K12 Physics Part_2

PHYSICS PART – I TEXTBOOK FOR CLASS XII o n T i l R b E u C p N re © e b o t t h s d e o n T i l R b E u C p N re © e b o t t h s d e PHYSICS PART – I TEXTBOOK FOR CLASS XII o n T i l R b E u C p N re © e b o t t h s d e ISBN 81-7450-631-4 First Edition December 2006 Pausa 1928 ALL RIGHTS RESERVED No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior permission of the publisher. This book is sold subject to the condition that it shall not, by way of trade, be lent, resold, hired out or otherwise disposed of without the publisher’s consent, in any form of binding or cover other than that in which it is published. The correct price of this publication is the price printed on this page, Any revised price indicated by a rubber stamp or by a sticker or by any other means is incorrect and should be unacceptable. Reprinted December 2007 Agrahayana 1929 December 2008 Pausa 1930 PD 360T RNB © National Council of Educational Research and Training, 2006 OFFICES OF THE PUBLICATION DEPARTMENT, NCERT Rs. 115.00 Printed on 80 GSM paper with NCERT watermark Published at the Publication Department by the Secretary, National Council of Educational Research and Training, Sri Aurobindo Marg, New Delhi 110 016 and printed at Prabhat Printing Press, D-23, Industrial Area Site-A, Mathura o n T i l R b E u C p N re © e b o t t NCERT Campus Sri Aurobindo Marg New Delhi 110 016 108, 100 Feet Road Hosdakere Halli Extension Banashankari III Stage Bangalore 560 085 Navjivan Trust Building P.O.Navjivan Ahmedabad 380 014 CWC Campus Opp. Dhankal Bus Stop Panihati Kolkata 700 114 CWC Complex Maligaon Guwahati 781 021 Phone : 011-26562708 Phone : 080-26725740 h s d e Phone : 079-27541446 Phone : 033-25530454 Phone : 0361-2674869 Publication Team Head, Publication Department Chief Production Officer Chief Editor : Peyyeti Rajakumar : Shiv Kumar : Shveta Uppal : Gautam Ganguly : R.N. Bhardwaj Chief Business Manager Assistant Editor Production Assistant : Prakash Veer Singh Cover, Layout and Illustrations Shweta Rao FOREWORD The National Curriculum Framework (NCF), 2005 recommends that children’s life at school must be linked to their life outside the school. This principle marks a departure from the legacy of bookish learning which continues to shape our system and causes a gap between the school, home and community. The syllabi and textbooks developed on the basis of NCF signify an attempt to implement this basic idea. They also attempt to discourage rote learning and the maintenance of sharp boundaries between different subject areas. We hope these measures will take us significantly further in the direction of a child-centred system of education outlined in the National Policy on Education (NPE), 1986. The success of this effort depends on the steps that school principals and teachers will take to encourage children to reflect on their own learning and to pursue imaginative activities and questions. We must recognise that, given space, time and freedom, children generate new knowledge by engaging with the information passed on to them by adults. Treating the prescribed textbook as the sole basis of examination is one of the key reasons why other resources and sites of learning are ignored. Inculcating creativity and initiative is possible if we perceive and treat children as participants in learning, not as receivers of a fixed body of knowledge. These aims imply considerable change in school routines and mode of functioning. Flexibility in the daily time-table is as necessary as rigour in implementing the annual calendar so that the required number of teaching days are actually devoted to teaching. The methods used for teaching and evaluation will also determine how effective this textbook proves for making children’s life at school a happy experience, rather than a source of stress or boredom. Syllabus designers have tried to address the problem of curricular burden by restructuring and reorienting knowledge at different stages with greater consideration for child psychology and the time available for teaching. The textbook attempts to enhance this endeavour by giving higher priority and space to opportunities for contemplation and wondering, discussion in small groups, and activities requiring hands-on experience. The National Council of Educational Research and Training (NCERT) appreciates the hard work done by the textbook development committee responsible for this book. We wish to thank the Chairperson of the advisory group in science and mathematics, Professor J.V. Narlikar and the Chief Advisor for this book, Professor A.W. Joshi for guiding the work of this committee. Several teachers contributed to the development of this textbook; we are grateful to their principals for making this possible. We are indebted to the institutions and organisations which have generously permitted us to draw upon their resour ces, material and personnel. We are especially grateful to the members of the National Monitoring Committee, appointed by the Department of Secondary and Higher Education, Ministry of Human Resource Development under the Chairpersonship of Professor Mrinal Miri and Professor G.P. Deshpande, for their valuable time and contribution. As an organisation committed to systemic reform and continuous improvement in the quality of its products, NCERT welcomes comments and suggestions which will enable us to undertake further revision and refinement. o n T i l R b E u C p N re © e b o t t h s d e New Delhi 20 December 2006 Director National Council of Educational Research and Training o n T i l R b E u C p N re © e b o t t h s d e TEXTBOOK DEVELOPMENT COMMITTEE CHAIRPERSON, ADVISORY GROUP FOR TEXTBOOKS IN SCIENCE AND MATHEMATICS J.V. Narlikar, Emeritus Professor, Inter-University Centre for Astronomy and Astrophysics (IUCAA), Ganeshkhind, Pune University Campus, Pune CHIEF ADVISOR A.W. Joshi, Honorary Visiting Scientist, National Centre for Radio Astrophysics (NCRA), Pune University Campus, Pune (Formerly Professor at Department of Physics, University of Pune) MEMBERS A.K. Ghatak, Emeritus Professor, Department of Physics, Indian Institute of Technology, New Delhi Alika Khare, Professor, Department of Physics, Indian Institute of Technology, Guwahati Anjali Kshirsagar, Reader, Department of Physics, University of Pune, Pune Anuradha Mathur, PGT , Modern School, Vasant Vihar, New Delhi B.K. Sharma, Professor, DESM, NCERT, New Delhi Gagan Gupta, Reader, DESM, NCERT, New Delhi Atul Mody, Lecturer (S.G.), VES College of Arts, Science and Commerce, Mumbai Chitra Goel, PGT, Rajkiya Pratibha Vikas Vidyalaya, Tyagraj Nagar, New Delhi H.C. Pradhan, Professor, Homi Bhabha Centre of Science Education (TIFR ), Mumbai N. Panchapakesan, Professor (Retd. ), Department of Physics and Astrophysics, University of Delhi, Delhi R. Joshi, Lecturer (S.G.), DESM, NCERT, New Delhi S.K. Dash, Reader, DESM, NCERT, New Delhi S. Rai Choudhary, Professor, Department of Physics and Astrophysics, University of Delhi, Delhi S.K. Upadhyay, PGT, Jawahar Navodaya Vidyalaya, Muzaffar Nagar V.H. Raybagkar, Reader, Nowrosjee Wadia College, Pune Vishwajeet Kulkarni, Teacher ( Grade I ), Higher Secondary Section, Smt. Parvatibai Chowgule College, Margao, Goa MEMBER-COORDINATOR V.P. Srivastava, Reader, DESM, NCERT, New Delhi S.N. Prabhakara, PGT, DM School, Regional Institute of Education (NCERT), Mysore o n T i l R b E u C p N re © e b o t t h s d e CONSTITUTION OF INDIA Part IV A (Article 51 A) Fundamental Duties – It shall be the duty of every citizen of India — (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and the National Anthem; (b) to cherish and follow the noble ideals which inspired our national struggle for freedom; (c) to uphold and protect the sovereignty, unity and integrity of India; (d) to defend the country and render national service when called upon to do so; (e) to promote harmony and the spirit of common brotherhood amongst all the people of India transcending religious, linguistic and regional or sectional diversities; to renounce practices derogatory to the dignity of women; (f) to value and preserve the rich heritage of our composite culture; (g) to protect and improve the natural environment including forests, lakes, rivers, wildlife and to have compassion for living creatures; (h) to develop the scientific temper, humanism and the spirit of inquiry and reform; (i) to safeguard public property and to abjure violence; (j) to strive towards excellence in all spheres of individual and collective activity so that the nation constantly rises to higher levels of endeavour and achievement; (k) who is a parent or guardian, to provide opportunities for education to his child or, as the case may be, ward between the age of six and fourteen years. o n T i l R b E u C p N re © e b o t t Fundamental Duties h s d e ACKNOWLEDGEMENTS The National Council of Educational Research and Training acknowledges the valuable contribution of the individuals and organisations involved in the development of Physics Textbook for Class XII. The Council also acknowledges the valuable contribution of the following academics for reviewing and refining the manuscripts of this book: Anu Venugopalan, Lecturer, School of Basic and Applied Sciences, GGSIP University, Delhi; A.K. Das, PGT, St. Xavier’s Senior Secondary School, Delhi; Bharati Kukkal, PGT, Kendriya Vidyalaya, Pushp Vihar, New Delhi; D.A. Desai, Lecturer (Retd.), Ruparel College, Mumbai; Devendra Kumar, PGT, Rajkiya Pratibha Vikas Vidyalaya, Yamuna Vihar, Delhi; I.K. Gogia, PGT , Kendriya Vidyalaya, Gole Market, New Delhi; K.C. Sharma, Reader, Regional Institute of Education (NCERT), Ajmer; M.K. Nandy, Associate Professor, Department of Physics, Indian Institute of Technology, Guwahati; M.N. Bapat, Reader , Regional Institute of Education (NCERT), Mysore; R. Bhattacharjee, Asstt. Professor, Department of Electronics and Communication Engineering, Indian Institute of Technology, Guwahati; R.S. Das, Vice-Principal (Retd.), Balwant Ray Mehta Senior Secondary School, Lajpat Nagar, New Delhi; Sangeeta D. Gadre, Reader, Kirori Mal College, Delhi; Suresh Kumar, PGT, Delhi Public School, Dwarka, New Delhi; Sushma Jaireth, Reader, Department of Women’s Studies, NCERT, New Delhi; Shyama Rath, Reader, Department of Physics and Astrophysics, University of Delhi, Delhi; Yashu Kumar, PGT , Kulachi Hans Raj Model School, Ashok Vihar, Delhi. The Council also gratefully acknowledges the valuable contribution of the following academics for the editing and finalisation of this book: B.B. Tripathi, Professor (Retd.), Department of Physics, Indian Institute of Technology, New Delhi; Dipan K. Ghosh, Professor, Department of Physics, Indian Institute of Technology, Mumbai; Dipanjan Mitra, Scientist, National Centre for Radio Astrophysics (TIFR), Pune; G.K. Mehta, Raja Ramanna Fellow, Inter-University Accelerator Centre, New Delhi; G.S. Visweswaran, Professor, Department of Electrical Engineering, Indian Institute of Technology, New Delhi; H.C. Kandpal, Head, Optical Radiation Standards, National Physical Laboratory, New Delhi; H.S. Mani, Raja Ramanna Fellow, Institute of Mathematical Sciences, Chennai; K. Thyagarajan, Professor, Department of Physics, Indian Institute of Technology, New Delhi; P.C. Vinod Kumar, Professor, Department of Physics, Sardar Patel University, Vallabh Vidyanagar, Gujarat; S. Annapoorni, Professor, Department of Physics and Astrophysics, University of Delhi, Delhi; S.C. Dutta Roy, Emeritus Professor, Department of Electrical Engineering, Indian Institute of Technology, New Delhi; S.D. Joglekar, Professor, Department of Physics, Indian Institute of Technology, Kanpur; V. Sundara Raja, Professor, Sri Venkateswara University, Tirupati. Special thanks are due to Hukum Singh, Professor and Head, DESM, NCERT for his support. The Council also acknowledges the support provided by the APC office and the administrative staff of the DESM; Deepak Kapoor, Incharge, Computer Station; Inder Kumar, DTP Operator; Mohd. Qamar Tabrez, Copy Editor; Ashima Srivastava, Proof Reader in shaping this book. The contributions of the Publication Department in bringing out this book are also duly acknowledged. o n T i l R b E u C p N re © e b o t t h s d e CONSTITUTION OF INDIA Part III (Articles 12 – 35) (Subject to certain conditions, some exceptions and reasonable restrictions) guarantees these Fundamental Rights Right to Equality • before law and equal protection of laws; • irrespective of religion, race, caste, sex or place of birth; • of opportunity in public employment; • by abolition of untouchability and titles. Right to Freedom • of expression, assembly, association, movement, residence and profession; • of certain protections in respect of conviction for offences; • of protection of life and personal liberty; • of free and compulsory education for children between the age of six and fourteen years; • of protection against arrest and detention in certain cases. Right against Exploitation • for prohibition of traffic in human beings and forced labour; • for prohibition of employment of children in hazardous jobs. o n Right to Freedom of Religion • freedom of conscience and free profession, practice and propagation of religion; • freedom to manage religious affairs; • freedom as to payment of taxes for promotion of any particular religion; • freedom as to attendance at religious instruction or religious worship in educational institutions wholly maintained by the State. Cultural and Educational Rights • for protection of interests of minorities to conserve their language, script and culture; • for minorities to establish and administer educational institutions of their choice. Right to Constitutional Remedies • by issuance of directions or orders or writs by the Supreme Court and High Courts for enforcement of these Fundamental Rights. T i l R b E u C p N re © e b o t t h s d e PREFACE It gives me pleasure to place this book in the hands of the students, teachers and the public at large (whose role cannot be overlooked). It is a natural sequel to the Class XI textbook which was brought out in 2006. This book is also a trimmed version of the textbooks which existed so far. The chapter on thermal and chemical effects of current has been cut out. This topic has also been dropped from the CBSE syllabus. Similarly, the chapter on communications has been substantially curtailed. It has been rewritten in an easily comprehensible form. Although most other chapters have been based on the earlier versions, several parts and sections in them have been rewritten. The Development Team has been guided by the feedback received from innumerable teachers across the country. In producing these books, Class XI as well as Class XII, there has been a basic change of emphasis. Both the books present physics to students without assuming that they would pursue this subject beyond the higher secondary level. This new view has been prompted by the various observations and suggestions made in the National Curriculum Framework (NCF), 2005. Similarly, in today’s educational scenario where students can opt for various combinations of subjects, we cannot assume that a physics student is also studying mathematics. Therefore, physics has to be presented, so to say, in a stand-alone form. As in Class XI textbook, some interesting box items have been inserted in many chapters. They are not meant for teaching or examinations. Their purpose is to catch the attention of the reader, to show some applications in daily life or in other areas of science and technology, to suggest a simple experiment, to show connection of concepts in different areas of physics, and in general, to break the monotony and enliven the book. Features like Summary, Points to Ponder, Exercises and Additional Exercises at the end of each chapter, and Examples have been retained. Several concept-based Exercises have been transferred from end-of-chapter Exercises to Examples with Solutions in the text. It is hoped that this will make the concepts discussed in the chapter more comprehensible. Several new examples and exercises have been added. Students wishing to pursue physics further would find Points to Ponder and Additional Exercises very useful and thoughtful. To provide resources beyond the textbook and to encourage eLearning, each chapter has been provided with some relevant website addresses under the title ePhysics. These sites provide additional materials on specific topics and also provide learners the opportunites for interactive demonstrations/ experiments. The intricate concepts of physics must be understood, comprehended and appreciated. Students must learn to ask questions like ‘why’, ‘how’, ‘how do we know it’. They will find almost always that the question ‘why’ has no answer within the domain of physics and science in general. But that itself is a learning experience, is it not? On the other hand, the question ‘how’ has been reasonably well answered by physicists in the case of most natural phenomena. In fact, with the understanding of how things happen, it has been possible to make use of many phenomena to create technological applications for the use of humans. For example, consider statements in a book, like ‘A negatively charged electron is attracted by the positively charged plate’, or ‘In this experiment, light (or electron) behaves like a wave’. You will realise that it is not possible to answer ‘why’. This question belongs to the domain of philosophy or metaphysics. But we can answer ‘how’, we can find the force acting, we can find o n T i l R b E u C p N re © e b o t t h s d e but three of them were generated by friends and well-wishers not part of any team. We welcome suggestions and comments from our valued users. o n xii T i l R b E u C p N re © e b o t t h s d e A. Sen of Pune. Prof. especially students and teachers. NCERT. and we can develop instruments which will use these phenomena to our advantage. Krishna Kumar. for his help from time-to-time. the Review Team and Editing Teams involved college and university teachers. 4 (Part I) and 9 (Part II). G. We are thankful to Dr P. scientists from national institutes and laboratories. along with a team of members. Ravindra. The Textbook Development Team. J OSHI Chief Advisor Textbook Development Committee . as well as higher secondary teachers. Professor Roopmanjari Ghosh of Delhi and Dr Rajesh B Khaparde of Mumbai for allowing us to use their box items. for entrusting us with the task of presenting this textbook as a part of the national effort for improving science education. W. Most box items were generated by members of one or the other team. respectively in Chapters 3. We are very thankful to the members of the Review and Editing Workshops to discuss and refine the first draft of the textbook. We also express our gratitude to Prof. We wish our young readers a happy journey into the exciting realm of physics. Joint Director. The feedback and critical look provided by higher secondary teachers in the various teams are highly laudable. NCERT.the wavelength of the photon (or electron). we can determine how things behave under different conditions. I also thank Prof. teachers from Indian Institutes of Technology.N. was always willing to help us in our endeavour in every possible way. Hukum Singh. Head. It has been a pleasure to work for these books at the higher secondary level. NCERT. Department of Education in Science and Mathematics. Director. 6 1.6 2.1 1.12 2.10 2.7 1.CONTENTS FOREWORD PREFACE CHAPTER ONE ELECTRIC CHARGES AND FIELDS 1.4 1.14 1.10 1.7 2.1 2.15 Introduction Electric Charges Conductors and Insulators Charging by Induction Basic Properties of Electric Charge Coulomb’s Law Forces between Multiple Charges Electric Field Electric Field Lines Electric Flux Electric Dipole Dipole in a Uniform External Field Continuous Charge Distribution Gauss’s Law Application of Gauss’s Law v xi CHAPTER TWO ELECTROSTATIC POTENTIAL AND CAPACITANCE 2.12 1.9 1.5 2.5 1.2 1.2 2.3 2.13 1.3 1.8 1.11 2.13 Introduction Electrostatic Potential Potential due to a Point Charge Potential due to an Electric Dipole Potential due to a System of Charges Equipotential Surfaces Potential Energy of a System of Charges Potential Energy in an External Field Electrostatics of Conductors Dielectrics and Polarisation Capacitors and Capacitance The Parallel Plate Capacitor Effect of Dielectric on Capacitance o n T i l R b E u C p N re © e b o t t h s d e 1 1 5 6 8 10 15 18 23 25 27 31 32 33 37 51 53 54 55 57 60 61 64 67 71 73 74 75 .9 2.4 2.11 1.8 2. 3 3.8 4.11 o n Introduction Magnetic Force Motion in a Magnetic Field Motion in Combined Electric and Magnetic Fields Magnetic Field due to a Current Element.7 3. the Ampere Torque on Current Loop.9 3.1 4.13 3. Power Combination of Resistors — Series and Parallel Cells.10 3.16 Van de Graaff Generator CHAPTER THREE CURRENT ELECTRICITY 3.4 3. emf.1 5.12 3.2.15 3.11 3.2 3.5 4.6 3. Internal Resistance Cells in Series and in Parallel Kirchhoff’s Laws Wheatstone Bridge Meter Bridge Potentiometer 78 80 83 CHAPTER FOUR MOVING CHARGES AND MAGNETISM 4.10 4.14 Combination of Capacitors 2.3 4.16 Introduction Electric Current Electric Currents in Conductors Ohm’s law Drift of Electrons and the Origin of Resistivity Limitations of Ohm’s Law Resistivity of various Materials Temperature Dependence of Resistivity Electrical Energy.8 3.2 4.9 4.2 xiv Introduction The Bar Magnet 173 174 .14 3.1 3.15 Energy Stored in a Capacitor 2.6 4. Magnetic Dipole The Moving Coil Galvanometer T i l R b E u C p N re © e b o t t h s d e 93 93 94 95 97 101 101 103 105 107 110 113 115 118 120 122 132 133 137 140 143 145 147 150 154 157 163 CHAPTER FIVE MAGNETISM AND MATTER 5. Biot-Savart Law Magnetic Field on the Axis of a Circular Current Loop Ampere’s Circuital Law The Solenoid and the Toroid Force between Two Parallel Currents.7 4.5 3.4 4. 7 Power in AC Circuit: The Power Factor 7.4 Introduction Displacement Current Electromagnetic Waves Electromagnetic Spectrum o n T i l R b E u C p N re © e b o t t h s d e 233 234 237 237 241 244 252 255 259 269 270 274 280 288 ANSWERS xv .4 6.6 6.4 5.9 Transformers CHAPTER EIGHT ELECTROMAGNETIC WAVES 8.7 6.3 6.2 6.8 LC Oscillations 7.6 5.9 6.5 5.3 8.2 AC Voltage Applied to a Resistor 7.1 Introduction 7.3 5.1 6.6 AC Voltage Applied to a Series LCR Circuit 7.7 Magnetism and Gauss’s Law The Earth’s Magnetism Magnetisation and Magnetic Intensity Magnetic Properties of Materials Permanent Magnets and Electromagnets 181 185 189 191 195 CHAPTER SIX ELECTROMAGNETIC INDUCTION 6.5.3 Representation of AC Current and Voltage by Rotating Vectors — Phasors 7.5 AC Voltage Applied to a Capacitor 7.2 8.1 8.5 6.10 Introduction The Experiments of Faraday and Henry Magnetic Flux Faraday’s Law of Induction Lenz’s Law and Conservation of Energy Motional Electromotive Force Energy Consideration: A Quantitative Study Eddy Currents Inductance AC Generator 204 205 206 207 210 212 215 218 219 224 CHAPTER SEVEN ALTERNATING CURRENT 7.4 AC Voltage Applied to an Inductor 7.8 6. Vol. at Cadarache in the South of France and will provide 500 MW of fusion power. to meet the needs of a growing world population. 11-12/ May-June 2006. h s d e . with abundant fuel resources. the Russian Federation and the USA. On earth. see Nuclear India.iter.in) Different stages in the evolution of the universe. the Republic of Korea.org and the Nobel Prize in Physics 2006) o n Cut away view of International Thermonuclear Experimental Reactor (ITER) device. the People’s Republic of China.org and http://www. The man in the bottom shows the scale. India is one of the seven full partners in the project. fusion research is aimed at demonstrating that this energy source can be used to produce electricity in a safe and environmentally benign way. For details of India’s role.COVER DESIGN (Adapted from http://nobelprize. Fusion is the energy source of the sun and the stars.dae. ITER is a joint international research and development project that aims to demonstrate the scientific and technical feasibility of fusion power.gov. Nov. issue available at Department of Atomic Energy (DAE) website mentioned above. Japan. ITER will be constructed in Europe. T i l R b E u C p N re © e b o t t BACK COVER (Adapted from http://www. 39. the others being the European Union (represented by EURATOM). New Delhi 110 016 and printed at Supreme Offset Press. mechanical.N. in any form of binding or cover other than that in which it is published. stored in a retrieval system or transmitted. 115. 2007 Agrahayana 1929 December 2008 Pausa 1930 PD 360T RNB © National Council of Educational Research and Training. National Council of Educational Research and Training. Bhardwaj Production Assistant : Prakash Veer Singh Cover.ISBN 81-7450-631-4 First Edition December. resold. NCERT Rs. by way of trade. be lent. 100 Feet Road Hosdakere Halli Extension Banashankari III Stage Bangalore 560 085 Navjivan Trust Building P. 2006 Pausa 1928 Reprinted December.O. New Delhi 110 017 o n T i l R b E u C p N re © e b o t t 108. photocopying. electronic. 2006 ALL RIGHTS RESERVED No part of this publication may be reproduced. K-5. Sri Aurobindo Marg.Navjivan Ahmedabad 380 014 CWC Campus Opp. Malvaya Nagar. hired out or otherwise disposed of without the publisher’s consent. This book is sold subject to the condition that it shall not. in any form or by any means. Dhankal Bus Stop Panihati Kolkata 700 114 CWC Complex Maligaon Guwahati 781 021 NCERT Campus Sri Aurobindo Marg New Delhi 110 016 Phone : 011-26562708 Phone : 080-26725740 h s d e Phone : 079-27541446 Phone : 033-25530454 Phone : 0361-2674869 Publication Team Head. recording or otherwise without the prior permission of the publisher. Publication Department Chief Production Officer Chief Editor : Peyyeti Rajakumar : Shiv Kumar : Shveta Uppal : Gautam Ganguly Chief Business Manager Assistant Editor : R. The correct price of this publication is the price printed on this page. Any revised price indicated by a rubber stamp or by a sticker or by any other means is incorrect and should be unacceptable. Layout and Illustrations Shweta Rao . OFFICES OF THE PUBLICATION DEPARTMENT.00 Printed on 80 GSM paper with NCERT watermark Published at the Publication Department by the Secretary. Many such pairs of materials were known which . This is precisely the topic we are going to discuss in this and the next chapter. Electrostatics deals with the study of forces.1 INTRODUCTION All of us have the experience of seeing a spark or hearing a crackle when we take off our synthetic clothes or sweater. 1.Chapter One ELECTRIC CHARGES AND FIELDS 1. Static means anything that does not move or change with time. We also experience a sensation of an electric shock either while opening the door of a car or holding the iron bar of a bus after sliding from our seat. around 600 BC. Have you ever tried to find any explanation for this phenomenon? Another common example of electric discharge is the lightning that we see in the sky during thunderstorms. This is almost inevitable with ladies garments like a polyester saree.2 ELECTRIC CHARGE Historically the credit of discovery of the fact that amber rubbed with wool or silk cloth attracts light objects goes to Thales of Miletus. You might have also heard that this is due to generation of static electricity. which were accumulated due to rubbing of insulating surfaces. fields and potentials arising from static charges. The reason for these experiences is discharge of electric charges through our body. particularly in dry weather. The name electricity is coined from the Greek word elektron meaning amber. Greece. In fact they remain stuck to the screen for a while. they repel each other [Fig. 1. The two strands FIGURE 1.edu/travoltage/HTML/ . pith balls and bits of papers. silk. If a plastic rod rubbed with fur is made to touch two small pith balls (now-a-days we can use polystyrene balls) suspended by silk or nylon thread. You will see that the strips get attracted to the screen. The glass rod repels the fur.1(a)].1(d)] and are also repelled by the rod. They also do not attract or repel other light objects as they did on being electrified. The property which differentiates the two kinds of charges is called the polarity of charge. after many careful studies by different scientists. Similarly.ucla. with unlike charges attract each other. 1. Now if the electrified glass rod is brought in contact with silk. These seemingly simple facts were established from years of efforts and careful experiments and their analyses. We say that the bodies like glass or plastic rods. they no longer attract each other. What can you conclude from these 2 observations? It just tells us that unlike charges acquired by the objects Interactive animation on simple electrostatic experiments: http://ephysics. also repel each other.1(e)]. Take them near a TV screen or computer monitor. Cut out long thin strips of white paper and lightly iron them. the charges acquired after rubbing are lost when the charged bodies are brought in contact. two plastic rods rubbed with cat’s fur repelled each other [Fig. It is said that the pith balls are electrified or are charged by contact. which the rods were rubbed.Physics on rubbing could attract light objects like straw. 1. with which it was rubbed.physics.1(b)] but attracted the fur. On the other hand. 1.1(f )]. It was observed that if two glass rods rubbed with wool or silk cloth are brought close to each other. fur and pith balls are electrified. 1. the plastic rod attracts the glass rod [Fig. The experiments on pith balls suggested that there are two kinds of electrification and we find that (i) like charges repel and (ii) unlike charges attract each other. This is true for any pair of objects that are rubbed to be electrified. A similar effect is found if the pith balls are touched with a glass rod rubbed with silk [Fig. When a glass rod is rubbed with silk.1 Rods and pith balls: like charges repel and of wool or two pieces of silk cloth. Thus. A dramatic observation is that a pith ball touched with glass rod attracts another pith ball touched with plastic rod [Fig. They acquire an electric charge on rubbing. then the balls repel each other [Fig. 1.1(c)] and repel the silk or wool with which the glass rod is rubbed. You can perform the following activity at home to experience such an effect. that there were only two kinds of an entity which is called the electric charge. The experiments also demonstrated that the charges are transferred from the rods to the pith balls on contact. However. the rod acquires one kind of charge and the silk acquires the second kind of charge. the glass rod and wool attracted each other. It was concluded. batteries. and a wide variety of the practical appliances used in daily life are based on the principles of this science. UNIFICATION OF ELECTRICITY AND MAGNETISM In olden days. If an object possesses an electric charge. the charge on glass rod or cat’s fur is called positive and that on plastic rod or silk is termed negative. electricity and magnetism were treated as separate subjects. In 1820 Danish scientist Oersted found that a compass needle is deflected by passing an electric current through a wire placed near the needle. By convention. charge flows on to the leaves and they diverge.Electric Charges and Fields neutralise or nullify each other’s effect. in the frame of reference where all the charges are at rest. the forces are purely electrical. cat’s fur. This field is called electromagnetism. Electromagnetic force is one of the fundamental forces of nature. Ampere and Faraday supported this observation by saying that electric charges in motion produce magnetic fields and moving magnets generate electricity. while magnetism described interactions of magnets. compass needles. Its effect is felt even when the distance between the interacting particles is very large because the force decreases inversely as the square of the distance between the interacting bodies. He also argued that light is electromagnetic in nature and its speed can be found by making purely electric and magnetic measurements. chemical force between atoms holding the matter together. We know that when we add a positive number to a negative number of the same magnitude. etc.. the sum is zero. You know that gravitational force is a long-range force. He claimed that the science of optics is intimately related to that of electricity and magnetism.2(a)]. lightning. Virtually every force that we can think of like friction. and even the forces describing processes occurring in cells of living organisms. It consists of a vertical metal rod housed in a box. Maxwell put forth four equations that play the same role in classical electromagnetism as Newton’s equations of motion and gravitation law play in mechanics. 1. The science of electricity and magnetism is the foundation for the modern technological civilisation. A simple apparatus to detect charge on a body is the gold-leaf electroscope [Fig. Most of the phenomena occurring around us can be described under electromagnetism. Electricity dealt with charges on glass rods. Although charged particles in motion exert both electric and magnetic forces. When it has no charge it is said to be neutral. iron filings. The degree of divergance is an indicator of the amount of charge. When a charged object touches the metal knob at the top of the rod. it is said to be electrified or charged. etc. telecommunication. Electric power. radio and television. with two thin gold leaves attached to its bottom end. Therefore the charges were named as positive and negative by the American scientist Benjamin Franklin. We will learn in this chapter that electric force is also as pervasive and is in fact stronger than the gravitational force by several orders of magnitude (refer to Chapter 1 of Class XI Physics Textbook). 3 . have its origin in electromagnetic force. This might have been the philosophy in naming the charges as positive and negative. The unification was achieved when the Scottish physicist Maxwell and the Dutch physicist Lorentz put forward a theory where they showed the interdependence of these two subjects. The separation is a rough measure of the amount of charge on the electroscope. 1. excess charge or deficit of charge. Take a large bottle that can hold this rod and a cork which will fit in the opening of the bottle. 1. Although normally the materials are electrically neutral. When we say that a body is charged. electroscope.2 Electroscopes: (a) The gold leaf iron it lightly with the mountain fold up. arising from the forces between charged particles. To electrify a neutral body. they do contain charges. On charging the curtain rod by touching the ball end with an electrified body.Physics Students can make a simple electroscope as follows [Fig. Fold a small. the adhesive force of glue. You know that all matter is made up of atoms and/or molecules. A body can thus be charged positively by 4 losing some of its electrons. Hold the strip by pinching it at the fold. we always refer to this experiment. A paper scale may be put inside the bottle in advance to measure the separation of leaves. Fold the strips into half so that you make a mark of fold. a body can be charged negatively . being less tightly bound in the atom. The same effect is seen in the leaf electroscope. charge is transferred to the curtain rod and the attached aluminium foil. Forces that hold the molecules together. This forms the leaves of your electroscope. Fit the cork in the bottle with about 5 cm of the ball end projecting above the cork. thin aluminium foil (about 6 cm in length) in the middle and attach it to the flattened end of the rod by cellulose tape. but their charges are exactly balanced. as shown electroscope. forces that hold atoms together in a solid. Make a hole in the cork sufficient to hold the curtain rod snugly. Similarly. forces associated with surface tension. Let us first try to understand why material bodies acquire charge. all are basically electrical in nature. are the charges which are transferred from one body to the other.3. It is therefore essential that we learn more about such a force. use the white paper strips we used for seeing the attraction of charged bodies. Both the halves of the foil get similar charge and therefore repel each other. Cut out a piece of length about 20 cm with the ball at one end and flatten the cut end. This shows that the strip has acquired charge on ironing. we need to add or remove one kind of FIGURE 1. Thus the electric force is all pervasive and it encompasses almost each and every field associated with our life. You would notice that the two halves move apart. (b) Schematics of a simple in Fig. Slide the rod through the hole in the cork with the cut end on the lower side and ball end projecting above the cork. The divergence in the leaves depends on the amount of charge on them.3 Paper strip charge. When you fold it into half. Open the strip and FIGURE 1.2(b)]: Take a thin aluminium curtain rod with ball ends fitted for hanging the curtain. both the halves have the same charge. To understand how the electroscope works. In solids. Hence they repel each other. some of the electrons. A thick metal plate is buried deep into the earth and thick wires are drawn from this plate. no transfer of charge will take place from the plastic rod to the pith ball. The first two carry * There is a third category called semiconductors. Why does the transfer of charge not take place from the rod to the ball? Some substances readily allow passage of electricity through them. We will find that the pith ball acquires a negative charge. but a metal article like spoon does not. Most substances fall into one of the two classes stated above*. Therefore. When we rub a glass rod with silk. When some charge is transferred to a conductor. Metals.Electric Charges and Fields by gaining electrons. the bodies get charged and that is why we have to stick to certain pairs of materials to notice charging on rubbing the bodies. it stays at the same place. that are transferred. They are called insulators. nylon. This process of sharing the charges with the earth is called grounding or earthing. if some charge is put on an insulator. 1. Also only the less tightly bound electrons in a material body can be transferred from it to another by rubbing. However. You will learn why this happens in the next chapter. Suppose we connect one end of a copper wire to a neutral pith ball and the other end to a negatively charged plastic rod. it shows signs of charging. which offer resistance to the movement of charges which is intermediate between the conductors and insulators. The electric wiring in our houses has three wires: live. wood offer high resistance to the passage of electricity through them. these are used in buildings for the purpose of earthing near the mains supply. If a similar experiment is repeated with a nylon thread or a rubber band. others do not. The charges on metal leak through our body to the ground as both are conductors of electricity. Most of the non-metals like glass. if a metal rod with a wooden or plastic handle is rubbed without touching its metal part.3 CONDUCTORS AND INSULATORS A metal rod held in hand and rubbed with wool will not show any sign of being charged. Thus the rod gets positively charged and the silk gets negatively charged. Those which allow electricity to pass through them easily are called conductors. some of the electrons from the rod are transferred to the silk cloth. plastic. is a very small fraction of the total number of electrons in the material body. porcelain. In contrast. When we bring a charged body in contact with the earth. Earthing provides a safety measure for electrical circuits and appliances. when a body is rubbed with another. all the excess charge on the body disappears by causing a momentary current to pass to the ground through the connecting conductor (such as our body). 5 . Also the number of electrons. No new charge is created in the process of rubbing. human and animal bodies and earth are conductors. it readily gets distributed over the entire surface of the conductor. They have electric charges (electrons) that are comparatively free to move inside the material. neutral and earth. This property of the materials tells you why a nylon or plastic comb gets electrified on combing dry hair or on rubbing. some of the negative charges on the rod are transferred to the pith ball and it also gets charged. (iii) Separate the spheres by a small distance while the glass rod is still held near sphere A. reside on the surfaces. The accumulated charges remain on the surface.4(d).Physics electric current from the power station and the third is earthed by connecting it to the buried metal plate. This is charging by induction. till the glass rod is held near the sphere. If the rod is removed. As the negative charge starts building up at the left surface of A. 1. has an excess of positive charge.4 CHARGING BY INDUCTION FIGURE 1. The positively charged glass rod does not lose any of its charge. They. However. The charges on spheres rearrange themselves as shown in Fig.4(e). 6 When we touch a pith ball with an electrified plastic rod. (iv) Remove the rod. refrigerator. When any fault occurs or live wire touches the metallic body. the metal spheres will each be equal and oppositely charged. (i) Bring two metal spheres. The charges on them get uniformly distributed over them. 1. A and B. The process is called induction of charge and happens almost instantly. 1. Thus the pith ball is charged by contact. Fig. However. supported on insulating stands. In this process. taking care that it does not touch the sphere. 1. has an excess of negative charge and the right surface of sphere B. The free electrons in the spheres are attracted towards the rod.4(c). The two spheres are found to be oppositely charged and attract each other.4(b) shows the equilibrium situation. Let us try to understand what could be happening by performing the following experiment. the charges are not acted by any outside force and they redistribute to their original neutral state. as shown in Fig. other electrons are repelled by these. TV are connected to the earth wire. equilibrium is reached under the action of force of attraction of the rod and the force of repulsion due to the accumulated charges. . The rods induce opposite charges on the near surfaces of the objects and similar charges move to the farther side of the object. contrary to the process of charging by contact. a similar effect takes place. Metallic bodies of the electric appliances such as electric iron. as shown in Fig. 1. as shown in Fig. This leaves an excess of positive charge on the rear surface of sphere B. In a short time. the charge flows to the earth without damaging the appliance and without causing any injury to the humans. in contact as shown in Fig. say A. separate the spheres quite apart. 1.4 Charging by induction.4(a). this would have otherwise been unavoidable since the human body is a conductor of electricity. The left surface of sphere A. (ii) Bring a positively charged rod near one of the spheres. therefore. It is then repelled by the plastic rod but is attracted by a glass rod which is oppositely charged.4(b). When electrified rods are brought near light objects. Now. why a electrified rod attracts light objects. 1. as shown. is a question we have still left unanswered. Both kinds of charges are bound in the metal spheres and cannot escape. not all of the electrons in the spheres have accumulated on the left surface of A. by bringing a positively charged rod near it. This process of charge distribution stops when the net force on the free electrons inside the metal is zero. the magnitude of force depends on the distance between the charges and in this case the force of attraction overweighs the force of repulsion. as shown in Fig. 1. being light.10. The positive charge continues to be held at the near end [Fig.1 7 .] The centres of the two types of charges are slightly separated. The near end becomes positively charged due to deficit of electrons.10 and 2.5(c). the metal sphere gets charged by the process of induction and the rod does not lose any of its charge. In this case the electrons will flow from the ground to the sphere when the sphere is connected to the ground with a wire. The positive charge will spread uniformly over the sphere as shown in Fig. Remove the electrified rod. 1. as shown in Fig. are pulled towards the rods.physicsclassroom. Example 1. Disconnect the sphere from the ground.Electric Charges and Fields [This happens even when the light object is not a conductor.com/mmedia/estatics/estaticTOC.html Interactive animation on charging a two-sphere system by induction: FIGURE 1. 1.5(d)]. We know that opposite charges attract while similar charges repel. http://www. However.1 How can you charge a metal sphere positively without touching it? Solution Figure 1. 1.5(b). Connect the sphere to the ground by a conducting wire.5 In this experiment. As a result the particles like bits of paper or pith balls. Bring a negatively charged rod close to the metallic sphere. the free electrons in the sphere move away due to repulsion and start piling up at the farther end. The electrons will flow to the ground while the positive charges at the near end will remain held there due to the attractive force of the negative charges on the rod. As the rod is brought close to the sphere.5(e). The mechanism for how this happens is explained later in Sections 1. Similar steps are involved in charging a metal sphere negatively by induction. Can you explain why? EXAMPLE 1.5(a) shows an uncharged metallic sphere on an insulating metal stand. Charge has magnitude but no direction. charges add up like real numbers or they are scalars like the mass of a body.5.e. –3. +2.5. If the sizes of charged bodies are very small as compared to the distances between them. +4 and –5. the total charge of a system containing five charges +1. we treat them as point charges.Physics 1. If a system contains n charges q1.3 Quantisation of charge Experimentally it is established that all free charges are integral multiples of a basic unit of charge denoted by e. charges may get redistributed but it is found that the total charge of the isolated system is always conserved. It is not possible to create or destroy net charge carried by any isolated system although the charge carrying particles may be created or destroyed in a process. we shall now describe some other properties of the electric charge. no new charges are either created or destroyed. Conservation of charge has been established experimentally. …. Thus charge q on a body is always given by 8 q = ne . When we rub two bodies. q3.2 Charge is conserved We have already hinted to the fact that when bodies are charged by rubbing. 1.5. A picture of particles of electric charge enables us to understand the idea of conservation of charge.. there is one difference between mass and charge. we shall follow it up in the next section. similar to the mass. All the charge content of the body is assumed to be concentrated at one point in space. Proper signs have to be used while adding the charges in a system. is (+1) + (+2) + (–3) + (+4) + (–5) = –1 in the same unit. qn. q2. i. due to interactions among the bodies. there is transfer of electrons from one body to the other. namely positive and negative and their effects tend to cancel each other. We shall tentatively assume that this can be done and proceed. the total charge of the system is obtained simply by adding algebraically q1 and q2 . If a system contains two point charges q1 and q2. what one body gains in charge the other body loses.5 BASIC PROPERTIES OF ELECTRIC CHARGE We have seen that there are two types of charges. in some arbitrary unit. then the total charge of the system is q1 + q2 + q3 + … + qn . Within an isolated system consisting of many charged bodies. For example. 1. Sometimes nature creates charged particles: a neutron turns into a proton and an electron.1 Additivity of charges We have not as yet given a quantitative definition of a charge. The proton and electron thus created have equal and opposite charges and the total charge is zero before and after the creation. However. Mass of a body is always positive whereas a charge can be either positive or negative. Here. 1. (see Chapter 2 of Class XI. very small because at the macroscopic level. charges of this large magnitude are seldom encountered and hence we use smaller units 1 µC (micro coulomb) = 10–6 C or 1 mC (milli coulomb) = 10–3 C. Part I). As many points very close to each other normally give an impression of a continuous line. the quantisation of charge has no practical consequence and can be ignored.Electric Charges and Fields where n is any integer. In electrostatics. a unit of charge is called a coulomb and is denoted by the symbol C. A dotted line viewed from a distance appears continuous to us but is not continuous in reality. their difference is also an integer. This basic unit of charge is the charge that an electron or proton carries. all the observable charges have to be integral multiples of e. the total amount of charge on the body is n 2 × e + n1 × (–e) = (n 2 – n1) e. if a body contains n1 electrons and n 2 protons. At this scale the fact that charge of a body can increase or decrease in units of e is not visible. This situation can be compared with the geometrical concepts of points and lines. In this system. The grainy nature of the charge is lost and it appears to be continuous.602192 × 10–19 C Thus. the value of the basic unit of charge is e = 1. In the International System (SI) of Units. there are about 6 × 1018 electrons in a charge of –1C. At this scale. say 1 µC.6 × 10–19 C. By convention. Thus. Physics Textbook . Thus. At the microscopic level. Since e = 1. many small charges taken together appear as a continuous charge distribution. where the charges involved are of the order of a few tens or hundreds of e. The quantisation of charge was first suggested by the experimental laws of electrolysis discovered by English experimentalist Faraday. we deal with charges of a few µC. at the macroscopic level. one coulomb is the charge flowing through a wire in 1 s if the current is 1 A (ampere). therefore charge on an electron is written as –e and that on a proton as +e. At the macroscopic level. the fact that charge can increase or decrease only in units of e is not very different from saying that charge can take continuous values. Since n1 and n2 are integers. It was experimentally demonstrated by Millikan in 1912. In terms of this definition. 9 . contains something like 1013 times the electronic charge.. positive or negative. Thus the charge on any body is always an integral multiple of e and can be increased or decreased also in steps of e. however. one deals with charges that are enormous compared to the magnitude of charge e. The step size e is. If the protons and electrons are the only basic charges in the universe. a charge of magnitude. i. The fact that electric charge is always an integral multiple of e is termed as quantisation of charge. A coulomb is defined in terms the unit of the electric current which you are going to learn in a subsequent chapter. There are a large number of situations in physics where certain physical quantities are quantised. the charge on an electron is taken to be negative.e. q1 q 2 10 . Therefore the number of molecules in one cup of water is (250/18) × 6.6 × 10–10 C/s) = 6. Thus. It was also used later by Cavendish to measure the very feeble gravitational force between two objects.2 1. however.. One coulomb is.02 × 1023.6 × 10–19 C = 1.Physics they can be counted.3 EXAMPLE 1. they appear in discrete lumps and quantisation of charge cannot be ignored.25 × 109 s = 6. Thus to collect a charge of one coulomb.02 × 1023 × 10 × 1. Example 1.6 COULOMB’S LAW Coulomb’s law is a quantitative statement about the force between two point charges. It is the scale involved that is very important.6 × 10–10 C. Therefore the charge given out in one second is 1. i. A cubic piece of copper of side 1 cm contains about 2. therefore. 10 electrons and 10 protons. When the linear size of charged bodies are much smaller than the distance separating them. from a body from which 109 electrons move out every second.e. one mole (= 6. It is equal to (250/18) × 6. The molecular mass of water is 18g. Coulomb measured the force between two point charges and found that it varied inversely as the square of the distance between the charges and was directly proportional to the product of the magnitude of the two charges and acted along the line joining the two charges. also important to know what is roughly the number of electrons contained in a piece of one cubic centimetre of a material. Each molecule of water contains two hydrogen atoms and one oxygen atom. Example 1.25 × 109 ÷ (365 × 24 × 3600) years = 198 years.02 × 1023 molecules) of water is 18 g. we will need approximately 200 years. to verify Newton’s Law of Gravitation. Hence the total positive and total negative charge has the same magnitude. It is.34 × 107 C. the magnitude of the force (F) between them is given by (1. a very large unit for many practical purposes.5 × 10 24 electrons.3 How much positive and negative charge is there in a cup of water? Solution Let us assume that the mass of one cup of water is 250 g. q2 are separated by a distance r in vacuum.6 × 10–19 × 109 C = 1. if two point charges q1. Thus. The time required to accumulate a charge of 1 C can then be estimated to be 1 C ÷ (1.2 If 109 electrons move out of a body to another body every second.1) r2 How did Coulomb arrive at this law from his experiments? Coulomb used a torsion balance* for measuring the force between two charged metallic F =k * A torsion balance is a sensitive device to measure force. how much time is required to get a total charge of 1 C on the other body? Solution In one second 109 electrons move out of the body. the size may be ignored and the charged bodies are treated as point charges. EXAMPLE 1. r = 1 m anticipated by Priestley and 9 also by Cavendish earlier.1). a simple mathematical statement. One coulomb is between unlike and like evidently too big a unit to be used. arrived in 1785 at the The unit of charge that results from this choice is called inverse square law relation. now known as Coulomb’s Putting this value of k in Eq. (1.Electric Charges and Fields spheres. (1.854 × 10–12 C2 N–1m–2 * Implicit in this is the assumption of additivity of charges and conservation: two charges (q/2 each) add up to make a total charge q. The choice of k determines the size of the unit charged spheres. a coulomb which we defined earlier in Section 1. The law had been q1 = q2 = 1 C. In the relation. How then could he discover a relation like Eq. The value of ε0 in SI units is F = ε 0 = 8. it has also been retired to a small estate to established down to subatomic level (r ~ 10–10 m). Coulomb Coulomb.1)? Coulomb thought of the following simple way: Suppose the charge on a metallic sphere is q. etc. so that Coulomb’s law is written as CHARLES AUGUSTIN DE COULOMB (1736 –1806) q1 q2 1 (1. When the separation between two spheres is much larger than the radius of each sphere.4. In 1776. q/4. However. While the original experiments returned to Paris and established it at a macroscopic scale. physicist. as a military engineer in was initially experimentally arrived at in the manner the West Indies. Coulomb varied the distance for a fixed pair of charges and measured the force for different separations. the value of k is about 9 × 10 . Repeating this process.1) is usually put as k = 1/4πε0 for later convenience. In fact. Coulomb discovered his law without knowing the He invented a torsion explicit magnitude of the charge. we can get charges q/2. he described above. the charge will spread over the two spheres. one uses smaller units like 1 mC or 1 µC. He thus 9 of charge. do his scientific research. we see that for law. Comparing forces for Coulomb (1736 – 1806) different pairs of charges at different distances.2) 4 π ε0 r2 ε0 is called the permittivity of free space . the charge on each sphere will be q/2*. to begin with. the charged spheres may be regarded as point charges. began his career Coulomb’s law. (1. (1. electrostatics. (1. k is so far arbitrary. In practice. distance of 1 m from another charge of the same Coulomb also found the magnitude in vacuum experiences an electrical force of inverse square law of force repulsion of magnitude 9 × 109 N.1). keeping Charles Augustin de the distance fixed for each pair. in magnetic poles. He then varied the charges in pairs. the charges on the spheres were unknown. forces of electric attraction Eq. 11 . If the sphere is put in contact with an identical uncharged sphere. 1 C is the charge that when placed at a published his results. In SI units. By symmetry. Eq. We can choose any positive or repulsion between small value of k. a French arrived at the relation. it is the other balance to measure the quantity of a force and used way round: Coulomb’s law can now be employed to it for determination of furnish a definition for a unit of charge.1). F = 9 × 10 N though Cavendish never That is. The constant k in Eq. 3). respectively (r12 = r21). • The force F12 on charge q1 due to charge q2.3) is valid for any sign of q1 and q2 whether positive or negative.6(b)]. F12 = q1 q2 1 ˆ12 = − F21 r 2 4 π ε0 r12 Thus. we define the unit vectors: FIGURE 1. We shall consider electrostatics in matter in the next chapter. Coulomb’s law agrees with the Newton’s third law. If q1 and q2 are of opposite signs. F21 is along – r ˆ 21(= r ˆ 12). ˆ21 = r r r21 ˆ12 = 12 . by simply interchanging 1 and 2..e. 1.1. as it should be for negative). If q1 and q2 are of the same sign (either both positive or both ˆ 21. the vector leading from 2 to 1 is denoted by r12: r12 = r1 – r2 = – r21 The magnitude of the vectors r21 and r12 is denoted by r21 and r12 .3) are relevant: • Equation (1. Equation (1. Thus.3) takes care of both cases correctly [Fig.3)] gives the force between two charges q1 and q2 in vacuum. The direction of a vector is specified by a unit vector along the vector.6(a)]. To denote the direction from 1 to 2 (or from 2 to 1).6 (a) Geometry and (b) Forces between charges. we do not have to write separate equations for the cases of like and unlike charges. If the charges are placed in matter or the intervening space has matter. r12 r21 ˆ21 = r ˆ12 r Coulomb’s force law between two point charges q1 and q2 located at r1 and r2 is then expressed as F21 = q1 q 2 1 ˆ21 r 2 4 π εo r21 (1. as expected for unlike charges. 12 . (1. Let the position vectors of charges q1 and q2 be r1 and r2 respectively [see Fig. which denotes repulsion. i. which denotes attraction. is obtained from Eq. We denote force on q1 due to q2 by F12 and force on q2 due to q1 by F21.Physics Since force is a vector. The two point charges q1 and q2 have been numbered 1 and 2 for convenience and the vector leading from 1 to 2 is denoted by r21: r21 = r2 – r1 In the same way. (1. • Coulomb’s law [Eq. the situation gets complicated due to the presence of charged constituents of matter. r . F21 is along r like charges.3) Some remarks on Eq. (1. it is better to write Coulomb’s law in the vector notation. the magnitude of force is | F| = 1 e2 = 8. Thus.11 ×10–31 kg = 2. The actual values of these forces between two protons inside a nucleus (distance between two protons is ~ 10-15 m inside a nucleus) are Fe ~ 230 N whereas FG ~ 1.3×10–8 N / 9. the acceleration that an electron will undergo is a = 2.html Interactive animation on Coulomb’s law: 1 e2 4 πε0 r 2 where the negative sign indicates that the force is attractive.9 × 10–34 N.3 × 10–8 N / 1. we can conclude that the effect of gravitational field is negligible on the motion of electron and it undergoes very large accelerations under the action of Coulomb force due to a proton.67 × 10–27 kg.67 × 10–27 kg = 1. the gravitational force is attractive in nature and the Coulomb force is repulsive .Electric Charges and Fields Example 1. me = 9. however the masses of an electron and a proton are different.5 × 1022 m/s2 Comparing this with the value of acceleration due to gravity. The value for acceleration of the proton is 2. (a) Compare the strength of these forces by determining the ratio of their magnitudes (i) for an electron and a proton and (ii) for two protons. it may be mentioned here that the signs of the two forces are different.edu/physlet_resources/bu_semester2/menu_semester2.11 × 10–31 kg) Solution (a) (i) The electric force between an electron and a proton at a distance r apart is: http://webphysics.davidson.4 × 1039 FG 4 πε0Gm pm e (ii) On similar lines. (b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are 1 Å (= 10-10 m) apart? (mp = 1. F = ma. FG = −G m p me Fe e2 = = 2. both have inverse-square dependence on the distance between the charges/masses. The (dimensionless) ratio of the two forces shows that electrical forces are enormously stronger than the gravitational forces.4 13 .987 × 109 Nm2/C2 × (1. the ratio of the magnitudes of electric force to the gravitational force between two protons at a distance r apart is : Fe e2 = = 1. The corresponding gravitational force (always attractive) is: Fe = − r2 where mp and me are the masses of a proton and an electron respectively. (b) The electric force F exerted by a proton on an electron is same in magnitude to the force exerted by an electron on a proton.4 × 1019 m/s2 EXAMPLE 1.6 ×10–19C)2 / (10–10m)2 4 πε 0 r 2 = 2. For two protons.4 Coulomb’s law for electrostatic force between two point charges and Newton’s law for gravitational force between two stationary point masses.3 × 1036 FG 4 πε 0Gm pm p However.3 × 10–8 N Using Newton’s second law of motion. What is the expected repulsion of A on the basis of Coulomb’s law? Spheres A and C and spheres B and D have identical sizes. Ignore the sizes of A and B in comparison to the separation between their centres.Physics Example 1. 1. 1.5 14 FIGURE 1. as shown in Fig.5 A charged metallic sphere A is suspended by a nylon thread. as shown in Fig. The resulting repulsion of A is noted (for example.0 cm between their centres.7(a). Spheres A and B are touched by uncharged spheres C and D respectively. Another charged metallic sphere B held by an insulating handle is brought close to A such that the distance between their centres is 10 cm. 1. EXAMPLE 1. by shining a beam of light and measuring the deflection of its shadow on a screen).7 .7(b).7(c). C and D are then removed and B is brought closer to A to a distance of 5. as shown in Fig. each sphere carries a charge q/2. the magnitude of the electrostatic force on each is EXAMPLE 1. the force on q1 due to q3. due to B. qn in vacuum. q3. To better understand the concept. .8 A system of (a) three charges (b) multiple charges. as shown in Fig. q2. 1. The force on one charge. 1. the magnitude of the electrostatic force on each is given by F = 1 qq ′ 4 πε 0 r 2 neglecting the sizes of spheres A and B in comparison to r. 15 . .. q3. F12 = F13 = 1 q1q 3 ˆ13 r 2 4 πε 0 r13 FIGURE 1. F12 is given by Eq. Now. What is the force on q1 due to q2.7 FORCES BETWEEN MULTIPLE CHARGES The mutual electric force between two charges is given by Coulomb’s law. q2 and q3. q3 can therefore be obtained by performing a vector addition of the forces due to each one of these charges... (1. if the force on q1 due to q2 is denoted by F12. The individual forces are unaffected due to the presence of other charges. is given by Thus. How to calculate the force on a charge where there are not one but several charges around? Consider a system of n stationary charges q1. if the separation between A and B is halved. When an identical but uncharged sphere C touches A. 1 q1q 2 ˆ12 r 2 4 πε 0 r12 In the same way. Recall that forces of mechanical origin add according to the parallelogram law of addition. remains unaltered. due to two other charges q2. Thus..8(a). This is termed as the principle of superposition. the charges redistribute on A and C and.Electric Charges and Fields Solution Let the original charge on sphere A be q and that on B be q ′ . qn? Coulomb’s law is not enough to answer this question.3) even though other charges are present.. say q1. Similarly.5 F′ = 1 (q /2)(q ′ /2) 1 (qq ′ ) = =F 2 4πε 0 4 πε 0 r 2 (r /2) Thus the electrostatic force on A.. consider a system of three charges q1. by symmetry. denoted by F13. Is the same true for forces of electrostatic origin? Experimentally it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges. after D touches B. the redistributed charge on each is q′/2. taken one at a time. At a distance r between their centres. 6 FIGURE 1.Physics which again is the Coulomb force on q1 due to q3. if we draw a perpendicular AD to the side BC.. = Example 1. 1.. + 12 n r ˆ1n   2 r r13 r1n  r12  q1 n qi ˆ1i ∑ r (1.e. .+ F1n = 1 4 πε 0  q1q 2  qq qq ˆ12 + 12 3 r ˆ13 + . is then given by the vector sum of the forces F12. qn.. q4. . What is the force on a charge Q (with the same sign as q ) placed at the centroid of the triangle. it is unaffected by the presence of the other charges q3. The total force F1 on the charge q1. qn.. q3 each equal to q at the vertices of an equilateral triangle of side l.8(b)...e. the force on q1 due to q2 is the same as given by Coulomb’s law. All of electrostatics is basically a consequence of Coulomb’s law and the superposition principle. as shown in Fig. even though other charge q2 is present. AD = AC cos 30º = ( 3 /2 ) l and the distance AO of the centroid O from A is (2/3) AD = ( 1/ 3 ) l.. Thus the total force F1 on q1 due to the two charges q2 and q3 is given as F1 = F12 + F13 = 1 q1q 2 1 q1q3 ˆ12 + ˆ13 r r 2 2 4 πε 0 r12 4 πε 0 r13 (1... q2.9? EXAMPLE 1. F1 = F12 + F13 + . . . as shown in Fig.. q2..6 Consider three charges q1. i. F13.. due to all other charges. 1.9 16 Solution In the given equilateral triangle ABC of sides of length l.4) The above calculation of force can be generalised to a system of charges more than three. The principle of superposition says that in a system of charges q1. By symmatry AO = BO = CO.. F1n: i...5) 4 πε0 i = 2 r12 i The vector sum is obtained as usual by the parallelogram law of addition of vectors. Therefore.Electric Charges and Fields Thus. where r It is clear also by symmetry that the three forces will sum to zero. q. 1.7 Consider the charges q. Suppose that the resultant force was non-zero but in some direction. the total force on Q = 4πε 2 ( r 0 l EXAMPLE 1. 3 4πε 0 3 Force F2 on Q due to charge q at B = 4πε 0 Force F1 on Q due to charge q at A = Qq along AO l2 Qq along BO l2 3 Qq Force F3 on Q due to charge q at C = 4πε 2 along CO 0 l 3 Qq The resultant of forces F 2 and F 3 is 4πε 2 along OA. What is the force on each charge? FIGURE 1. 17 . where r ˆ 2 is a The total force F2 on charge q at B is thus F2 = F r unit vector along AC.7 ˆ 2. By the parallelogram law. the total force F1 on the charge q at A is given by ˆ1 where r ˆ1 is a unit vector along BC. = 0.10. as shown in Fig. as shown in Fig.6 ˆ is the unit vector along OA. 1. Example 1. and –q placed at the vertices of an equilateral triangle.10 Solution The forces acting on charge q at A due to charges q at B and –q at C are F12 along BA and F13 along AC respectively. F1 = F r The force of attraction or repulsion for each pair of charges has the same magnitude F = q2 4 π ε0 l 2 EXAMPLE 1.10. Consider what would happen if the system was rotated through 60º about O. by the 0 l 3 Qq ˆ−r ˆ) parallelogram law. the FIGURE 1. i.6) ˆ = r/r. we say that the charge Q produces an electric field everywhere in the surrounding. Eq. Thus. at the origin O. the field there acts on it and produces a force.7) 4 πε 0 r 2 Note that the charge q also exerts an equal and opposite force on the charge Q. the early scientists introduced the concept of field. In order to answer such questions. the electric field due to a charge Q is numerically equal to the force exerted by it. (b) due to a charge –Q. it experiences a force F equal to the charge q multiplied by the electric field E at the location of q. is a unit vector from the origin to the point r. . The electric field produced by the charge Q at a point r is given as E (r) = 1 Q 1 Q ˆ= ˆ r r 2 4 πε 0 r 4 πε 0 r 2 (1.. If we denote the position of charge q by the vector r.e.6) where r specifies the value of the electric field for each value of the position vector r. where OP = r. The electrostatic force between the charges Q and q can be looked upon as an interaction between charge q and the electric field of Q and vice versa. then what is left in the surrounding? Is there nothing? If there is nothing at the point P.8) defines the SI unit of electric field as N/C*. If we place another point charge q at a point P. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law. then how does a force act when we place the charge q at P.11 Electric electric field due to a charge Q at a point in space may be defined field (a) due to a as the force that a unit positive charge would experience if placed charge Q.Physics ˆ . When another charge q is brought at some point P. Thus. We may ask the question: If charge q is removed. F1 + F2 + F3 = 0 The result is not at all surprising. we can infer that if q is unity.8) Equation (1. where n ˆ is Similarly the total force on charge –q at C is F3 = 3 F n the unit vector along the direction bisecting the ∠BCA. as 1 Qq ˆ F= r (1.7 1. EXAMPLE 1. (1. It is interesting to see that the sum of the forces on the three charges is zero.8 ELECTRIC FIELD Let us consider a point charge Q placed in vacuum.8). Thus.(1. then the charge Q will exert a force on q as per Coulomb’s law. We obtain the force F exerted by a charge Q on a charge q. Some important remarks may be made here: (i) From Eq. According to this. The proof is left to you as an exercise. The word “field” signifies how some distributed quantity (which could be a scalar or a vector) varies with position. The effect of the charge has been incorporated in the existence of the electric field. 18 * An alternate unit V/m will be introduced in the next chapter. F(r) = q E(r) (1. q2. at each point.9) A practical way to get around the problem (of keeping Q undisturbed in the presence of q) is to hold Q to its location by unspecified forces! This may look strange but actually this is what happens in practice. For different positions of the charge q all over the space. which is producing the electric field. Like the electric field at a point in space due to a single charge. The force F on the charge q due to the charge Q depends on the particular location of charge q which may take any value in the space around the charge Q. in other words.8.. Note that the source charge Q must remain at its original location. the electric field E due to Q is also dependent on the space coordinate r. if a charge q is brought at any point around Q. if the source charge is negative. (iv) Since the magnitude of the force F on charge q due to charge Q depends only on the distance r of the charge q from charge Q. However... we get different values of electric field E. The magnitude of electric field E due to a point charge is thus same on a sphere with the point charge at its centre. qn with position vectors r1.. r2. the electric field will be directed radially outwards from the charge. 19 .. is independent of q.1 Electric field due to a system of charges Consider a system of charges q1. (ii) Note that the electric field E due to Q. rn relative to some origin O. the electric field vector. which tests the effect of a source charge. Thus. is called a source charge and the charge q. Q itself is bound to experience an electrical force due to q and will tend to move. so the ratio F/q does not depend on q. electric field at a point in space due to the system of charges is defined to be the force experienced by a unit test charge placed at that point. We can use Coulomb’s law and the superposition principle to determine this field at a point P denoted by position vector r. . without disturbing the original positions of charges q1.. A way out of this difficulty is to make q negligibly small. q2. points radially inwards. it has a spherical symmetry. though defined operationally in terms of some test charge q. The charge Q. the magnitude of its electric field E is same.. (iii) For a positive charge. .15). The field exists at every point in three-dimensional space. Thus at equal distances from the charge Q. When we are considering the electric force on a test charge q due to a charged planar sheet (Section 1. qn. is called a test charge.. This is because F is proportional to q. the charges on the sheet are held to their locations by the forces due to the unspecified charged constituents inside the sheet. On the other hand..Electric Charges and Fields at that point. the magnitude of the electric field E will also depend only on the distance r. 1. . The force F is then negligibly small but the ratio F/q is finite and defines the electric field:  F E = lim   q →0  q  (1. for any system of charges.Physics Electric field E1 at r due to q1 at r1 is given by E1 = 1 q1 ˆ1P r 2 4 πε 0 r1P ˆ1P is a unit vector in the direction from q1 to P. By the superposition principle. Electric field is a vector field. 1.12) E(r) = E1 (r) + E2 (r) + … + En(r) = 1 q1 1 q2 1 qn ˆ + ˆ2P + . the electric field E at r due to the system of charges is (as shown in Fig.2 Physical significance of electric field You may wonder why the notion of electric field has been introduced here at all. where r and r1P is the distance between q1 and P. but not really necessary. q4. . The term field in physics generally refers to a quantity that is defined at every point in space and may vary from point to point. After all.10) E is a vector quantity that varies from one point to another point in space and is determined from the positions of the source charges. Similar expressions hold good for fields E3. qn. + ˆnP r r r 2 1P 2 4 πε 0 r1P 4 πε 0 r2P 4 πε 0 rn2P 1 4πε 0 ˆi P ∑ r 2i r i =1 i P n E(r) = q (1. En due to charges q3. the effect of any motion of q1 on q2 cannot 20 . the speed of light. .... Suppose we consider the force between two distant charges q1. E4. 1. Electric field is a characteristic of the system of charges and is independent of the test charge that you place at a point to determine the field.8... Thus. the measurable quantity is the force on a charge which can be directly determined using Coulomb’s law and the superposition principle [Eq. ˆ2P is a unit vector in the direction from q2 to P where r and r 2P is the distance between q 2 and P. (1. emerges only when we go beyond electrostatics and deal with timedependent electromagnetic phenomena... however. Why then introduce this intermediate quantity called the electric field? For electrostatics. Electric field is an elegant way of characterising the electrical environment of a system of charges. The true physical significance of the concept of electric field. In the same manner.5)].12 Electric field at a point due to a system of charges is the vector sum of the electric fields at the point due to individual charges. the concept of electric field is convenient. q2 in accelerated motion. electric field E2 at r due to q2 at r2 is E2 = 1 q2 ˆ2P r 4 πε 0 r22P FIGURE 1.. Now the greatest speed with which a signal or information can go from one point to another is c. since force is a vector quantity. Electric field at a point in the space around a system of charges tells you the force a unit positive test charge would experience if placed at that point (without disturbing the system). The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance [Fig. They have an independent dynamics of their own. not merely mathematical constructs.5 × 10–2 m. the time required by the electron to fall through a distance h is given by t e = 2h = ae 2h me eE For e = 1. The acceleration of the electron is ae = eE/me where me is the mass of the electron. The time of fall for the proton is ap = eE/mp EXAMPLE 1. Compute the time of fall in each case. so the negatively charged electron experiences a downward force of magnitude eE where E is the magnitude of the electric field.13(b)]. i. leaves behind propagating electromagnetic fields transporting energy.13 Solution In Fig.6 × 10–19C. 1.0 × 104 N C–1. It is precisely here that the notion of electric field (strictly.13 (b). and the positively charged proton experiences a downward force of magnitude eE .Electric Charges and Fields arise instantaneously. turned on briefly and switched off.8 An electron falls through a distance of 1.11 × 10–31 kg. There will be some time delay between the effect (force on q2) and the cause (motion of q1). h = 1. 1. they evolve according to laws of their own. The acceleration of the proton is where mp is the mass of the proton. Example 1. Thus.13(a) the field is upward. Thus. Starting from rest. 1.0 × 104 N C–1 [Fig. which then propagate with the speed c. reach q2 and cause a force on q2. te = 2. The notion of field elegantly accounts for the time delay. 1. even though electric and magnetic fields can be detected only by their effects (forces) on charges. mp = 1.. me = 9. The concept of field was first introduced by Faraday and is now among the central concepts in physics. FIGURE 1. a source of timedependent electromagnetic fields.13(a)]. electromagnetic field) is natural and very useful. Contrast the situation with that of ‘free fall under gravity’. they are regarded as physical entities. They can also transport energy. the field is downward. The field picture is this: the accelerated motion of charge q1 produces electromagnetic waves.67 × 10–27 kg.9 × 10–9s In Fig.8 21 .5 cm in a uniform electric field of magnitude 2. E = 2.e. 9 × 1012 m s –2 which is enormous compared to the value of g (9. of magnitude +10–8 C and –10–8 C.9 The electric field vector E2A at A due to the negative charge q2 points towards the right and has the same magnitude. the effect of acceleration due to gravity can be ignored in this example. the heavier particle (proton) takes a greater time to fall through the same distance.9 Two point charges q1 and q2.3 × 10 –7 s Thus. Thus. Hence the magnitude of the total electric field EA at A is EA = E1A + E2A = 7. The acceleration of the electron is even greater.8 = 1. 22 .1 m apart. This is in basic contrast to the situation of ‘free fall under gravity’ where the time of fall is independent of the mass of the body. respectively.05 m)2 EXAMPLE 1.0 × 104 N C −1 ) 1.Physics tp = 2h = ap 2h m p eE = 1. the acceleration due to gravity.6 × 10−19 C) × (2.2 × 104 N C–1 EA is directed toward the right.67 × 10 −27 kg = EXAMPLE 1.8 m s –2) . are placed 0. B and C shown in Fig.6 × 104 N C–1 (0. 1. FIGURE 1. let us calculate the acceleration of the proton in the given electric field: ap = eE mp (1. Calculate the electric fields at points A.14 Solution The electric field vector E1A at A due to the positive charge q1 points towards the right and has a magnitude E1A = (9 × 109 Nm 2 C-2 ) × (10−8 C) = 3. To see if this is justified.14. Example 1. Note that in this example we have ignored the acceleration due to gravity in calculating the time of fall. In this figure. an infinite number of lines can be drawn in any region. 1. We thus get many field lines.6 × 104 N C–1 (0. Let the point charge be placed at the origin. Have we lost the information about the strength or magnitude of the field now. It is a vector quantity and can be represented as we represent vectors. The magnitude of each electric field vector at point C.14. so the density of field lines is more near the charge and the lines are closer. . resulting in well-separated lines.2 × 104 N C–1 EB is directed towards the left. the force acting on a unit positive charge.10 m)2 The directions in which these two vectors point are indicated in Fig. The resultant of these two vectors is π π + E 2 cos = 9 × 103 N C–1 3 3 EC points towards the right.e. due to charge q1 and q2 is E1C = E 2C = (9 × 109 Nm 2 C –2 ) × (10 −8 C) = 9 × 103 N C–1 (0. E is strong near the charge.9 ELECTRIC FIELD LINES We have studied electric field in the last section. i. all pointing outwards from the point charge. Connect the arrows pointing in one direction and the resulting figure represents a field line. because it was contained in the length of the arrow? No. EC = E1 cos EXAMPLE 1.15 Field of a point charge. Now the magnitude of the field is represented by the density of field lines. Draw vectors pointing along the direction of the electric field with their lengths proportional to the strength of the field at each point. FIGURE 1.05 m)2 The electric field vector E2B at B due to the negative charge q2 points towards the right and has a magnitude E 2B = (9 × 109 Nm 2 C –2 ) × (10 −8 C) = 4 × 103 N C–1 (0. Since the magnitude of electric field at a point decreases inversely as the square of the distance of that point from the charge. Figure 1.Electric Charges and Fields The electric field vector E1B at B due to the positive charge q1 points towards the left and has a magnitude E1B = (9 × 109 Nm 2 C –2 ) × (10−8 C) = 3..15 m)2 The magnitude of the total electric field at B is EB = E1B – E2B = 3. the vector gets shorter as one goes away from the origin. each arrow indicates the electric field. placed at the tail of that arrow. But the number of lines is not 23 important. In fact.9 1. Let us try to represent E due to a point charge pictorially. Another person may draw more lines. always pointing radially outward.15 shows such a picture. the field gets weaker and the density of field lines is less. Away from the charge. in three-dimensions the solid angle* subtended by a small perpendicular plane area ∆S . Figure 1. We started by saying that the field lines carry information about the direction of electric field at different points in space.16. This term is somewhat misleading. the number of field lines crossing the enclosing area remains constant. The solid angle ∆Ω of the cone is defined to be equal 2 to ∆S/R . The picture of field lines was invented by Faraday to develop an intuitive non. let us try to relate the area with the solid angle. at a distance r . We can imagine two equal and small elements of area placed at points R and S normal to the field lines there. Since n and ∆Ω are common. especially in case of magnetic fields. The picture shows that the field at R is stronger than at S. can be written as ∆Ω = ∆S/r2. The more appropriate term is field lines (electric or magnetic) that we have adopted in this book. An electric field line is. cutting unit area element is therefore n/( r12 ∆Ω) at 2 P1 andn/( r2 ∆Ω) at P2 . Consider the intersection of the given cone with a sphere of radius R.e. To understand the dependence of the field lines on the area. a generalization of angle to three dimensions. Since the electric field decreases as the square of the distance from a point charge and the area enclosing the charge increases as the square of the distance. the strength of the field clearly has a 1/r 2 dependence. closeness) of the field lines at different points indicates distance and its relation to the the relative strength of electric field at those points. lines crowd where the field is strong and are spaced apart where it is weak.16 shows a set of field lines. Likewise.16 Dependence of Having drawn a certain set of field lines. one has to consider the number of lines per unit cross-sectional area. Then the angle subtended by ∆l at O can be approximated as ∆θ = ∆l/r.. 24 . 1. Recall how a (plane) angle is defined in two-dimensions.. We draw the figure on the plane of paper. In Fig. i. The number of lines (say n) cutting these area elements are the same. FIGURE 1. or rather the solid angle subtended by an area element. Electric field lines are thus a way of pictorially mapping the electric field around a configuration of charges. Let a small transverse line element ∆l be placed at a distance r from a point O. perpendicular to the lines. Faraday called them lines of force. whatever may be the distance of the area from the charge. where ∆S is the area on the sphere cut out by the cone.mathematical way of visualizing electric fields around charged configurations. respectively. The field number of field lines. respectively. The number of field lines in our picture cutting the area elements is proportional to the magnitude of field at these points. the relative density electric field strength on the (i. in general. The number of field lines.e. So if one wishes to estimate the density of field lines. in twodimensions but we live in three-dimensions. * Solid angle is a measure of a cone. the element of area subtending the solid angle 2 ∆Ω is r12 ∆Ω at P1 and an element of area r2 ∆Ω at P2. for two points P1 and P2 at distances r1 and r2 from the charge. We know that in a given solid angle the number of radial field lines is the same.Physics It is the relative density of lines in different regions which is important. out of the surface dS is v. (ii) In a charge-free region. In the picture of electric field lines described above. the field lines are in 3-dimensional space. If the normal to the surface is not parallel to the direction of flow of liquid.) (iv) Electrostatic field lines do not form any closed loops. The field lines of a single positive charge are radially outward while those of a single negative charge are radially inward. we define an analogous quantity and call it electric flux. 1. placed normal to the field at a point is a measure of the strength of electric field at that point. Figure 1. i. though the figure shows them only in a plane. The field lines follow some important general properties: (i) Field lines start from positive charges and end at negative charges. the field at the point of intersection will not have a unique direction.10 ELECTRIC FLUX Consider flow of a liquid with velocity v. If there is a single charge. As mentioned earlier. 25 .. An arrow on the curve is obviously necessary to specify the direction of electric field from the two possible directions indicated by a tangent to the curve. which is absurd. a dipole.17 shows the field lines around some simple charge configurations. but makes an angle θ with it. while those around the configuration of two equal and opposite charges (q. (iii) Two field lines can never cross each other. electric field lines can be taken to be continuous curves without any breaks..17 Field lines due to some simple charge configurations. (If they did. We should however note that there is no flow of a physically observable quantity unlike the case of liquid flow. Therefore the flux going ˆ dS. This follows from the conservative nature of electric field (Chapter 2). we saw that the number of field lines crossing a unit area. n For the case of the electric field. the projected area in a plane perpendicular to v is v dS cos θ. –q). in a direction normal to the surface. to v.e. a curve in three dimensions. they may start or end at infinity. A field line is a space curve. The field lines around a system of two positive charges ( q .e. i. The rate of flow of liquid is given by the volume crossing the area per unit time v dS and represents the flux of liquid flowing across the plane. through a small flat surface dS. This means that if FIGURE 1.Electric Charges and Fields a curve drawn in such a way that the tangent to it at each point is in the direction of the net field at that point. show clearly the mutual attraction between the charges. q ) give a vivid pictorial description of their mutual repulsion. 1. Clearly. If you hold it normal to the flow. The projection of the area element normal to E is ∆S cosθ. as seen before. the amount of water flowing through a ring will naturally depend on how you hold the ring. n We now come to the definition of electric flux.e. Thus. Electric flux ∆φ through an area element ∆S is defined by ∆φ = E. It has a magnitude and also a direction. But a normal can point in two directions. the number of field lines crossing ∆ S is proportional to E ∆S cosθ.19. This shows that an area element should be treated as a vector. E times the * It will not be proper to say that the number of field lines is equal to E ∆S. the number of field lines crossing the area element will be smaller. Notice one ambiguity here. in a stream. inclination θ between E and n orientation. this convention is very simple. How to associate a vector to the area of a curved surface? We imagine dividing the surface into a large number of very small area elements. For a closed surface.18 Dependence of flux on the through it than if you hold it with some other ˆ. Which direction do we choose as the direction of the vector associated with the area element? This problem is resolved by some convention appropriate to the given context. The vector associated with every area element of a closed surface is taken to be in the direction of the outward normal. field lines will be parallel to ∆S and will not cross it at all (Fig.18). Thus the direction of a planar area vector is along its normal. as explained before. Now suppose we tilt the area element by angle θ. How to specify the direction of a planar area? Clearly. a matter of how many field lines we choose to draw. For example. Notice we could look at the expression E ∆S cosθ in two ways: E (∆S cosθ ) i. n which. When θ = 90°. maximum water will flow FIGURE 1. 1. the normal to the plane specifies the orientation of the plane. The angle θ here is the angle between E and ∆S.Physics we place a small planar element of area ∆S normal to E at a point. 26 . is proportional to the number of field lines cutting the area element. What is physically significant is the relative number of field lines crossing a given area at different points. For the case of a closed surface.∆S = E ∆S cosθ (1. Thus. Each small area element may be treated as planar and a vector associated with it. θ is the angle between E and the outward normal to the area element. This is the convention used in Fig. The direction of an area element is along its normal. The number of field lines is after all.. the area element vector ∆S at a point on a closed ˆ where ∆S is the magnitude of the area element and surface equals ∆S n ˆ is a unit vector in the direction of outward normal at that point.19 Convention for defining normal ˆ and ∆S. with the convention stated already.11) FIGURE 1. The orientation of area element and not merely its magnitude is important in many contexts. the number of field lines crossing it is proportional * to E ∆S . component of E along the normal to the area element times the magnitude of the area element.e. Since the charge q and –q are separated by some distance. Also where p E +q = q ˆ p 4 π ε 0 (r − a )2 [1. the electric fields due to them. the direction from –q to q is said to be the direction of the dipole. as shown in Fig. on a plane perpendicular to the dipole axis through its centre. the fields due to q and –q nearly cancel out. and (ii) when it is in the equatorial plane of the dipole. However. at distances much larger than the separation of the two charges forming a dipole (r >> 2a).11 ELECTRIC DIPOLE An electric dipole is a pair of equal and opposite point charges q and –q. The total charge of the electric dipole is obviously zero. The mid-point of locations of –q and q is called the centre of the dipole. by the parallelogram law of vectors.12) is written as an integral. i. when added. Thus. separated by a distance 2a. i. in principle. the total flux φ through a surface S is φ ~ Σ E. The line connecting the two charges defines a direction in space. 1. By convention.13(a)] 4 πε 0 (r + a )2 ˆ is the unit vector along the dipole axis (from –q to q).11.∆S (1. The electric field due to a dipole therefore falls off. This does not mean that the field of the electric dipole is zero. 1.e. calculate the flux at each element and add them up. at large distance. to calculate the total flux through any given surface. or E⊥ ∆S.11) can be used.12) The approximation sign is put because the electric field E is taken to be constant over the small area element. (1. The electric field at any general point P is obtained by adding the electric fields E–q due to the charge –q and E+q due to the charge q. Then q ˆ E −q = − p [1. This is mathematically exact only when you take the limit ∆S → 0 and the sum in Eq. faster than like 1/r 2 (the dependence on r of the field due to a single charge q). (1..Electric Charges and Fields projection of area normal to E. The basic definition of electric flux given by Eq. The results are simple for the following two cases: (i) when the point is on the dipole axis. All we have to do is to divide the surface into small area elements.. (i) For points on the axis Let the point P be at distance r from the centre of the dipole on the side of the charge q.13(b)] 27 .20(a). These qualitative ideas are borne out by the explicit calculation as follows: 1. do not exactly cancel out.1 The field of an electric dipole The electric field of the pair of charges (–q and q) at any point in space can be found out from Coulomb’s law and the superposition principle. The unit of electric flux is N C–1 m2. The components along the dipole axis ˆ . p is the dipole moment vector of magnitude p = q × 2a and directed from –q to q.20 Electric field of a dipole at (a) a point on the axis. it depends on the product qa. the components normal to the dipole axis cancel away.21) .16(a)] [1. (b) a point on the equatorial plane of the dipole. the electric field of a dipole at large distances takes simple forms: At a point on the dipole axis E=− E= 2p 4 πεo r 3 p 4 πεo r 3 (r >> a) (1.18) 4 π εo r 3 From Eqs.16(b)] and are equal. it is clear that the dipole field at large distances does not involve q and a separately. The total electric field is opposite to p ˆ E = – ( E + E ) cosθ p +q –q =− 2q a ˆ p 4 π εo (r 2 + a 2 )3 / 2 (1. This suggests the definition of dipole moment. We have add up. The dipole moment vector p of an electric dipole is defined by ˆ p = q × 2a p (1.15) and (1.17) At large distances (r >> a).Physics The total field at P is E = E + q + E −q = = 1  q  1 ˆ −  p 2 4 π ε 0  (r − a ) (r + a )2  4ar q ˆ p 4 π εo ( r 2 − a 2 )2 4qa ˆ p 4 πε 0r 3 (1. this reduces to 2qa ˆ p (r >> a ) (1. 1.15) (ii) For points on the equatorial plane The magnitudes of the electric fields due to the two charges +q and –q are given by E +q = E –q = FIGURE 1.19) that is.14) For r >> a E= (r >> a) (1. it is a vector whose magnitude is charge q times the separation 2a (between the pair of charges q. In terms of p. Clearly. –q) and the direction is along the line from –q to q.20) At a point on the equatorial plane 28 E=− (r >> a) (1. (1. The directions of E +q and E –q are as shown in Fig.20(b). 1 q 2 4 πε 0 r + a 2 1 q 4 πε 0 r 2 + a 2 [1.18). EXAMPLE 1. H2O.11. 1.10 Two charges ±10 µ C are placed 5. Therefore they have a permanent electric dipole moment. CO2 and CH4 are of this type of molecules. However. Eqs. Further. the charge q approaches infinity in such a way that the product p = q × 2a is finite.20) and (1. Such molecules are called polar molecules.2 Physical significance of dipoles In most molecules.21(b). Such a dipole is referred to as a point dipole. Various materials give rise to interesting properties and important applications in the presence or absence of electric field. even in the absence of an electric field. the magnitude and the direction of the dipole field depends not only on the distance r but also on the angle between the position vector r and the dipole moment p. the centres of positive charges and of negative charges* lie at the same place.21) are exact. 1. and (b) a point Q. (1. as shown in Fig. they develop a dipole moment when an electric field is applied.Electric Charges and Fields Notice the important point that the dipole field at large distances falls off not as 1/r 2 but as1/r 3. 1. Therefore. Example 1.10 FIGURE 1. is an example of this type. as shown in Fig.0 mm apart. true for any r. For a point dipole. Determine the electric field at (a) a point P on the axis of the dipole 15 cm away from its centre O on the side of the positive charge. the centres of negative charges and of positive charges do not coincide.21 * Centre of a collection of positive point charges is defined much the same way as the centre of mass: rcm = ∑ qi ri i . But in some molecules. Water molecules.21(a). their dipole moment is zero. We can think of the limit when the dipole size 2a approaches zero. ∑ qi i 29 . 15 cm away from O on a line passing through O and normal to the axis of the dipole. 6 × 105 N C–1 (15)3 × 10 −6 m 3 along the dipole moment direction AB.854 × 10 −12 C2 N −1 m −2 ) = 3.99 × 106 N C–1 along BQ Field at Q due to charge –10 µC at A = 10−5 C −12 2 −1 −2 10 −5 C 1 4 π (8. we can expect to get approximately the same result by directly using the formula for dipole field at a point on the normal to the axis of the dipole: . from – q to q ). × 1 [15 + (0.854 × 10 −12 C N 2 −1 m ) −2 × 1 (15 − 0.25)2 × 10 −4 m 2 = 4.25)2 ] × 10 −4 m 2 = 3. Here.13 × 106 N C–1 along BP Field at P due to charge –10 µC = 1 10 –5 C × (15 + 0.854 × 10 −12 C2 N −1 m −2 ) × [152 + (0. 2a distance apart. we can expect to get approximately the same result as above by directly using the formula for electric field at a far-away point on the axis of a dipole. For a dipole consisting of charges ± q. (b) Field at Q due to charge + 10 µC at B = 4 π (8.25)2 ] × 10 −4 m 2 2 Clearly. E = 2 × 5 × 10 −8 C m 4π (8. As in (a). In this example.854 × 10 −12 C N m ) 2 −1 −2 × 1 = 2.. the ratio OP/OB is quite large (= 60). Thus. Therefore.e. which is close to the result obtained earlier.7 × 105 N C–1 along BP.854 × 10 C N m ) = 3. The direction of electric field on the dipole axis is always along the direction of the dipole moment vector (i.99 × 106 N C–1 along QA.86 × 106 N C–1 along PA The resultant electric field at P due to the two charges at A and B is = 2.33 × 105 N C–1 along BA.25)2 × 3. the resultant electric field at Q due to the two charges at A and B is EXAMPLE 1.99 × 106 N C –1 along BA 30 = 1.10 =2× 0. the components of these two forces with equal magnitudes cancel along the direction OQ but add up along the direction parallel to BA. p =10–5 C × 5 × 10–3 m = 5 × 10–8 C m Therefore.25)2 × 10 −4 m 2 4π (8. the electric field at a distance r from the centre on the axis of the dipole has a magnitude E = 2p 4 πε 0r 3 (r/a >> 1) where p = 2a q is the magnitude of the dipole moment.25 15 2 + (0.Physics Solution (a) Field at P due to charge +10 µC = 10 −5 C 4π (8. it has not been induced by E. A comb run through dry hair attracts pieces of paper. The magnitude of p × E is also p E sinθ and its direction is normal to the paper. It is easily seen that when p is parallel to E.10 = 1.23 Electric force on a dipole: (a) E parallel to p. This brings us to a common observation in frictional electricity. since E is uniform. the net force on the dipole is in the direction of decreasing field.22) τ =p×E This torque will tend to align the dipole with the field E. the charges are separated. 1.23 is self-explanatory. so let us consider the simpler situations when p is parallel to E or antiparallel to E. In addition there will. In either case. What then explains the attractive force? Taking the clue from the preceding FIGURE 1. (b) E antiparallel to p. in general. we mean that p exists irrespective of E. as shown in Fig.22. The direction of electric field in this case is opposite to the direction of the dipole moment vector. 31 .33 × 105 N C–1. but there is a net force on the dipole if E is not uniform. Again the result agrees with that obtained before. coming out of it. When the net force is zero. The net force on the dipole is zero. as we know. coming out of it. In general. When p is antiparallel to E.) There is a force qE on q and a force –qE on –q. But the paper is not charged.22 Dipole in a uniform electric field. the net torque is zero. acquires charge through friction. (1. the torque is zero.12 DIPOLE IN A UNIFORM EXTERNAL FIELD Consider a permanent dipole of dipole moment p in a uniform external field E. What happens if the field is not uniform? In that case. The comb. When p is aligned with E. each force multiplied by the arm of the couple (perpendicular distance between the two antiparallel forces). resulting in a torque on the dipole. The general case is involved.854 × 10 −12 C2 N –1 m –2 ) EXAMPLE 1. (By permanent dipole. Figure 1.Electric Charges and Fields E = p 4 π ε0 r 3 (r/a >> 1) = 1 5 × 10 −8 C m × (15)3 × 10−6 m 3 4 π (8. the force depends on the orientation of p with respect to E. Its magnitude equals the magnitude of FIGURE 1. the net force will evidently be non-zero. However. Magnitude of torque = q E × 2 a sinθ = 2 q a E sinθ Its direction is normal to the plane of the paper. the torque (couple) is independent of the origin. 1. Thus. be a torque on the system as before. so the forces act at different points. the dipole has a net force in the direction of increasing field. it is impractical to work in terms of discrete charges and we need to work with continuous charge distributions.23) ∆S We can do this at different points on the conductor and thus arrive at a continuous function σ.. In this situation. The units for λ are C/m. The units for σ are C/m2. on the surface of a charged conductor. charge distribution is discontinuous. the electric field due to the comb is not uniform. Similar considerations apply for a line charge distribution and a volume charge distribution. We then define a surface charge density σ at the area element by ∆Q (1.13 CONTINUOUS CHARGE DISTRIBUTION We have so far dealt with charge configurations involving discrete charges q1. The surface charge density σ so defined ignores the quantisation of charge and the discontinuity in charge distribution at the microscopic level*. is large microscopically but small macroscopically. i. ∆V ) chosen is small on the macroscopic scale but contains a very large number of microscopic constituents. For example. . however. ∆Q (1. 1. σ represents macroscopic surface charge density.Physics discussion.. it is impractical to specify the charge distribution in terms of the locations of the microscopic charged constituents.. Further. The notion of continuous charge distribution is similar to that we adopt for continuous mass distribution in mechanics.e. surface and volume charge densities. and ∆Q is the charge contained in that line element. the element (∆l.25) ∆V where ∆Q is the charge included in the macroscopically small volume element ∆V that includes a large number of microscopic charged constituents. One reason why we restricted to discrete charges is that the mathematical treatment is simpler and does not involve calculus. The linear charge density λ of a wire is defined by σ= FIGURE 1. q2. It is more feasible to consider an area element ∆S (Fig.. includes a large number of microscopic charged constituents. ∆S. called the surface charge density. qn. The units for ρ are C/m3. When we refer to ρ= 32 * At the microscopic level. In each case. . The volume charge density (sometimes simply called charge density) is defined in a similar manner: λ = ∆Q (1. it is easily seen that the paper should move in the direction of the comb! 1. is a smoothed out average of the microscopic charge density over an area element ∆S which. which in a sense. induces a net dipole moment in the direction of field. For many purposes. because they are discrete charges separated by intervening space where there is no charge.24 Definition of linear.24) ∆l where ∆l is a small line element of wire on the macroscopic scale that. however. as said before.24) on the surface of the conductor (which is very small on the macroscopic scale but big enough to include a very large number of electrons) and specify the charge ∆Q on that element. the charged comb ‘polarizes’ the piece of paper. discrete or continuous or part discrete and part continuous. using Coulomb’s law and the superposition principle. which encloses a point charge q at its centre. since the normal to a sphere at every point is ˆ have along the radius vector at that point. In short. The charge density ρ may vary from point to point.26) 4 πε0 r' 2 ˆ ′ is a where r′ is the distance between the charge element and P. Divide the charge distribution into small volume elements of size ∆V. for simplicity. The flux through an area element ∆S is ˆ i ∆S r (1. The unit vector r the area element. Suppose a continuous charge distribution in space has a charge density ρ. but we omit that discussion here. 1. r′. electric field can be determined for any charge distribution. it is a function of r. (1.14 GAUSS’S LAW As a simple application of the notion of electric flux. In a strict mathematical method. The total flux through the sphere is obtained by adding up flux through all the different area elements: ∆φ = q ∆S FIGURE 1.27) 4 πε 0 all ∆V r' 2 Note that ρ. The field due to a continuous charge distribution can be obtained in much the same way as for a system of discrete charges.Electric Charges and Fields the density of a liquid.e. Electric field due to the charge ρ∆V is given by Coulomb’s law: ρ ∆V 1 ˆ' r (1.10). consider any general point P (inside or outside the distribution) with position vector R (Fig. Divide the sphere into small area elements. Now. We regard it as a continuous fluid and ignore its discrete molecular constitution. The charge in a volume element ∆V is ρ∆V.24). 1. let us consider the total flux through a sphere of radius r.29) enclosing a point charge q at its centre.. the total electric field due to the charge distribution is obtained by summing over electric fields due to different volume elements: ∆E = ρ ∆V 1 ˆ' Σ r (1.25 Flux through a sphere (1. Eq. i. and r unit vector in the direction from the charge element to P. we are referring to its macroscopic density. By the superposition principle. Now.28) 4 πε 0 r 2 where we have used Coulomb’s law for the electric field due to a single ˆ is along the radius vector from the centre to charge q. the area element ∆S and r the same direction.25. ∆φ = E i ∆ S = 4 πε 0 r 2 since the magnitude of a unit vector is 1. Therefore. E≅ 1. q 33 . as shown in Fig. Choose any convenient origin O and let the position vector of any point in the charge distribution be r. we should let ∆V→0 and the sum then becomes an integral. r ˆ ′ all can vary from point to point. Physics φ= Σ q 4 π ε0 r 2 all ∆S ∆S Since each area element of the sphere is at the same distance r from the charge. so by definition of flux. Thus. The great significance of Gauss’s law Eq. the outward normal to 2 is along E while the outward normal to 1 is opposite to E.31). Let us note some important points regarding this law: (i) Gauss’s law is true for any closed surface. the total area of the sphere. represents only the total charge inside S.31)] is due to all the charges. The term q on the right side of Gauss’s law.31). We state Gauss’s law without proof: Electric flux through a closed surface S = q/ε0 (1. (1. where φ1 and φ2 represent the flux through the surfaces 1 and 2 (of circular cross-section) of the cylinder and φ3 is the flux through the curved cylindrical part of the closed surface. 1. the total flux is zero.26. Here the electric field is uniform and we are considering a closed cylindrical surface. φ3 = 0. Therefore. both inside and outside S. We can see that explicitly in the simple situation of Fig. φ1 = –E S1. The total flux φ through the surface is φ = φ1 + φ2 + φ3. the electric field [whose flux appears on the left side of Eq. with its axis parallel to the uniform field E.30) is a simple illustration of a general result of electrostatics called Gauss’s law. whenever you find that the net electric flux through a closed surface is zero. is that it is true in general. 34 . we conclude that the total charge contained in the closed surface is zero. equals 4πr 2. no matter what its shape or size.26 Calculation of the flux of uniform electric field through the surface of a cylinder. as expected by Gauss’s law. however. (ii) The term q on the right side of Gauss’s law.31) q = total charge enclosed by S. φ= × 4 πr 2 = ε0 (1. The law implies that the total electric flux through a closed surface is zero if no charge is enclosed by the surface. The charges may be located anywhere inside the surface. Further. Thus. includes the sum of all charges enclosed by the surface. (1. Eq. FIGURE 1. Now the normal to the surface 3 at every point is perpendicular to E. and not only for the simple cases we have considered above. (iii) In the situation when the surface is so chosen that there are some charges inside and some outside. Thus.30) Equation (1. φ2 = +E S2 S1 = S2 = S where S is the area of circular cross-section. φ= q 4 πεo r q 4 πε 0 r 2 2 all ∆S Σ ∆S = q 4 πε 0 r 2 q S Now S. (1. 11 = –ELa2 φR= ER. (vi) Finally. Now the magnitude of the electric field at the left face is EL = αx1/2 = αa1/2 (x = a at the left face). The corresponding fluxes are ˆ L =E ∆S cosθ = –E ∆S. take care not to let the Gaussian surface pass through any discrete charge. ∆S = ∆S E L ⋅ n L L L L EXAMPLE 1. the angle between E and ∆ S is ± π/2. However. 1.1 m. Any violation of Gauss’s law will indicate departure from the inverse square law. This is because electric field due to a system of discrete charges is not well defined at the location of any charge. Gauss’s law is based on the inverse square dependence on distance contained in the Coulomb’s law. Ey = Ez = 0. = ERa2 Net flux through the cube since θ = 0° 35 .Electric Charges and Fields (iv) The surface that we choose for the application of Gauss’s law is called the Gaussian surface.27 Solution (a) Since the electric field has only an x component. the field grows without any bound. You may choose any Gaussian surface and apply Gauss’s law. Therefore.27 are Ex = αx1/2.∆S is separately zero for each face of the cube except the two shaded ones. in which α = 800 N/C m1/2. FIGURE 1. (As you go close to the charge. for faces perpendicular to x direction. the flux φ = E. Calculate (a) the flux through the cube. the Gaussian surface can pass through a continuous charge distribution.∆S = ER ∆S cosθ = ER ∆S. This is facilitated by the choice of a suitable Gaussian surface.11 The electric field components in Fig.) However. The magnitude of electric field at the right face is ER = α x1/2 = α (2a)1/2 (x = 2a at the right face). (v) Gauss’s law is often useful towards a much easier calculation of the electrostatic field when the system has some symmetry. Example 1. and (b) the charge within the cube. since θ = 180° φ = E . Assume that a = 0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the x-axis so that one face is at x = +10 cm and the other is at x = –10 cm (Fig.28 36 (d) The net charge within the cylinder can be found by using Gauss’s law which gives q = ε0 φ = 3. 1. Therefore.Physics = φR + φL = ERa2 – ELa2 = a2 (ER – EL) = αa2 [(2a)1/2 – a1/2] = αa5/2 ( 2 –1 EXAMPLE 1. the outward flux is ii ∆S φL= E. the flux out of the side of the cylinder is zero.14 N m2 C–1 EXAMPLE 1.854 × 10–12 C = 9.05 N m C 2 ) ( 2 –1 ) –1 (b) We can use Gauss’s law to find the total charge q inside the cube.57 + 0 = 3. since ˆ ii ∆S = – ∆S = + 200 × π (0. Therefore.854 × 10–12 C = 2.12 An electric field is uniform.12 FIGURE 1. Therefore. (b) For any point on the side of the cylinder E is perpendicular to ∆S and hence E.1)5/2 = 1.∆S = + 1.05)2 = + 1. Example 1.05 × 8.57 N m2 C–1 On the right face.14 × 8.∆S = 0.11 = 800 (0.28). and uniform with the same magnitude but in the negative x direction for negative x.57 N m2 C–1. (a) What is the net outward flux through each flat face? (b) What is the flux through the side of the cylinder? (c) What is the net outward flux through the cylinder? (d) What is the net charge inside the cylinder? Solution (a) We can see from the figure that on the left face E and ∆S are parallel.∆S = – 200 ˆ = + 200 ∆S. and in the positive x direction for positive x.27 × 10–12 C. It is given that E = 200 ˆ i N/C for x > 0 and E = –200 ˆ i N/C for x < 0. q = 1. (c) Net outward flux through the cylinder φ = 1. We have φ = q/ε0 or q = φε0.78 × 10–11 C . E and ∆S are parallel and therefore φR = E.57 + 1. The wire is obviously an axis of symmetry. where l is the length of the cylinder. inward if λ < 0). At the cylindrical part of the surface. The electric fields produced by the two elements of the pair when summed give a resultant electric field which is radial (the components normal to the radial vector cancel).29. The direction of electric field at every point must be radial (outward if λ > 0.15 APPLICATIONS OF GAUSS’S LAW The electric field due to a general charge distribution is. as shown.29 (a) Electric field due to an infinitely long thin straight wire is radial.29(b). however.Electric Charges and Fields 1. 37 . This is true for any such pair and hence the total field at any point P is radial. 1. the summation (or integration) involved in this equation cannot be carried out to give electric field at every point in space. In practice. it is possible to obtain the electric field in a simple way using the Gauss’s law. For some symmetric charge configurations. (1. and its magnitude is constant. Since the field is everywhere radial. Finally. P ′. electric field does not depend on the position of P along the length of the wire. 1.15. E is normal to the surface at every point. since it depends only on r. flux through the two ends of the cylindrical Gaussian surface is zero. Suppose we take the radial vector from O to P and rotate it around the wire. as seen above. This is best understood by some examples. To calculate the field. FIGURE 1.27). 1. The surface area of the curved part is 2πrl. since the wire is infinite. The points P. imagine a cylindrical Gaussian surface. P′′ so obtained are completely equivalent with respect to the charged wire. In short. Consider a pair of line elements P1 and P2 of the wire. the electric field is everywhere radial in the plane cutting the wire normally. (b) The Gaussian surface for a long thin wire of uniform linear charge density. except for some special cases. and its magnitude depends only on the radial distance r.1 Field due to an infinitely long straight uniformly charged wire Consider an infinitely long thin straight wire with uniform linear charge density λ. This is clear from Fig. This implies that the electric field must have the same magnitude at these points. given by Eq. as shown in the Fig. e. only the two faces 1 and 2 will contribute to the flux. 1. Therefore FIGURE 1. we cannot take E to be normal to the curved part of the cylindrical Gaussian surface. as A = A a negative or positive.. Note that when we write a vector A as a scalar multiplied by a unit ˆ . Therefore. 38 Therefore by Gauss’s law. Without this assumption. The direction of A will be the same as that of the unit ˆ if A > 0 and opposite to a ˆ if A < 0. the electric field will not depend on y and z coordinates and its direction at every point must be parallel to the x-direction. However. E = 2πε r 0 λ Vectorially. therefore.32) is approximately true for electric field around the central portions of a long wire. E is directed outward if λ is positive and inward if λ is negative.Physics Flux through the Gaussian surface = flux through the curved cylindrical part of the surface = E × 2πrl The surface includes charge equal to λ l. as shown.30).32) ˆ is the radial unit vector in the plane normal to the wire passing where n through the point. By symmetry. It can be vector.) As seen from the figure. We can take the Gaussian surface to be a rectangular parallelepiped of cross sectional area A. The charge enclosed by the closed surface is σA. A ≥ 0 . Further. the electric field E is due to the charge on the entire wire.2 Field due to a uniformly charged infinite plane sheet Let σ be the uniform surface charge density of an infinite plane sheet (Fig.15. do not contribute to the total flux. electric field lines are parallel to the other faces and they. E at any point is given by E= λ 2πε 0r ˆ n (1. .∆S through both the surfaces are equal and add up. the assumption that the wire is infinitely long is crucial. i. When we want to restrict to vector a non-negative values. (1. The unit vector normal to surface 1 is in –x direction while the unit vector normal to surface 2 is in the +x direction. 1.e. Eq.30 Gaussian surface for a the net flux through the Gaussian surface is 2 EA. Also note that though only the charge enclosed by the surface (λl ) was included above. flux E. We take the x-axis normal to the given plane. the scalar A is an algebraic number. where the end effects may be ignored. we use the symbol A and call it the modulus of A . uniformly charged infinite plane sheet. Gauss’s law then gives E × 2πrl = λl/ε0 i. (A cylindrical surface will also do.. Thus. The Gaussian surface is again a sphere through P centred at O. passing through P. we take the Gaussian surface to be a sphere of radius r and with centre O.15. therefore. E= σ ˆ n 2ε 0 (1. The charge enclosed is σ × 4 π R 2. E= q 4 πε 0 r 2 ˆ r (1. E = σ R2 q = ε0 r 2 4 π ε0 r 2 where q = 4 π R2 σ is the total charge on the spherical shell. The situation has obvious spherical symmetry. To calculate E at P. (b) r < R.Electric Charges and Fields 2 EA = σA/ε0 or. (1. Thus.33) is approximately true in the middle regions of the planar sheet. can depend only on r (the radial distance from the centre of the shell to the point) and must be radial (i.31). This. 39 . 1. (ii) Field inside the shell: In Fig.3 Field due to a uniformly charged thin spherical shell Let σ be the uniform surface charge density of a thin spherical shell of radius R (Fig. has the same magnitude E and is along the radius vector at each point. 1. (That is what we mean by spherical symmetry. the point P is inside the shell.. Thus for points outside the shell. (i) Field outside the shell: Consider a point P outside the shell with radius vector r. Vectorially.e. E and ∆S at every point are parallel and the flux through each element is E ∆S. the field due to a uniformly charged shell is as if the entire charge of the shell is concentrated at its centre. All points on this sphere are equivalent relative to the given charged configuration.34) FIGURE 1. away from the ends. Note that the above application of the Gauss’ law has brought out an additional fact: E is independent of x also.33) ˆ is a unit vector normal to the plane and going away from it. For a finite large planar sheet. is exactly the field produced by a charge q placed at the centre O. Eq. The field at any point P. however. the flux through the Gaussian surface is E × 4 π r 2. outside or inside. 1. The electric field is directed outward if q > 0 and inward if q < 0.31(b).31 Gaussian surfaces for a point with (a) r > R. where n E is directed away from the plate if σ is positive and toward the plate if σ is negative.) The electric field at each point of the Gaussian surface. By Gauss’s law 2 E × 4 π r2 = ε 4 π R 0 σ Or. Summing over all ∆S. E = σ/2ε0 Vectorically. along the radius vector). 32 Solution The charge distribution for this model of the atom is as shown in Fig. For this model. The obvious Gaussian surface is a spherical surface centred at the nucleus..32. the field due to a uniformly charged thin shell is zero at all points inside the shell*. is E × 4 π r 2. The atom as a whole is neutral. E = 0 (r < R ) (1. This is because * Compare this with a uniform mass shell discussed in Section 8. Its direction is along (or opposite to) the radius vector r from the origin to the point P. no matter what the direction of r. This important result is a direct consequence of Gauss’s law which follows from Coulomb’s law. since we must have 4 πR3 ρ = 0 – Ze 3 3 Ze or ρ = − 4 π R 3 EXAMPLE 1. calculated as before. Because of the spherical symmetry of the charge distribution. The experimental verification of this result confirms the 1/r2 dependence in Coulomb’s law.Physics The flux through the Gaussian surface.5 of Class XI Textbook of Physics. namely. 1. However.13 To find the electric field E(r) at a point P which is a distance r away from the nucleus.13 An early model for an atom considered it to have a positively charged point nucleus of charge Ze. We consider two situations. the Gaussian surface encloses no charge. surrounded by a uniform density of negative charge up to a radius R. Gauss’s law then gives E × 4 π r2 = 0 i. r < R and r > R.e. what is the electric field at a distance r from the nucleus? FIGURE 1. The total negative charge in the uniform spherical charge distribution of radius R must be –Z e. in this case. 40 . we use Gauss’s law.35) that is. the magnitude of the electric field E(r) depends only on the radial distance. Example 1. (i) r < R : The electric flux φ enclosed by the spherical surface is φ = E (r ) × 4 π r 2 where E (r ) is the magnitude of the electric field at r. since the atom (nucleus of charge Z e + negative charge) is neutral. This immediately gives us the negative charge density ρ. that is. and has the same magnitude at all points on the surface. however. 4 π ε0  r 2 R 3  r < R The electric field is directed radially outward. both cases give the same result: E = 0. In particular. This system is unchanged if (a) translated parallel to the y-z plane in any direction. it must be parallel to the x-direction. E (r ) × 4 π r 2 = 0 or E (r ) = 0. The direction. (ii) r > R: In this case. in this example. As the system is unchanged under such symmetry operation. By using rotation symmetry around the x -axis. the electric field E must be unchanged. 0). Consider. q = Z e + 4 πr3 ρ 3 Substituting for the charge density ρ obtained earlier. E (r ) = Z e 1 r  −  . Similarly translational symmetry along the z-axis shows that the electric field at two point (0. 0) as at (0. is opposite of each other on either side of the sheet. z1) and (0. y1. z2) must be the same. 0. The charge q enclosed by the Gaussian surface is the positive nuclear charge and the negative charge within the sphere of radius r. so must its properties be. independent of the x-coordinate. (b) rotated about the x-axis through any angle. we have q = Ze−Ze r3 R3 Gauss’s law then gives.e. 0. 41 . y2. EXAMPLE 1. Try to think of a symmetry now which will tell you that the magnitude of the electric field is a constant. the total charge enclosed by the Gaussian spherical surface is zero since the atom is neutral. r > R At r = R. Consideration of these symmetries helps one arrive at results much faster than otherwise by a straightforward calculation.. i. Compare this with the effort needed to arrive at this result by a direct calculation using Coulomb’s law. It thus turns out that the magnitude of the electric field due to a uniformly charged infinite conducting sheet is the same at all points in space.13 ON SYMMETRY OPERATIONS In Physics. we often encounter systems with various symmetries. Translation symmetry along the y-axis shows that the electric field must be the same at a point (0.Electric Charges and Fields the field at any point on the spherical Gaussian surface has the same direction as the normal to the surface there. Thus. we can conclude that E must be perpendicular to the y-z plane. from Gauss’s law. for example an infinite uniform sheet of charge (surface charge density σ) along the y-z plane. but no creation or destruction of charge. the unit of charge is coulomb. the charge on a glass rod rubbed with silk is positive. . 4. there is a transfer of electric charge from one body to another. the mobile charges are electrons.. that on a plastic rod rubbed with fur is then negative. Proton and electron have charges +e. Additivity of electric charges means that the total charge of a system is the algebraic sum (i. ±1. k (q1q2 ) ˆ21 r 2 r21 1 4πε 0 ˆ21 is a unit vector in the direction from q1 to q2 and k = where r 6. Conductors allow movement of electric charge through them. In metals. and that like charges repel and unlike charges attract. ±2.854 × 10–12 C2 N–1 m–2 The approximate value of k is k = 9 × 109 N m2 C–2 The ratio of electric force and gravitational force between a proton and an electron is k e2 ≅ 2. molecules and bulk matter. q = n e. F21 = force on q2 due to q1 = 3.. where n = 0. the force on any charge.e.. ±3. is 42 . Mathematically. respectively. q2. Coulomb’s Law: The mutual electrostatic force between two point charges q1 and q2 is proportional to the product q1q2 and inversely proportional to the square of the distance r21 separating them. 5. –e.4 × 1039 G m em p 7. Conservation of electric charges means that the total charge of an isolated system remains unchanged with time. Superposition Principle: The principle is based on the property that the forces with which two charges attract or repel each other are not affected by the presence of a third (or more) additional charge(s). additivity and conservation. the sum taking into account proper signs) of all individual charges in the system. one can infer that there are two types of charges in nature. say q1.e. q3. insulators do not. For an assembly of charges q1. in electrolytes both positive and negative ions are mobile... 2. In SI units. Electric charge has three basic properties: quantisation. is the constant of proportionality. Electric and magnetic forces determine the properties of atoms. For macroscopic charges for which n is a very large number. . Quantisation of electric charge means that total charge (q) of a body is always an integral multiple of a basic quantum of charge (e) i..Physics SUMMARY 1. From simple experiments on frictional electricity. The experimental value of the constant ε0 is ε0 = 8. By convention.. This means that when bodies are charged through friction.. quantisation of charge can be ignored. it is radially outwards from q. The flux ∆φ of electric field E through a small area element ∆S is given by ∆φ = E. they crowd near each other in regions of strong electric field and are far apart where the electric field is weak. 14. 4 πε o r 3 Dipole electric field on the axis at a distance r from the centre: E = 2 pr 4 πε 0 (r 2 − a 2 )2 for r >> a ≅ 2p 4 πε 0r 3 The 1/r 3 dependence of dipole electric fields should be noted in contrast to the 1/r 2 dependence of electric field due to a point charge. In regions of constant electric field.. (iii) Electrostatic field lines start at positive charges and end at negative charges —they cannot form closed loops. An electric field line is a curve drawn in such a way that the tangent at each point on the curve gives the direction of electric field at that point. Some of the important properties of field lines are: (i) Field lines are continuous curves without any breaks. 12. if q is positive. 8. The relative closeness of field lines indicates the relative strength of electric field at different points. An electric dipole is a pair of equal and opposite charges q and –q separated by some distance 2a.e. the force on q1 due to q3. Field of an electric dipole in its equatorial plane (i. Like Coulomb force. the force is given by the Coulomb’s law for two charges stated earlier. 13. For each pair. 11. 43 . electric field also satisfies superposition principle. which can be considered planar for sufficiently small ∆S. In a uniform electric field E. a dipole experiences a torque τ given by τ =p×E but experiences no net force. 9. Electric field due to a point charge q has a magnitude |q|/4πε0r 2. and so on. the field lines are uniformly spaced parallel straight lines.Electric Charges and Fields the vector sum of the force on q1 due to q2. and radially inwards if q is negative. (ii) Two field lines cannot cross each other. The electric field E at a point due to a charge configuration is the force on a small positive test charge q placed at the point divided by the magnitude of the charge. the plane perpendicular to its axis and passing through its centre) at a distance r from the centre: E= −p 1 4 πε o (a 2 + r 2 )3 / 2 for r >> a ≅ −p . 10. Its dipole moment vector p has magnitude 2qa and is in the direction of the dipole axis from –q to q.∆S The vector area element ∆S is ˆ ∆S = ∆S n ˆ is normal to the where ∆S is the magnitude of the area element and n area element. Gauss’s law: The flux of electric field through any closed surface S is 1/ε0 times the total charge enclosed by S. The electric field outside the shell is as though the total charge is concentrated at the centre. The field is zero at all points inside the shell Physical quantity Vector area element Electric field Electric flux Dipole moment Symbol ∆S E Dimensions [L 2] [MLT–3A–1] [ML3 T–3A–1] [LTA] Unit m2 V m–1 Vm Cm Remarks ˆ ∆S = ∆S n φ p ∆ φ = E . 15. (ii) Infinite thin plane sheet of uniform surface charge density σ E= σ 2 ε0 ˆ n ˆ is a unit vector normal to the plane.Physics ˆ is taken to be the direction For an area element of a closed surface. where n (iii) Thin spherical shell of uniform surface charge density σ E= q ˆ r 4 πε 0 r 2 (r ≥ R ) E=0 (r < R ) where r is the distance of the point from the centre of the shell and R the radius of the shell. The same result is true for a solid sphere of uniform volume charge density. The law is especially useful in determining electric field E. outward on either side. when the source distribution has simple symmetry: (i) Thin infinitely long straight wire of uniform linear charge density λ E= λ 2 πε 0 r ˆ n where r is the perpendicular distance of the point from the wire and ˆ is the radial unit vector in the plane normal to the wire passing n through the point. q is the total charge of the shell: q = 4πR2σ. by convention. n of outward normal.∆ S Vector directed from negative to positive charge Charge density linear surface λ σ ρ [L–1 TA] [L –2 C m–1 Cm –2 Charge/length Charge/area Charge/volume TA] 44 volume [L–3 TA] C m–3 . there is no analogous law on quantisation of mass. one can have conserved vector quantity (e. four-body. This is precisely the size of the nucleus. but is not invariant for frames of reference in relative motion. interestingly. kinetic energy is a scalar under rotation. 10.. 4. the value of k is no longer arbitrary. is too big a unit for electric effects. i. 9. angular momentum of an isolated system). The rather large value of k. 1 C = 1 A s. It says two things: force on one charge due to another charge is unaffected by the presence of other charges.Electric Charges and Fields POINTS TO PONDER You might wonder why the protons. can be a dominating and more pervasive force in nature. 3. 5. For a surface charge distribution. charge is a scalar. Thus while 1 ampere is a unit of reasonable size for magnetic effects. what is defined is the unit of current (A) via its magnetic effect (Ampere’s law) and the unit of charge (coulomb) is simply defined by (1C = 1 A s). This is how gravity. This is not always true for every scalar. A quantity may be scalar but not conserved (like kinetic energy in an inelastic collision). it is also invariant for frames of reference in relative motion.e. Conservation of total charge of an isolated system is a property independent of the scalar nature of charge noted in point 6. etc. The range of distance where this force is effective is. 7. i. In SI units. it is approximately 9 × 109 N m2 C–2. electric field is discontinuous across the surface. Why do they not fly away? You will learn that there is a third kind of a fundamental force. all carrying positive charges. called the strong force which holds them together. or equated with the law of addition of vectors. Superposition principle should not be regarded as ‘obvious’. In this case. due to the laws of quantum mechanics. 6. Also the electrons are not allowed to sit on top of the protons. and there are no additional three-body. For continuous volume charge distribution. For example. The additive property of charge is not an ‘obvious’ property. inside the nucleus. Coulomb force and gravitational force follow the same inverse-square law. Quantisation of electric charge is a basic (unexplained) law of nature. Conservation refers to invariance in time in a given frame of reference. 2. forces which arise only when there are more than two charges.. The electric field due to a discrete charge configuration is not defined at the locations of the discrete charges.. however. the large size of the unit of charge (1C) from the point of view of electric effects arises because (as mentioned in point 3 already) the unit of charge is defined in terms of magnetic forces (forces on current–carrying wires) which are generally much weaker than the electric forces. 8. allowing possibility of cancellation of electric forces. however. despite being a much weaker force.e. 1. This gives the atoms their structure as they exist in nature. On the other hand. 45 . very small ~10-14 m. But gravitational force has only one sign (always attractive). Charge is not only a scalar (or invariant) under rotation. are compactly residing inside the nucleus. it is defined at any point in the distribution. It is related to the fact that electric charge has no direction associated with it.g. while Coulomb force can be of both signs (attractive and repulsive). The constant of proportionality k in Coulomb’s law is a matter of choice if the unit of charge is to be defined using Coulomb’s law. 7 1.8 µC in air is 0. and qD = –5 µC are located at the corners of a square ABCD of side 10 cm. That is.10 1. Calculate the magnitude of the torque acting on the dipole.e.9 1. Look up a Table of Physical Constants and determine the value of this ratio. qC = 2 µC.3 1. Explain how this observation is consistent with the law of conservation of charge. What does the ratio signify? (a) Explain the meaning of the statement ‘electric charge of a body is quantised’. what is the force experienced by the test charge? A system has two charges qA = 2.6 1.5 1. a field line cannot have sudden breaks.1 1.0.11 46 1. An electric dipole is the simplest example of this fact. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first? Check that the ratio ke2/G memp is dimensionless.2 N. (a) Estimate the number of electrons transferred (from which to which?) (b) Is there a transfer of mass from wool to polythene? (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm.5 × 10–9 C is placed at this point. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1. A similar phenomenon is observed with many other pairs of bodies. its field falls off faster than 1/r 2.4 1. respectively.5 × 10–7 C and qB = –2. What is the force on a charge of 1 µC placed at the centre of the square? (a) An electrostatic field line is a continuous curve. The electric field due to a charge configuration with total charge zero is not zero. +15 cm). (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.Physics 11. Four point charges qA = 2 µC. typical of field due to a single charge.8 1.4 µC due to another small sphere of charge – 0. EXERCISES 1. What are the total charge and electric dipole moment of the system? An electric dipole with dipole moment 4 × 10–9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 NC–1.12 .5 × 10–7 C located at points A: (0. large scale charges? When a glass rod is rubbed with a silk cloth.. but for distances large compared to the size of the configuration. charges appear on both.2 What is the force between two small charged spheres having charges of 2 × 10–7C and 3 × 10–7C placed 30 cm apart in air? The electrostatic force on a small sphere of charge 0. What is the mutual force of 1. qB = –5 µC. 0. A polythene piece rubbed with wool is found to have a negative charge of 3 × 10–7 C. –15 cm) and B: (0. Why not? (b) Explain why two field lines never cross each other at any point? Two point charges qA = 3 µC and qB = –3 µC are located 20 cm apart in vacuum. 33 shows tracks of three charged particles in a uniform electrostatic field. A third sphere of the same size but uncharged is brought in contact with the first. What is the new force of repulsion between A and B? Figure 1. Give the signs of the three charges. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis? What is the net flux of the uniform electric field of Exercise 1.16 1.34 47 . 1. and the distance between them is halved? Suppose the spheres A and B in Exercise 1. as shown in Fig.Electric Charges and Fields electrostatic repulsion if the charge on each is 6.13 1.5 × 10–7 C? The radii of A and B are negligible compared to the distance of separation.18 Consider a uniform electric field E = 3 × 103 î N/C.) FIGURE 1. (b) What is the force of repulsion if each sphere is charged double the above amount. and finally removed from both.12 have identical sizes.17 1.14 FIGURE 1.33 1. then brought in contact with the second.34.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes? Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8. Which particle has the highest charge to mass ratio? 1. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero.0 × 103 Nm2/C.15 1. could you conclude that there were no charges inside the box? Why or Why not? A point charge +10 µC is a distance 5 cm directly above the centre of a square of side 10 cm. 55 × 104 NC–1 in Millikan’s oil drop experiment. the plates have surface charge densities of opposite signs and of magnitude 17.23 1.25 An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.60 × 10–19 C). (b) What is the total electric flux leaving the surface of the sphere? An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm.81 m s–2. Estimate the radius of the drop.0 × 10–22 C/m2. (b) in the outer region of the second plate. On their inner faces.0 µC/m2.26 48 . If the electric field 20 cm from the centre of the sphere is 1. how much flux would pass through the surface? (b) What is the value of the point charge? A conducting sphere of radius 10 cm has an unknown charge. What is the net electric flux through the surface? A point charge causes an electric flux of –1.0 cm on edge. 1.26 g cm–3. (g = 9.19 A point charge of 2. and (c) between the plates? 1. (a) Find the charge on the sphere. e = 1.Physics 1. (a) If the radius of the Gaussian surface were doubled. The density of the oil is 1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 µC is at the centre of a cubic Gaussian surface 9.5 × 103 N/C and points radially inward.20 1. Two large.22 1. Calculate the linear charge density. What is E: (a) in the outer region of the first plate. Which among the curves shown in Fig.0 cm radius centred on the charge. what is the net charge on the sphere? A uniformly charged conducting sphere of 2. thin metal plates are parallel and close to each other.4 m diameter has a surface charge density of 80.24 ADDITIONAL EXERCISES 1.21 1.35 cannot possibly represent electrostatic field lines? 1. where n ˆ is the Show that the electric field in the hole is (σ/2ε0) n unit vector in the outward normal direction. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. however. 1. What are the force and torque experienced by a system having a total dipole moment equal to 10–7 Cm in the negative z-direction ? (a) A conductor A with a cavity as shown in Fig. 1. Show that the total charge on the outside surface of A is Q + q [Fig. 1. FIGURE 1. the so called ‘up’ quark (denoted by u) of charge + (2/3) e. electric field is along the z-direction throughout. A proton and a neutron consist of three quarks each.36 1.35 1.Electric Charges and Fields FIGURE 1. 49 . and σ is the surface charge density near the hole. Two types of quarks. [Hint: Use Coulomb’s law directly and evaluate the necessary integral. Show that the entire charge must appear on the outer surface of the conductor. Suggest a possible way.) Suggest a possible quark composition of a proton and neutron.36(a) is given a charge Q. and the ‘down’ quark (denoted by d) of charge (–1/3) e. at the rate of 105 NC–1 per metre.29 A hollow charged conductor has a tiny hole cut into its surface. Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. not constant but increases uniformly along the positive z-direction. (Quarks of other types have also been found which give rise to different unusual varieties of matter.36(b)]. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A.28 In a certain region of space.] It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. together with electrons build up ordinary matter.27 1.31 ˆ .30 1. The magnitude of electric field is. me = 9.5 cm is 9.6 × 10–19 C.Physics 1. A particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed vx (like particle 1 in Fig. where E = 0) of the configuration. 1.34 50 .33 1. Suppose that the particle in Exercise in 1.) 1. If E between the plates separated by 0. The length of plate is L and an uniform electric field E is maintained between the plates.1 × 10–31 kg. (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.0 × 106 m s–1. where will the electron strike the upper plate? (|e|=1.e.33 is an electron projected with velocity vx = 2. Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics. Show that the vertical deflection of the particle at the far edge of the plate is 2 qEL2/(2m vx ).1 × 102 N/C. Show that the equilibrium of the test charge is necessarily unstable..33). A small test charge is placed at a null point (i.32 (a) Consider an arbitrary electrostatic field configuration. Thus. the notion of potential energy was introduced. like the potential energy of a mass in a gravitational field. This is not surprising. First. Consider an electrostatic field E due to some charge configuration. Coulomb force between two (stationary) charges is also a conservative force. The sum of kinetic and potential energies is thus conserved. consider the field E due to a charge Q placed at the origin. since both have inverse-square dependence on distance and differ mainly in the proportionality constants – the masses in the gravitational law are replaced by charges in Coulomb’s law. When an external force does work in taking a body from a point to another against a force like spring force or gravitational force. for simplicity. With reference . gaining kinetic energy and losing an equal amount of potential energy.1 INTRODUCTION In Chapters 6 and 8 (Class XI). Forces of this kind are called conservative forces. Now. Spring force and gravitational force are examples of conservative forces. When the external force is removed.Chapter Two ELECTROSTATIC POTENTIAL AND d e h CAPACITANCE T lis o n R b E u C p N re © e b o t t 2. imagine that we bring a test charge q from a point R to a point P against the repulsive force on it due to the charge Q. we can define electrostatic potential energy of a charge in an electrostatic field. that work gets stored as potential energy of the body. the body moves. Second. potential energy difference (2. we assume that the test charge q is so small that it does not disturb the original configuration. we apply an external force Fext just enough to moved from the point R to the counter the repulsive electric force FE (i. It means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other. –WRP .1. For definiteness.) Therefore. this work done increases its potential energy by an amount equal to potential energy difference between points R and P.. work done by external forces in moving a charge q from R to P is o n 52 T i l R b E u C p N re © e b o t t WRP = g h s d e (2. (2. The concept of the potential energy would not be meaningful if the work depended on the path. 2. i. point P against the repulsive force on it by the charge Q (> 0) This means there is no net force on or acceleration of placed at the origin. and gets fully stored in the form of potential energy of the charge q. let us take Q. If the external force is removed on reaching P. the electric force will take the charge away from Q – the stored energy (potential energy) at P is used to provide kinetic energy to the charge q in such a way that the sum of the kinetic and potential energies is conserved. the charge q when it is brought from R to P. . At every point in electric field. Thus.2) (Note here that this displacement is in an opposite sense to the electric force and hence work done by electric field is negative. a particle with charge q possesses a certain electrostatic potential energy. namely the charge Q at the origin (or else. we can define electric potential energy difference between two points as the work required to be done by an external force in moving (without accelerating ) charge q from one point to another for electric field of any arbitrary charge configuration.Physics to Fig. In this situation.e. This is the fundamental characteristic of a conservative force. i.e. q > 0. Thus. We omit this proof here. we keep Q fixed at the origin by some unspecified force). The path-independence of work done by an electrostatic field can be proved using the Coulomb’s law. it is brought with infinitesimally slow constant speed. Two remarks may be made here.1 A test charge q (> 0) is R to P. this will happen if Q and q are both positive or both negative. Two important comments may be made at this stage: (i) The right side of Eq..e.2) depends only on the initial and final positions of the charge.1) = g This work done is against electrostatic repulsive force and gets stored as potential energy. First. Fext= –FE). in bringing the charge q from FIGURE 2. work done by the external force is the negative of the work done by the electric force. This work is obviously proportional to q. was not due to any exceptional property of animal tissues but was also generated whenever any wet body was sandwiched between dissimilar metals. Volta established that the animal electricity observed by Luigi Galvani. convenient to divide the work by the amount of charge q. (2.2) defines potential energy difference in terms of the physically meaningful quantity work. Note. respectively. since this will not change the potential energy difference: Put it differently. (2. 2. where E is the electric field at that point due to the given charge configuration. 1737–1798. In other words. in experiments with frog muscle tissue placed in contact with dissimilar metals. With this choice. If. Potential energy of charge q at a point (in the presence of field due to any charge configuration) is the work done by the external force (equal and opposite to the electric force) in bringing the charge q from infinity to that point. since the force at any point is q E . that it is not the actual value of potential but the potential difference that is physically significant. We define potential energy of a test charge q in terms of the work done on the charge q.1). therefore. (2.Electrostatic Potential and Capacitance (ii) Equation (2. It is. We can always add an arbitrary constant a to potential energy at every point. From Eq. there is a freedom in choosing the point where potential energy is zero. so that the resulting quantity is independent of q.3) (2.3) provides us with a definition of potential energy of a charge q at any point. or battery.2 ELECTROSTATIC POTENTIAL Consider any general static charge configuration. we get from Eq. professor at Pavia.4) implies: Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V ) at that point. Clearly. it is only the difference of potential energy that is significant. h s d e COUNT ALESSANDRO VOLTA (1745 –1827) where VP and VR are the electrostatic potentials at P and R. 53 . This led him to develop the first voltaic pile .2) Since the point P is arbitrary.4) Count Alessandro Volta (1745 – 1827) Italian physicist. potential energy so defined is undetermined to within an additive constant. Eq. A convenient choice is to have electrostatic potential energy zero at infinity. as before. (2. we choose the potential to be zero at infinity. consisting of a large stack of moist disks of cardboard (electrolyte) sandwiched between disks of metal (electrodes). we get: Work done by external force in bringing a unit positive charge from point R to P = VP – V R o n T i l R b E u C p N re © e b o t t (2.What this means is that the actual value of potential energy is not physically significant. as before. Eq. work done per unit test charge is characteristic of the electric field associated with the charge configuration. if we take the point R at infinity. This leads to the idea of electrostatic potential V due to a given charge configuration. Consider a point charge Q at the origin (Fig. Total work done (W) by the external force is obtained by integrating Eq. the electrostatic force on a unit positive charge is FIGURE 2. For that we must calculate the work done in bringing a unit positive test charge from infinity to the point P. we choose a convenient path – along the radial direction from infinity to the point P.3). take Q to be positive. the work done against the repulsive force on the test charge is positive. Since work done is independent of the path. against the repulsive force of charge Q (Q > 0).5) where is the unit vector along OP′. and depends only on its initial and final positions. We wish to determine the potential at any point P with position vector r from the origin. by definition is the potential at P due to the charge Q (2. For Q > 0.7) This. To obtain the work done per unit test charge. The qualifying remarks made earlier regarding potential energy also apply to the definition of potential. (2. is the potential at P due to the charge Q.2 Work done on a test charge q by the electrostatic field due to any given charge configuration is independent of the path. the electrostatic potential (V ) at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point.6) The negative sign appears because for ∆ r ′ < 0. o n 54 T i l R b E u C p N re © e b o t t 2. 2. ∆ W is positive .3 POTENTIAL DUE TO A r P OINT CHARGE h s d e (2. For definiteness. At some intermediate point P′ on the path. obtain the work done dW in bringing it from infinity to the point and determine the ratio dW/ dq. we should take an infinitesimal test charge dq.8) . Also. Work done against this force from r′ to r′ + ∆ r′ is (2. FIGURE 2. the external force at every point of the path is to be equal and opposite to the electrostatic force on the test charge at that point.3 Work done in bringing a unit positive test charge from infinity to the point P. (2.6) from r¢= ∞ to r¢= r.Physics In other words. 1 (a) Calculate the potential at a point P due to a charge of 4 × 10–7C located 9 cm away.8) is consistent with the choice that potential at infinity be zero. we -2 of (Q/4πe0) m ] (black curve) for a point charge Q.5). We also saw that the electric field of a dipole at a point with position vector r depends not just on the magnitude r. so that the electrostatic force and the FIGURE 2. 55 .4) shows how the electrostatic potential ( 1/r ) and the electrostatic field ( 1/r 2 ) varies with r. For Q < 0. an electric dipole consists of two charges q and –q separated by a (small) distance 2a .e. work done will be path independent.8) is true for any sign of the charge Q.. (2. Further. note that Eq.] Finally. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point P is positive. It is characterised by a dipole moment vector p whose magnitude is q × 2a and which points in the direction from – q to q (Fig. but also on the angle between r and p. (b) Hence obtain the work done in bringing a charge of 2 × 10–9 C from infinity to the point P.4 Variation of potential V with r [in units of displacement (from infinity to P) are (Q/4πe0) m-1] (blue curve) and field with r [in units in the same direction. Example 2. the force on a unit positive test charge is attractive. The work done corr esponding to the later will be zero. V < 0. [This is as it should be. 2. 2. i. Figure (2. Its total charge is zero. though we considered Q > 0 in its derivation.4 POTENTIAL o n –5 T i l R b E u C p N re © e b o t t EXAMPLE 2. work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative.1 h s d e DUE TO AN ELECTRIC DIPOLE As we learnt in the last chapter. since for Q < 0. Any arbitrary infinitesimal path can be resolved into two perpendicular displacements: One along r and another perpendicular to r.Electrostatic Potential and Capacitance Equation (2. Does the answer depend on the path along which the charge is brought? Solution (a) = 4 × 104 V (b) = 8 × 10 J No. 11) (2. Using Eqs. at large distance. h s d e (2.12) Using the Binomial theorem and retaining terms upto the first order in a/r . we obtain.9) (2. respectively. cosq cosq Similarly. the potential due to the dipole is the sum of potentials due to the charges q and –q FIGURE 2. we take the origin at the centre of the dipole.13(b)] (2. electrostatic potential also follows the superposition principle. [2.9) and (2.5 Quantities involved in the calculation of potential due to a dipole. now. determine the electric potential due to a dipole and contrast it with the potential due to a single charge. (2. Now we know that the electric field obeys the superposition principle. by geometry.13) and p = 2qa.Physics the field falls off.14) . we get Now. We. As before.13(a)] [2. Thus. o n 56 T i l R b E u C p N re © e b o t t Now. Since potential is related to the work done by the field. p cos q = g where r1 and r2 are the distances of the point P from q and – q. not as 1/r 2 (typical of field due to a single charge) but as 1/r 3.10) We take r much greater than a ( r >> a ) and retain terms only upto the first order in a/r (2. 17) o n T i l R b E u C p N re © e b o t t DUE TO A h s d e SYSTEM OF CHARGES FIGURE 2.15): (i) The potential due to a dipole depends not just on r but also on the angle between the position vector r and the dipole moment vector p. rn relative to some origin (Fig. negative sign for q = π. qn with position vectors r1.8) and (2. respectively. where r2P and r3P are the distances of P from charges q2 and q3. (2. (r >> a) (2.) (ii) The electric dipole potential falls off. so that higher order terms in a/ r are negligible.15) Equation (2.. however. the potential V 2 at P due to q2 and V3 due to q3 are given by .) 2. approximately true only for distances large compared to the size of the dipole. By the superposition principle.16) (Positive sign for q = 0. at large distance. … . + Vn (2. π ) is given by (2.Electrostatic Potential and Capacitance where r ˆ is the unit vector along the position vector OP. That is. Eq.5 for graphs of 1/r 2 versus r and 1/ r versus r. The potential V1 at P due to the charge q1 is where r1P is the distance between q1 and P.15) is. the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges V = V1 + V2 + . For a point dipole p at the origin. The important contrasting features of electric potential of a dipole from that due to a single charge are clear from Eqs. as indicated. The electric potential of a dipole is then given by g . (You can refer to the Fig. (It is. q2 . keeping q fixed. potential on the dipole axis (q = 0.. drawn there in another context. characteristic of the potential due to a single charge. (2. as 1/r 2 . and so on for the potential due to other charges. r2. 2.5 POTENTIAL Consider a system of charges q1 .…. however. not as 1/r. 2. axially symmetric about p.15). if you rotate the position vector r about p .15) is. exact.6).6 Potential at a point due to a system of charges is the sum of potentials due to individual charges.) The potential in the equatorial plane (q = π/2) is zero. the points corresponding to P on the cone so generated will have the same potential as at P. From Eq. (2. Similarly. 57 . as before. therefore.7 Let P be the required point on the x-axis where the potential is zero. If x lies on the extended line OA. the negative charge is taken to be on the right side of the origin (Fig.18) If we have a continuous charge distribution characterised by a charge density r (r). integrate) over all such contributions.Physics (2. which is [2. That is. Thus. We then determine the potential due to each volume element and sum (strictly speaking . and thus determine the potential due to the entire distribution. the electric field outside the shell is as if the entire charge is concentrated at the centre. the potential outside the shell is given by o n 58 T i l R b E u C p N re © e b o t t –8 where q is the total charge on the shell and R its radius. (Ther e is no possibility of potentials due to the two charges adding up to zero for x < 0. This implies (Section 2. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.2 Two charges 3 × 10 C and –2 × 10 C are located 15 cm apart. the required condition is . we divide it.19(b)] h s [2. We have seen in Chapter 1 that for a uniformly charged spherical shell.6) that potential is constant inside the shell (as no work is done in moving a charge inside the shell). Solution Let us take the origin O at the location of the positive charge. equals its value at the surface. we have x x where x is in cm. obviously x must be positive. 2. The line joining the two charges is taken to be the x -axis. EXAMPLE 2.) If x lies between O and A. –8 FIGURE 2. The electric field inside the shell is zero.19(a)] d e Example 2.7). and. into small volume elements each of size ∆v and carrying a charge r ∆v. If x is the x-coordinate of P.2 which gives x = 9 cm. E.8 (a) and (b) show the field lines of a positive and negative point charge respectively. work done by the field is negative.3 59 .3 Figures 2. VB – VA . Due to force of repulsion on the negative charge. Thus. (VP – VQ ) is positive. equipotential surfaces: d e EXAMPLE 2.Electrostatic Potential and Capacitance which gives x = 45 cm Thus. (P.)B and hence sign of potential ener gy differences is positive. Example 2. Also V B is less negative T i l R b E u C p N re © e b o t t FIGURE 2.2 h s http://sol.sci.8 EXAMPLE 2. Thus. Similarly. velocity decreases and hence the kinetic energy decreases in going from B to A. In moving a small positive charge fr om Q to P. (e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A? Solution (a) As (b) (c) (d) (e) than V A . work has to be done by an external agency against the electric field. (a) Give the signs of the potential differ ence VP – VQ. electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge.html Electric potential.edu/~jfalward/electricpotential/electricpotential. In moving a small negative charge from B to A work has to be done by the external agency. VB > V A or ( VB – VA) is positive. (c) Give the sign of the work done by the field in moving a small positive charge from Q to P. A small negative charge will be attracted towards positive charge. (d) Give the sign of the work done by the external agency in moving a small negative charge from B to A. o n . Therefore. The negative charge moves from higher potential energy to lower potential energy.E. VP > V Q. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity. A and B.uop. (b) Give the sign of the potential energy difference of a small negative charge between the points Q and P.)A > (P. It is positive. and (b) electric field lines are radial. it would have non-zero component along the surface. Clearly. For a uniform electric field E. Equipotential surfaces for (a) a dipole and (b) two identical positive charges are shown in Fig. o n 60 T i l R b E u C p N re © e b o t t This shows that V is a constant if r is constant . FIGURE 2. the potential is given by Eq. therefore. This is true in general: for any charge configuration. be normal to the equipotential surface at every point. But this is in contradiction to the definition of an equipotential surface: there is no potential difference between any two points on the surface and no work is required to move a test charge on the surface.11 Some equipotential surfaces for (a) a dipole. Now the electric field lines for a single charge q are radial lines starting from or ending at the charge. h s d e FIGURE 2. For a single charge q. The electric field must. If the field were not normal to the equipotential surface.8): FIGURE 2.10 Equipotential surfaces for a uniform electric field.10). starting from the charge if q > 0. (2.9 For a single charge q (a) equipotential surfaces are spherical surfaces centred at the charge. depending on whether q is positive or negative. the equipotential surfaces are planes normal to the x -axis. planes parallel to the y. the electric field at every point is normal to the equipotential surface passing through that point. Thus.. To move a unit test charge against the direction of the component of the field. The proof of this statement is simple.e. equipotential surface through a point is normal to the electric field at that point. . 2.6 EQUIPOTENTIAL S URFACES An equipotential surface is a surface with a constant value of potential at all points on the surface. Equipotential surfaces offer an alternative visual picture in addition to the picture of electric field lines around a charge configuration. work would have to be done.z plane (Fig. i. along the x -axis. (b) two identical positive charges.Physics 2.11. 2. say. equipotential surfaces of a single point charge are concentric spherical surfaces centred at the charge. There is no external field against which work needs to be done. δl is the perpendicular distance of the surface A from P.20) (2. (ii) Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.6.12) with potential values V and V + δ V. Let P be a point on the surface B. h s d e OF A S YSTEM OF CHARGES where r1P is the distance of a point P in space from the location of q1. 2. The work done in this process is |E |d l.12 From the potential to the field. From the definition of potential. This work equals the potential difference VA–VB. Let us calculate the work done (externally) in building up this configuration.Electrostatic Potential and Capacitance 2.1 Relation between field and potential Consider two closely spaced equipotential surfaces A and B (Fig. first the charge q1 is brought from infinity to the point r1 . Suppose. |E |d l = V (V +dV )= –dV i.20) as We thus arrive at two important conclusions concerning the relation between electric field and potential: (i) Electric field is in the direction in which the potential decreases steepest. dV = – |dV|. This charge produces a potential in space given by o n T i l R b E u C p N re © e b o t t (2. where δV is the change in V in the direction of the electric field E . 2. work done in bringing charge q2 from infinity to the point r2 is q2 times the potential at r2 due to q1 : work done on q2 = 61 . Imagine that a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field. This means that we consider the charges q1 and q2 initially at infinity and determine the work done by an external agency to bring the charges to the given locations. |E |= Since dV is negative.21) FIGURE 2.e. so work done in bringing q1 from infinity to r1 is zero. Thus.7 POTENTIAL E NERGY Consider first the simple case of two charges q1and q2 with position vector r1 and r2 relative to some origin.. we can rewrite Eq (2. work done in this step is h s d e (2. with the notation given in the figure. 2 at r3 (2. (2. Since electrostatic force is conservative. r3. is independent of the manner in which the configuration is assembled.26). the electrostatic force is attractive.22) is easily generalised for a system of any number of point charges. q2 and q3 located at r1. electrostatic force is repulsive and a positive amount of work is needed to be done against this force to bring the charges from infinity to a finite distance apart. this work gets stored in the form of potential energy of the system. (2. In other words. If q1q2 > 0.22). More generally. a negative amount of work is needed for the reverse path (from infinity to the present locations). the potential energy of a system of two charges q1 and q2 is FIGURE 2. since for like charges (q1q2 > 0). (2.26) Again. respectively.22) o n 62 T i l R b E u C p N re © e b o t t Obviously. As before. r2. if q2 was brought first to its present location and q1 brought later. (2.13 Potential energy of a system of charges q 1 and q2 is directly proportional to the product of charges and inversely to the distance between them. the path independence of work done).14 Potential energy of a system of three charges is given by Eq. (2. so the potential energy is negative. This is as expected. Let us calculate the potential energy of a system of three charges q1. is unaltered whatever way the charges are brought to the specified locations.25)].23) The charges q1 and q2 produce a potential. Equation (2. (2. For unlike charges (q1 q2 < 0). (2. . no work is required.23) and Eq. the potential energy expression.Physics where r12 is the distance between points 1 and 2.22) is true for any sign of q1and q2. a positive amount of work is needed against this force to take the charges from the given location to infinity. Next we bring q2 from infinity to r2. Eq. In that case.26). potential energy is positive. Thus. because of the conservative nature of the electrostatic force (or equivalently. The potential energy FIGURE 2. Eq. because of path-independence of work for electrostatic force. the potential energy U would be the same.24) Work done next in bringing q3 from infinity to the point r3 is q3 times V1. which at any point P is given by (2. To bring q1 first from infinity to r1. the final expression for U.25) The total work done in assembling the charges at the given locations is obtained by adding the work done in different steps [Eq. Equation (2. 4 Four charges are arranged at the corners of a square ABCD of side d. and –q are brought to B. Suppose. respectively. C and D. and then the charges – q. This is given by (charge at C) × (potential at C due to charges at A and B) (iv) Work needed to bring –q to D when + q at A.–q at B. This is given by (charge at B) × (electrostatic potential at B due to charge +q at A) (iii) Work needed to bring charge +q to C when + q is at A and –q is at B.15 h s d e EXAMPLE 2. +q. This is given by (charge at D) × (potential at D due to charges at A. as shown in Fig. B and C) o n T i l R b E u C p N re © e b o t t FIGURE 2. and +q at C. the four charges being held fixed at its corners. and not on how they ar e put together.(a) Find the work required to put together this arrangement. Example 2. (b) A charge q0 is brought to the centre E of the square. 2. How much extra work is needed to do this? Solution (a) Since the work done depends on the final arrangement of the charges. and not the way the state is achieved. C and D. (ii) Work needed to bring –q to B when + q is at A.Electrostatic Potential and Capacitance is characteristic of the present state of configuration.4 63 . The total work needed can be calculated in steps: (i) Work needed to bring charge + q to A when no charge is present elsewhere: this is zero. we calculate work needed for one way of putting the charges at A.15. B. first the charge +q is brought to A. but often they are unknown or unspecified. Hence no work is required to bring any charge to point E. By definition. By definition.1 Potential energy of a single charge In Section 2. Note again that we are interested in determining the potential energy of a given charge q (and later. We assume that the charge q does not significantly affect the sources producing the external field. (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same. in fact.7. The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D. or the external sources are held fixed by other unspecified forces. we ask a related but a distinct question.and the potential energy of the system of those charges was determined. What is the potential energy of a charge q in a given field? This question was. and not how they are assembled. B.7. In this section.8 POTENTIAL E NERGY IN AN EXTERNAL FIELD 2.4 The work done depends only on the arrangement of the charges. The external electric field E and the corresponding external potential V may vary from point to point. C and D). The total work required is o n 64 T i l R b E u C p N re © e b o t t EXAMPLE 2.1 and 2. Even if q is finite.The external sources may be known. . the source of the electric field was specified – the charges and their locations . B.8. (ii).Physics Add the work done in steps (i). this is the total electrostatic energy of the charges. The external field E is not produced by the given charge(s) whose potential energy we wish to calculate. But here we address this question again to clarify in what way it is different from the discussion in Section 2. we are not interested in the potential energy of the sources producing the external electric field. E is produced by sources external to the given charge(s). V at a point P is the work done in bringing a unit positive charge from infinity to the point P. what is specified is the electric field E or the electrostatic potential V due to the external sources.) (b) The extra work necessary to bring a charge q0 to the point E when the four charges are at A. the starting point that led us to the notion of the electrostatic potential (Sections 2. (iii) and (iv). The main difference is that we are now concerned with the potential energy of a charge (or charges) in an external field.2). This is true if q is very small. h s d e 2. C and D is q 0 × (electrostatic potential at E due to the charges at A. its influence on the external sources may still be ignored in the situation when very strong sources far away at infinity produce a finite field E in the region of interest. a system of charges) in the external field. work done in bringing a charge q from infinity to the point P in the external field is qV. In this step. we can write: Potential energy of q at r in an external field = qV ( r) (2. 0) and (9 cm.6 × 10–19J. work is done not only against the external field E but also against the field due to q1.27) and (2. i. (2. using Eq.2 Potential energy of a system of two charges in an external field Next.6 × 10–7J). (2. XI Physics Part I. This unit of energy is defined as 1 electron volt or 1eV. nuclear and particle physics.Electrostatic Potential and Capacitance (We continue to take potential at infinity to be zero.27) where V(r) is the external potential at the point r. This work is stored in the form of potential energy of q.1.28) (2.6 × 10–13J. Next.5 (a) Determine the electrostatic potential energy of a system consisting of two charges 7 µC and –2 µ C (and with no external field) placed at (–9 cm.] 2. We have made use of Eqs. Work done in this step is q1 V(r1). (b) How much work is required to separate the two charges infinitely away from each other? 65 . If the point P has position vector r relative to some origin. By the superposition principle for fields. 1 eV=1. The units based on eV are most commonly used in atomic. Table 6.. respectively. 1 GeV = 109eV = 1. 0.6 × 10–19 J. [This has already been defined on Page 117. Work done on q2 against the external field = q2 V (r2) Work done on q2 against the field due to q1 where r12 is the distance between q1 and q2 . in an external field? First. Potential energy of the system = the total work done in assembling the configuration o n T i l R b E u C p N re © e b o t t (2. it would gain energy of q∆ V = 1.6×10–19 C is accelerated by a potential difference of ∆V = 1 volt.8. (1 keV = 103eV = 1.29) EXAMPLE 2. Thus.5 h s d e Example 2. we calculate the work done in bringing the charge q1 from infinity to r1. we consider the work done in bringing q2 to r2 .e.22). 0) respectively.27). if an electron with charge q = e = 1.6 × 10–16 J. we ask: what is the potential energy of a system of two charges q1 and q2 located at r1and r2. 0. 1 MeV = 106eV = 1.6 × 10–10J and 1 TeV = 1012eV = 1.) Thus. we add up the work done on q2 against the two fields (E and that due to q1): Work done in bringing q2 to r2 Thus. 31) This work is stored as the potential energy of the system. Similar to other potential energies.16. in a uniform electric field.5 and the net electrostatic energy is h s d e 2. As seen in the last chapter. A natural choice is to take q0 = π / 2.8. there is a freedom in choosing the angle where the potential energy U is taken to be zero. Suppose an external torque t ext applied in such a manner that it just neutralises this torque and rotates it in the plane of paper from angle q0 to angle q1 at an infinitesimal angular speed and without angular acceleration.7 J. o n T i l R b E u C p N re © e b o t t E XAMPLE 2.3 Potential energy of a dipole in an external field Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform electric field E. g (2. (b) W = U2 – U1 = 0 – U = 0 – (–0.32) 66 .7) = 0.Physics (c) Suppose that the same system of charges is now placed in an external electric field E = A (1/r 2). We find. 2. We can then associate potential energy U (q ) with an inclination q of the dipole. FIGURE 2.7 J. there is the energy of interaction of the two charges with the exter nal electric field. In addition.16 Potential energy of a dipole in a uniform external field. the dipole experiences no net force. as shown in Fig. The amount of work done by the external torque will be given by (2. What would the electrostatic energy of the configuration be? Solution (a) = – 0.) We can then write. but experiences a torque t given by t = p×E (2. (c) The mutual interaction energy of the two charges remains unchanged. (Α n explanation for it is provided towards the end of discussion. A = 9 × 105 C m–2.30) which will tend to rotate it (unless p is parallel or is antiparallel to E). Thus. The potential energy expression then reads (2.. they collide with each other and with the ions. (2. Ui = –pE cos q = –6×10 ×10 cos 0° = –6 J –6 Final potential energy (when q = 60°). but o n T i l R b E u C p N re © e b o t t EXAMPLE 2. Conductors contain mobile charge carriers. These electrons are free within the metal but not free to leave the metal. r1 and r2 denote the position vectors of + q and –q. q [V (r1 ) – V (r2)]=0. (2. Uf = –6 × 10 × 106 cos 60° = –3 J Change in potential energy = –3 J – (–6J) = 3 J So. there is loss in potential energy. We thus obtain. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. We apply Eq.32).e.33) Here. Example 2. In electrolytic conductors. In this case.6 A molecule of a substance has a permanent electric –29 dipole moment of magnitude 10 C m.34) and it then reduces to Eq. assume 100% polarisation of the sample.Electrostatic Potential and Capacitance This expression can alternately be understood also from Eq.34) h s d e OF CONDUCTORS 67 . A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 106 V m–1.29). 2. This must be the energy released by the substance in the form of heat in aligning its dipoles. In an external electric field. the work done against the external field E in bringing +q and – q are equal and opposite and cancel out. (q) differs from U (q ) by a quantity which is just a constant We note that U¢ for a given dipole. (2. In metallic conductors. the outer (valence) electrons part away from their atoms and are free to move. we can drop the second term in Eq. i. We can now understand why we took q0=π/2. Since a constant is insignificant for potential energy. For simplicity. these charge carriers are electrons.29) to the present system of two charges +q and –q. p = 6 × 10 × 10 C m –6 = 6 × 10 C m –6 6 Initial potential energy. The displacement parallel to the force is 2a cosq. the charge carriers are both positive and negative ions.9 ELECTROSTATICS Conductors and insulators were described briefly in Chapter 1. dipole moment of each molecules = 10–29 C m As 1 mole of the substance contains 6 × 1023 molecules. 23 –29 total dipole moment of all the molecules. (2. [ V(r1 )–V (r2 )] = –E × 2a cosq .6 g (2. they drift against the direction of the field. The free electrons form a kind of ‘gas’. Solution Here. Now. In a metal. The positive ions made up of the nuclei and the bound electrons remain held in their fixed positions. and move randomly in different directions. the potential difference between positions r1 and r2 equals the work done in bringing a unit positive charge against field from r2 to r1. The direction of the field is suddenly changed by an angle of 60º. Electrostatic field is zero inside a conductor . the volume v can be made vanishingly small. This means there is no net charge at any point inside the conductor. when there is no current inside or on the surface of the conductor. the free charge carriers would experience force and drift. Since E = 0 inside the conductor and has no tangential component on the surface.Physics the situation in this case is more involved – the movement of the charge carriers is affected both by the external electric field as also by the so-called chemical forces (see Chapter 3). h s d e 3. But the surface S can be made as small as you like. Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface This follows from results 1 and 2 above. by Gauss’s law. the electric field is zero everywhere inside the conductor. Inside a conductor. There may also be an external electrostatic field. electrostatic field must be normal to the surface at every point If E were not normal to the surface. and any excess charge must reside at the surface. E should have no tangential component. This follows from the Gauss’s law. Let us note important results regarding electrostatics of conductors.. If the conductor is charged. That is. neutral or charged. electrostatic field is zero. there is no potential difference between any two points inside or on the surface of the conductor. A conductor has free electrons. We shall restrict our discussion to metallic solid conductors. therefore. the excess charge can reside only on the surface in the static situation. In the static situation. the result. there is no net charge enclosed by S. When the conductor is charged. . field is zero even at the surface. Thus the total electric flux through S is zero. Free charges on the surface of the conductor would then experience force and move.) See result 5. Hence. The interior of a conductor can have no excess charge in the static situation A neutral conductor has equal amounts of positive and negative charges in every small volume or surface element. Hence. i. it would have some non-zero component along the surface. o n 68 T i l R b E u C p N re © e b o t t 2. On the closed surface S bounding the volume element v. Thus electrostatic field at the surface of a charged conductor must be normal to the surface at every point. (For a conductor without any surface charge density. At the surface of a charged conductor. no work is done in moving a small test charge within the conductor and on its surface. This fact can be taken as the defining property of a conductor. electrostatic field is zero Consider a conductor. 4. the free charges have so distributed themselves that the electric field is zero everywhere inside. In the static situation.e. Consider any arbitrary volume element v inside a conductor. As long as electric field is not zero. 1. In the static situation. But the vanishing of electric field in the (charge-free) cavity of a conductor is. with no charges inside the cavity. We have proved a simple case of this result already: the electric field inside a charged spherical shell is zero. electric field is normal to the surface inward.35) for electric field at the surface of a char ged conductor.36) h s d e FIGURE 2. Electrostatic shielding Consider a conductor with a cavity. the contribution to the total flux through the pill box comes only from the outside (circular) cross-section of the pill box.Electrostatic Potential and Capacitance electric field normal to the surface exists. This equals ± EdS (positive for s > 0. 2. E may be considered constant and E and dS are parallel or antiparallel.17. The charge enclosed by the pill box is sdS. this means potential will be different for the surface and a point just outside the surface. as shown in Fig. the electrostatic field is zero. negative for s < 0). but this constant may differ from one conductor to the other.35) where σ is the surface charge density and to the surface in the outward direction. whatever be the size and shape of the cavity and whatever be the charge on the conductor and the external fields in which it might be placed. Eq. as mentioned above. The proof of the result for the shell makes use of the spherical symmetry of the shell (see Chapter 1). (2. we get the vector relation. is a unit vector normal To derive the result. choose a pill box (a short cylinder) as the Gaussian surface about any point P on the surface. Electric field at the surface of a charged conductor (2. just outside. In a system of conductors of arbitrary size. shape and charge configuration. a very general result. By Gauss’s law EdS = E= Including the fact that electric field is normal to the surface. It has a small area of cross section d S and negligible height. the field is normal to the surface with magnitude E. A related result is that even if the conductor o n T i l R b E u C p N re © e b o t t (2. Thus. 69 . for s < 0. 5. (2.35). Just inside the surface. each conductor is characterised by a constant value of potential. For s > 0. which is true for both signs of s. 6. electric field is normal to the surface outward.17 The Gaussian surface (a pill box) chosen to derive Eq. The pill box is partly inside and partly outside the surface of the conductor. since over the small area dS. A remarkable result is that the electric field inside the cavity is zero. Figure 2.18 are omitted here. but we note their important implication. Why is this necessary? (c) Vehicles carrying inflammable materials usually have metallic ropes touching the ground during motion. a paper does not conduct electricity. Example 2.7 (a) A comb run through one’s dry hair attracts small bits of paper. (There are no charges placed in the cavity. This is known as electrostatic shielding. Why? What happens if the hair is wet or if it is a rainy day? (Remember. The proofs of the results noted in Fig. The molecules in the paper gets polarised by the charged comb.) (b) Or dinary rubber is an insulator.Physics is charged or charges are induced on a neutral conductor by an external field. any cavity in a conductor remains shielded from outside electric influence: the field inside the cavity is always zero . Why? (d) A bird perches on a bare high power line. 2. Why? Solution (a) This is because the comb gets charged by friction.) o n 70 T i l R b E u C p N re © e b o t t EXAMPLE 2. and nothing happens to the bird. The effect can be made use of in protecting sensitive instruments from outside electrical influence. Whatever be the charge and field configuration outside.19 Some important electr ostatic properties of a conductor. or if it is rainy day. friction between hair and the comb reduces.18 The electric field inside a cavity of any conductor is zero. . A man standing on the ground touches the same line and gets a fatal shock.7 h s d e FIGURE 2. resulting in a net force of attraction. FIGURE 2. But special rubber tyr es of aircraft are made slightly conducting.19 gives a summary of the important electrostatic properties of a conductor. all charges reside only on the outer surface of a conductor with cavity. The comb does not get charged and thus it will not attract small bits of paper. If the hair is wet. All charges reside only on the outer surface of a conductor with cavity. this free movement of charges is not possible. Examples of non-polar molecules are oxygen (O 2 ) and hydrogen (H 2 ) molecules which. o n T i l R b E u C p N re © e b o t t h s d e . however. Unlike in a conductor.21 Some examples of polar external field). It only reduces it. In a non-polar molecule. Such molecules have a and non-polar molecules. This happens until. It turns out that the external field induces dipole moment by stretching or re-orienting molecules of the dielectric.Electrostatic Potential and Capacitance (b) To enable them to conduct charge (produced by friction) to the ground.7 2.9 what happens when a conductor is placed in an external electric field. The free charge carriers move and charge distribution in the conductor adjusts itself in such a way that the electric field due to induced charges opposes the external field within the conductor.20 Difference in behaviour surface of the dielectric which produce a field that of a conductor and a dielectric opposes the external field. in an external electric field. they have no (or negligible number of ) charge carriers. (c) Reason similar to (b). In contrast to conductors.10 DIELECTRICS AND POLARISATION Dielectrics are non-conducting substances. the two fields cancel each other and the net electrostatic field in the conductor is zero. in the static situation. The collective effect of all the molecular dipole moments is net charges on the FIGURE 2. we need to look at the charge distribution of a dielectric at the molecular level. a polar molecule is one in which the centres of positive and negative charges are separated (even when there is no FIGURE 2. because of their symmetry. as too much of static electricity accumulated may result in spark and result in fire. The molecules of a substance may be polar or non-polar. An ionic molecule such as HCl or a molecule of water 71 (H2O) are examples of polar molecules. On the other hand. (d) Current passes only when there is difference in potential. E XAMPLE 2. The molecule then has no permanent (or intrinsic) dipole moment. the centres of positive and negative charges coincide. In a dielectric. permanent dipole moment. have no dipole moment. To understand the effect. The extent of the effect depends on the nature of the dielectric. Recall from Section 2. the opposing field so induced does not exactly cancel the external field. The displacement stops when the external force on the constituent charges of the molecule is balanced by the restoring force (due to internal fields in the molecule). FIGURE 2. but we shall not pursue that here. the moment in an external electric field. but for a different reason. but generally the alignment effect is more important for polar molecules. the individual dipole moments tend to align with the field. The dipole moment per unit volume is called polarisation and is denoted by P..e. the positive and negative charges of a nonpolar molecule are displaced in opposite directions. (Substances for which this assumption is true are called linear isotropic dielectrics. the dielectric is polarised. When summed over all the molecules. The question is: how does the polarised dielectric modify the original external field inside it? Let us consider.) The induced dipole moments of different molecules add up giving a net dipole moment of the dielectric in the presence of the external field. for simplicity. Thus 72 . (a) Non-polar different permanent dipoles are oriented molecules. whether polar or non-polar.22 A dielectric develops a net dipole In the absence of any external field. randomly due to thermal agitation. there is then a net dipole moment in the direction of the external field. It is possible to relate ce to the molecular properties of the substance. There may be. Thus in either case. The field causes a uniform polarisation P of the dielectric. A dielectric with polar molecules also develops a net dipole moment in an external field. The extent of polarisation depends on the relative strength of two mutually opposite factors: the dipole potential energy in the external field tending to align the dipoles with the field and thermal energy tending to disrupt the alignment. in addition. We consider only the simple situation when the induced dipole moment is in the direction of the field and is proportional to the field strength. o n T i l R b E u C p N re © e b o t t h s d e (2. so the total dipole moment is zero. i. The non-polar molecule thus develops an induced dipole moment. a dielectric develops a net dipole moment in the presence of an external field. For linear isotropic dielectrics. The dielectric is said to be polarised by the external field. When an external field is applied. (b) Polar molecules.Physics In an external electric field. a rectangular dielectric slab placed in a uniform external field E 0 parallel to two of its faces. the ‘induced dipole moment’ effect as for non-polar molecules.37) where ce is a constant characteristic of the dielectric and is known as the electric susceptibility of the dielectric medium. Anywhere inside the dielectric.) Now. The unbalanced charges are the induced charges due to the external field. is the charge on one of the conductors – the total charge of the capacitor is zero. there is evidently a net charge density. That is. as stated above. reduced from the case when no dielectric is present.23. say sp and –s p.38) The constant C is called the capacitance of the capacitor.Electrostatic Potential and Capacitance every volume element ∆ v of the slab has a dipole moment P ∆v in the direction of the field. and potentials V1 and V2. in fact. at the surfaces of the dielectric normal to the electric field. However. The total field in the dielectric is. the positive charge of one dipole sits close to the negative charge of the adjacent dipole. Usually. The volume element ∆v is macroscopically small but contains a very large number of molecular dipoles.23 A uniformly polarised dielectric amounts to induced surface charge density. the field produced by these surface charges opposes the external field. say Q1 and Q 2. This is. though this. the electric field will also be doubled at every point. h s d e CAPACITANCE (2. with potential difference V = V1 – V2 between them. We shall consider only this kind of charge configuration of the capacitor. Q is called the charge of the capacitor. (Even a single conductor can be used as a capacitor by assuming the other at infinity. The capacitance C depends only on the 73 . potential difference V is the work done per unit positive charge in taking a small test charge from the conductor 2 to 1 against the field. in practice. As seen in Fig 2. Clearly. say doubled.24 A system of two conductors Consequently. but no volume charge density. if the charge on the capacitor is. the positive ends of the dipoles remain unneutralised at the right surface and the negative ends at the left surface. thereby. (This follows from the direct proportionality between field and charge implied by Coulomb’s law and the superposition principle. V is also proportional to Q. The electric field in the region between the conductors is proportional to the charge Q. We should note that the surface charge density ±s p arises from bound (not free charges) in the dielectric.11 CAPACITORS A capacitor is a system of two conductors separated by an insulator (Fig. because. 2. Thus the polarised dielectric is equivalent to two charged surfaces with induced surface charge densities.) The conductors may be so charged by connecting them to the two terminals of a battery. The conductors have charges. 2. C is independent of Q or V. the two conductors have charges Q and – Q. the ratio Q/V is a constant: o n T i l R b E u C p N re © e b o t t AND FIGURE 2. and separated by an insulator forms a capacitor. FIGURE 2. the volume element ∆v has no net charge (though it has net dipole moment).24). This is of practical importance. for a capacitor to store a large amount of charge without leaking. for air it is about 3 × 106 Vm–1. the most common units are its sub-multiples 1 µF = 10–6 F. there is a limit to the amount of charge that can be stored on a given capacitor without significant leaking. size. In practice. etc.25). For a separation between conductors of the order of 1 cm or so. a capacitor is a key element of most ac circuits with important functions. A strong electric field can ionise the surrounding air and accelerate the charges so produced to the oppositely charged plates. The two plates have charges Q and –Q . V is small for a given Q. Equation (2. as described in Chapter 7. Thus. this field corresponds to a potential difference of 3 × 104 V between the conductors. The maximum electric field that a dielectric medium can withstand without break-down (of its insulating property) is called its dielectric strength. Besides its use in storing charge. a farad is a very big unit. A capacitor with fixed capacitance is symbolically shown as -|| --.] The SI unit of capacitance is 1 farad (=1 coulomb volt-1) or 1 F = 1 C V –1. 2.Physics geometrical configuration (shape.25 T i l R b E u C p N re © e b o t t h s d e 2. we can use the result on electric field by an infinite plane sheet of uniform surface charge density (Section 1. the electric field in different regions is: Outer region I (region above the plate 1). In other words. The parallel plate capacitor. Using Eq. separation) of the system of two conductors. [As we shall see later.39) . while the one with variable capacitance is shown as .33). o n FIGURE 2. the charge of the capacitor leaks away due to the reduction in insulating power of the intervening medium. Let A be the area of each plate and d the separation between them. The effect of a dielectric medium between the plates is discussed in the next section. 1 nF = 10–9 F. Put differently. This means a capacitor with large capacitance can hold large amount of charge Q at a relatively small V.38) shows that for large C . Plate 1 has surface charge density s = Q /A and plate 2 has a surface charge density –s .15). it also depends on the nature of the insulator (dielectric) separating the two conductors. 74 (2. thereby neutralising the charge on the capacitor plates. High potential difference implies strong electric field around the conductors.12 THE PARALLEL PLATE CAPACITOR A parallel plate capacitor consists of two large plane parallel conducting plates separated by a small distance (Fig. its capacitance should be high enough so that the potential difference and hence the electric field do not exceed the break-down limits. 1 pF = 10–12 F. We first take the intervening medium between the plates to be vacuum. Since d is much smaller than the linear dimension of the plates (d2 << A). (1. at least partly. ] However. By the same token.magnet. depends only on the geometry of the system. For plates with finite area.10. As before. Now for uniform electric field. as expected. the electric fields due to the two charged plates add up. giving (2.Electrostatic Potential and Capacitance Outer region II (region below the plate 2).45) h s http://micro. Thus. o n T i l R b E u C p N re © e b o t t (2. d = 1 mm. we have two large plates. say 1 cm: which is a plate about 30 km in length and breadth! 2. we get (You can check that if 1F= 1C V –1 = 1C (NC–1m)–1 = 1 C2 N–1m–1. potential difference is simply the electric field times the distance between the plates.42) The capacitance C of the parallel plate capacitor is then = which. each of area A.41) The direction of electric field is from the positive to the negative plate. capacitors in action Interactive Java tutorial d e OF DIELECTRIC ON CAPACITANCE 75 .40) In the inner region between the plates 1 and 2. (2.35). The field lines bend outward at the edges – an effect called ‘fringing of the field’. When there is vacuum between the plates. The charge on the plates is ±Q. (2. as remarked earlier. and the field there is given by Eq.edu/electromag/java/capacitance/ Factors affecting capacitance. [E and s are related by Eq. this will not be true near the outer boundaries of the plates. that is.41). Another way of seeing the ‘bigness’ of 1F is to calculate the area of the plates needed to have C = 1F for a separation of.) This shows that 1F is too big a unit in practice. for d2 << A. For typical values like A = 1 m2. separated by a distance d.44) (2.43) (2. (2. these effects can be ignored in the regions sufficiently far from the edges. let us see how the capacitance of a parallel plate capacitor is modified when a dielectric is present. (2.13 EFFECT With the understanding of the behavior of dielectrics in an external field developed in Section 2. s will not be strictly uniform on the entire plate. the electric field is localised between the two plates and is uniform throughout. corresponding to the charge density ±s (with s = Q/A ).fsa. 53) is called the dielectric constant of the substance.46) and (2.Physics and the potential difference V0 is V0 = E0d The capacitance C 0 in this case is (2.49).46) Consider next a dielectric inserted between the plates fully occupying the intervening region. is then (2. as explained in Section 2.. From Eqs. e 0 is called the permittivity of the vacuum. (s – sp ) is proportional to s and we can write where K is a constant characteristic of the dielectric. Clearly. (2. the effect is equivalent to two charged sheets (at the surfaces of the dielectric normal to the field) with surface charge densities s p and –sp. with dielectric between the plates.49) For linear dielectrics. the dielectric constant of a substance is the factor (>1) by which the capacitance increases from its vacuum value. when the dielectric is inserted fully between the plates of a capacitor. (2. we expect sp to be proportional to E0.e. Though we arrived at . The dielectric is polarised by the field and.52) For vacuum K = 1 and e = e 0. That is.51) The product e K is called the permittivity of the medium and is 0 denoted by e e = e0 K (2. The electric field in the dielectric then corresponds to the case when the net surface charge density on the plates is ±(s – sp ).48) (2.47) (2.54) Thus. to s . As remarked before.50) The capacitance C.10. o n 76 T i l R b E u C p N re © e b o t t so that the potential difference across the plates is h s d e (2. i. 51) (2. We then have (2. K > 1. from Eq. it is clear that K is greater than 1. Thus. The dimensionless ratio (2. Thus. (2. as seen in above equation. If the dielectric is now inserted. Since P is in the same direction as E . Above equation is general. ELECTRIC DISPLACEMENT We have introduced the notion of dielectric constant and arrived at Eq.54). E and D are parallel. without giving the explicit relation between the induced charge density sp and the polarisation P.8 A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4) d. It is a vector quantity. We take without proof the result that g where is a unit vector along the outward normal to the surface. induced charge density is positive and at the left surface. true for any shape of the dielectric. For a dielectric medium. the corresponding role is taken up by D. all the three vectors P. in fact. it is negative. Putting the equation for electric field in vector form g g or (e0 E + P) g =s The quantity e 0 E + P is called the electric displacement and is denoted by D. The potential difference will then be o n T i l R b E u C p N re © e b o t t EXAMPLE 2. E is related to the free charge density s . How is the capacitance changed when the slab is inserted between the plates? Solution Let E 0 = V0/ d be the electric field between the plates when there is no dielectric and the potential difference is V0. D = e0 K E and P = D –e0E = e0 (K –1)E This gives for the electric susceptibility ce defined in Eq. For the slab in Fig.Electrostatic Potential and Capacitance Eq. (2. (2. it holds good for any type of capacitor and can. The ratio of the magnitudes of D and E is Thus. be viewed in general as a definition of the dielectric constant of a substance. 2. Thus at the right surface. D = e0 E + P.8 h s d e 77 . where d is the separation of the plates. D g =s . The significance of D is this : in vacuum.37) ce = e0 (K –1) Example 2. When a dielectric medium is present. as guessed already in our qualitative discussion before. it is D not E that is directly related to free charge density s. the electric field in the dielectric will be E = E0/ K. P is along at the right surface and opposite to at the left surface.54) for the case of a parallel plate capacitor.23. We can combine several capacitors of capacitance C 1. Two simple possibilities are discussed below. Charge would flow until the net charge on both C 1 and C 2 is zero and there is no electric field in the conductor connecting C 1 and C 2 . The left plate of C 1 and the right plate of C2 are connected to two terminals of a battery and have charges Q and –Q . (2. in the series combination. (2.26 Combination of two capacitors in series.26 shows capacitors C1 and C2 combined in series. . Cn to obtain a system with some effective capacitance C .14.57) with Eq. This would result in an electric field in the conductor connecting C 1and C 2. It then follows that the right plate of C 1 has charge –Q and the left plate of C 2 has charge Q. o n T i l R b E u C p N re © e b o t t 2.55) h s d e (2. the net charge on each capacitor would not be zero.56).56) Now we can regard the combination as an effective capacitor with charge Q and potential difference V. charges on the two plates (±Q) are the same on each capacitor. respectively.1 Capacitors in series V = V1 + V2 = i. Figure 2.. The effective capacitance of the combination is (2.…. The total potential drop V across the combination is the sum of the potential drops V 1 and V2 across C1 and C2. The capacitance thus increases 2.14 COMBINATION OF CAPACITORS FIGURE 2. respectively.8 The potential difference decreases by the factor (K + 3)/K while the free charge Q0 on the plates remains unchanged. C 2.58) 78 FIGURE 2. .57) We compare Eq. The effective capacitance depends on the way the individual capacitors are combined.27 Combination of n capacitors in series. Thus. (2. and obtain (2. If this was not so.e.Physics E XAMPLE 2. CV = C 1V + C 2V + . (2.59) Following the same steps as for the case of two capacitors.) o n T i l R b E u C p N re © e b o t t E XAMPLE 2. In this case.29 h s d e FIGURE 2..64) C = C1 + C2 The general formula for effective capacitance C for parallel combination of n capacitors [Fig.66) i.28 (b)] follows similarly.61) Q1 = C 1V.. 79 . Cn Example 2.67) C = C1 + C2 + . (2. (Note. the same potential difference is applied across both the capacitors.62) Q = Q1 + Q 2 and potential difference V.60) 2.63). + Q n (2. Equation (2. But the plate charges (±Q1) on capacitor 1 and the plate charges (±Q 2) on the capacitor 2 are not necessarily the same: (2.. the char ge on a capacitor is the charge on the plate with higher potential. we get the general formula for effective capacitance of a series combination of n capacitors: (2. (b) n capacitors.55).29.Electrostatic Potential and Capacitance The proof clearly goes through for any number of capacitors arranged in a similar way. generalises to (2. Determine (a) the equivalent capacitance of the network and (b) the charge on each capacitor. from Eq. as shown in Fig.63) Q = CV = C1 V + C 2V The effective capacitance C is.14. (2. CnV which gives (2....65) Q = Q 1 + Q2 + .2 Capacitors in parallel Figure 2. (2. for n capacitors arranged in series.9 FIGURE 2.28 Parallel combination of (a) two capacitors. equal and opposite to the charge on the plate with lower potential..e. Q2 = C2V The equivalent capacitor is one with charge (2. 2.28 (a) shows two capacitors arranged in parallel. 2.9 A network of four 10 µF capacitors is connected to a 500 V supply. say Q . Work in charging the capacitor may be done in this step (d W¢ ). At this stage. Next imagine that a small from Q ¢to Q¢ + δ Q¢ . is a system of two conductors with charge Q and –Q . (b) Clearly. since the potential difference across AB is Q/ C1. resulting in charge Q ′ on viewed as stored in the energy of conductor 1 increasing to Q¢ + d Q¢ . conductor 2 has charge –Q at the end (Fig 2. the potential difference V¢between conductors 1 to 2 is Q¢ FIGURE 2. conductor 1 gets charge Q.30 (a) Work done in a small /C . o n T i l R b E u C p N re © e b o t t .9 Also. C1.15 ENERGY STORED IN A CAPACITOR 80 (2. The effective capacitance C′ of these three capacitors is given by For C1 = C2 = C3 = 10 µF. vanishingly small) amount of charge. across CD is Q/ C3 . the charge on each of the capacitors. (b) Total work done charge d Q¢is transferred from conductor 2 to 1. Q ¢ / C4 = 500 V. C2 and C3 are connected in series. C¢= (10/3) µF. the equivalent capacitance C of the network is C = C¢+ C4 = µF =13. we have E XAMPLE 2. Consider the intermediate situation when the conductors 1 and 2 have charges Q¢and –Q¢respectively. so that at the end.68) . In transferring positive charge from conductor 2 to conductor 1. consider initially two uncharged conductors 1 and 2. Imagine next a process of transferring charge from conductor 2 to conductor 1 bit by bit.30 ). C2 and C3 is the same. To calculate the total work done. and h s d e 2. since at any stage conductor 1 is at a higher potential than conductor 2. work will be done externally. Now. across BC is Q / C2. The network has C¢and C4 connected in parallel. as we have seen above. is given by electric field between the plates.e. we first calculate the work done in a small step involving transfer of an infinitesimal (i.. This gives for the given value of the capacitances. from the figure. Let the charge on C4 be Q¢ . Thus. C1. where C is the step of building charge on conductor 1 capacitance of the system. To determine the energy stored in this configuration.3µ F A capacitor.Physics Solution (a) In the given network. By charge conservation. 75) . The total work done (W ) is the sum of the small work (d W ) over the very large number of steps involved in building the charge Q¢from zero to Q. this stored-up energy is released.68) by integration This is not surprising since integration is nothing but summation of a large number of small terms. It is possible to view the potential energy of the capacitor as ‘stored’ in the electric field between the plates.69) are identical because the term of second order in d Q¢ . We can write the final result.74) and (2.74) h s d e 81 . i. dQ¢2 /2C. (2. = = (2.. For the same reason. Eq. the final result for potential energy [Eq.70) The same result can be obtained directly from Eq.72) in different ways (2.68) and (2. (2. since dQ ¢is arbitrarily small.69) Equations (2. is negligible. (2. Energy stored in the capacitor = The surface charge density s is related to the electric field E between the plates. (2.71) (2. a parallel plate capacitor [of area A(of each plate) and separation d between the plates]. Eq.72) (2. we get Energy stored in the capacitor U= (2.75) From Eqs.73)] is independent of the manner in which the charge configuration of the capacitor is built up. this work is stored in the form of potential energy of the system. To see this. consider for simplicity.76) o n T i l R b E u C p N re © e b o t t (2.e.Electrostatic Potential and Capacitance Since d Q¢can be made as small as we like.68) can be written as (2. (2.73) Since electrostatic force is conservative. (2. When the capacitor discharges. 10 . (2. EXAMPLE 2. 2.5 × 10 –6 J (b) In the steady situation. the two capacitors have their positive plates at the same potential. How much electrostatic energy is stored by the capacitor? (b) The capacitor is disconnected from the battery and connected to another 900 pF capacitor [Fig. u =(1/2)e E2 (2. Example 2. During this period. the result on energy density of an electric field is. The charge on each capacitor is then Q¢ = CV ¢ . and their negative plates at the same potential. The total energy of the system is Thus in going from (a) to (b). though no charge is lost. Let the common potential difference be V′ . Eq (2.31(b)].31 h s d e Solution (a) The charge on the capacitor is –12 –8 Q = CV = 900 × 10 F × 100 V = 9 × 10 C The energy stored by the capacitor is 2 = (1/2) CV = (1/2) QV = (1/2) × 9 × 10–8C × 100 V = 4. If we define energy density as energy stored per unit volume of space. Energy is lost during this time in the form of heat and electromagnetic radiation.77) 0 Though we derived Eq. the final energy is only half the initial energy.31(a)]. By charge conservation. This implies V¢ = V/2. in fact. 2.77) for the case of a parallel plate capacitor.76) shows that Energy density of electric field. a transient current flows from the first capacitor to the second. very general and holds true for electric field due to any configuration of charges. Where has the remaining energy gone? There is a transient period before the system settles to the situation (b). Q¢ = Q/2.Physics Note that Ad is the volume of the region between the plates (where electric field alone exists).10 (a) A 900 pF capacitor is charged by 100 V battery [Fig. What is the electrostatic energy stored by the system? o n 82 T i l R b E u C p N re © e b o t t FIGURE 2. html h s Van de Graaff generator. on which we place a charge Q. We see that.com/emotor/vdg. the field outside the sphere is just that of a point charge Q at the centre.16 VAN DE GRAAFF GENERATOR This is a machine that can build up high voltages of the order of a few million volts.html http://www. protons. and inside it is constant. 2. the charge q on the former o n T i l R b E u C p N re © e b o t t (2. namely the value at the radius R . We thus have: Potential inside conducting spherical shell of radius R carrying charge Q = constant (2. Taking both charges q and Q into account we have for the total potential V and the potential difference the values Assume now that q is positive.32. the inner sphere is always at a higher potential: the difference V (r )–V (R ) is positive. independent of the amount of charge Q that may have accumulated on the larger sphere and even if it is positive. carrying some charge q.80) FIGURE 2.coe. The resulting large electric fields are used to accelerate charged particles (electrons. This charge spreads itself uniformly all over the sphere.ufrj. The potential due to Q is constant upto radius R and so cancels out in the difference! This means that if we now connect the smaller and larger sphere by a wire. principle and demonstration: d e 83 . The potential due to this new charge clearly has the following values at the radii indicated: Potential due to small sphere of radius r carrying charge q at surface of small sphere at large shell of radius R. So the potential outside is that of a point charge. Suppose we have a large spherical conducting shell of radius R . and place it at the centre. let us suppose that in some way we introduce a small sphere of radius r.14. The principle underlying the machine is as follows. as shown in Fig. ions) to high energies needed for experiments to probe the small scale structure of matter. while the field inside the sphere vanishes. As we have seen in Section 1.Electrostatic Potential and Capacitance 2.78) Now. http://amasce.br/~acmg/myvdg.79) (2. into the large one.32 Illustrating the principle of the electrostatic generator. one at the centre of the shell. Thus positive charge is transferred to the shell. o n 84 T i l R b E u C p N re © e b o t t SUMMARY 1. Thus.33. where it spreads out uniformly on the outer surface. to the top. at least until we reach the breakdown field of air. The potential (Eq. 3. we can in this way keep piling up larger and larger amount of charge on the latter. provided we are somehow able to introduce the small charged sphere into the larger one. 2. Potential at a point is arbitrary to within an additive constant. is wound around two pulleys – one at ground level. voltage differences of as much as 6 or 8 million volts (with respect to ground) can be built up. potential at a point with position vector r due to a point charge Q placed at the origin is given is given by The electrostatic potential at a point with position vector r due to a point dipole of dipole moment p placed at the origin is g . If potential at infinity is chosen to be zero. height several meters above the ground on an insulating column. sprayed on to it by a brush at ground level. A schematic diagram of the van de Graaff generator is given in Fig. There it transfers its positive charge to another conducting brush connected to the large shell. 2. 2.Physics will immediately flow onto the matter. It is a machine capable of building up potential difference of a few million volts. even though the charge Q may be quite large.78) at the outer sphere would also keep rising. Potential at a point is the work done per unit charge (by an external agency) in bringing a charge from infinity to that point. In this way. like rubber or silk. which is the difference in potential energy of charge q between the final and initial points. h s d e Electrostatic for ce is a conservative force. since it is the potential difference between two points which is physically significant. A long narrow endless belt insulating material. This belt is kept continuously moving by a motor driving the lower pulley. and fields close to the breakdown field of air which is about 3 × 106 V/m. The natural tendency is for positive charge to move from higher to lower potential.33 Principle of construction shell (of few metres radius) is supported at a of Van de Graaf f generator. Work done by an exter nal force (equal and opposite to the electrostatic force) in bringing a charge q from a point R to a point P is VP – VR . A large spherical conducting FIGURE 2. This is the principle of the van de Graaff generator. It continuously carries positive charge. 5. r 2 is given by where r12 is distance between q1 and q2 . the potential at a point P is given by the superposition principle where r1P is the distance between q1 and P. rn. The potential energy of a charge q in an external potential V( r) is qV ( r). q2 . r2 . .. 10. C is determined purely geometrically. q2 at r1. gives rise to a field in the opposite direction..E . In a cavity within a conductor (with no charges).Electrostatic Potential and Capacitance The result is true also for a dipole (with charges –q and q separated by 2 a) for r >> a . The unit of capacitance is farad:. For a point charge. by the shapes. C= where A is the area of each plate and d the separation between them. concentric spheres centered at a location of the charge are equipotential surfaces. For a parallel plate capacitor (with vacuum between the plates). Charges in a conductor can reside only at its surface. .. Electrostatics field E is zero in the interior of a conductor. sizes and relative positions of the two conductors. The potential energy of a dipole moment p in a uniform electric field E is –p. qn with position vectors r1 . E is in the direction of the steepest decrease of potential. This effect. called polarisation. surface and s is the surface charge density. 8. 1 F = 1 C V –1. 7. C = KC 0 where K is the dielectric constant of the insulating substance. Its capacitance is defined by C = Q/V. o n T i l R b E u C p N re © e b o t t h s d e 85 . where Q and –Q are the charges on the two conductors and V is the potential difference between them. 6. the capacitance C increases from its value C 0 when there is no medium (vacuum). Potential energy stored in a system of charges is the work done (by an external agency) in assembling the charges at their locations. just outside the surface of a charged conductor. 4. An equipotential surface is a surface over which potential has a constant value. Consequently. E is nor mal to the surface given by where is the unit vector along the outward normal to the 9. the electric field is zero. The net electric field inside the dielectric and hence the potential difference between the plates is thus reduced. The electric field E at a point is perpendicular to the equipotential surface through the point. Potential energy of two charges q 1. the electric field due to the charged plates induces a net dipole moment in the dielectric. Potential is constant within and on the surface of a conductor. A capacitor is a system of two conductors separated by an insulator.. For a charge configuration q1 . as and so on.. If the medium between the plates of a capacitor is filled with an insulating substance (dielectric). It is zero inside and outside. Electric field is discontinuous across the surface of a spherical charged shell. h s d e φ or V Potential difference is physically significant C [M–1 L –2 T–4 A 2] [L –2 F P AT] C m-2 Dipole moment per unit volume K [Dimensionless] POINTS TO PONDER 1. . 13. For capacitors in the series combination. A capacitor is so configured that it confines the electric field lines within a small region of space. But if there is a force on a charge. 12. C 2. how can it be at rest? Thus. the oscillations are damped and the dipole eventually aligns with E. which can be used for accelerating charged particles. 2. the total capacitance C is given by In the parallel combination. Only if there is a dissipative mechanism. C 3 . with charge Q and voltage V is Physical quantity Potential Capacitance Polarisation Dielectric constant o n 86 T i l R b E u C p N re © e b o t t Symbol Dimensions [M1 L 2 T–3 A–1] Unit V Remark The electric energy density (energy per unit volume) in a region with 2 electric field is (1/2)e0 E . A Van de Graaff generator consists of a large spherical conducting shell (a few metre in diameter). By means of a moving belt and suitable brushes.. The energy U stored in a capacitor of capacitance C . even though field may have considerable strength. Thus. The torque p × E on a dipole causes it to oscillate about E . 4. are individual capacitances.Physics 11. Electric potential is. the total capacitance C is: C = C 1 + C2 + C3 + . 3. the potential difference between the two conductors of a capacitor is small. charge is continuously transferred to the shell and potential difference of the order of several million volts is built up. Electrostatics deals with forces between charges at rest. however continuous across the surface.. it should be understood that each charge is being kept at rest by some unspecified force that opposes the net Coulomb force on the charge. where C1 . when we are talking of electrostatic force between charges.. equal to q/4πe0 R at the surface.. Electrostatic Potential and Capacitance 5. 6. Potential due to a charge q at its own location is not defined – it is infinite. In the expression q V ( r) for potential energy of a charge q , V ( r) is the potential due to external charges and not the potential due to q. As seen in point 5, this expression will be ill-defined if V ( r) includes potential due to a charge q itself. A cavity inside a conductor is shielded from outside electrical influences. It is worth noting that electrostatic shielding does not work the other way round; that is, if you put charges inside the cavity, the exterior of the conductor is not shielded from the fields by the inside charges. 7. 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. A regular hexagon of side 10 cm has a charge 5 µ C at each of its vertices. Calculate the potential at the centre of the hexagon. Two charges 2 µC and –2 µ C are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface? A spherical conductor of radius 12 cm has a charge of 1.6 × 10–7C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere? A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6? Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply? Thr ee capacitors of capacitances 2 pF, 3 pF and 4 pF ar e connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply. In a parallel plate capacitor with air between the plates, each plate –3 2 has an area of 6 × 10 m and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor? o n T i l R b E u C p N re © e b o t t EXERCISES h s d e 87 Physics 2.9 Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected. A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor? A 600pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process? 2.10 2.11 o n 88 T i l R b E u C p N re © e b o t t 2.12 2.13 2.14 2.15 2.16 g ADDITIONAL EXERCISES A charge of 8 mC is located at the origin. Calculate the work done in taking a small charge of –2 × 10–9 C from a point P (0, 0, 3 cm) to a point Q (0, 4 cm, 0), via a point R (0, 6 cm, 9 cm). A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube. Two tiny spheres carrying charges 1.5 µC and 2.5 µC are located 30 cm apart. Find the potential and electric field: (a) at the mid-point of the line joining the two charges, and (b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain. (a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by h s d e 2.17 2.18 where is a unit vector normal to the surface at a point and s is the surface charge density at that point. (The direction of is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is s / e0. (b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.] A long charged cylinder of linear charged density l is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders? In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å: Electrostatic Potential and Capacitance (a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton. (b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)? (c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation? If one of the two electrons of a H2 molecule is removed, we get a + hydrogen molecular ion H + 2. In the ground state of an H2, the two protons are separated by roughly 1.5 Å, and the electron is roughly 1 Å from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy. Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions. 2.19 2.20 2.21 2.22 Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a), respectively. (a) What is the electrostatic potential at the points (0, 0, z) and (x, y , 0) ? (b) Obtain the dependence of potential on the distance r of a point from the origin when r/ a >> 1. (c) How much work is done in moving a small test charge from the point (5,0,0) to (–7,0,0) along the x -axis? Does the answer change if the path of the test charge between the same points is not along the x -axis? Figure 2.34 shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on r for r/ a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge). 2.23 An electrical technician requires a capacitance of 2 µF in a circuit across a potential differ ence of 1 kV. A large number of 1 µF capacitors are available to him each of which can withstand a potential differ ence of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors. What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of µF or less. However, electr olytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.] o n T i l R b E u C p N re © e b o t t FIGURE 2.34 h s d e 2.24 89 Physics 2.25 Obtain the equivalent capacitance of the network in Fig. 2.35. For a 300 V supply, determine the charge and voltage acr oss each capacitor. o n 90 T i l R b E u C p N re © e b o t t FIGURE 2.35 2.26 2.27 2.28 2.29 The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. (a) How much electrostatic energy is stored by the capacitor? (b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates. A 4 µF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 µ F capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation? Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (½) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor ½. A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports (Fig. 2.36). Show h s d e FIGURE 2.36 Electrostatic Potential and Capacitance that the capacitance of a spherical capacitor is given by 2.30 2.31 where r1 and r 2 are the radii of outer and inner spheres, respectively. A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µ C. The space between the concentric spheres is filled with a liquid of dielectric constant 32. (a) Determine the capacitance of the capacitor. (b) What is the potential of the inner sphere? (c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller. Answer carefully: (a) Two large conducting spheres carrying charges Q1 and Q 2 are brought close to each other. Is the magnitude of electr ostatic force between them exactly given by Q1 Q2/4πe0r 2, where r is the distance between their centres? (b) If Coulomb’s law involved 1/r3 dependence (instead of 1/r 2), would Gauss’s law be still true ? (c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point? (d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical? (e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there? (f ) What meaning would you give to the capacitance of a single conductor? (g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6). A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 µ C. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends). 2.32 2.33 2.34 A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm–1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF? Describe schematically the equipotential surfaces corresponding to (a) a constant electric field in the z-direction, (b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction, o n T i l R b E u C p N re © e b o t t h s d e 91 Physics a single positive charge at the origin, and a uniform grid consisting of long equally spaced parallel charged wires in a plane. In a Van de Graaff type generator a spherical metal shell is to be a 15 × 106 V electrode. The dielectric strength of the gas surrounding 7 –1 the electrode is 5 × 10 Vm . What is the minimum radius of the spherical shell requir ed? (You will lear n from this exercise why one cannot build an electrostatic generator using a very small shell which requires a small charge to acquire a high potential.) A small sphere of radius r1 and charge q 1 is enclosed by a spherical shell of radius r2 and charge q 2. Show that if q1 is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2 on the shell is. Answer the following: (a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field –1 is about 100 Vm . Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!) (b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of 2 area 1m . Will he get an electric shock if he touches the metal sheet next morning? (c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged? (d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm–1 at its surface in the downward direction, corresponding to a surface charge density = –10–9 C m–2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually all over the globe pump an equal amount of negative charge on the earth.) (c) (d) 2.35 2.36 o n 92 T i l R b E u C p N re © e b o t t 2.37 h s d e Chapter Three CURRENT ELECTRICITY o n T i l R b E u C p N re © e b o t t 3.1 INTRODUCTION 3.2 ELECTRIC CURRENT h s d e In Chapter 1, all charges whether free or bound, were considered to be at rest. Charges in motion constitute an electric current. Such currents occur naturally in many situations. Lightning is one such phenomenon in which charges flow from the clouds to the earth through the atmosphere, sometimes with disastrous results. The flow of charges in lightning is not steady, but in our everyday life we see many devices where charges flow in a steady manner, like water flowing smoothly in a river. A torch and a cell-driven clock are examples of such devices. In the present chapter, we shall study some of the basic laws concerning steady electric currents. Imagine a small area held normal to the direction of flow of charges. Both the positive and the negative charges may flow forward and backward across the area. In a given time interval t, let q+ be the net amount (i.e., forward minus backward) of positive charge that flows in the forward direction across the area. Similarly, let q– be the net amount of negative charge flowing across the area in the forward direction. The net amount of charge flowing across the area in the forward direction in the time interval t, then, is q = q+– q –. This is proportional to t for steady current Physics and the quotient (3.1) is defined to be the current across the area in the forward direction. (If it turn out to be a negative number, it implies a current in the backward direction.) Currents are not always steady and hence more generally, we define the current as follows. Let ∆Q be the net charge flowing across a crosssection of a conductor during the time interval ∆t [i.e., between times t and (t + ∆t )]. Then, the current at time t across the cross-section of the conductor is defined as the value of the ratio of ∆ Q to ∆t in the limit of ∆t tending to zero, o n 94 T i l R b E u C p N re © e b o t t 3.3 ELECTRIC CURRENTS IN In SI units, the unit of current is ampere. An ampere is defined through magnetic effects of currents that we will study in the following chapter. An ampere is typically the order of magnitude of currents in domestic appliances. An average lightning carries currents of the order of tens of thousands of amperes and at the other extreme, currents in our nerves are in microamperes. h s d e (3.2) CONDUCTORS An electric charge will experience a force if an electric field is applied. If it is free to move, it will thus move contributing to a current. In nature, free charged particles do exist like in upper strata of atmosphere called the ionosphere . However, in atoms and molecules, the negatively charged electrons and the positively charged nuclei are bound to each other and are thus not free to move. Bulk matter is made up of many molecules, a gram of water, for example, contains approximately 1022 molecules. These molecules are so closely packed that the electrons are no longer attached to individual nuclei. In some materials, the electrons will still be bound, i.e., they will not accelerate even if an electric field is applied. In other materials, notably metals, some of the electrons are practically free to move within the bulk material. These materials, generally called conductors, develop electric currents in them when an electric field is applied. If we consider solid conductors, then of course the atoms are tightly bound to each other so that the current is carried by the negatively charged electrons. There are, however, other types of conductors like electrolytic solutions where positive and negative charges both can move. In our discussions, we will focus only on solid conductors so that the current is carried by the negatively charged electrons in the background of fixed positive ions. Consider first the case when no electric field is present. The electrons will be moving due to thermal motion during which they collide with the fixed ions. An electron colliding with an ion emerges with the same speed as before the collision. However, the direction of its velocity after the collision is completely random. At a given time, there is no preferential direction for the velocities of the electrons. Thus on the average, the Current Electricity number of electrons travelling in any direction will be equal to the number of electrons travelling in the opposite direction. So, there will be no net electric current. Let us now see what happens to such a piece of conductor if an electric field is applied. To focus our thoughts, imagine the conductor in the shape of a cylinder of radius R (Fig. 3.1). Suppose we now take two thin circular discs FIGURE 3.1 Charges +Q and –Q put at the ends of a dielectric of the same radius and put of a metallic cylinder. The electrons will drift positive charge +Q distributed over one disc because of the electric field created to and similarly –Q at the other disc. We attach neutralise the charges. The current thus will stop after a while unless the charges +Q the two discs on the two flat surfaces of the and – Q are continuously replenished. cylinder. An electric field will be created and is directed from the positive towards the negative charge. The electrons will be accelerated due to this field towards +Q. They will thus move to neutralise the charges. The electrons, as long as they are moving, will constitute an electric current. Hence in the situation considered, there will be a current for a very short while and no current thereafter. We can also imagine a mechanism where the ends of the cylinder are supplied with fresh charges to make up for any charges neutralised by electrons moving inside the conductor. In that case, there will be a steady electric field in the body of the conductor. This will result in a continuous current rather than a current for a short period of time. Mechanisms, which maintain a steady electric field are cells or batteries that we shall study later in this chapter. In the next sections, we shall study the steady current that results from a steady electric field in conductors. 3.4 OHM’S LAW A basic law regarding flow of currents was discovered by G.S. Ohm in 1828, long before the physical mechanism responsible for flow of currents was discovered. Imagine a conductor through which a current I is flowing and let V be the potential difference between the ends of the conductor. Then Ohm’s law states that V∝I or, V = R I where the constant of proportionality R is called the resistance of the conductor. The SI units of resistance is ohm, and is denoted by the symbol Ω. The resistance R not only depends on the material of the conductor but also on the dimensions of the conductor. The dependence of R on the dimensions of the conductor can easily be determined as follows. Consider a conductor satisfying Eq. (3.3) to be in the form of a slab of length l and cross sectional area A [Fig. 3.2(a)]. Imagine placing two such identical slabs side by side [Fig. 3.2(b)], so that the length of the combination is 2l . The current flowing through the combination is the same as that flowing through either of the slabs. If V is the potential difference across the ends of the first slab, then V is also the potential difference across the ends of the second slab since the second slab is o n T i l R b E u C p N re © e b o t t (3.3) h s d e FIGURE 3.2 Illustrating the relation R = r l/ A for a rectangular slab of length l and area of cross-section A . 95 Physics identical to the first and the same current I flows through both. The potential difference across the ends of the combination is clearly sum of the potential difference across the two individual slabs and hence equals 2 V. The current through the combination is I and the resistance of the combination R C is [from Eq. (3.3)], GEORG SIMON OHM (1787–1854) since V/I = R, the resistance of either of the slabs. Thus, doubling the length of a conductor doubles the resistance. In general, then resistance is proportional to length, Georg Simon Ohm (1787– 1854) German physicist, professor at Munich. Ohm was led to his law by an analogy between the conduction of heat: the electric field is analogous to the temperature gradient, and the electric current is analogous to the heat flow. o n 96 T i l R b E u C p N re © e b o t t Combining Eqs. (3.5) and (3.7), we have and hence for a given conductor (3.5) Next, imagine dividing the slab into two by cutting it lengthwise so that the slab can be considered as a combination of two identical slabs of length l , but each having a cross sectional area of A/2 [Fig. 3.2(c)]. For a given voltage V across the slab, if I is the current through the entire slab, then clearly the current flowing through each of the two half-slabs is I/2. Since the potential difference across the ends of the half-slabs is V, i.e., the same as across the full slab, the resistance of each of the half-slabs R1 is (3.6) h s d e (3.4) Thus, halving the area of the cross-section of a conductor doubles the resistance. In general, then the resistance R is inversely proportional to the cross-sectional area, (3.7) (3.8) (3.9) where the constant of proportionality r depends on the material of the conductor but not on its dimensions. r is called resistivity . Using the last equation, Ohm’s law reads (3.10) Current per unit area (taken normal to the current), I /A, is called current density and is denoted by j. The SI units of the current density are A/m2. Further, if E is the magnitude of uniform electric field in the conductor whose length is l, then the potential difference V across its ends is El. Using these, the last equation reads Current Electricity E l = jr l or, E = j r (3.11) The above relation for magnitudes E and j can indeed be cast in a vector form. The current density, (which we have defined as the current through unit area normal to the current) is also directed along E, and is also a vector j (º j E/E). Thus, the last equation can be written as, E = jr or, j=s E (3.12) (3.13) where s ≡1/r is called the conductivity. Ohm’s law is often stated in an equivalent form, Eq. (3.13) in addition to Eq.(3.3). In the next section, we will try to understand the origin of the Ohm’s law as arising from the characteristics of the drift of electrons. 3.5 D RIFT OF E LECTRONS R ESISTIVITY As remarked before, an electron will suffer collisions with the heavy fixed ions, but after collision, it will emerge with the same speed but in random directions. If we consider all the electrons, their average velocity will be zero since their directions are random. Thus, if there are N electrons and the velocity of the ith electron (i = 1, 2, 3, ... N ) at a given time is vi, then (3.14) Consider now the situation when an electric field is present. Electrons will be accelerated due to this field by (3.15) where –e is the charge and m is the mass of an electron. Consider again the ith electron at a given time t. This electron would have had its last collision some time before t, and let ti be the time elapsed after its last collision. If vi was its velocity immediately after the last collision, then its velocity V i at time t is since starting with its last collision it was accelerated (Fig. 3.3) with an acceleration given by Eq. (3.15) for a time interval ti . The average velocity of the electrons at time t is the average of all the V i’s. The average of v i’s is zero [Eq. (3.14)] since immediately after any collision, the direction of the velocity of an electron is completely random. The collisions of the electrons do not occur at regular intervals but at random times. Let us denote by τ , the average time between successive collisions. Then at a given time, some of the electrons would have spent o n T i l R b E u C p N re © e b o t t AND THE O RIGIN OF h s d e (3.16) FIGURE 3.3 A schematic picture of an electron moving from a point A to another point B through repeated collisions, and straight line travel between collisions (full lines). If an electric field is applied as shown, the electron ends up at point B′ (dotted lines). A slight drift in a direction opposite the electric field is visible. 97 Physics time more than τ and some less than τ . In other words, the time t i in Eq. (3.16) will be less than τ for some and more than τ for others as we go through the values of i = 1, 2 ..... N . The average value of t i then is τ (known as relaxation time). Thus, averaging Eq. (3.16) over the N -electrons at any given time t gives us for the average velocity vd This last result is surprising. It tells us that the electrons move with an average velocity which is independent of time, although electrons are accelerated. This is the phenomenon of drift and the velocity v d in Eq. (3.17) is called the drift velocity. Because of the drift, there will be net transport of charges across any area perpendicular to E. Consider a planar area A, located inside the conductor such that the normal to the area is parallel to E FIGURE 3.4 Current in a metallic (Fig. 3.4). Then because of the drift, in an infinitesimal conductor. The magnitude of current amount of time ∆t , all electrons to the left of the area at density in a metal is the magnitude of charge contained in a cylinder of unit distances upto |vd| ∆t would have crossed the area. If area and length vd. n is the number of free electrons per unit volume in the metal, then there are n ∆ t |vd |A such electrons. Since each electron carries a charge –e, the total charge transported across this area A to the right in time ∆ t is –ne A|v d|∆ t. E is directed towards the left and hence the total charge transported along E across the area is negative of this. The amount of charge crossing the area A in time ∆t is by definition [Eq. (3.2)] I ∆t , where I is the magnitude of the current. Hence, o n 98 T i l R b E u C p N re © e b o t t Substituting the value of |vd | from Eq. (3.17) I = |j |A Hence, from Eqs.(3.19) and (3.20), h s d e (3.17) (3.18) (3.19) By definition I is related to the magnitude |j | of the current density by (3.20) (3.21) The vector j is parallel to E and hence we can write Eq. (3.21) in the vector form (3.22) Comparison with Eq. (3.13) shows that Eq. (3.22) is exactly the Ohm’s law, if we identify the conductivity s as o n T i l R b E u C p N re © e b o t t m –3 h s d e EXAMPLE 3.Current Electricity (3.18) vd = (I/ neA) Now. A = 1. Note that the drift speed of electrons is much –5 smaller.1 × 10–3 m s –1 = 1.6 × 10–19 C. The density of copper is 9. Since 6. Solution (a) The direction of drift velocity of conduction electrons is opposite to the electric field direction. about 10 times the typical ther mal speed at ordinary temperatures. (b) Compare the drift speed obtained above with. I = 1. extremely small.5 u. electrons drift in the direction of increasing potential.23) We thus see that a very simple picture of electrical conduction reproduces Ohm’s law. and its atomic mass is 63.0 × 10–7m2. (ii) An electric field travelling along the conductor has a speed of an electromagnetic wave. 99 . namely equal to 3. (i) thermal speeds of copper atoms at ordinary temperatures. We have. A cubic metre 3 23 of copper has a mass of 9.5 A. e = 1. This figure indicates the random vibrational speeds of copper atoms in a conductor. this is about 2 × 10 m/s.0 × 10 kg. We shall. The drift speed is.5 A.e. independent of E . (ii) speed of propagation of electric field along the conductor which causes the drift motion. The density of conduction electrons.5 g. discuss the limitations of Ohm’s law.1 * See Eq.23) of Chapter 13 from Class XI book. n is equal to the number of atoms per cubic metre (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one). (3. smaller by a factor of 10–11. i. Example 3.0 × 10–7 m2 carrying a current of 1. in comparison. made assumptions that t and n are constants. in the next section.1 (a) Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.. where k B is the Boltzmann typically of the order of 2 constant. The drift speed vd is given by Eq. Assume that each copper atom contributes roughly one conduction electron. For copper at 300 K.0 × 103 kg/m3.0 × 10 8 m s–1 (You will learn about this in Chapter 8).0 × 10 copper atoms have a mass of 63.5 × 10 28 which gives. = 8.1 mm s–1 (b) (i) At a temperature T . the thermal speed* of a copper atom of 2 mass M is obtained from [<(1/2) Mv > = (3/2) k BT ] and is thus . of course. (13. vd = 1. However. (b) Each ‘free’ electron does accelerate. these can be both positive and negative ions. in an electrolyte. From Eq. the paths are. An important quantity is the mobility m defined as the magnitude of the drift velocity per unit electric field: (3. 29 –3 ~10 m . in general. and the electron’s charge is small. how can we still obtain large amounts of current in a conductor? (d) When electrons drift in a metal from lower to higher potential. in an ionised gas. we have vd = . does it mean that all the ‘free’ electrons of the metal are moving in the same direction? (e) Are the paths of electrons straight lines between successive collisions (with the positive ions of the metal) in the (i) absence of electric field. But force should cause acceleration.2 Solution (a) Electric field is established throughout the circuit.17). Establishment of a current does not have to wait for electrons from one end of the conductor travelling to the other end.24) The SI unit of mobility is m2/Vs and is 104 of the mobility in practical units (cm2/Vs). (c) Simple. It loses its drift speed after collision but starts to accelerate and increases its drift speed again only to suffer a collision again and so on. the paths are straight lines. increasing its drift speed until it collides with a positive ion of the metal.5. the electron drift speed is estimated to be only a –1 few mm s for currents in the range of a few amperes? How then is current established almost the instant a circuit is closed? (b) The electron drift arises due to the force experienced by electrons in the electric field inside the conductor. h s d e 3. Why then do the electrons acquire a steady average drift speed? (c) If the electron drift speed is so small. (e) In the absence of electric field. Mobility is positive. therefore. they are electrons and positive charged ions. these mobile charge carriers are electrons. (ii) presence of electric field? o n 100 T i l R b E u C p N re © e b o t t EXAMPLE 3.1 Mobility As we have seen.Physics Example 3. electrons acquire only a drift speed.2 (a) In Example 3. because the electron number density is enormous. (d) By no means. conductivity arises from mobile charge carriers. in the presence of electric field. The drift velocity is superposed over the large random velocities of electrons. On the average. it does take a little while for the current to reach its steady value.1. (3. almost instantly (with the speed of light) causing at every point a local electron drift. In metals. curved. does not produce a current of the same magnitude as I in the opposite direction (Fig. (3. (3. the direction of V keeping its magnitude fixed.3) are actually widely used in electronic circuits. there is more than one value of V for the same current I (Fig.Current Electricity Hence. 3. in a diode which we will study in Chapter 14. The deviations broadly are one or more of the following FIGURE 3.7 RESISTIVITY VARIOUS MATERIALS 101 The resistivities of various common materials are listed in Table 3. i. 3. if I is the current for a certain V.5 The dashed line types: represents the linear Ohm’s (a) V ceases to be proportional to I (Fig. In V versus current I for a good other words.5).6 Characteristic curve of a diode. FIGURE 3. The solid line is the voltage (b) The relation between V and I depends on the sign of V. 3. semiconductors and insulators .7). law.6). Materials and devices not obeying Ohm’s law in the form of Eq. Note the different scales for negative and positive values of the voltage and current. 3. there do exist materials and devices used in electric circuits where the proportionality of V and I does not hold. The materials are classified as conductors.1. In this and a few subsequent chapters.25) where t is the average collision time for electrons. o n T i l R b E u C p N re © e b o t t OF h s d e FIGURE 3. A material exhibiting such behaviour is GaAs.e. for example.. This happens. 3.7 Variation of current versus voltage for GaAs.6 LIMITATIONS OF OHM’S LAW Although Ohm’s law has been found valid over a large class of materials. then reversing conductor. (c) The relation between V and I is not unique. however. we will study the electrical currents in materials that obey Ohm’s law. 5 × 105 1010 – 10 14 1013 – 10 16 ~1014 ~1016 Temperature coefficient of resistivity. The resistivities of semiconductors are also affected by presence of small amount of impurities. Fe.6 × 10–8 10 × 10–8 11 × 10–8 98 × 10–8 ~100 × 10–8 48 × 10–8 3.0005 – 0.0041 0. Cr) Manganin (alloy) Semiconductors Carbon (graphite) Germanium Silicon Insulators Pure Water Glass Hard Rubber NaCl Fused Quartz o n T i l R b E u C p N re © e b o t t 1. nichrome or similar ones.0045 0. rubber and plastics having resistivities 1018 times greater than metals or more.. viz. constantan. TABLE 3. in an increasing order of their values.05 – 0. These.6 × 10–8 1.46 2300 2. however. a (°C) –1 h s d e 0.7 × 10–8 2. At the other end are insulators like ceramic.07 102 Commercially produced resistors for domestic use or in laboratories are of two major types: wire bound resistors and carbon resistors.0043 0.7 × 10–8 5. This last feature is exploited in use of semiconductors for electronic devices.0068 0.0065 0. have resistivities characteristically decreasing with a rise in temperature. These resistances are typically in the range of a fraction of an ohm to a few hundred ohms. . Wire bound resistors are made by winding the wires of an alloy. r (Ω m) at 0°C SOME MATERIALS Conductors Silver Copper Aluminium Tungsten Iron Platinum Mercury Nichrome (alloy of Ni.0009 0. Metals have low resistivities in the range of 10–8 Ωm to 10–6 Ωm.1 R ESISTIVITIES OF Material Resistivity. In between the two a re the semiconductors.0004 0.5 × 10–5 0.0039 0. The choice of these materials is dictated mostly by the fact that their resistivities are relatively insensitive to temperature.Physics depending on their resistivities. manganin.002 × 10–3 – 0. Over a limited range of temperatures. 3. the resistance value is 36 × 104 Ω. 3. Carbon resistors are compact. TABLE 3. The third band indicates the decimal multiplier (as listed in Table 3. Different materials do not exhibit the same dependence on temperatures.2. with a tolerence value of 5%. (b) (47 × 10 Ω ) ± 5%. α is called the temperature co-efficient of resistivity. o n T i l R b E u C p N re © e b o t t OF h s d e 5 10 20 FIGURE 3. The last band stands for tolerance or possible variation in percentage about the indicated values. yellow and gold. Sometimes. rT = r0 [1 + α (T– T0)] (3. and from Eq. Carbon resistors are small in size and hence their values are given using a colour code.26). the dimension of α is (Temperature)–1. For example.2). the resistivity of a metallic conductor is approximately given by.8).8 T EMPERATURE D EPENDENCE R ESISTIVITY The resistivity of a material is found to be dependent on the temperature. this last band is absent and that indicates a tolerance of 20% (Fig. The first two bands from the end indicate the first two significant figures of the resistance in ohms.2 RESISTOR COLOUR CODES Colour Black Brown Red Orange Yellow Green Blue Violet Gray White Gold Silver No colour Number 0 1 2 3 4 5 6 7 8 9 Multiplier 1 10 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 –1 10 –2 Tolerance (%) The resistors have a set of co-axial coloured rings on them whose significance are listed in Table 3. inexpensive and thus find extensive use in electronic circuits. blue. (3.26) where rT is the resistivity at a temperature T and r0 is the same at a reference temperature T0 . if the four colours are orange. 103 .8 Colour coded resistors (a) (22 × 102 Ω ) ± 10%.Current Electricity Resistors in the higher range are made mostly from carbon. that is not too large. 11 Temperature dependence of resistivity for a typical semiconductor. We can qualitatively understand the temperature dependence of resistivity. As we increase temperature. however. Unlike metals. For insulators and semiconductors. 3. FIGURE 3. Some materials like Nichrome (which is an alloy of nickel.Physics For metals. average speed of the electrons.23). iron and chromium) exhibit a very weak dependence of resistivity with temperature (Fig. In a metal. n increases with temperature. . a is positive and values of a for some metals at T0 = 0°C are listed in Table 3. From this equation. (3. At temperatures much lower than 0°C.10 Resistivity rT of nichrome as a function of absolute temperature T.9 Resistivity r T of copper as a function of temperature T . increases resulting in more frequent collisions.1.11. A typical dependence is shown in Fig.9). The relation of Eq. where the graph can be approximated as a straight line. however. o n 104 T i l R b E u C p N re © e b o t t FIGURE 3. thus decreases with temperature.27) r thus depends inversely both on the number n of free electrons per unit volume and on the average time t between collisions. n is not dependent on temperature to any appreciable extent and thus the decrease in the value of t with rise in temperature causes r to increase as we have observed. 3. resistivity of a material is given by (3. Equation (3. the graph. 3. in the light of our derivation of Eq. can be used approximately over a limited range of T around any reference temperature T0. These materials are thus widely used in wire bound standard resistors since their resistance values would change very little with temperatures. which act as the carriers of current. Manganin and constantan have similar properties. deviates considerably from a straight line (Fig. The average time of collisions t.26) implies that a graph of rT plotted against T would be a straight line. the resistivities of semiconductors decrease with increasing temperatures.10).23) so that for such materials.26) thus. r decreases with temperature. (3. This increase more than compensates any decrease in t in Eq.(3. h s d e FIGURE 3. In a few seconds. Example 3. Solution When the current through the element is very small. a steady state will be reached when temperature will rise no further. o n = T i l R b E u C p N re © e b o t t = 820 °C h s d e EXAMPLE 3. the current settles.9 ELECTRICAL ENERGY.23 Ω and Rt = 5. When a negligibly small current passes through it. after a few seconds. we get T2 – T1 that is.4 The resistance of the platinum wire of a platinum resistance thermometer at the ice point is 5 Ω and at steam point is 5. Solution R 0 = 5 Ω . But due to heating effect of the current.3 Ω . The resistance R 2 at the steady temperature T 2 is R2 Using the relation R 2 = R1 [1 + a ( T2 – T 1)] with a = 1. When the toaster is connected to the supply. to a steady value of 2. POWER Consider a conductor with end points A and B.70 × 10 °C . When the toaster is connected to a 230 V supply.3 An electric toaster uses nichrome for its heating element. T2 = (820 + 27. the steady temperature of the heating element (when heating effect due to the current equals heat loss to the surroundings) is 847 °C.39 Ω. When the ther mometer is inserted in a hot bath. the resistance of the platinum wire is 5. heating effects can be ignored and the temperature T1 of the element is the same as room temperature.68 A.795 Ω. and both the r esistance of the element and the current drawn will achieve steady values. This will cause an increase in resistance and a slight decrease in current.795 Ω Now. its initial current will be slightly higher than its steady value of 2.Current Electricity Example 3. R 100 = 5.68 A. Calculate the temperature of the bath. in which a current I is flowing from A to B.0 °C) is found to be 75.0) °C = 847 °C Thus.70 × 10–4 °C–1. is 1. the temperature will rise.4 = 345. its resistance at room temperature (27. The electric potential at A and B are denoted by V (A) 105 . What is the steady temperature of the nichrome element? The temperature coefficient of resistance of nichrome averaged over the temperature range –4 –1 involved.3 EXAMPLE 3.65 °C 3. e. It is this power which heats up.33) as the power loss (“ohmic loss”) in a conductor of resistance R carrying a current I . seen earlier that on the average.31) The energy dissipated per unit time is the power dissipated P = ∆ W/∆ t and we have. their kinetic energy would also change so that the total energy is unchanged.32) h s d e (3. ∆W = I V∆ t (3. Thus. ∆K = – ∆U pot P = I 2 R = V 2/R (3. the energy gained by the charges thus is shared with the atoms. it is Q V(B)..32) and (3. the coil of an electric bulb to incandescence. Thus.3.29) .33). (3.30) Thus. During collisions. by definition. an amount of charge ∆ Q = I ∆t travels from A to B. change in its potential energy ∆ Upot is ∆U pot = Final potential energy – Initial potential energy = ∆ Q[(V (B) – V (A)] = – ∆Q V = –I V ∆t < 0 (3. FIGURE 3. charge carriers do not move with acceleration but with a steady drift velocity. We have. radiating out heat and light. Where does the power come from? As we have reasoned before. This is because of the collisions with ions and atoms during transit. in case charges were moving freely through the conductor under the action of electric field.12 Heat is produced in the show the dependence of the power dissipated in a resistor R which is connected across resistor R on the current through it and the voltage the terminals of a cell. their kinetic energy would increase as they move. (3. Conservation of total energy would then imply that. In the simple circuit shown with a cell (Fig. The energy dissipated in the resistor R comes from across it.Physics and V (B) respectively. we need an external source to keep a steady current through the conductor. we get ∆K = I V∆t > 0 (3. the conductor heats up. Equation (3. Eqs. it is the chemical energy of the cell which supplies this power for as long as it can. Since current is flowing from A to B. in an actual conductor. however. The potential energy of the charge at A.12). V (A) > V (B) and the potential difference across AB is V = V(A) – V(B) > 0. It is clearly this source which must supply this power. an amount of energy dissipated as heat in the conductor during the time interval ∆t is. In a time interval ∆ t.28) If charges moved without collisions through the conductor. P=IV Using Ohm’s law V = IR. i. the chemical energy of the electrolyte. was Q V (A) and similarly at B. for example. The expressions for power.33) has an important application to power transmission. which o n T i l R b E u C p N re © e b o t t that is. Electrical power is transmitted 106 from power stations to homes and factories. The atoms vibrate more vigorously. If V is the voltage across R and I the current through it. 3. Using electricity at such voltages is not safe and hence at the other end. the power wasted in the connecting wires is inversely proportional to V 2. 3. If a third resistor is joined with the series combination of the two (Fig. Two resistors are said to be in series if only one of their end points is joined (Fig. then P=VI (3. One obviously wants to minimise the power loss in the transmission cables connecting the power stations to homes and factories. which is wasted is Pc with Pc = I 2 Rc from Eq. To reduce Pc.15 Two resistors R 1 and R 2 connected in parallel.35) h s d e RESISTORS – SERIES AND FIGURE 3. we can extend this definition to series combination of any number of resistors. via transmission cables. (3.13). a device called a transformer lowers the voltage to a value suitable for use. R3. to which a power P is to be delivered via transmission cables having a resistance R c to be dissipated by it finally. to drive a device of power P . FIGURE 3. Two or more resistors are said to be in parallel if one end of all the resistors is joined together and similarly the other ends joined together (Fig. Thus. 3. these wires carry current at enormous values of V and this is the reason for the high voltage danger signs on transmission lines — a common sight as we move away from populated areas. Clearly. We shall see now how this can be achieved.14). The power dissipated in the connecting wires. Consider a device R.15).Current Electricity may be hundreds of miles away. 107 .13 A series combination of two resistor R1 and R2.10 COMBINATION PARALLEL The current through a single resistor R across which there is a potential difference V is given by Ohm’s law I = V/R. R 2.32). Resistors are sometimes joined together and there are simple rules for calculation of equivalent resistance of such combination. FIGURE 3. The transmission cables from power stations are hundreds of miles long and their resistance R c is considerable.34) The connecting wires from the power station to the device has a finite resistance R c.14 A series combination of three resistors R1. then all three are said to be in series. o n T i l R b E u C p N re © e b o t t OF (3. 3. this means that the same current I flows through R1 and R2. The currents I. R2 . we would have. R 2 and R 3. 3.38) (3. By Ohm’s law: Potential difference across R1 = V1 = I R 1. Ohm’s law applied to R2 gives (3. (3. FIGURE 3. The charge that flows in at A from the left flows out partly through R 1 and partly through R 2. I 1. I 2 shown in the figure are the rates of flow of charge at the points indicated.36) Consider now the parallel combination of two resistors (Fig.15).16). Since current measures the rate of flow of charge.43) If the combination was replaced by an equivalent resistance R eq.. . and Potential difference across R2 = V2 = I R2 . by Ohm’s law (3.Physics Consider two resistors R1 and R2 in series.. ∴ I = I 1 + I2 = Hence. which by Ohm’s law is Req = ( R1 + R 2) o n 108 T i l R b E u C p N re © e b o t t V = I R1 + I R2 + I R3 = I (R1+ R2+ R3). then similarly This obviously can be extended to a series combination of any number n of resistors R1 . 3.. If we had three resistors connected in series. . V = V1+ V2 = I ( R1 + R2) This is as if the combination had an equivalent resistance Req .40) I = I1 + I 2 The potential difference between A and B is given by the Ohm’s law applied to R 1 V = I1 R 1 (3.45) We can easily see how this extends to three resistors in parallel (Fig. Hence.39) Req = R1 + R2 + .37) (3. The potential difference V across the combination is V1+V 2. The charge which leaves R1 must be entering R2.41) Also.42) V = I2 R 2 (3. + Rn h s d e (3.44) (3. Rn... The equivalent resistance Req is (3. Hence.16 Parallel combination of three resistors R 1. . R2 and R3 are in parallel and hence we can replace them by an equivalent between point B and C with or.17 A combination of three resistors R 1.50) (3. R2.46) (3. R 2 and R3. Consider for example. . .48) An equivalent resistance Req that replaces the combination.17). the circuit in Fig. (3. R2 . R 2 and R3. R 1 and are in equivalent resistance . V = I3 R 3 So that I = I 1 + I2 + I3 = (3.Current Electricity Exactly as before I = I 1 + I2 + I3 and applying Ohm’s law to R1 . V = I2 R 2.49) h s d e If the voltage between A and C is V . where there are three resistors R1. The equivalent resistance of n resistors R 1. .52) in series (3. would be such that and hence We can reason similarly for any number of resistors in parallel. R2 and R3 we get. V = I1 R1 . R 3 are in parallel with an .54) (3. the current I is given by FIGURE 3. series with an equivalent resistance 109 . The circuit now has R1 and and hence their combination can be replaced by an equivalent resistance with o n T i l R b E u C p N re © e b o t t (3.51) (3.53) (3.47) (3.R n is These formulae for equivalent resistances can be used to find out currents and voltages in more complicated circuits. so that the potential difference between P and N is V+ – (–V– ) = V+ + V – . called the internal resistance. the electrodes exchange charges with the electrolyte.55) Note that ε is. Electrical negative electrodes in an open circuit. internal resistances of cells in the circuit may be neglected when the current I is such that e >> I r.Physics 3. A potential difference between P and N.57) 110 Note the negative sign in the expression (I r ) for the potential difference between A and B. As explained before. whereas through R. the electrolyte has the same potential throughout. across the electrolyte the same current flows through the electrolyte FIGURE 3. emf e is the potential difference between the positive and electrode. Thus h s d e (3. Clearly. consider a resistor R connected across the cell (Fig. The internal resistance of dry cells.18 (a) Sketch of but from N to P. Now. is used because of historical reasons.18). called the positive (P) and the negative (N). Consider first the negative terminal N. actually. Similarly. The actual values of the internal resistances of cells vary from cell to cell. They are immersed in an electrolytic solution. however. and was given at a time when the phenomenon was not understood properly. marked as B in the figure. it flows from P to N. The name emf. however. when no current is connections to the cell are flowing through the cell.e. INTERNAL RESISTANCE e = V++V – > 0 (3. Basically a cell has two electrodes.56) – referring to the N Thus.11 CELLS. . as shown in Fig. The positive electrode has a potential difference V+ (V+ > 0) between itself and the electrolyte solution immediately adjacent to it marked A in the figure. + referring to P and =e (3. is much higher than the common electrolytic cells. a steady current is maintained because current flows from N to P through the electrolyte. and B are points in the V = Potential difference between P and A electrolyte typically close to + Potential difference between A and B P and N. When there is no current. a potential difference and not a force. To understand the significance of e. 3. This is because the current I flows from B to A in the electrolyte. i. the negative electrode develops a negative potential – (V– ) (V– 0) relative to the electrolyte adjacent to it. (b) the symbol for + Potential difference between B and N a cell. The situation when R is infinite so that I = V/ R = 0.. In that case the potential difference between P and N is V = V ++ V– – I r o n T i l R b E u C p N re © e b o t t = e– I r We have already mentioned that a simple device to maintain a steady current in an electric circuit is the electrolytic cell. an electrolyte cell with The electrolyte through which a current flows has a finite positive terminal P and resistance r. made at P and N. 3. This difference is called the electromotive force (emf) of the cell and is denoted by e. In practical calculations. where V is the gap between the electrodes is exaggerated for clarity. Dipped in the solution. If however R is finite. A current I flows across R from C to D. I is not zero.18. EMF. This can amount to millions of volts and eventually the electrical resistance in the air breaks down and lightning flash begins and thousands of amperes of current flows. These charged particles get separated by updrifts in the clouds and gravity. The ground is positively charged at the time of thunderstorm development. (3.2 × 10–9 C/m2 . This electric field is not noticeable in daily life. which corresponds to a surface charge density of –1. It was believed to be the great weapon of Gods.58) The maximum current that can be drawn from a cell is for R = 0 and / r. o n T i l R b E u C p N re © e b o t t CHARGES IN CLOUDS h s d e 111 . in most cells the maximum allowed current is it is I max = e much lower than this to prevent permanent damage to the cell. leading to dipole structure. The surface charge density at ground is negative. which is of the order of picoampere / square metre. Over land the average electric field is about 120 V/m. An equal positive charge exists in the atmosphere.58). we get I R = e– I r Or. The electric field is of the order of 105 V/m. The separation of charges produce tremendous amount of electrical potential within the cloud as well as between the cloud and ground. including our bodies. (3.Current Electricity We also observe that since V is the potential difference across R. fracture and break apart. During fair weather also there is charge in the atmosphere. In the lower atmosphere the current is weaker and is maintained by thunderstorm.59) (3. Also cosmic and radioactive radiations ionise air into positive and negative ions and air becomes (weakly) electrically conductive. There are ice particles in the clouds. Sometimes a very weak positive charge is found near the base of the cloud. The average peak power per stroke is about 1012 watts. the total negative charge amount to about 600 kC. which grow. A lightning flash is composed of a series of strokes with an average of about four and the duration of each flash is about 30 seconds. The upper portion of the cloud becomes positively charged and the middle negatively charged. Atmospheric electricity arises due to the separation of electric charges. is conductor compared to air.57) and (3. the electric field is directed downward. The reason why it is not noticed is that virtually everything. The fair weather electric field arises due to the existence of a surface charge density at ground and an atmospheric conductivity as well as due to the flow of current from the ionosphere to the earth’s surface. we have from Ohm’s law V=I R Combining Eqs. The smaller particles acquire positive charge and the larger ones negative charge. But today the phenomenon of lightning can be explained scientifically by elementary principles of physics. In the ionosphere and magnetosphere strong electric current is generated from the solarterrestrial interaction. However. Over the entire earth’s surface. In olden days lightning was considered as an atmospheric flash of supernatural origin. collide. I1 = I2. 3. I1 × 4 = I2 × 4 that is. (b) The total current I in the circuit is h s d e Consider the resistors between A and B. the 12 Ω and 6 Ω resistors in parallel are equivalent to a resistor of [(12 × 6)/(12 + 6)] Ω = 4 Ω. I4 = A EXAMPLE 3. I1 = I2 = 1 A that is. which is otherwise obvious from the symmetry of the two arms. Current in 1 Ω resistor between B and C would be 2 A.Physics Example 3. (c) Obtain the voltage drops V AB. VBC and V CD. . consider the resistances between C and D. that is. I4 = 2I3 But.e. The equivalent resistance R of the network is obtained by combining these resistors (2 Ω and 4 Ω) with 1 Ω in series. i.19 Solution (a) The network is a simple series and parallel combination of resistors.5 that is. But I1 + I2 = I = 2 A. as shown in Fig. In the same way. I3 + I4 = I = 2 A Thus. (b) Obtain the current in each resistor. that is. and I4 in the 6 Ω resistor. Thus. while the current in the 6 Ω resistor is (4/3) A. This can also be obtained by multiplying the total current between A and B by the equivalent resistance between A and B. First the two 4Ω resistors in parallel are equivalent to a resistor = [(4 × 4)/(4 + 4)] Ω = 2 Ω. current in each 4 Ω resistor is 1 A. I3 × 12 = I4 × 6. If I1 is the current in one of the 4 Ω resistors and I2 the current in the other.. R = 2 Ω + 4 Ω + 1 Ω = 7 Ω. If I3 is the current in the 12 Ω resistor. o n 112 T i l R b E u C p N re © e b o t t FIGURE 3.5 A network of resistors is connected to a 16 V battery with internal resistance of 1Ω . (c) The voltage drop across AB is VAB = I1 × 4 = 1 A × 4 Ω = 4 V. I3 = A.19: (a) Compute the equivalent resistance of the network. the current in the 12 Ω resistor is (2/3) A. Now. And like resistors. 3. Then V (A) – V (B) is the potential difference between the positive and negative terminals of the first cell.12 CELLS IN Like resistors. while its emf is 16 V.20). B and C shown in Fig. r1.61) Hence. the combination can be considered as one cell of emf eeq and an internal resistance req.62) 113 . replace a combination of cells by an equivalent cell. 3. the potential difference between the terminals A and C of the combination is (3. V CD = 2 A × 4 Ω = 8 V Note that the total voltage drop across AD is 4 V + 2 V + 8 V = 14 V. Similarly. the voltage drop across CD is V CD = 12 Ω × I3 = 12 Ω × A = 8 V.5 (3. V (B). This can alternately be obtained by multiplying total current between C and D by the equivalent resistance between C and D. Thus. the terminal voltage of the battery is 14 V . FIGURE 3. (3.20. The loss of the voltage (= 2 V) is accounted for by the internal resistance 1 Ω of the battery [2 A × 1 Ω = 2 V]. For connections across A and C.Current Electricity V AB = 2 A × 2 Ω = 4 V The voltage drop across BC is V BC = 2 A × 1 Ω = 2 V Finally. 3.57) and hence. r2 their internal resistances. r2 are their internal resistances.60) (3. Let V (A). e 2 are the emf’s of the two cells and r1. V (C) be the potentials at points A. cells can be combined together in an electric circuit. for calculating currents and voltages in a circuit.20 Two cells of emf’s e1 and e2 in the series. o n T i l R b E u C p N re © e b o t t SERIES AND IN P ARALLEL h s d e E XAMPLE 3. one can. that is. We have already calculated it in Eq. where one terminal of the two cells is joined together leaving the other terminal in either cell free. e1. Consider first two cells in series (Fig. respectively. when the current leaves each cell from the positive electrode. I1 and I2 are the currents leaving the positive electrodes of the cells. the potential difference across its combination can be terminals is V (B 1) – V (B 2). h s d e (3. (3. Since as much charge flows in as out.68) Points B 1 and B 2 are connected exactly similarly to the second cell.20. and (ii) The equivalent internal resistance of a series combination of n cells is just the sum of their internal resistances. 3. we get eeq = e1 + e 2 and req = r1 + r2 (3.72) V = eeq – I req .66). Next.74). At the point B1. (3. (3. I1 and I 2 flow in whereas the current I flows out. If in the combination. Eq.63) (3.3. as in Eq. respectively. (3.67) I = I1 + I 2 FIGURE 3.69) (3. of emf eeq and internal resistance r eq. V is given by. the emf of the cell enters the expression for eeq with a negative sign. Hence.71) If we want to replace the combination by a single cell.73) and (3. Hence considering the second cell. o n 114 T i l R b E u C p N re © e b o t t Combining the last three equations Hence.21). consider a parallel combination of the cells (Fig.70) (3.57) replaced by one cell of emf eeq and internal resistances req whose values are given in Eqs. For connections Let V (B1) and V (B2) be the potentials at B1 and B 2. If instead we connect the two negatives.65) In Fig.Physics If we wish to replace the combination by a single cell between A and C of emf e eq and internal resistance req . from Eq. between B 1 and B 2. This is so.21 Two cells in parallel. we would have (3.64) (3. considering the first cell.66) 1 2 The rule for series combination clearly can be extended to any number of cells: (i) The equivalent emf of a series combination of n cells is just the sum of their individual emf’s. we also have (3. across A and C. the Then. we have (3. the current leaves any cell from the negative electrode. we had connected the negative electrode of the first to the positive electrode of the second. we would have VAC = eeq – I req Comparing the last two equations.61) would change to VBC = –e2– Ir2 and we will get eeq = e –e (e1 > e2) (3. V = V (P) – V (N) = e – I r o n T i l R b E u C p N re © e b o t t (3. .13 KIRCHHOFF’S RULES Electric circuits generally consist of a number of resistors and cells interconnected sometimes in a complicated way. (3.76) Gustav Robert Kirchhoff (1824 – 1887) German physicist.. the actual current in the resistor is in the direction of the arrow. professor at Heidelberg and at Berlin. (3.78) (3.76) can be extended easily. This will tell us the potential difference. say I . called Kirchhoff’s rules. and a directed arrow to indicate that a current I flows along the resistor in the direction indicated.. The formulae we have derived earlier for series and parallel combinations of resistors are not always sufficient to determine all the currents and potential differences in the circuit. r n respectively. If I turns out to be negative.e and of internal resistances 1 n r 1.. cell or some other source of electrical power) the positive and negative electrodes are labelled as well as a directed arrow with a symbol for the current flowing through the cell. the combination is equivalent to a single cell of emf e eq and internal resistance req . Given a circuit. are very useful for analysis of electric circuits.75) and (3. Eqs. we start by labelling currents in each resistor by a symbol.21). Mainly known for his development of spectroscopy. (3. so that the currents I1. h s d e 115 GUSTAV ROBERT KIRCHHOFF (1824 – 1887) . I2 flow out of positive terminals. such that L L 3. he also made many important contributions to mathematical physics.. his first and second rules for circuits.74) We can put these equations in a simpler way. we had joined the positive terminals together and similarly the two negative ones. If ultimately I is determined to be positive. If there an n cells of emf e . . for each source (i.75) and (3. If the negative terminal of the second is connected to positive terminal of the first. Similarly. among them. the current actually flows in a direction opposite to the arrow. .73) (3.e. Two rules.76) would still be valid with e 2 → –e2 Equations (3. connected in parallel.Current Electricity The last two equations should be the same and hence (3.77) (3.75) In Fig. 57) between the positive terminal P and the negative terminal N. I here is the current flowing from N to P through the cell]. (3. Thus starting with any point if we come back to the same point. leaving is I1 + I 2 and current entering is I3.22). At point at which charge flows into the junction).6 A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1 Ω (Fig. we now state the rules and the proof: (a) Junction rule: At any junction. Determine the equivalent resistance of the network and the current along each edge of the cube. while labelling the current I through the cell one goes from P to N. .22 At junction a the current of charges at any junction or at any point in a line. Thus.79) Having clarified labelling. the total change must be zero. the total current flowing in. If. the sum of the currents entering the junction is equal to the sum of currents leaving the junction (Fig. o n 116 T i l R b E u C p N re © e b o t t EXAMPLE 3. In a closed loop. current leaving h and by junction rule (b) Loop rule: The algebraic sum of changes in that will also be I 1.23). there is no accumulation FIGURE 3. must equal h current entering is I1.6 FIGURE 3. This rule is also obvious. we do come back to the starting point and hence the rule.23 h s d e Example 3.Physics [Eq. 3.22). 3. since electric potential is dependent on the location of the point. 3. There is only one the total current flowing out. This applies equally well if instead of a junction of several lines. the loop rules give –30I1 – resistors and cells in the loop is zero (Fig. 41 I3 + 45 = 0 and –30I1 + 21 I2 – 80 = 0. we consider a point in a line. For the loops ‘ahdcba’ potential around any closed loop involving and ‘ahdefga’. (which is the rate The junction rule says I3 = I1 + I2. then of course V = e+ I r (3. The proof of this rule follows from the fact that when currents are steady. Thus. e= The equivalent resistance R eq of the network is For R = 1 Ω.hawaii. the great power of Kirchhoff’s rules has not been very apparent in Example 3. and apply Kirchhoff’s second rule: –IR – (1/2)IR – IR + e = 0 where R is the resistance of each edge and e the emf of battery. This will be illustrated in Example 3. B and D.Current Electricity Solution The network is not reducible to a simple series and parallel combinations of r esistors.24 h s http://www. Further. i.23.24. It should be noted that because of the symmetry of the network. and only by application of Kirchhoff’s rules to junctions and closed loops (as many as necessary to solve the unknowns in the network) can we handle the problem. a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network. there will be no such simplification due to symmetry. The paths AA ′.7 FIGURE 3. using Kirchhoff’s first rule and the symmetry in the problem.7. I. In a general network. Next take a closed loop. at the corners A′. o n T i l R b E u C p N re © e b o t t EXAMPLE 3. 3. In this manner. the current in each must be the same.phys. however. Example 3. the total current (= 3I ) in the network is 3I = 10 V/(5/6) Ω = 12 A. There is.6 EXAMPLE 3. 3. the incoming current I must split equally into the two outgoing branches. say. ABCC′EA. say.edu/~teb/optics/java/kirch3/ Similation for application of Kirchhoff ís rules: d e 117 .e. the current in all the 12 edges of the cube are easily written down in terms of I. Thus. Req = (5/6) Ω and for e = 10 V.7 Determine the current in each branch of the network shown in Fig.. AD and AB are obviously symmetrically placed in the network. I = 4 A The current flowing in each edge can now be read off from the Fig.6. A. a galvanometer G (which is a device to detect currents) is connected.14 WHEATSTONE BRIDGE As an application of Kirchhoff’s rules consider the circuit shown in Fig.80(c)] that is. the total voltage drop over the closed loop BADEB equal to zero. the above values of currents satisfy the second rule for every closed loop of the network. we get 10 – 4I2– 2 (I2 + I3) – I1 = 0 [3. This (i. 7I1– 6I2 – 2I3 = 10 For the closed loop ABCA. In this case. we assume that the cell has no internal resistance. is called the galvanometer arm. b. Of special interest. the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. CA : A. I3 = A o n 118 T i l R b E u C p N re © e b o t t AB : A.80 a. For example.e. DEB : BC : A AD : CD : 0 A. that is. 10 – 4(I1– I2) + 2(I2 + I3 – I1) – I1 = 0 [3. Kirchhoff’s second rule for the closed loop ADCA gives. the Kirchhoff’s junction rule applied to junctions D and B (see the figure) . which is called the Wheatstone bridge. Between the other two vertices. A h s d e The currents in the various branches of the network are EXAMPLE 3.. such that there is no current through G.Physics Solution Each branch of the network is assigned an unknown current to be determined by the application of Kirchhoff’s rules. I2 and I3 which can be found by applying the second rule of Kirchhoff to three different closed loops.7 It is easily verified that Kirchhoff’s second rule applied to the remaining closed loops does not provide any additional independent equation. we get 5 – 2 (I2 + I3 ) – 2 (I2 + I 3 – I1) = 0 [3. We then have three unknowns I1. shown as BD in the figure. The bridge has four resistors R 1. In general there will be currents flowing across all the resistors as well as a current Ig through G.80(b)] that is. B and D. AC) is called the battery arm. We can easily get the balance condition. c) are three simultaneous equations in three unknowns.25. To reduce the number of unknowns at the outset. Across one pair of diagonally opposite points (A and C in the figure) a source is connected. This line. as required by Kirchhoff’s second rule. I1 + 6I2 + 2 I3 =10 For the closed loop BCDEB. R2.80(a)] that is. These can be solved by the usual method to give I1 = 2. 2I1 – 4I 2 – 4I3 = –5 Equations (3. I2 = A. 3. For simplicity. is the case of a balanced bridge where the resistors are such that Ig = 0. 3. R3 and R4.5A. 26 FIGURE 3. I4 = I 2 I2 R 4 + 0 – I 1 R 3 = 0 From Eq. (3. The bridge then is balanced. CD = 5Ω. and from the balance condition the value of the unknown resistance R 4 is given by. Example 3. which we insert in the fourth arm.83(a)] EXAMPLE 3.25 h s d e 119 . 3.82) whereas from Eq. (3. we obtain. BC = 10Ω . (3. upon using I 3 = I1 . and DA = 60Ω .Current Electricity immediately gives us the relations I1 = I 3 and I2 = I4.81) –I 1 R1 + 0 + I2 R2 = 0 and the second loop gives. Hence.83(b)] A practical device using this principle is called the meter bridge. Next. The Wheatstone bridge and its balance condition provide a practical method for determination of an unknown resistance.82). [3. It will be discussed in the next section. we go on varying R3 till the galvanometer shows a null deflection.81). we apply Kirchhoff’s loop rule to closed loops ADBA and CBDC. The first loop gives (Ig = 0) (3. Let us suppose we have an unknown resistance. o n T i l R b E u C p N re © e b o t t [3.26) have the following resistances: AB = 100Ω . R 4 is thus not known. Keeping known resistances R 1 and R2 in the first and second arm of the bridge.8 FIGURE 3.8 The four arms of a Wheatstone bridge (Fig. we obtain the condition This last equation relating the four resistors is called the balance condition for the galvanometer to give zero or null deflection. we obtain. Physics A galvanometer of 15Ω resistance is connected across BD. R is a resistance to be the galvanometer is connected to a ‘jockey’. 3.27. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC. [3.84b) by 10 20I1 – 60Ig – 10I2 = 0 From Eqs.84(a)] Considering the mesh BCDB.5Ig h s [3. 60I2 + 5 (I2 + Ig ) = 10 13I2 + Ig = 2 65I2 + 5Ig = 10 Multiplying Eq.84d) and (3. The metallic strip has two gaps across which resistors can be connected. knife-edge which can slide over the wire to make electrical connection.84a) we have 63Ig – 2I2 = 0 I2 = 31. we have 100I1 + 15Ig – 60I2 = 0 or 20I1 + 3Ig – 12I 2= 0 [3. The other end of FIGURE 3. (3. we have 10 (I1 – Ig) – 15Ig – 5 (I2 + Ig ) = 0 10I1 – 30Ig –5I2 = 0 2I1 – 6Ig – I2 = 0 The meter bridge is shown in Fig.5Ig ) + Ig = 2 410.84(b)] d e [3. we get 13 (31.8 Substituting the value of I2 into Eq.87 mA. One end of a galvanometer is connected to the metallic strip midway between the two gaps. It consists of a wire of length 1 m and of uniform cross sectional area stretched taut and clamped between two thick metallic strips bent at right angles. (3. R is an unknown resistance whose value we want to determine.15 METER BRIDGE . as shown. The end points where the wire is clamped are connected to a cell through a key.84(c)].84(d)] [3.84(c)] [3. The jockey measured and S is a standard is essentially a metallic rod whose one end has a resistance. 3.5 Ig = 2 Ig = 4. Solution Considering the mesh BADB. Across the other gap. It is 120 connected across one of the gaps.84(e)] EXAMPLE 3. Wire AC is 1 m long. we connect a o n T i l R b E u C p N re © e b o t t Considering the mesh ADCEA.27 A meter bridge. BC.86 Ω.88) Substituting the value of R/ S from Eq. (3.9 which gives S = 13. a distance l cm from the end A. we get g o n T i l R b E u C p N re © e b o t t (3. The four resistances of the bridge at the balance point then are R . 121 . i. Eq. Determine the values of R and S. the unknown resistance R is known in terms of the standard known resistance S by (3. The jockey is connected to some point D on the wire.Current Electricity standard known resistance S. It can be shown that the percentage error in R can be minimised by adjusting the balance point near the middle of the bridge. The balance condition. S. The jockey can be moved along the wire. we get After S is connected in parallel with a resistance of 12 Ω . the null point is found at a distance of 33. we would get various values of l1. Rcm l1 and Rcm(100– l1). once we have found out l1. The portion DC of the wire similarly has a resistance Rcm (100-l ).87) h s d e EXAMPLE 3. the resistance across the gap changes from S to Seq. Solution From the first balance point. If the jockey is moved along the wire. S. where and hence the new balance condition now gives (3. Using the value of R / S above. and calculate R each time.e. [3.) Example 3. The portion AD of the wire has a resistance Rcm l. where Rcm is the resistance of the wire per unit centimetre.85) Thus.5 Ω . An error in measurement of l1 would naturally result in an error in R. when l1 is close to 50 cm.7 cm from A..86) By choosing various values of S.87). The four arms AB. the null point occurs at 51. then there will be one position where the galvanometer will show no current. we get R = 6.9 cm.83(a)] gives (3. R cm l and Rcm (100-l )] obviously form a Wheatstone bridge with AC as the battery arm and BD the galvanometer arm. 3.27). If now a resistance of 12Ω is connected in parallel with S. Let the distance of the jockey from the end A at the balance point be l= l1. (This requires a suitable choice of S.9 In a metre bridge (Fig. DA and CD [with resistances R. if another emf e is balanced against l2 (AN2) 2 (3. the cell sends a current (I ) through the resistance box (R). In practice one of the cells is chosen as a standard cell whose emf is known to a high degree of accuracy. 3. Figure 3.16 POTENTIOMETER This is a versatile instrument. A current I flows through the wire which can be varied by a variable resistance (rheostat. Then. Since the wire is uniform. If V is the terminal potential difference of the cell and balance is obtained at length l2 (AN2). The emf of the other cell is then easily calculated from Eq.93(b)] 122 . the potential difference between A and any point at a distance l from A is FIGURE 3.93(a)] When key K2 is closed. G is a galvanometer and R a variable resistance (rheostat). R) in the circuit. o n T i l R b E u C p N re © e b o t t f l1 + 0 – e1 = 0 f l2 + 0 – e2 = 0 From the last two equations l l where f is the potential drop per unit length. 3 are terminals of a two way key (a) circuit for comparing emfs of two cells.28 (a) shows an application of the potentiometer to compare the emf of two cells of emf e 1 and e 2 . It is basically a long piece of uniform wire.Physics 3. 3. We can also use a potentiometer to measure internal resistance of a cell [Fig. the wires run from A to C. (Fig. there is no deflection in the galvanometer. at a distance l1 from A. (3. (3. 1. balance is obtained at length l1 (AN1). Consider first a position of the key where 1 and 3 are connected so that the galvanometer is connected to e 1 . 2. The jockey is moved along the wire till at a point N1. We can apply Kirchhoff’s loop rule to the closed loop AN1G31A and get. In actual design.28). V = f l2 [3. The small vertical portions are the thick metal strips connecting the various sections of the wire. The points marked 1. In the figure. For this the cell (emf e ) whose internal resistance (r ) is to be determined is connected across a resistance box through a key K2.90) Similarly.92) This simple mechanism thus allows one to compare the emf’s of any two sources. as shown in the figure. With key K2 open.28 A potentiometer.91) h s d e (3.89) (3.28 (b)].92). the wire is sometimes cut in several pieces placed side by side and connected at the ends by thick metal strip. 2. e = f l1 [3. sometimes a few meters in length across which a standard cell is connected. 3 form a two way key. (b) circuit for determining internal resistance of a cell. the total resistance between A and B.. we have e /V = l1/l 2 But.29 h s d e EXAMPLE 3. will be given by the following expression: The total resistance between A and C will be sum of resistance between A and B and B and C.94(b)] e/V = (r+R )/R From Eq. 3.94(b)] we have (R+r )/R = l1 /l 2 (3.e. V 1 = I R1 = 123 . R 1. This gives [3. [3.94(a)] and [3.94(a)] [3.Current Electricity So.29). Solution While the slide is in the middle of the potentiometer only half of its resistance (R 0/2) will be between the points A and B. (3. A voltage V is supplied to the potentiometer. As such it is unaffected by the internal resistance of the source. R 1 + R 0/2 ∴ The current flowing through the potentiometer will be o n T i l R b E u C p N re © e b o t t FIGURE 3.10 A resistance of R Ω draws current from a potentiometer. i. e = I (r + R ) and V = IR. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.95) we can find the internal resistance of a given cell. The potentiometer has a total r esistance R 0 Ω (Fig. Example 3. The potentiometer has the advantage that it draws no current from the voltage source being measured.95) Using Eq.10 The voltage V1 taken from the potentiometer will be the product of current I and resistance R 1. Hence. say. 10 or V1 . Metals have low resistivity. we must have a closed circuit in which an external agency moves electric charge from lower to higher potential energy. it is the voltage difference between the two terminals of a source in open circuit. i. Electrical resistivity of substances varies over a very wide range. V ∝ I or V = RI. ionic crystals and electrolytic liquids. Current density j gives the amount of charge flowing per second per unit area normal to the flow. in some cases. The work done per unit charge by the source in taking the charge from lower to higher potential energy (i. To maintain a steady current. Insulators like glass and rubber have 1022 to 1024 times greater resistivity. where n is the number density (number per unit volume) of charge carriers each of charge q. o n 124 T i l R b E u C p N re © e b o t t SUMMARY 1.e.e.. 5. 3. . of the source. the carriers of current are electrons.. Ohm’s law: The electric current I flowing through a substance is proportional to the voltage V across its ends. we have a E XAMPLE 3. j = nq vd Current through a given area of a conductor is the net charge passing per unit time through the area. and vd is the drift velocity of the charge carriers. In most substances. from one terminal of the source to the other) is called the electromotive force. The unit of resistance is ohm: 1Ω = 1 V A–1. the magnitude of the current I through the area is nev d A. 2. 7. positive and negative ions carry the electric current. in the range of 10–8 Ω m to 10–6 Ω m. 4. or emf. called resistivity is a pr operty of the material and depends on temperature and pressure. Semiconductors like Si and Ge lie roughly in the middle range of resistivity on a logarithmic scale. h s d e where r.Physics Substituting for R1. The resistance R of a conductor depends on its length l and constant cross-sectional area A through the relation. where R is called the resistance of the substance. 6. For electrons q = – e. If j is normal to a cross-sectional area A and is constant over the area. for example. Note that the emf is not a force. If such collisions occur on an average at a time interval t. R2 . Since the method involves a condition of no current flow. R4 as shown in the text. 12. one obtains The proportionality between the force eE on the electrons in a metal due to the external field E and the drift velocity vd (not acceleration) can be understood. the sum of currents entering the junction must equal the sum of currents leaving it. 125 . The Wheatstone bridge is an arrangement of four resistances – R 1. In the temperature range in which resistivity increases linearly with temperature. The null-point condition is given by o n T i l R b E u C p N re © e b o t t h s d e using which the value of one resistance can be determined. 15. internal resistance of a cell and compare emf’s of two sources.. A rectifier combines features (a) and (b). the voltage Vext across R is given by + where r is the internal resistance of the source.. and Ohm’s law.. 11. The potentiometer is a device to compare potential differences. knowing the other three resistances. (b) the relation between V and I depends on the sign of V for the same absolute value of V . This gives 9. 10. but it is not a fundamental law of nature. (a) Total resistance R of n resistors connected in series is given by Vext = IR = R = R1 + R2 +. the device can be used to measure potential difference.. vd = at = eEt/m where a is the acceleration of the electron. I = nevd A. (b) Loop Rule: The algebraic sum of changes in potential around any closed loop must be zero. (c) The relation between V and I is non-unique. Ohm’s law is obeyed by many substances. if we assume that the electrons suffer collisions with ions in the metal. the temperature coefficient of resistivity a is defined as the fractional increase in resistivity per unit increase in temperature. Kirchhoff’s Rules – (a) Junction Rule: At any junction of circuit elements. An example of (a) is when r increases with I (even if temperature is kept fixed). It fails if (a) V depends on I non-linearly. 14. + Rn (b) Total resistance R of n resistors connected in parallel is given by 13. GaAs shows the feature (c). Using E = V/l. When a source of emf e is connected to an external resistance R .Current Electricity 8. R 3. which deflect them randomly. The Ohm’s law asserts that the plot of I versus V is linear i. 2.Physics Physical Quantity Electric current Charge Voltage. Current is a scalar although we represent current with an arrow. Currents do not obey the law of vector addition. The assertion that V = IR is a statement of Ohm’s law is not true. The current I through an area of cross-section is given by the scalar product of two vectors: I = j .. Homogeneous conductors like silver or semiconductors like pure germanium or germanium containing impurities obey Ohm’s law within some range of electric field values.e. Equation E = r j leads to another statement of Ohm’s law.. ∆S where j and ∆ S are vectors.e. . R is independent of V . This equation defines resistance and it may be applied to all conducting devices whether they obey Ohm’s law or not. 3. i. The motion 4. If the field becomes too strong. Motion of conduction electrons in electric field E is the sum of (i) motion due to random collisions and (ii) that due to E. A resistor obeys Ohm’s law while a diode does not. Refer to V-I curves of a resistor and a diode as drawn in the text. Electric potential difference Electromotive force Resistance Resistivity Electrical conductivity Electric field Drift speed Symbol I Q. q V Dimensions [A] [T A] [M L T A ] [M L T A ] [M L T A ] [M L T A ] [M –1 3 –3 –2 2 –3 –2 2 –3 –1 2 –3 –1 Unit A C V V Ω Ωm S Remark SI base unit Work/charge Work/charge R = V/ I R = rl/A e R r s L T A] –3 3 2 s = 1/r –1 Relaxation time Current density Mobility o n 126 T i l R b E u C p N re © e b o t t E [M L T –1 –3 A ] –1 Vm Electric force charge eEτ vd = m vd [L T ] [T] [L ms s –1 h s d e t j –2 A] A m–2 current/area m [M L3 T –4 A–1 ] m 2 V –1s –1 POINTS TO PONDER 1. That current is a scalar also follows from it’s definition. a conducting material obeys Ohm’s law when the resistivity of the material does not depend on the magnitude and direction of applied electric field. there are departures from Ohm’s law in all cases. and the total current drawn from the battery. At room temperature (27.Current Electricity due to random collisions averages to zero and does not contribute to vd (Chapter 11. and its resistance is measured to be 5. obtain the potential drop across each r esistor. vd . What is the temperatur e of the element if the resistance is found to be 117 Ω . thus is only due to applied electric field on the electron.0 Ω .1 Ω at 27. What is the resistivity of the material at the temperature of the experiment? A silver wire has a resistance of 2. v+ ~ 0 which gives r=0 j = r– v Thus.7 Ω at 100 °C. r+ = – r– Further.5 A. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 20 V and negligible internal resistance. and 3 Ω are combined in series. If the inter nal resistance of the battery is 0.70 × 10–4 °C–1. 3. What is the total resistance of the combination? (b) If the combination is connected to a battery of emf 12 V and negligible internal resistance. the relation j = r v does not apply to the total current charge density. what is the maximum current that can be drawn from the battery? A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor.2 A which settles after a few seconds to o n T i l R b E u C p N re © e b o t t EXERCISES h s d e 127 . Determine the temperature coefficient of resistivity of silver.7 3. Bending or reorienting the wire does not change the validity of Kirchhoff’s junction rule.2 3. and a resistance of 2.5 °C.1 3. A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.4 Ω. The relation j = r v should be applied to each type of charge carriers separately. 2 Ω.0 × 10–7 m2. 4 Ω and 5 Ω are combined in parallel. 6.0 °C) the resistance of a heating element is 100 Ω.6 3. If the current in the circuit is 0.8 The storage battery of a car has an emf of 12 V.4 3. given that the temperature coefficient of the material of the resistor is 1.3 3. In a conducting wire. determine the current through each resistor. 5.5 3. Textbook of Class XI). what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed? (a) Three resistors 1 Ω. the total current and charge density arises from both positive and negative charges: j = r+ v + + r– v– r = r+ + r– Now in a neutral wire carrying electric current. (a) Three resistors 2 Ω. A heating element using nichrome connected to a 230 V supply draws an initial current of 3. Kirchhoff’s junction rule is based on conservation of charge and the outgoing currents add up and are equal to incoming current at a junction. How long does an electron take to drift from one end of a wire 3. 3.8 A.5 × 10 m . Determine the current in each branch of the network shown in Fig. the balance point is found to be at 39.0 cm. 14 The earth’s surface has a negative surface charge density of 10–9 C m–2 .Physics a steady value of 2.0 × 10 m and it is carrying a current of 3.30 h s d e 3.1 is 8.30: 3.10 3.27]. when the resistor Y is of 12. 13 (a) In a metre bridge [Fig. (c) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current? A storage battery of emf 8.5 Ω is being charged by a 120 V dc supply using a series r esistor of 15.0 °C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.5 cm fr om the end A . Determine the resistance of X. ADDITIONAL EXERCISES 3.9 o n 128 T i l R b E u C p N re © e b o t t FIGURE 3. a cell of emf 1. 3. What is the steady temperature of the heating element if the room temperature is 27. Why are the connections between resistors in a Wheatstone or meter bridge made of thick copper strips? (b) Determine the balance point of the bridge above if X and Y are interchanged.0 m long to its other end? The –6 2 area of cross-section of the wire is 2.12 3.0 cm length of the wire.5 Ω .25 V gives a balance point at 35.0 V and internal resistance 0.0 A.11 3. If there were no mechanism of sustaining atmospheric electric .5 Ω. The potential difference of 400 kV between the top of the atmosphere and the surface results (due to the low conductivity of the lower atmosphere) in a current of only 1800 A over the entire globe.70 × 10–4 °C–1. what is the emf of the second cell? The number density of free electrons in a copper conductor 28 –3 estimated in Example 3. If the cell is replaced by another cell and the balance point shifts to 63. What is the terminal voltage of the battery during charging? What is the purpose of having a series resistor in the charging circuit? In a potentiometer arrangement. 6 0.6 118.19 3. how much time (roughly) would be required to neutralise the earth’s surface? (This never happens in practice because there is a mechanism to replenish electric charges.87 11.0 6. 6 kV must have a very large internal resistance.2 78.5 138. (Radius of earth = 6.8 15. Why? Choose the correct alternative: (a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.8 98.7. Which of these quantities is constant along the conductor: current. how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance? (b) Given the resistances of 1 Ω.0 4.) (a) Six lead-acid type of secondary cells each of emf 2.31.7 19.94 7.20 Answer the following questions: (a) A steady current flows in a metallic conductor of non-uniform cross-section.Current Electricity field. amber) is greater than that of a metal by a factor of the order of (1022/1023). 3 Ω. give examples of elements which do not obey Ohm’s law. say. one of aluminium and the other of copper have the same resistance. of Cu = 8. 2 Ω .37 × 106 m.8 1.) What conclusion can you draw from the following observations on a resistor made of alloy manganin? Current A 0. how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω. Relative density of Al = 2.17 3.4 0.0 7. 3. namely the continual thunderstorms and lightning in different parts of the globe). (b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.0 2. rCu = 1.7 39.0 Voltage V 59.5 Ω. (d) The resistivity of a typical insulator (e.9 V and a large inter nal resistance of 380 Ω. electric field.0 Voltage V 3. (rAl = 2.0 3.63 × 10–8 Ω m.15 3. (iv) (6/11) Ω? (c) Determine the equivalent resistance of networks shown in Fig.16 3. (a) Given n resistors each of resistance R . (c) A low voltage supply from which one needs high currents must have very low internal resistance. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car? Two wires of equal length.g.9.0 8.0 V and internal resistance 0.2 158.0 5. o n T i l R b E u C p N re © e b o t t h s d e 129 .2 0. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. (iii) 6 Ω.18 3. current density.015 Ω are joined in series to provide a supply to a resistance of 8.72 × 10–8 Ω m. What are the current drawn from the supply and its terminal voltage? (b) A secondary cell after long use has an emf of 1. drift speed? (b) Is Ohm’s law universally applicable for all conducting elements? If not.. (c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature.4 Current A 3. Why? (d) A high tension (HT) supply of. 22 Figure 3.40 Ω maintaining a potential drop across the resistor wire AB.32 Determine the current drawn from a 12V supply with internal resistance 0. h s d e 3. a very high resistance of 600 kΩ is put in series with it. A standard cell which maintains a constant emf of 1.33 shows a potentiometer with a cell of 2.3 cm length of the wire. To ensure very low currents drawn from the standard cell. turns out to be at 82. FIGURE 3.5 Ω by the infinite network shown in Fig.02 V (for very moderate currents upto a few mA) gives a balance point at 67. The standard cell is then replaced by a cell of unknown emf e and the balance point found similarly.3 cm length of the wire.Physics FIGURE 3.0 V and internal resistance 0.33 (a) What is the value e ? (b) What purpose does the high resistance of 600 kΩ have? .21 o n 130 T i l R b E u C p N re © e b o t t FIGURE 3. which is shorted close to the balance point. Each resistor has 1Ω resistance.31 3.32. 3. 35 shows a 2. Determine the value of X . When a resistor of 9.5 Ω is used in the external circuit of the cell. how will you modify the circuit? Figure 3.34 FIGURE 3.0 Ω is found to be 58. Determine the internal resistance of the cell.24 Figure 3.23 3.35 h s d e 131 .3 cm.3 cm. say of the order of a few mV (such as the typical emf of a thermo-couple)? If not. the balance point shifts to 64.0V instead of 2.5 cm.8 cm length of the potentiometer wire.0 V potentiometer used for the determination of internal resistance of a 1. The balance point with a standard resistor R = 10. The balance point of the cell in open circuit is 76. o n T i l R b E u C p N re © e b o t t FIGURE 3.Current Electricity (c) Is the balance point affected by this high resistance? (d) Is the balance point affected by the internal resistance of the driver cell? (e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1. while that with the unknown resistance X is 68.0V? (f ) Would the circuit work well for determining an extremely small emf.5 V cell. What might you do if you failed to find a balance point with the given cell of emf e ? 3.34 shows a potentiometer circuit for comparison of two resistances. .Physics Chapter Four MOVING CHARGES AND MAGNETISM 4. in 1820. The deflection increases on increasing the current or bringing the needle closer to the wire. that it was realised that they were intimately related*.1(b)]. the laws obeyed by electricity and magnetism were unified and formulated by 132 * See the box in Chapter 1.4. He investigated this phenomenon. Page 3. This situation is depicted in Fig. the Danish physicist Hans Christian Oersted noticed that a current in a straight wire caused a noticeable deflection in a nearby magnetic compass needle. Following this there was intense experimentation. It is noticeable when the current is large and the needle sufficiently close to the wire so that the earth’s magnetic field may be ignored. 4. Oersted concluded that moving charges or currents produced a magnetic field in the surrounding space. Reversing the direction of the current reverses the orientation of the needle [Fig. However. it was only about 200 years ago. Iron filings sprinkled around the wire arrange themselves in concentric circles with the wire as the centre [Fig. During a lecture demonstration in the summer of 1820.1(c)].1 INTRODUCTION Both Electricity and Magnetism have been known for more than 2000 years. He found that the alignment of the needle is tangential to an imaginary circle which has the straight wire as its centre and has its plane perpendicular to the wire. In 1864.1(a). 4. Moving Charges and Magnetism James Maxwell who then realised that light was electromagnetic waves. Radio waves were discovered by Hertz, and produced by J.C.Bose and G. Marconi by the end of the 19th century. A remarkable scientific and technological progress has taken place in the 20th century. This is due to our increased understanding of electromagnetism and the invention of devices for production, amplification, transmission and detection of electromagnetic waves. FIGURE 4.1 The magnetic field due to a straight long current-carrying wire. The wire is perpendicular to the plane of the paper. A ring of compass needles surrounds the wire. The orientation of the needles is shown when (a) the current emerges out of the plane of the paper, (b) the current moves into the plane of the paper. (c) The arrangement of iron filings around the wire. The darkened ends of the needle represent north poles. The effect of the earth’s magnetic field is neglected. In this chapter, we will see how magnetic field exerts forces on moving charged particles, like electrons, protons, and current-carrying wires. We shall also learn how currents produce magnetic fields. We shall see how particles can be accelerated to very high energies in a cyclotron. We shall study how currents and voltages are detected by a galvanometer. In this and subsequent Chapter on magnetism, we adopt the following convention: A current or a field (electric or magnetic) emerging out of the plane of the paper is depicted by a dot ( ). A current or a field going into the plane of the paper is depicted by a cross ( ⊗ )*. Figures. 4.1(a) and 4.1(b) correspond to these two situations, respectively. HANS CHRISTIAN OERSTED (1777–1851) 4.2 MAGNETIC FORCE 4.2.1 Sources and fields Before we introduce the concept of a magnetic field B, we shall recapitulate what we have learnt in Chapter 1 about the electric field E. We have seen that the interaction between two charges can be considered in two stages. The charge Q, the source of the field, produces an electric field E, where Hans Christian Oersted (1777–1851) Danish physicist and chemist, professor at Copenhagen. He observed that a compass needle suffers a deflection when placed near a wire carrying an electric current. This discovery gave the first empirical evidence of a connection between electric and magnetic phenomena. * A dot appears like the tip of an arrow pointed at you, a cross is like the feathered tail of an arrow moving away from you. 133 Physics E=Q / (4πε0)r2 (4.1) ˆ is unit vector along r, and the field E is a vector where r field. A charge q interacts with this field and experiences a force F given by ˆ / (4πε0) r 2 F = q E =qQ r (4.2) As pointed out in the Chapter 1, the field E is not just an artefact but has a physical role. It can convey energy and momentum and is not established instantaneously but takes finite time to propagate. The concept of a field was specially stressed by Faraday and was incorporated by Maxwell in his unification of electricity and magnetism. In addition to depending on each point in space, it can also vary with time, i.e., be a function of time. In our discussions in this chapter, we will assume that the fields do not change with time. The field at a particular point can be due to one or more charges. If there are more charges the fields add vectorially. You have already learnt in Chapter 1 that this is called the principle of superposition. Once the field is known, the force on a test charge is given by Eq. (4.2). Just as static charges produce an electric field, the currents or moving charges produce (in addition) a magnetic field, denoted by B (r), again a vector field. It has several basic properties identical to the electric field. It is defined at each point in space (and can in addition depend on time). Experimentally, it is found to obey the principle of superposition: the magnetic field of several sources is the vector addition of magnetic field of each individual source. HENDRIK ANTOON LORENTZ (1853 – 1928) Hendrik Antoon Lorentz (1853 – 1928) Dutch theoretical physicist, professor at Leiden. He investigated the relationship between electricity, magnetism, and mechanics. In order to explain the observed effect of magnetic fields on emitters of light (Zeeman effect), he postulated the existence of electric charges in the atom, for which he was awarded the Nobel Prize in 1902. He derived a set of transformation equations (known after him, as Lorentz transformation equations) by some tangled mathematical arguments, but he was not aware that these equations hinge on a new concept of space and time. 4.2.2 Magnetic Field, Lorentz Force Let us suppose that there is a point charge q (moving with a velocity v and, located at r at a given time t ) in presence of both the electric field E (r) and the magnetic field B (r). The force on an electric charge q due to both of them can be written as F = q [ E (r) + v × B (r)] ≡ Felectric +Fmagnetic (4.3) 134 This force was given first by H.A. Lorentz based on the extensive experiments of Ampere and others. It is called the Lorentz force. You have already studied in detail the force due to the electric field. If we look at the interaction with the magnetic field, we find the following features. (i) It depends on q, v and B (charge of the particle, the velocity and the magnetic field). Force on a negative charge is opposite to that on a positive charge. (ii) The magnetic force q [ v × B ] includes a vector product of velocity and magnetic field. The vector product makes the force due to magnetic Moving Charges and Magnetism field vanish (become zero) if velocity and magnetic field are parallel or anti-parallel. The force acts in a (sideways) direction perpendicular to both the velocity and the magnetic field. Its direction is given by the screw rule or right hand rule for vector (or cross) product as illustrated in Fig. 4.2. (iii) The magnetic force is zero if charge is not moving (as then |v|= 0). Only a moving charge feels the magnetic force. The expression for the magnetic force helps us to define the unit of the magnetic field, if one takes q, F and v, all to be unity in the force FIGURE 4.2 The direction of the magnetic ˆ , where θ equation F = q [ v × B] =q v B sin θ n force acting on a charged particle. (a) The is the angle between v and B [see Fig. 4.2 (a)]. force on a positively charged particle with The magnitude of magnetic field B is 1 SI unit, velocity v and making an angle θ with the when the force acting on a unit charge (1 C), magnetic field B is given by the right-hand moving perpendicular to B with a speed 1m/s, rule. (b) A moving charged particle q is is one newton. deflected in an opposite sense to –q in the presence of magnetic field. Dimensionally, we have [B] = [F/qv] and the unit of B are Newton second / (coulomb metre). This unit is called tesla ( T ) named after Nikola Tesla (1856 – 1943). Tesla is a rather large unit. A smaller unit (non-SI) called gauss (=10–4 tesla) is also often used. The earth’s magnetic field is about 3.6 × 10–5 T. Table 4.1 lists magnetic fields over a wide range in the universe. TABLE 4.1 ORDER OF MAGNITUDES OF MAGNETIC FIELDS IN A VARIETY OF PHYSICAL SITUATIONS Physical situation Surface of a neutron star Typical large field in a laboratory Near a small bar magnet On the earth’s surface Human nerve fibre Interstellar space Magnitude of B (in tesla) 10 8 1 10 –2 10 –5 10 –10 10 –12 4.2.3 Magnetic force on a current-carrying conductor We can extend the analysis for force due to magnetic field on a single moving charge to a straight rod carrying current. Consider a rod of a uniform cross-sectional area A and length l. We shall assume one kind of mobile carriers as in a conductor (here electrons). Let the number density of these mobile charge carriers in it be n. Then the total number of mobile charge carriers in it is nAl. For a steady current I in this conducting rod, we may assume that each mobile carrier has an average 135 Physics drift velocity vd (see Chapter 3). In the presence of an external magnetic field B, the force on these carriers is: F = (nAl )q vd × B where q is the value of the charge on a carrier. Now nqvd is the current density j and |(nq vd )|A is the current I (see Chapter 3 for the discussion of current and current density). Thus, F = [(nqe vd )Al ] × B = [ jAl ] × B = I1 × B (4.4) where l is a vector of magnitude l, the length of the rod, and with a direction identical to the current I. Note that the current I is not a vector. In the last step leading to Eq. (4.4), we have transferred the vector sign from j to l. Equation (4.4) holds for a straight rod. In this equation, B is the external magnetic field. It is not the field produced by the current-carrying rod. If the wire has an arbitrary shape we can calculate the Lorentz force on it by considering it as a collection of linear strips dlj and summing F= ∑ Idl j j ×B This summation can be converted to an integral in most cases. ON PERMITTIVITY AND PERMEABILITY In the universal law of gravitation, we say that any two point masses exert a force on each other which is proportional to the product of the masses m1, m2 and inversely proportional to the square of the distance r between them. We write it as F = Gm1m2/r 2 where G is the universal constant of gravitation. Similarly in Coulomb’s law of electrostatics we write the force between two point charges q1, q2, separated by a distance r as F = kq1q2/r 2 where k is a constant of proportionality. In SI units, k is taken as 1/4πε where ε is the permittivity of the medium. Also in magnetism, we get another constant, which in SI units, is taken as µ/4π where µ is the permeability of the medium. Although G, ε and µ arise as proportionality constants, there is a difference between gravitational force and electromagnetic force. While the gravitational force does not depend on the intervening medium, the electromagnetic force depends on the medium between the two charges or magnets. Hence while G is a universal constant, ε and µ depend on the medium. They have different values for different media. The product εµ turns out to be related to the speed v of electromagnetic radiation in the medium through εµ =1/ v 2. Electric permittivity ε is a physical quantity that describes how an electric field affects and is affected by a medium. It is determined by the ability of a material to polarise in response to an applied field, and thereby to cancel, partially, the field inside the material. Similarly, magnetic permeability µ is the ability of a substance to acquire magnetisation in magnetic fields. It is a measure of the extent to which magnetic field can penetrate matter. EXAMPLE 4.1 136 Example 4.1 A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B (Fig. 4.3). What is the magnitude of the magnetic field? Moving Charges and Magnetism FIGURE 4.3 Solution From Eq. (4.4), we find that there is an upward force F, of magnitude IlB,. For mid-air suspension, this must be balanced by the force due to gravity: m g = I lB B= = mg Il EXAMPLE 4.1 0.2 × 9.8 = 0.65 T 2 × 1.5 Note that it would have been sufficient to specify m/l, the mass per unit length of the wire. The earth’s magnetic field is approximately 4 × 10–5 T and we have ignored it. http://www.phys.hawaii.edu/~teb/optics/java/partmagn/index.html Charged particles moving in a magnetic field. Interactive demonstration: Example 4.2 If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig. 4.4), which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge). FIGURE 4.4 EXAMPLE 4.2 Solution The velocity v of particle is along the x-axis, while B, the magnetic field is along the y-axis, so v × B is along the z-axis (screw rule or right-hand thumb rule). So, (a) for electron it will be along –z axis. (b) for a positive charge (proton) the force is along +z axis. 4.3 MOTION IN A MAGNETIC FIELD We will now consider, in greater detail, the motion of a charge moving in a magnetic field. We have learnt in Mechanics (see Class XI book, Chapter 6) that a force on a particle does work if the force has a component along (or opposed to) the direction of motion of the particle. In the case of motion 137 Physics of a charge in a magnetic field, the magnetic force is perpendicular to the velocity of the particle. So no work is done and no change in the magnitude of the velocity is produced (though the direction of momentum may be changed). [Notice that this is unlike the force due to an electric field, q E, which can have a component parallel (or antiparallel) to motion and thus can transfer energy in addition to momentum.] We shall consider motion of a charged particle in a uniform magnetic field. First consider the case of v perpendicular to B. The perpendicular force, q v × B, acts as a centripetal force and produces a circular motion perpendicular to the magnetic field. The particle will describe a circle if v and B are perpendicular to each other (Fig. 4.5). If velocity has a component along B, this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to B is as before a circular one, thereby producing a helical motion (Fig. 4.6). You have already learnt in earlier classes (See Class XI, Chapter 4) that if r is the radius of the circular path of a particle, then a force of m v2 / r, acts perpendicular to the path towards the centre of the circle, and is called the centripetal force. If the velocity v is perpendicular to the magnetic field B, the magnetic force is perpendicular to both v and B and acts like a centripetal force. It has a magnitude q v B. Equating the two expressions for centripetal force, m v 2/r = q v B, which gives r = m v / qB FIGURE 4.6 Helical motion FIGURE 4.5 Circular motion (4.5) 138 for the radius of the circle described by the charged particle. The larger the momentum, the larger is the radius and bigger the circle described. If ω is the angular frequency, then v = ω r. So, ω = 2π ν = q B/ m [4.6(a)] which is independent of the velocity or energy . Here ν is the frequency of rotation. The independence of ν from energy has important application in the design of a cyclotron (see Section 4.4.2). The time taken for one revolution is T= 2π/ω ≡ 1/ν. If there is a component of the velocity parallel to the magnetic field (denoted by v||), it will make the particle move along the field and the path of the particle would be a helical one (Fig. 4.6). The distance moved along the magnetic field in one rotation is called pitch p. Using Eq. [4.6 (a)], we have p = v||T = 2πm v|| / q B [4.6(b)] The radius of the circular component of motion is called the radius of the helix. Moving Charges and Magnetism Example 4.3 What is the radius of the path of an electron (mass 9 × 10-31 kg and charge 1.6 × 10–19 C) moving at a speed of 3 ×107 m/s in a magnetic field of 6 × 10 –4 T perpendicular to it? What is its frequency? Calculate its energy in keV. ( 1 eV = 1.6 × 10–19 J). Solution Using Eq. (4.5) we find r = m v / (qB ) = 9 ×10–31 kg × 3 × 107 m s–1 / ( 1.6 × 10–19 C × 6 × 10–4 T ) = 26 × 10–2 m = 26 cm ν = v / (2 πr) = 2×106 s–1 = 2×106 Hz =2 MHz. E = (½ )mv 2 = (½ ) 9 × 10–31 kg × 9 × 1014 m2/s2 = 40.5 ×10–17 J –16 ≈ 4×10 J = 2.5 keV. EXAMPLE 4.3 HELICAL MOTION OF CHARGED PARTICLES AND AURORA BOREALIS In polar regions like Alaska and Northern Canada, a splendid display of colours is seen in the sky. The appearance of dancing green pink lights is fascinating, and equally puzzling. An explanation of this natural phenomenon is now found in physics, in terms of what we have studied here. Consider a charged particle of mass m and charge q, entering a region of magnetic field B with an initial velocity v. Let this velocity have a component vp parallel to the magnetic field and a component vn normal to it. There is no force on a charged particle in the direction of the field. Hence the particle continues to travel with the velocity vp parallel to the field. The normal component vn of the particle results in a Lorentz force (vn × B) which is perpendicular to both vn and B. As seen in Section 4.3.1 the particle thus has a tendency to perform a circular motion in a plane perpendicular to the magnetic field. When this is coupled with the velocity parallel to the field, the resulting trajectory will be a helix along the magnetic field line, as shown in Figure (a) here. Even if the field line bends, the helically moving particle is trapped and guided to move around the field line. Since the Lorentz force is normal to the velocity of each point, the field does no work on the particle and the magnitude of velocity remains the same. During a solar flare, a large number of electrons and protons are ejected from the sun. Some of them get trapped in the earth’s magnetic field and move in helical paths along the field lines. The field lines come closer to each other near the magnetic poles; see figure (b). Hence the density of charges increases near the poles. These particles collide with atoms and molecules of the atmosphere. Excited oxygen atoms emit green light and excited nitrogen atoms emits pink light. This phenomenon is called Aurora Borealis in physics. 139 Physics 4.4 MOTION FIELDS IN COMBINED ELECTRIC AND MAGNETIC 4.4.1 Velocity selector You know that a charge q moving with velocity v in presence of both electric and magnetic fields experiences a force given by Eq. (4.3), that is, F = q (E + v × B) = FE + FB We shall consider the simple case in which electric and magnetic fields are perpendicular to each other and also perpendicular to the velocity of the particle, as shown in Fig. 4.7. We have, ˆ = –qB ˆ FE = qE = qE ˆ j, FB = qv × B, = q v ˆ i × Bk j ˆ Therefore, F = q ( E – vB ) j . ( ) FIGURE 4.7 Thus, electric and magnetic forces are in opposite directions as shown in the figure. Suppose, we adjust the value of E and B such that magnitudes of the two forces are equal. Then, total force on the charge is zero and the charge will move in the fields undeflected. This happens when, E (4.7) B This condition can be used to select charged particles of a particular velocity out of a beam containing charges moving with different speeds (irrespective of their charge and mass). The crossed E and B fields, therefore, serve as a velocity selector . Only particles with speed E /B pass undeflected through the region of crossed fields. This method was employed by J. J. Thomson in 1897 to measure the charge to mass ratio (e/m) of an electron. The principle is also employed in Mass Spectrometer – a device that separates charged particles, usually ions, according to their charge to mass ratio. qE = qvB or v = http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=33.0 4.4.2 Cyclotron The cyclotron is a machine to accelerate charged particles or ions to high energies. It was invented by E.O. Lawrence and M.S. Livingston in 1934 to investigate nuclear structure. The cyclotron uses both electric and magnetic fields in combination to increase the energy of charged particles. As the fields are perpendicular to each other they are called crossed fields. Cyclotron uses the fact that the frequency of revolution of the charged particle in a magnetic field is independent of its energy. The particles move most of the time inside two semicircular disc-like metal containers, D1 and D2, which are called dees as they look like the letter D. Figure 4.8 shows a schematic view of the cyclotron. Inside the metal boxes the particle is shielded and is not acted on by the electric field. The magnetic field, however, acts on the particle and makes it go round in a circular path inside a dee. Every time the particle moves from one dee to another it is acted upon by the electric field. The sign of the electric field is changed alternately in tune with the circular motion of the particle. This ensures that the particle is always accelerated by the electric field. Each time the acceleration increases the energy of the particle. As energy 140 Cyclotron Interactive demonstration: Moving Charges and Magnetism increases, the radius of the circular path increases. So the path is a spiral one. The whole assembly is evacuated to minimise collisions between the ions and the air molecules. A high frequency alternating voltage is applied to the dees. In the sketch shown in Fig. 4.8, positive ions or positively charged particles (e.g., protons) are released at the centre P. They move in a semi-circular path in one of the dees and arrive in the gap between the dees in a time interval T/2; where T, the period of revolution, is given by Eq. (4.6), T = 1 νc = 2πm qB qB (4.8) 2 πm This frequency is called the cyclotron frequency for obvious reasons and is denoted by νc . The frequency νa of the applied voltage is adjusted so that the polarity of the dees is reversed in the same time that it takes the ions to complete one half of the revolution. The requirement νa = νc is called the resonance condition. The phase of the supply is adjusted so that when the positive ions arrive at the edge of D1, D2 is at a lower potential and the ions are accelerated across the gap. Inside the dees the particles travel in a region free of the electric field. The increase in their kinetic energy is qV each time they cross from one dee to another (V refers to the voltage across the dees at that time). From Eq. (4.5), it is clear that the radius of their path goes on increasing each time their kinetic energy increases. The ions are repeatedly accelerated across the dees until they have the required energy to have a radius approximately that of the dees. They are then deflected by a magnetic field and leave the system via an exit slit. From Eq. (4.5) we have, or ν c = qBR (4.9) m where R is the radius of the trajectory at exit, and equals the radius of a dee. Hence, the kinetic energy of the ions is, v= q 2B2R2 1 mv 2 = 2 2m (4.10) The operation of the cyclotron is based on the fact that the time for one revolution of an ion is independent of its speed or radius of its orbit. The cyclotron is used to bombard nuclei with energetic particles, so accelerated by it, and study FIGURE 4.8 A schematic sketch of the cyclotron. There is a source of charged particles or ions at P which move in a circular fashion in the dees, D1 and D2, on account of a uniform perpendicular magnetic field B. An alternating voltage source accelerates these ions to high speeds. The ions are eventually ‘extracted’ at the exit port. 141 Physics the resulting nuclear reactions. It is also used to implant ions into solids and modify their properties or even synthesise new materials. It is used in hospitals to produce radioactive substances which can be used in diagnosis and treatment. Example 4.4 A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons? If the radius of its ‘dees’ is 60 cm, what is the kinetic energy (in MeV) of the proton beam produced by the accelerator. (e =1.60 × 10–19 C, mp = 1.67 × 10–27 kg, 1 MeV = 1.6 × 10–13 J). Solution The oscillator frequency should be same as proton’s cyclotron frequency. Using Eqs. (4.5) and [4.6(a)] we have B = 2π m ν/q =6.3 ×1.67 × 10–27 × 107 / (1.6 × 10–19) = 0.66 T Final velocity of protons is v = r × 2π ν = 0.6 m × 6.3 ×107 = 3.78 × 107 m/s. E = ½ mv 2 EXAMPLE 4.4 = 1.67 ×10–27 × 14.3 × 1014 / (2 × 1.6 × 10–13) = 7 MeV. ACCELERATORS IN INDIA India has been an early entrant in the area of accelerator- based research. The vision of Dr. Meghnath Saha created a 37" Cyclotron in the Saha Institute of Nuclear Physics in Kolkata in 1953. This was soon followed by a series of Cockroft-Walton type of accelerators established in Tata Institute of Fundamental Research (TIFR), Mumbai, Aligarh Muslim University (AMU), Aligarh, Bose Institute, Kolkata and Andhra University, Waltair. The sixties saw the commissioning of a number of Van de Graaff accelerators: a 5.5 MV terminal machine in Bhabha Atomic Research Centre (BARC), Mumbai (1963); a 2 MV terminal machine in Indian Institute of Technology (IIT), Kanpur; a 400 kV terminal machine in Banaras Hindu University (BHU), Varanasi; and Punjabi University, Patiala. One 66 cm Cyclotron donated by the Rochester University of USA was commissioned in Panjab University, Chandigarh. A small electron accelerator was also established in University of Pune, Pune. In a major initiative taken in the seventies and eighties, a Variable Energy Cyclotron was built indigenously in Variable Energy Cyclotron Centre (VECC), Kolkata; 2 MV Tandem Van de Graaff accelerator was developed and built in BARC and a 14 MV Tandem Pelletron accelerator was installed in TIFR. This was soon followed by a 15 MV Tandem Pelletron established by University Grants Commission (UGC), as an inter-university facility in Inter-University Accelerator Centre (IUAC), New Delhi; a 3 MV Tandem Pelletron in Institute of Physics, Bhubaneshwar; and two 1.7 MV Tandetrons in Atomic Minerals Directorate for Exploration and Research, Hyderabad and Indira Gandhi Centre for Atomic Research, Kalpakkam. Both TIFR and IUAC are augmenting their facilities with the addition of superconducting LINAC modules to accelerate the ions to higher energies. Besides these ion accelerators, the Department of Atomic Energy (DAE) has developed many electron accelerators. A 2 GeV Synchrotron Radiation Source is being built in Raja Ramanna Centre for Advanced Technologies, Indore. The Department of Atomic Energy is considering Accelerator Driven Systems (ADS) for power production and fissile material breeding as future options. 142 Moving Charges and Magnetism 4.5 MAGNETIC FIELD DUE BIOT-SAVART LAW TO A CURRENT ELEMENT, All magnetic fields that we know are due to currents (or moving charges) and due to intrinsic magnetic moments of particles. Here, we shall study the relation between current and the magnetic field it produces. It is given by the Biot-Savart’s law. Figure 4.9 shows a finite conductor XY carrying current I. Consider an infinitesimal element dl of the conductor. The magnetic field dB due to this element is to be determined at a point P which is at a distance r from it. Let θ be the angle between dl and the displacement vector r. According to Biot-Savart’s law, the magnitude of the magnetic field dB is proportional to the current I, the element length |dl|, and inversely proportional to the square of the distance r. Its direction* is perpendicular to the plane containing dl and r . Thus, in vector notation, dB ∝ Idl×r r3 FIGURE 4.9 Illustration of the Biot-Savart law. The current element I dl produces a field dB at a distance r. The ⊗ sign indicates that the field is perpendicular to the plane of this page and directed into it. = µ0 I d l × r 4π r3 [4.11(a)] where µ0/4 π is a constant of proportionality. The above expression holds when the medium is vacuum. The magnitude of this field is, dB = µ0 I dl sin θ [4.11(b)] 4π r2 where we have used the property of cross-product. Equation [4.11 (a)] constitutes our basic equation for the magnetic field. The proportionality constant in SI units has the exact value, µ0 = 10 −7 Tm/A [4.11(c)] 4π We call µ0 the permeability of free space (or vacuum). The Biot-Savart law for the magnetic field has certain similarities as well as differences with the Coulomb’s law for the electrostatic field. Some of these are: (i) Both are long range, since both depend inversely on the square of distance from the source to the point of interest. The principle of superposition applies to both fields. [In this connection, note that the magnetic field is linear in the source I dl just as the electrostatic field is linear in its source: the electric charge.] (ii) The electrostatic field is produced by a scalar source, namely, the electric charge. The magnetic field is produced by a vector source I d l. * The sense of dl×r is also given by the Right Hand Screw rule : Look at the plane containing vectors dl and r. Imagine moving from the first vector towards second vector. If the movement is anticlockwise, the resultant is towards you. If it is clockwise, the resultant is away from you. 143 Physics (iii) The electrostatic field is along the displacement vector joining the source and the field point. The magnetic field is perpendicular to the plane containing the displacement vector r and the current element I d l. (iv) There is an angle dependence in the Biot-Savart law which is not present in the electrostatic case. In Fig. 4.9, the magnetic field at any point in the direction of dl (the dashed line) is zero. Along this line, θ = 0, sin θ = 0 and from Eq. [4.11(a)], |dB| = 0. There is an interesting relation between ε0, the permittivity of free space; µ , the permeability of free space; and c, the speed of light in vacuum: 0 ε 0 µ 0 = ( 4 πε 0 )  1 1 1   µ0   = 2 10 −7 = = 8 2  4π    9 × 109   (3 × 10 ) c ( ) We will discuss this connection further in Chapter 8 on the electromagnetic waves. Since the speed of light in vacuum is constant, the product µ0ε0 is fixed in magnitude. Choosing the value of either ε0 or µ0, fixes the value of the other. In SI units, µ0 is fixed to be equal to 4π × 10–7 in magnitude. Example 4.5 An element ∆ l = ∆x ˆ i is placed at the origin and carries a large current I = 10 A (Fig. 4.10). What is the magnetic field on the y-axis at a distance of 0.5 m. ∆x = 1 cm. FIGURE 4.10 Solution |dB | = µ0 I dl sin θ 4π r2 [using Eq. (4.11)] Tm A dl = ∆x = 10−2 m , I = 10 A, r = 0.5 m = y, µ0 / 4 π = 10 −7 θ = 90° ; sin θ = 1 10 −7 × 10 × 10 −2 = 4 × 10–8 T 25 × 10 −2 The direction of the field is in the +z-direction. This is so since, dB = EXAMPLE 4.5 ˆ i׈ j = y ∆x k dl × r = ∆x ˆ i×yˆ j = y ∆x ˆ ( ) We remind you of the following cyclic property of cross-products, ˆ ˆ; ˆ ˆ =ˆ ˆ ׈ i׈ j=k j×k i; k i=ˆ j 144 Note that the field is small in magnitude. 11. 4. For Fig.11. This is shown in Fig. dB = µ0 Idl 4π x + R2 2 ( ) (4. The evaluation entails summing up the effect of infinitesimal current elements (I dl) mentioned in the previous section. 4. vacuum). Consider a conducting element dl of the loop. the d B ⊥ component due to d l is cancelled by the contribution due to the diametrically opposite dl element. The net contribution along x-direction can be obtained by integrating dBx = dB cos θ over the loop.14) From Eqs. The x-axis is the axis of the loop. we shall evaluate the magnetic field due to a circular coil along its axis. The magnitude dB of the magnetic field due to dl is given by the Biot-Savart law [Eq. they cancel out and we obtain a null result. It has an x-component dBx and a component perpendicular to x-axis. the element d l in Fig.11 depicts a circular loop carrying a steady current I.6 MAGNETIC FIELD CURRENT LOOP ON THE AXIS OF A CIRCULAR In this section. For example. 4. Figure 4.11. Shown are the magnetic field dB (due to a line element dl ) and its components along and perpendicular to the axis.12) FIGURE 4. For example. Further. dB = µ 0 I dl × r r3 4π (4. any element of the loop will be perpendicular to the displacement vector from the element to the axial point. 4.14). 4. shown in Fig. Let x be the distance of P from the centre O of the loop. We assume that the current I is steady and that the evaluation is carried out in free space (i. We wish to calculate the magnetic field at the point P on this axis. The loop is placed in the y-z plane with its centre at the origin O and has a radius R. When the components perpendicular to the x-axis are summed over. only the x-component survives. (4.11 Magnetic field on the axis of a current carrying circular loop of radius R. we shall use the Biot-Savart law to calculate the magnetic field due to a circular loop.e.Moving Charges and Magnetism In the next section. 4. dB⊥. cos θ = R (x + R 2 )1/ 2 2 (4.13) The direction of dB is shown in Fig.13) and (4. Thus.11(a)].11 is in the y-z plane whereas the displacement vector r from dl to the axial point P is in the x-y plane. It is perpendicular to the plane formed by dl and r. 4. Hence |dl × r|=r dl. Thus.. dB x = µ0 Idl 4π R (x 2 + R2 ) 3/2 145 .11. Now r 2 = x 2 + R 2 . Physics The summation of elements d l over the loop yields 2 π R . B = Bx ˆ i= µ0 I R 2 ( ) (4. Here x = 0. 4.13 .13(b)? 146 EXAMPLE 4.6 A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2. 4. B0 = µ0 I ˆ i FIGURE 4. and we obtain. The upper side of the loop may be thought of as the north pole and the lower side as the south pole of a magnet. The direction of the field is given by the right-hand thumb rule described in the text.16) 2R The magnetic field lines due to a circular wire form closed loops and are shown in Fig. Thus. the magnetic field at P due to entire circular loop is ˆ i 3/2 (4. The right-hand thumb gives the direction of the magnetic field.0 cm as shown in Fig. the circumference of the loop. (a) What is the magnetic field due to the straight segments? (b) In what way the contribution to B from the semicircle differs from that of a circular loop and in what way does it resemble? (c) Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Fig. we may obtain the field at the centre of the loop. Consider the magnetic field B at the centre of the arc.15) 2 x 2 + R2 As a special case of the above result.12.6 FIGURE 4. 4.13(a). Example 4. The direction of the magnetic field is given by (another) right-hand thumb rule stated below: Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current.12 The magnetic field lines for a current loop. 7 Consider a tightly wound 100 turn coil of radius 10 cm. We consider the boundary to be made up of a number of small line elements. We shall assume that. The number of turns N = 100. The surface has current passing through it. B= EXAMPLE 4.14 passing through the surface. All such products are added together.17(a)] proves sufficient.. The integral is taken over the closed loop coinciding with the boundary C of the surface. The magnitude of the magnetic field is. Straight segments do not contribute to |B|. Therefore. We take the value of the tangential component of the magnetic field. dl × r are all parallel to each other (into the plane of the paper). Let the fingers of the right-hand be curled in the sense the boundary is traversed in the loop integral “B. 4.17(a)] where I is the total current through the surface. Thus B is 1. (c) Same magnitude of B but opposite in direction to that in (b). Consider one such element of length dl.7 AMPERE’S CIRCUITAL LAW There is an alternative and appealing way in which the Biot-Savart law may be expressed. Example 4. Ampere’s law states that this integral is equal to µ0 times the total current FIGURE 4. given by the right-hand rule. we may take each circular element to have the same radius R = 10 cm = 0. The sum then tends to an integral. i. (b) For all segments of the semicircular arc. dl × r = 0. in such cases. it is possible to choose the loop (called an amperian loop) such that at each point of the loop. a much simplified version of Eq. Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. All such contributions add up in magnitude. For several applications. Bt. We consider the limit as the lengths of elements get smaller and their number gets larger.d l.7 µ0 NI 2R = 4π × 10 –7 × 102 × 1 −4 = 2π × 10 −4 = 6.d l].14).Moving Charges and Magnetism Solution (a) dl and r for each element of the straight segments are parallel.9 × 10–4 T normal to the plane of the paper going into it. [4. ∫ Bidl = µ 0 I [4. The relation above involves a sign-convention. either 147 .28 × 10 T 2 × 10 –1 4. carrying a current of 1 A.6 EXAMPLE 4. Ampere’s circuital law considers an open surface with a boundary (Fig. What is the magnitude of the magnetic field at the centre of the coil? Solution Since the coil is tightly wound. Then the direction of the thumb gives the sense in which the current I is regarded as positive.1 m. at this element and multiply it by the length of that element dl [Note: Btdl=B.e. It tends to blow up only when we come very close to the wire. in Fig.17(b)] When there is a system with a symmetry such as for a straight infinite current-carrying wire in Fig. Whenever there is symmetry. Let Ie be the current enclosed by the loop. This is unlike the electrostatic field lines which originate from positive charges and end at negative charges.1(c). or (ii) B is normal to the loop. The field is directly proportional to the current and inversely proportional to the distance from the (infinitely long) current source. [4. Ampere was a child prodigy who mastered advanced mathematics by the age of 12. 4. (i) It implies that the field at every point on a circle of radius r. or (iii) B vanishes.18) The above result for the infinite wire is interesting from several points of view. mathematician and chemist who founded the science of electrodynamics. He hypothesised that all magnetic phenomena are due to circulating electric currents. Eq. Now. The expression for the magnetic field of a straight wire provides a theoretical justification to Oersted’s experiments. BL =µ0Ie [4. much the same way Gauss’ law helps in determination of the electric field.15. Andre Ampere (1775 – 1836) Andre Marie Ampere was a French physicist. He carried out a large series of experiments to explore the relationship between current electricity and magnetism. the magnetic field possesses what is called a cylindrical symmetry. the solutions simplify. the iron filings form concentric circles. (4. (with the wire along the axis). the field due to it at a nonzero distance is not infinite. Thus. (ii) The field direction at any point on this circle is tangential to it.9 below. (iii) Another interesting point to note is that even though the wire is infinite.Physics (i) B is tangential to the loop and is a non-zero constant B. 2πr. the lines of constant magnitude of magnetic field form concentric circles. Ampere was humble and absentminded. In other words. B = µ0 I/ (2πr) (4.17 (b)]. These lines called magnetic field lines form closed loops. the Ampere’s law enables an easy evaluation of the magnetic field. ANDRE AMPERE (1775 –1836) 148 . Notice now. We find that the magnetic field at a distance r outside the wire is tangential and given by B × 2πr = µ0 I. This is exhibited in the Example 4. Then. He once forgot an invitation to dine with the Emperor Napoleon. is same in magnitude. let L be the length (part) of the loop for which B is tangential. The field that normally can depend on three coordinates depends only on one: r. B. 4.17) reduces to. The boundary of the loop chosen is a circle and magnetic field is tangential to the circumference of the circle. Ampere grasped the significance of Oersted’s discovery. He died of pneumonia at the age of 61. These investigations culminated in 1827 with the publication of the ‘Mathematical Theory of Electrodynamic Phenomena Deduced Solely from Experiments’. His gravestone bears the epitaph: Tandem Felix (Happy at last). for the left hand side of Eq. The law gives. Ampere’s circuital law is not new in content from Biot-Savart law. For this loop. in the interior (current or charge). is a circle concentric with the cross-section.Moving Charges and Magnetism (iv) There exists a simple rule to determine the direction of the magnetic field due to a long wire.8 L =2πr * Note that there are two distinct right-hand rules: One which gives the direction of B on the axis of current-loop and the other which gives direction of B for a straight conducting wire. The Amperian loop. This rule.19(a)] 1 (r > a) r (b) Consider the case r < a. Fingers and thumb play different roles in the two. The following example will help us understand what is meant by the term enclosed current. We also note that Ampere’s circuital law holds for steady currents which do not fluctuate with time. L =2πr Ie = Current enclosed by the loop = I The result is the familiar expression for a long straight wire B (2π r) = µ0I B= B∝ µ0 I 2πr [4.15 shows a long straight wire of a circular cross-section (radius a) carrying steady current I. is: Grasp the wire in your right hand with your extended thumb pointing in the direction of the current. EXAMPLE 4. called the right-hand rule*. labelled 2. Both.15 Solution (a) Consider the case r > a . For this loop. taking the radius of the circle to be r. Both relate the magnetic field and the current. FIGURE 4. and both express the same physical consequences of a steady electrical current. Ampere’s and Gauss’s law relate a physical quantity on the periphery or boundary (magnetic or electric field) to another physical quantity. Ampere’s law is to Biot-Savart law. Your fingers will curl around in the direction of the magnetic field. Calculate the magnetic field in the region r < a and r > a. 149 .8 Figure 4. The Amperian loop is a circle labelled 1. Example 4. The current I is uniformly distributed across this cross-section. what Gauss’s law is to Coulomb’s law. the source. namely. Since the current distribution is uniform. The synchrotron uses a combination of both to generate the high magnetic fields required. The television uses the solenoid to generate magnetic fields needed. the current enclosed is. 4. . there exists a large number of situations of high symmetry where the law can be conveniently applied. but is less than this value. B (2 π r ) = µ0  µ I  B =  0 2r  2 πa  I r2 a2 [4. In both.16 EXAMPLE 4. it may not always facilitate an evaluation of the magnetic field in every case. it cannot be applied to extract the simple expression B = µ0I/2R [Eq.16) shows a plot of the magnitude of B with distance r from the centre of the wire. This example possesses the required symmetry so that Ampere’s law can be applied readily. For example. (4.8 Figure (4. We shall use it in the next section to calculate the magnetic field produced by two commonly used and very useful magnetic systems: the solenoid and the toroid. for the case of the circular loop discussed in Section 4. It should be noted that while Ampere’s circuital law holds for any loop. solenoid and toroid.16)] for the field at the centre of the loop.19(b)] B∝r (r < a) FIGURE 4.8 THE SOLENOID AND THE TOROID 150 The solenoid and the toroid are two pieces of equipment which generate magnetic fields. we come across a situation of high symmetry where Ampere’s law can be conveniently applied. The direction of the field is tangential to the respective circular loop (1 or 2) and given by the right-hand rule described earlier in this section.  πr2  Ir 2 Ie = I  2 =  πa  a2 Using Ampere’s law. However.Physics Now the current enclosed I e is not I.6. Figure 4.18 represents this idealised picture. In Fig. We show a section of this solenoid in an enlarged manner in Fig. Only the exterior semi-circular part is shown.17 (a) The magnetic field due to a section of the solenoid which has been stretched out for clarity.1 The solenoid We shall discuss a long solenoid. strong and along the axis of the solenoid. The field outside the solenoid approaches zero. (b) The magnetic field of a finite solenoid.17(b). Figure 4. So each turn can be regarded as a circular loop.17 displays the magnetic field lines for a finite solenoid. 4. As the solenoid is made longer it appears like a long cylindrical metal sheet. By long solenoid we mean that the solenoid’s length is large compared to its radius.18 The magnetic field of a very long solenoid. it is clear from the circular loops that the field between two neighbouring turns vanishes. We shall assume that the field outside is zero.17(a). In Fig. The field inside becomes everywhere parallel to the axis. 151 . FIGURE 4. Figure 4.8. Notice how the circular loops between neighbouring turns tend to cancel. It consists of a long wire wound in the form of a helix where the neighbouring turns are closely spaced. 4. we see that the field at the interior mid-point P is uniform. The net magnetic field is the vector sum of the fields due to all the turns. 4. The field at the exterior mid-point Q is weak and moreover is along the axis of the solenoid with no perpendicular or normal component. We consider a rectangular Amperian loop abcd to determine the field.17(a).17(b) shows the entire finite solenoid with its magnetic field. FIGURE 4. Enamelled wires are used for winding so that turns are insulated from each other.Moving Charges and Magnetism 4. 19(b) shows a sectional view of the toroid. The solenoid is commonly used to obtain a uniform magnetic field.8. The circular areas bounded by loops 2 and 3 both cut the toroid: so that each turn of current carrying wire is cut once by the loop 2 and twice by the loop 3. L = 2π r1. B h = µ0I (n h) B = µ0 n I (4. 4. so Ie = 0. Let the magnetic field inside the solenoid be B. Let the magnetic field along loop 1 be B1 in magnitude. the magnetic field should be tangential to each of them and constant in magnitude for a given loop. Once again we employ Ampere’s law in the form of Eq. Figure 4. the field component is zero. 4. The direction of the magnetic field inside is clockwise as per the right-hand thumb rule for circular loops. the relevant length of the Amperian loop is. Along transverse sections bc and ad.17 (b)] BL = µ0Ie. By symmetry. We shall see in the next chapter that a large field is possible by inserting a soft iron core inside the solenoid. the loop encloses no current. L = h.Physics Consider a rectangular Amperian loop abcd. these two sections make no contribution. However. Ie = I (n h). Let the magnetic field along loop 3 be B3.17(a)]. From Ampere’s circuital law [Eq. We find. The field B inside the toroid is constant in magnitude for the ideal toroid of closely wound turns. Then in Ampere’s circuital law [Eq. L = 2π r. B (2πr) = µ0NI . B1 (2 π r1) = µ0(0). and B3 = 0. we see that the current coming out of the plane of the paper is cancelled exactly by the current going into it.20) The direction of the field is given by the right-hand rule. The enclosed current is. from the sectional cut. FIGURE 4. Thus. It can be viewed as a solenoid which has been bent into a circular shape to close on itself. (b) A sectional view of the toroid.17 (a)]. Along cd the field is zero as argued above.2 The toroid The toroid is a hollow circular ring on which a large number of turns of a wire are closely wound.19(a) carrying a current I.19 (a) A toroid carrying a current I. Thus. the magnetic field at any point P in the open space inside the toroid is zero. 4. Ie= 0. 4. We shall now consider the magnetic field at S. Thus. We shall now show that magnetic field at Q is likewise zero. then the total number of turns is nh. Let n be the number of turns per unit length. It is shown in Fig. The magnetic field can be obtained at an arbitrary distance r from the centre O of the toroid by Ampere’s circuital law. where I is the current in the solenoid. The current enclosed Ie is (for N turns of toroidal coil) N I. Let the field along ab be B. B1 = 0 152 Thus. We shall see that the magnetic field in the open space inside (point P) and exterior to the toroid (point Q) is zero. The dashed lines labelled 1. Thus. 2 and 3 are shown by dashed lines. [4. Three circular Amperian loops 1. Once again from Ampere’s law L = 2 π r3. 2 and 3 are three circular Amperian loops. However. but does not change much during one circular orbit.3 (see also the box on helical motion of charged particles earlier in this chapter) that orbits of charged particles are helical. The particles will never touch the sides of the container.iter. (4. In reality the turns of the toroidal coil form a helix and there is always a small magnetic field external to the toroid. Such magnetic bottles are of great use in confining the high energy plasma in fusion experiments.e. Then B= µ0 NI N = 2πr n = (average) perimeter of the toroid × number of turns per unit length and thus. It has a horizontal component against the forward motion. The plasma will destroy any other form of material container because of it’s high temperature. enclosed in an evacuated container (see figure below where we have not shown the container). We consider two solenoids at a distance from each other. If the magnetic field is non-uniform. for achieving controlled fusion.org. (4. being set up in France. Let r be the average radius of the toroid and n be the number of turns per unit length. B = µ0 n I.22) i. Charged particles moving in the region between the two solenoids will start with a small radius.21) to make the comparison easier with the solenoid result given in Eq. We re-express Eq. you may visit http://www. Toroids are expected to play a key role in the tokamak. 153 .21) 2πr We shall now compare the two results: for a toroid and solenoid. [See the direction of F as the particle approaches coil 2 in the figure. Such an arrangement will act like magnetic bottle or magnetic container.Moving Charges and Magnetism (4.20). MAGNETIC CONFINEMENT We have seen in Section 4. The radius will increase as field decreases and the radius will decrease again as field due to the second solenoid takes over. of which India is a collaborating nation.] This makes the particles turn back when they approach the solenoid. (4. then the radius of the helix will decrease as it enters stronger magnetic field and the radius will increase when it enters weaker magnetic fields. For details of ITER collaboration and the project.. an equipment for plasma confinement in fusion power reactors. There is an international collaboration called the International Thermonuclear Experimental Reactor (ITER). the result for the solenoid! In an ideal toroid the coils are circular. Another useful container is a toroid. The solenoids act as a mirror or reflector. we have learnt that an external magnetic field will exert a force on a current-carrying conductor. the same magnetic field Ba at all points along up by conductor ‘a’ at conductor ‘b’.5 m and radius r = 0.4). The right-hand rule tells us that the direction of this field is downwards (when the conductors are placed horizontally). . Eq.20 shows two long parallel conductors a FIGURE 4. Thus. we can use the long solenoid formula. This follows from the Lorentz force formula.01 m. namely.19(a)] or from Ampere’s circuital law. on the shape and size of the conductors as well as the distances between the conductors.28 × 10–3 T n = EXAMPLE 4. The magnitude of this force is given by Eq. the conductor ‘b’. In this section. Ba is the magnetic field set produces. We label this force as Fba.e.20) B = µ0n I = 4π × 10–7 × 103 × 5 = 6. l >> a . the force on a segment L of ‘b’ due to ‘a’. which will perhaps help us to appreciate Ampere’s painstaking work.. Ba = µ0 I a 2πd 154 The conductor ‘b’ carrying a current Ib will experience a sideways force due to the field Ba. Further.9 FORCE BETWEEN TWO PARALLEL CURRENTS.Physics Example 4. What is the magnitude of the magnetic field inside the solenoid? Solution The number of turns per unit length is. It carries a current of 5 A. 500 = 1000 turns/m 0.5 The length l = 0. l/a = 50 i. [4. THE AMPERE We have learnt that there exists a magnetic field due to a conductor carrying a current which obeys the Biot-Savart law. Thus. respectively. Its magnitude is given by Eq.9 4. Figure 4. Ampere studied the nature of this magnetic force and its dependence on the magnitude of the current. The conductor ‘a’ currents Ia and Ib and separated by a distance d. (4. The direction of this force is towards the conductor ‘a’ (Verify this).9 A solenoid of length 0. Hence.5 m has a radius of 1 cm and is made up of 500 turns. we shall take the simple example of two parallel currentcarrying conductors. (4. it is logical to expect that two current-carrying conductors placed near each other will exert (magnetic) forces on each other.20 Two long straight and b separated by a distance d and carrying (parallel) parallel conductors carrying steady currents I a and I b. In the period 1820-25. This definition of the ampere was adopted in 1946. namely. Then. This rule is the opposite of what we find in electrostatics. holds even for the case of time-dependent situations with electromagnetic fields. which is one of the seven SI base units. Fba = –Fab (4. Like (same sign) charges repel each other. provided the momentum carried by fields is also taken into account. would produce on each of these conductors a force equal to 2 × 10–7 newtons per metre of length. It is equal in magnitude to Fba. parallel conductors of negligible cross-section. straight. Let fba represent the magnitude of the force Fba per unit length. Thus. can now be defined in terms of the ampere. from Eq. the coulomb. One can show that oppositely directed currents repel each other. and antiparallel currents repel. It is a theoretical definition.Moving Charges and Magnetism Fba = Ib L Ba = µ0 I a I b 2πd L (4. on a segment of length L of ‘a’ due to the current in ‘b’. We have seen from above that currents flowing in the same direction attract each other.23) It is of course possible to compute the force on ‘a’ due to ‘b’. 155 . Thus. and directed towards ‘b’. This. When a steady current of 1A is set up in a conductor. Thus. Newton’s third law may not hold for forces between charges and/or conductors.24) Note that this is consistent with Newton’s third Law. Parallel currents attract. we have shown that the Biot-Savart law and the Lorentz force yield results in accordance with Newton’s third Law*. In practice one must eliminate the effect of the earth’s magnetic field and substitute very long wires by multiturn coils of appropriate geometries. The ampere is the value of that steady current which. (4.25) The above expression is used to define the ampere (A). An instrument called the current balance is used to measure this mechanical force. f ba = µ0 I a I b 2 πd (4. From considerations similar to above we can find the force Fab. when maintained in each of the two very long. at least for parallel conductors and steady currents. * It turns out that when we have time-dependent currents and/or charges in motion. The SI unit of charge. but like (parallel) currents attract each other. and placed one metre apart in vacuum. the quantity of charge that flows through its cross-section in 1s is one coulomb (1C). however. An essential consequence of the Newton’s third law in mechanics is conservation of momentum of an isolated system.23). Keep your face away from mercury vapours as they are poisonous. θ = 90° Hence. Take some mercury in a dish and adjust the spring such that the tip is just above the mercury surface. . you may require some small adjustments to get a good effect. we can increase the effective length of the parallel current and by using mercury. It would be like pulling a wire by a string going over a pulley to which a 50 mg weight is attached. . Take a soft spring whose natural period of oscillations is about 0. The displacement of the wire would be quite unnoticeable. (b) south to north? Solution F = I l × B F = IlB sinθ The force per unit length is f = F/l = I B sinθ (a) When the current is flowing from east to west. tick. Thus for 5 A current in each wire at a separation of 1cm. The spring shrinks with a jerk. because of the smallness of the factor µ. we can make the displacement of even a few mm observable very dramatically. Do not inhale mercury vapours for long. . Example 4. A very long straight conductor is carrying a steady current of 1A.10 The horizontal component of the earth’s magnetic field at a certain place is 3. Take the DC current source. Hence it is difficult to demonstrate attraction or repulsion between currents. the spring relaxes and tries to come back to its original position. the circuit is completed through mercury. the force per metre would be 5 × 10–4 N.Physics ROGET’S SPIRAL FOR ATTRACTION BETWEEN PARALLEL CURRENTS Magnetic effects are generally smaller than electric effects. Hang it vertically and attach a pointed tip to its lower end. the tip comes out of mercury (just by a mm or so). connect one of its terminals to the upper end of the spring. You will also need a constant-current supply giving a constant current of about 5 A. and dip the other terminal in mercury. With the use of a soft spring. If the tip of the spring touches mercury. .5 – 1s. and the cycle continues with tick. the circuit is broken. f=IB = 1 × 3 × 10–5 = 3 × 10–5 N m–1 156 EXAMPLE 4. the tip again touches mercury establishing a current in the circuit. Let the tip be adjusted so that it just touches the mercury surface. Let the DC source be put off to begin with. Switch on the constant current supply. In the beginning. tick.10 . which is about 50 mg weight.0 ×10–5 T and the direction of the field is from the geographic south to the geographic north. As a consequence. the current stops. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west. the force between currents is rather small. and watch the fascinating outcome. as shown in the figure here. a a + IbB = I (ab ) B 2 2 =IAB = IbB (4. MAGNETIC DIPOLE 4. The field exerts no force on the two arms AD and BC of the loop.Moving Charges and Magnetism This is larger than the value 2×10–7 Nm–1 quoted in the definition of the ampere. The direction of the force is downwards. (b) The couple acting on the coil. This is illustrated in Fig. It is perpendicular to the arm AB of the loop and exerts a force F1 on it which is directed into the plane of the loop.21(b) shows a view of the loop from the AD end.26) where A = ab is the area of the rectangle.21 (a) A rectangular current-carrying coil in uniform magnetic field. This torque is (in magnitude). Hence it is important to eliminate the effect of the earth’s magnetic field and other stray fields while standardising the ampere. The magnetic moment m points downwards. τ = F1 a a + F2 2 2 FIGURE 4. It does not experience a net force. the net force on the loop is zero. θ = 0o f=0 Hence there is no force on the conductor. We next consider the case when the plane of the loop. but makes an angle with it.10).1 Torque on a rectangular current loop in a uniform magnetic field We now show that a rectangular loop carrying a steady current I and placed in a uniform magnetic field experiences a torque. It shows that the torque on the loop tends to rotate it anti-clockwise.10 4. We first consider the simple case when the rectangular loop is placed such that the uniform magnetic field B is in the plane of the loop. Its magnitude is. F2 = I b B = F1 Thus. F1 = I b B Similarly it exerts a force F2 on the arm CD and F2 is directed out of the plane of the paper.10. There is a torque on the loop due to the pair of forces F1 and F2. EXAMPLE 4. 4.10 TORQUE ON CURRENT LOOP. We take the angle between the field and the normal to 157 . is not along the magnetic field.21(a). This direction may be obtained by the directional property of cross product of vectors. (b) When the current is flowing from south to north. Figure 4. This behaviour is analogous to that of electric dipole in a uniform electric field (Section 1. The torque τ is along the axis and tends to rotate the coil anticlockwise. 27) can be expressed by one expression τ = m×B (4. This makes the forces collinear and the net force and torque zero.Physics the coil to be angle θ (The previous case corresponds to θ = π/2). 4. we see that the torque τ vanishes when m is either parallel or antiparallel to the magnetic field B. Figure 4. (4. τ = F1 FIGURE 4. a a sin θ + F2 sin θ 2 2 = I ab B sin θ = I A B sin θ (4. The torques in Eqs.28) where the direction of the area vector A is given by the right-hand thumb rule and is directed into the plane of the paper in Fig.28).22 illustrates this general case. Then as the angle between m and B is θ . The torque is. The forces on the arms BC and DA are equal. resulting in no net force or torque. which connects the centres of mass of BC and DA. Figure 4.26) and (4. This is because the perpendicular distance between the forces of the couple has decreased. the dimensions of the magnetic moment are [A][L2] and its unit is Am2. As is clear from Eq. The forces on arms AB and CD are F1 and F2. with magnitude.22(b) is a view of the arrangement from the AD end and it illustrates these two forces constituting a couple. opposite. From Eq.27) As θ 0.21.27) can be expressed as vector product of the magnetic moment of the coil and the magnetic field. The forces F1 and F2 acting on the arms AB and CD are indicated.29) This is analogous to the electrostatic case (Electric dipole of dipole moment pe in an electric field E). This indicates a state of equilibrium as there is no torque on the coil (this also applies to any object with a magnetic moment m). We define the magnetic moment of the current loop as. When m and B are parallel the 158 . (4. They too are equal and opposite. (4. The magnitude of the torque on the loop is. (4.22 (a) The area vector of the loop ABCD makes an arbitrary angle θ with the magnetic field.29).26) and (4. m=IA (4. Being collinear along the axis they cancel each other. however. the perpendicular distance between the forces of the couple also approaches zero. F1 = F2 = I b B But they are not collinear! This results in a couple as before. (b) Top view of the loop. and act along the axis of the coil. Eqs. less than the earlier case when plane of loop was along the magnetic field. B= m = N I A = N I π r2 = 100 × 3. with m=NIA (4. (d) From Newton’s second law.11 A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.16) dt µ0 NI 2R Here.2 4 × 10 −5 × 10 = (using π × 3.2 = 10) 2 × 10 −1 2 × 10 −1 = 2 × 10–3 T The direction is given by the right-hand thumb rule. Eq. θ = π/2 (or 90º).14 × 10–2 = 10 A m2 The direction is once again given by the right hand thumb rule.29). A uniform magnetic field of 2T in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. (b) The magnetic moment is given by Eq. The coil rotates through an angle of 90º under the influence of the magnetic field.Moving Charges and Magnetism equilibrium is a stable one. (a) What is the field at the centre of the coil? (b) What is the magnetic moment of this coil? The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. (4. still holds. Hence. Solution dω = m B sin θ (a) From Eq. initial torque τi = 0. I = 3.1 m. dω dω dθ dω ω = = dt dθ dt dθ EXAMPLE 4.2 A. From chain rule. θ = 0. Thus.1 kg m2. I ω dω = m B sin θ dθ 159 .2 A. If the loop has N closely wound turns. the equilibrium is unstable as any rotation produces a torque which increases with the amount of rotation. The presence of this torque is also the reason why a small magnet or any magnetic dipole aligns itself with the external magnetic field. and R = 0. final torque τf = m B = 10 × 2 = 20 N m.2 × 3. Thus. B= 4π × 10−7 × 102 × 3. (4. the expression for torque. N = 100. (c) τ = m × B [from Eq. When they are antiparallel.11 Using this. (c) What are the magnitudes of the torques on the coil in the initial and final position? (d) What is the angular speed acquired by the coil when it has rotated by 90º? The moment of inertia of the coil is 0.30) Example 4. Finally.29)] = m B sin θ Initially.30). (4. Any small rotation of the coil produces a torque which brings it back to its original position. I where I is the moment of inertia of the coil. (4. This attractive property of magnets made it difficult for them to move around. The name lodesto ne (or loadsto ne ) g iven to a naturally occurring ore of ironmagnetite means leading stone. distant galaxies. A thin long piece of a magnet. The word magnet is derived from the name of an island in Greece called magnesia where mag netic ore deposits were found. men and beasts all are permeated throug h and through with a host of magnetic fields from a variety of sources. A similar effect was observed when it was placed on a piece of cork which was then allowed to float in still water. pointed in the north-south direction. when suspended freely. Caravans crossing the Gobi desert also employed magnetic needles. The directional property of magnets was also known since ancient times. A Chinese legend narrates the tale of the victory of the emperor Huang-ti about four thousand years ago.Chapter Five MAGNETISM AND MATTER 5. The earth magnetism predates human evolution. The technological exploitation of this property is generally credited to the Chinese. Chinese texts dating 400 BC mention the use of magnetic needles for navigation on ships. Vast.1 INTRODUCTION Magnetic phenomena are universal in nature. Shepherds on this island complained that their wooden shoes (which had nails) at times stayed struck to the ground. the tiny invisible atoms. as early as 600 BC. which he owed to his craftsmen (whom . Their iron-tipped rods were similarly affected. This is an artist magnetism are: of the earliest known compasses. it points in the north-south direction.1 is an artist description of this chariot. In the previous chapter we have learned that moving charges or electric currents produce magnetic fields. Einstein was fascinated. In the present chapter. and to defeat them. there is an attractive force between the north pole of one magnet and the south pole of the other. dia-. (iii) There is a repulsive force when north poles ( or south poles ) of two magnets are brought close together. We begin with a description of a bar magnet and its behaviour in an external magnetic field. which was made in the early part of the nineteenth century is credited to Oersted. The figure swiveled around so that the finger of the statuette on it always pointed south. We conclude with a section on electromagnets and permanent magnets. (i) The earth behaves as a magnet with the magnetic field pointing approximately from the geographic south to the north. and played endlessly with it. 174 . This discovery. Conversely. and ferromagnetism. among others. Unlike electric charges. If a bar magnet is broken into two halves. mounted on the chariot always points Some of the commonly known ideas regarding south. We describe Gauss follow it up with an account of the earth how materials can be classified on the basis of their magnetic properties.1 The arm of the statuette as a subject in its own right. (iv) We cannot isolate the north.2 THE BAR MAGNET One of the earliest childhood memories of the famous physicist Albert Einstein was that of a magnet gifted to him by a relative. 5. (ii) When a bar magnet is freely suspended. thousands of years old. we get two similar bar mag nets with somewhat weaker properties.Physics nowadays you would call engineers). The tip which points to the geographic north is called the north pole and the tip which points to the geographic south is called the south pole of the magnet. or south pole of a magnet. Fig ure 5. we take a look at magnetism FIGURE 5. Ampere. (v) It is possible to make magnets out of iron and its alloys. These built a chariot on which they placed a magnetic figure with arms ou tstretched. Biot and Savart. We describe para-. isolated magnetic north and south poles known as magnetic monopoles do not exist. With this chariot. He wondered how the magnet could affect objects such as nails or pins placed away from it and not in any way connected to it by a spring or string. Huang-ti to attack the enemy from the rear in thick fog. 5. The curves labelled i and ii are closed Gaussian surfaces. begin from a positive charge and end on the negative charge or escape to infinity. the South pole . The pattern of iron filings suggests that the magnet has two poles similar to the positive and neg ative charg e of an electric dipole.3 The field lines of (a) a bar magnet. respectively. This nomenclature is avoided since it can be confusing. This is FIGURE 5. (ii) The tangent to the field line at a given point represents the direction of the net magnetic field B at that point.3(c). Electric field filings surrounding a bar magnet. For comparison refer to the Chapter 1. (b) a current-carrying finite solenoid and (c) electric dipole. one pole is designated the North pole and the other. The lines of an electric dipole are also displayed in Fig.Magnetism and Matter We begin our study by examining iron filings sprinkled on a sheet of glass placed over a short bar magnet. Unli ke electrostatics the fi eld lines i n magnetism do not indicate the direction of the force on a (moving) charge. 175 . Their magnetic field lines. When suspended freely.2. * In some textbooks the magnet ic field l ines are called magnetic lines of force. This is unlike the electric dipole where these field lines is a magnetic dipole. 5. A similar pattern of iron filings is observed around a current carrying solenoid. 5. the field lines are very similar. As mentioned in the introductory section. The arrangement of iron filings is shown in Fig. At large distances. Figure 1. 5.17(d).2. FIGURE 5.1 The magnetic field lines The pattern of iron filings permits us to plot the magnetic field lines*. properties are: The pattern suggests (i) The magnetic field lines of a magnet (or a solenoid) form continuous that the bar magnet closed loops. these poles point approximately towards the geog raphic north and south poles.3.2 The shown both for the bar-magnet and the current-carrying solenoid in arrangement of iron Fig. The magnetic field pattern mimics lines are a visual and intuitive realisation of the magnetic field. 3(a). we have explained how a current loop acts as a magnetic dipole (Section 4. Let I be the current in the solenoid. consider a circular element of thickness dx of the solenoid at a distance x from its centre. the magnitude of the field at point P due to the circular 176 element is . Recall that the magnetic dipole moment m associated with a current loop was defined to be m = NI A where N is the number of turns in the loop.13). and radius a. (iv) The magnetic field lines do not intersect. 5. for if they did. One can test this analogy by moving a small compass needle in the neig hbou rhood of a bar mag net and a current-carrying finite solenoid and noting that the deflections of the needle are similar in both cases. We can evaluate the axial field at a point P. 5. 5. We shall demonstrate finite solenoid in order to demonstrate its similarity that at large distances this axial field to that of a bar magnet. (4. at a distance r from the centre O of the solenoid. It consists of n d x turns. the direction of the magnetic field would not be unique at the point of intersection. the stronger is the magnitude of the magnetic field B. Let its length be 2l moment m of the needle. We mentioned Ampere all magnetic phenomena can be explained in terms of circulating currents.10). emerging from one face of the solenoid and entering into the other face. 4. To make this analog y more firm we calculate the axial field of a finite solenoid FIGURE 5. In Fig.2 Bar magnet as an equivalent solenoid In the previous chapter. I the current and A the area vector (Eq.4(a) consists of arrangement may be used to determine either B or the magnetic n turns per unit length. The resemblance of magnetic field lines for a bar magnet and a solenoid suggest that a bar magnet may be thought of as a large number of circulating currents in analogy with a solenoid. This gives us an idea of the magnetic field direction at various points in space. From Eq. 5.4 (a).4 Calculation of (a) The axial field of a depicted in Fig. Cutting a bar magnet in half is like cutting a solenoid. To do this. We get two smaller solenoids with weaker magnetic properties. (b) A magnetic needle resembles that of a bar magnet.Physics (iii) The larger the number of field lines crossing per unit area. The field lines remain continuous. The Let the solenoid of Fig. B is larger around region ii than in region i . In Section 4. One way is to place a small magnetic compass needle at various positions and note its orientation. in a uniform magnetic field B. One can plot the magnetic field lines in a variety of ways.2.30).6 of the previous chapter we have calculated the magnetic field on the axis of a circular current loop. x )2 + a 2 ] 3 2 » r3 and B = m0 n I a 2 2r 3 -l ò dx l (5. The field strength due to qm at a distance r from it is given by m0 qm/4pr 2 .e. dx [(r ..e. magnetic monopoles do not exist. Note that the range of x is from l to + l . in a manner analogous to that of an electric dipole (Chapter 1). 5. r >> a and r >> l . 5.2) This is also the far axial magnetic field of a bar magnet which one may obtain experimentally. i.x )2 + a 2 ]3 / 2 2 This integration can be done by trigonometric substitu tions. Thus. The torque on the needle is [see Eq.1) 2 r3 Note that the magnitude of the magnetic moment of the solenoid is. both for the axial and the equatorial case. This is done by placing a small compass needle of known magnetic moment m and moment of inertia I and allowing it to oscillate in the magnetic field.4(b). the magnetic field lines gives us an approximate idea of the magnetic field B. t =m B (5.Magnetism and Matter dB = m0n dx I a 2 2[(r . This exercise. However. The method is simple and appealing.29)].3) 177 . Thus. Some textbooks assign a magnetic charge (also called pole strength) +qmto the north pole and qm to the south pole of a bar magnet of length 2l . Then the denominator is approximated by B= m 0nIa 2 ò l -l [(r . The magnetic moment of a bar magnet is thus equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field. however. Thus. and magnetic moment qm(2l ).2. Consider the far axial field of the solenoid.x )2 + a 2 ] 3 2 The magnitude of the total field is obtained by summing over all the elements x = l to x = + l . m = n (2 l ) I (pa 2 ) area).. and we have avoided this approach for that reason.3 The dipole in a uniform magnetic field The pattern of iron filings. a bar magnet and a solenoid produce similar magnetic fields. is not necessary for our purpose. The magnetic field due to the bar magnet is then obtained. i. (4. We may at times be required to determine the magnitude of B accurately. B= = m0 n I 2l a 2 m0 2m 4p r 3 (5. This arrangement is shown in Fig. 70 = 0.mgB We have emphasised in Chapter 2 that the zero of potential energy can be fixed at one zero means fixing the zero of potential energy at q = 90”.5 ´ 10-6 6.6) shows that potential energy is minimum (= mB) at q = 0” (most stable position) and maximum (= +mB) at q = 180” (most unstable position).70 s.6) = .5) m T2 An expression for magnetic potential energy can also be obtained on lines similar to electrostatic potential energy.1 In Fig.4) 4 p2 I or (5.67s 10 From Eq. when the needle is perpendicular to the field. The square of the angular frequency is w 2 = mB/I and the time period is. the magnetic needle has magnetic moment 6. i. d 2q = . For small values of q in radians.1 B= 4 p2 I mT 2 = 4 ´ (3.mB sin q dt 2 Negative sign with mB sinq implies that restoring torque is in opposition to deflecting torque.4(b).. It performs 10 complete oscillations in 6. The magnetic potential energy Um is given by B= U m = ò t (q )dq = ò mB sin q = -mB cos q (5. 2 dt I This represents a simple harmonic motion.67) 178 = 0.14)2 ´ 7.e.5) EXAMPLE 5.5 10 kg m2. in equilibrium I d 2q » mB q dt 2 d2q mB =q or.Physics In magnitude t = mB sinq Here t is restoring torque and q is the angle between m and B.01 T . Equation (5. Example 5. Wha t is th e magnitude of the magnetic field? Solution The time period of oscillation is. I T = 2p I mB (5.7 10 Am2 and moment of inertia I = 7. 5. T = 6. we approximate sin q » q and get Therefore.7 ´ 10 ´ (0. (5. 3 179 . In a bar magnet the intensity of the magnetic field is the strongest at the two ends (poles) and weakest at the central region. m = 160 0.3 (a) What happens if a bar magnet is cut into two pieces: (i) transverse to its length. In this case.016 Nm. A repulsive force in some situation establishes that both are magnetised. Determine the current flowing through the solenoid.40 A m2 (1/2) (b) From Eq. Solution (a) From Eq.3). one of which is definitely known to be magnetised. If it is always attractive.016 = m (800 10 T) 2/800 = 0. therefore. (b) No force if the field is uniform. however. (5. Why? (c) Must every magnetic configuration have a north pole and a south pole? What about the field due to a toroid? (d) Two identical looking iron bars A and B are given.40 104/(1000 2) = 2A Example 5. to see which EXAMPLE 5. Thus. how does one ascertain which one? [Use nothing else but the bars A and B. The iron nail experiences a nonuniform field due to the bar magnet.40 A m2 0. t = m B sin q. hence sin q =1/2. the most stable position is q = 0” and the most unstable position is q = 180”.) How would one ascertain whether or not both are magnetised? If only one is magnetised. each with a north and south pole.2 A short bar magnet placed with its axis at 30” with an external field of 800 G experiences a torque of 0. (4. Work done is given by W = U m (q = 180°) . ms = 0. (5. it experiences both force and torque. (d) Try to bring different ends of the bars closer. experiences a force of attraction in addition to a torque. This is not so for a toroid or even for a straight infinite conductor.U m (q = 0°) EXAMPLE 5. From part (a). then one of them is not magnetised.Magnetism and Matter Example 5. (ii) along its length? (b) A magnetised needle in a uniform magnetic field experiences a torque but no net force.] Solution (a) In either case. There is induced magnetic moment in the nail. q = 30”.30). An iron nail near a bar magnet. (a) What is the magnetic moment of the magnet? (b) What is the work done in moving it from its most stable to most unstable position? (c) The bar magnet is replaced by a solenoid of cross-sectional area 2 10 m2 and 1000 turns. This fact may be used to determine whether A or B is the magnet.6). but of the same magnetic moment. The net force is attractive because the induced south pole (say) in the nail is closer to the north pole of magnet than induced north pole.40 = 1000 I 2 10 I = 0. one gets two magnets.40 800 10 = 0.2 = 2 m B = 2 0. True only if the source of the field has a net nonzero magnetic moment. (c) Not necessarily. (We do not know which one.064 J (c) From Eq. ms = NIA. 4 Solution From Eq. (5.4 What is the magnitude of the equatorial and axial fields due to a bar magnet of length 5. A experiences no force.2). (5. A) and lower one of its ends. Table 5. If you do not notice any change from the end to the middle of B. the axial field (BA) of a bar magnet for r >> l is: BA = 4 p r3 m0 2m (5. EXAMPLE 5. by making the following replacements: E ®B .E Magnetism m m0 m / 4p r 3 m0 2m / 4p r 3 m B .4 The electrostatic analog Comparison of Eqs. sugg ests that magnetic field at large distances due to a bar magnet of magnetic moment m can be obtained from the equation for electric field due to an electric dipole of dipole moment p. where l is the size of the magnet: BE = - m0 m 4 pr 3 (5.2. 4 pe 0 4p In particular.B m0 Example 5.7) BE = m0m 4pr3 = 10 -7 ´ 0. and then on the middle of B. pick up one. (5. If you notice that in the middle of B. the same as in Example 5.4 ( 0.8).4 ´ 10-7 T . (5.3 one of the two bars is a magnet.2) in the vector form. TABLE 5. p m 1 ® 0 m.2 ´ 10 T 180 m 0 2m From Eq.8) is just Eq.5) 3 = 10-7 ´ 0.0 cm at a distance of 50 cm from its mid-point? The magnetic moment of the bar magnet is 0. B). we can write down the equatorial field (BE) of a bar magnet at a distance r.8) Equation (5.40 A m2.2. (5. B A = 4p r 3 = 6. (say. 5. then B is magnetised.Physics EXAMPLE 5. then A is magnetised.1 THE DIPOLE ANALOGY Electrostatics Dipole moment Equatorial Field for a short dipole Axial Field for a short dipole External Field: torque External Field: Energy 1/e0 p /4pe0 r 3 2p/4pe0 r 3 p E .1 summarises the analogy between electric and magnetic dipoles.125 = 3.4 -7 0.3) and (5. for r >> l .6) with the corresponding equations for electric dipole (Chapter 1).7) Likewise. first on one of the ends of the other (say. (a) In which configuration the system is not in equilibrium? (b) In which configuration is the system in (i) stable. we studied Gauss see that for a closed surface represented by i .5 Figure 5. Hence Q3 is stable. This is consistent with the fact that no net charge is enclosed by the surface. Equilibrium is stable when mQ is parallel to BP. and (ii) unstable equilibrium? (c) Which configuration corresponds to the lowest potential energy among all the configurations shown? FIGURE 5. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q.5 shows a small magnetised needle P placed at a point O. Thus. the number of lines leaving the surface is equal to the number of lines entering it. (5.5 Solution Potential energy of the configuration arises due to the potential energy of one dipole (say. Use the result that the field due to P is given by the expression [Eqs. For instan ce for th e con figura tion Q3 for wh ich Q is along the perpendicular bisector of the dipole P. PQ4 (unstable) (c) PQ6 EXAMPLE 5. for the closed surface ii .Magnetism and Matter Example 5. (a) PQ1 and PQ2 (b) (i) PQ3.5 5. (ii) PQ5. The arrow shows the direction of its magnetic moment. the magnetic moment of Q is parallel to the magnetic field at the position 3. 181 . and unstable when it is anti-parallel to BP. However. Q) in the magnetic field due to other (P).3 MAGNETISM AND GAUSS S LAW In Chapter 1.8)]: BP = BP = m 0 mP 4p r 3 (on the normal bisector) m0 2 m P (on the axis) 4p r 3 where mP is the magnetic moment of the dipole P. there is a net outward flux.7) and (5. PQ6 (stable). since it does include a net (positive) charge. in the same figure. This is true for any closed surface. and in his work he anticipated major mathematical development of later times. There are no sources or sinks of B. Some of them may describe electrostatic field lines correctly.9) D S¢ . å Eg DS = EXAMPLE 5. en gin eerin g. Then.D S. Thus. We divide S into many small area elements and calcu late the individual flux through each. a stron omy and even land surveying. Both cases visually demonstrate that the number of magnetic field lines leaving the surface is balanced by the number of lines entering it.3(a) or Fig . 5. The ne t magnetic flux is zero for both the surfaces . His ma th ema tica l th eory of curved surfa ce laid the foundation for the later work of Riemann. fB = ’ all ’ å Df B = ’ all ’ å Bg DS = 0 (5. 5. Gauss The net magnetic flux through any closed surface is zero . The magnetic flux through S is defined as D f B = B. FIGURE 5. Along with Wilhelm Welser. Point out which ones. physics.6 Consider a small vector area element D S of a closed surface S as in Fig. Examine the Gaussian surfaces represented by i or ii in Fig 5.6 Example 5. Compare this where with the Gauss in that case is given by q e0 where q is the electric charge enclosed by the surface. KARL FRIEDRICH GAUSS (1777 1855) Ka rl Fri e dri c h Ga u ss (17 77 He was a child prodigy and was gifted in ma thematics. Point out what is wrong with them. the net flux f B is.3(b). 182 .Physics The situation is radically different for magnetic fields which are continuous and form closed loops. 5.6. the simplest magnetic element is a dipole or a current loop.7 show magnetic field lines (thick lines in the figure) wrongly. where B is the field at D S. he built the first electric telegra ph in 1833. The properties of numbers fascinated him.6 Many of the diagrams given in Fig. The difference between the Gauss electrostatics is a reflection of the fact that isolated magnetic poles (also called monopoles) are not known to exist. All magnetic phenomena can be explained in terms of an arrangement of dipoles and/or current loops. represent electric field of a long positively charged wire.Magnetism and Matter FIGURE 5. EXAMPLE 5. The field lines shown.6 183 .. Th e correct magnetic field lines a re circling the straight conductor. Magnetic field lines can never emanate from a point.e. Over any closed surface. in fact.7 Solution (a) Wrong. i. as shown in figure. pictorially as many field lines should seem to enter the surface as the number of lines leaving it. the net flux of B must always be zero. as described in Chapter 4. neither in empty space. for clarity of figure. the entire region enclosed by th e windin gs contains magnetic field.7 (a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). There is further error in the figure. This is impossible for a magnetic field. These are field lines outside and inside a bar magnet. (g) Wrong. electrostatic field lines can never form closed loops. Do the magnetic field lines also represent the lines of force on a moving charged particle at every point? (b) Magnetic field lines can be entirely confined within the core of a toroid. (e) Right. By contrast. how would the Gauss magnetism be modified? (d) Does a bar magnet exert a torque on itself due to its own field? Does one element of a current-carrying wire exert a force on another element of the same wire? (e) Magnetic field arises due to charges in motion. The magnetic force is always normal to B (remember magnetic force = qv B). Not all field lines emanate out of a north pole (or converge into a south pole). the net flux of the field is zero. Some fringing of lines is inevitable. 184 EXAMPLE 5.7 EXAMPLE 5. such a thing violates Ampere eventually to form closed loops. in fact. All the field lines seem to emanate out of the shaded plate. Can a system have magnetic moments even though its net charge is zero? Solution (a) No. Magnetic field lines between two pole pieces cannot be precisely straight at the ends. nor when the loop encloses charges. The difference between Fig. Actually. Around both the N-pole. Ampere field lines. These field lines cannot possibly represent a magnetic field.Physics (b) Wrong. It is misleading to call magnetic field lines as lines of force. Why? (c) If magnetic monopoles existed. Magnetostatic field lines can never form closed loops around empty space. [5. Magnetic field lines (like electric field lines) can never cross each other. since each loop encloses a region across which a current passes. Note. only a few field lines within the toroid have been sh own. Ma gnetic lines a re completely confined within a toroid. (d) Wrong. Example 5. Look at the upper region. The given field lines. Otherwise. and the S-pole. The net flux through a surface surrounding the shaded plate is not zero. because otherwise the direction of field at the point of intersection is ambiguous. Note carefully the direction of field lines inside. (c) Right.6 . A closed loop of static magnetic field line must enclose a region across which a current is passing. (f ) Wrong. but not within a straight solenoid. Field lines due to a solenoid at its ends and outside cannot be so completely straight and confined. show the electrostatic field lines around a positively charged upper plate and a negatively charged lower plate. Nothing wrong here in field lines forming closed loops.7(e) and (f )] should be carefully grasped. But the flux of field B through any closed surface must always be zero. Originally the magnetic field was thought of as arising from a giant bar magnet placed approximately along the axis of rotation of the earth and deep in the interior.3” with respect to the later. the flux through the cross-section at each end would be non-zero. the mean of the magnetic moments due to various current loops may not be zero. The magnetic field is now thought to arise due to electrical currents produced by convective motion of metallic fluids (consisting mostly of molten iron and nickel) in the outer core of the earth.22” E in the Antarctica. Likewise.Magnetism and Matter (b) If field lines were entirely confined between two ends of a straight solenoid. a place somewhere in north Canada. the pole near the geographic south pole is called http://www.] (d) No. Yet. However. EXAMPLE 5. But there is a force (or torque) on an element of the same wire.74” S. The axis of the dipole does not coincide with the axis of rotation of the earth but is presently titled by approximately 11. S If monopoles existed.) (e) Yes. [Ana logous to Ga uss ò S Bgds = m0qm where qm is th e (monopole) magnetic charge enclosed by S . The magnetic south pole is at 79. The pole near the geographic north pole of the earth is called the north magnetic pole . There is no force or torque on an element due to the field produced by that element itself.noaa. The strength of the earth its value being of the order of 10 T.7 5. this simplistic picture is certainly not correct. the magnetic poles are located where the magnetic field lines due to the dipole enter or leave the earth. (For the special case of a straight wire.gov/seg/geomag/ Geomagnetic field frequently asked questions 185 . this difficulty is absent because it has no (c) Gauss B through any closed surface is always zero Ñ ò B gds = 0 . The average of the charge in the system may be zero.4 THE EARTH S MAGNETISM Earlier we have referred to the magnetic field of the earth.74” N and a longitude of 71. This is known as the dynamo effect.ngdc. What causes the earth to have a magnetic field is not clear. The magnetic field lines of the earth resemble that of a (hypothetical) magnetic dipole located at the centre of the earth. 108. We will come across such examples in connection with paramagnetic material where atoms have net dipole moment through their net charge is zero. In this way of looking at it. this force is zero. The location of the north magnetic pole is at a latitude of 79.8” W. For a toroid. the right hand side would be equal to the monopole (magn etic cha rge) qm enclosed by S. the line of longitude towards the north pole being the direction of true north. Thus. we take the radius of 106)3 Am 2 EXAMPLE 5. the field lines go into the earth at the north magnetic pole (Nm ) and come out from the south magnetic pole (Sm ). then. which is free to swing horizontally.4.4 ´ 106 )3 =4 m0 / 4 p 102 = 1. is the angle between the true geographic north and free to move in horizontal plane. If one looks at the magnetic field lines of the earth (Fig.4 10 23 2 (m0/4p = 10 ) This is close to the value 8 A m quoted in geomagnetic texts. Hence. (5. m = 4 ´ 10-5 ´ (6.9 A magnetic needle This.4 G = 4 the earth 6. FIGURE 5. This plane wou ld intersect the su rface of the earth in a longitude like circle. one can define magnetic meridian of a place as the vertical plane which passes through the imaginary line joining the magnetic north and the south poles. in reality. 186 near the equator.8 We are given that BE ~ 0. one sees that unlike in the case of a bar magnet. north-south The declination is greater at higher latitudes and smaller direction.Physics the south magnetic pole . Since the line joining the magnetic poles is titled with respect to the geographic axis of the earth. In a similar way. 5. The vertical plane containing the longitude circle and the axis of rotation of the earth is called the geographic meridian.4 106 m.1 Magnetic declination and dip Consider a point on the earth the longitude circle determines the geographic north-south direction.8 The earth 0. Estimate the earth Solution From Eq. BE = m0m 4pr3 10 T. This angle is called the points toward the magnetic magnetic declination or simply declination (Fig.8 The earth as a giant magnetic dipole. For r. 5. The convention arose because the magnetic north was the direction to which the north pole of a magnetic needle pointed. Example 5. There is some confusion in the nomenclature of the poles. 5. FIGURE 5.05 10 22 (6. the north magnetic pole behaves like the south pole of a bar magnet inside the earth and vice versa. it being .7). the equatorial magnetic field is.8). would then lie in the magnetic meridian and the north pole of the needle would point towards the magnetic north pole. A magnetic needle. the north shown by a compass needle. the magnetic meridian at a point makes angle with the geographic meridian.9). the north pole of a magnet was so named as it was the no rth se eking pole .4 G. The declination in India is small. Representing the verticle component by ZE. viz. Also shown are the declination. dip is the angle that the total magnetic field BE of the earth makes with the surface of the earth. Thus.10(a)] HE = BE cosI which gives.11 The earth magnetic field. 5. at both these places a magnetic needle shows the true north quite accurately. Figure 5. tan I = ZE HE [5. This is known as the angle of dip (also known as inclination). we have ZE = BE sinI [5. FIGURE 5. its horizontal and vertical components. FIGURE 5.10(c)] 187 . The angle between BE and the horizontal component HE is the angle of dip.10 The circle is a section through the earth containing the magnetic meridian. If a magnetic needle is perfectly balanced about a horizontal axis so that it can swing in a plane of the magnetic meridian. The plane is a section through the earth.10). To describe the magnetic field of the earth at a point on its surface. Likewise in most of the southern hemisphere. the angle of dip or the inclination I and the horizontal component of the earth HE. There is one more quantity of interest.Magnetism and Matter 0”41¢ E at Delhi and 0”58¢ W at Mumbai. we need to specify three quantities. I. the declination D. HE and ZE. the needle would make an angle with the horizontal (Fig. The total magnetic field at P can be resolved into a horizontal component H E and a vertical component ZE. I. Thus. The angle that BE makes with HE is the angle of dip.11 shows the magnetic meridian plane at a point P on the surface of the earth. In most of the northern hemisphere.. BE. the north pole of the dip needle tilts downwards. These are known as the element of the earth . D and the inclination or angle of dip.10(b)] [5. the south pole of the dip needle tilts downwards. rendering it useless as a direction finder. At the magnetic poles such a needle will point straight down. The needle of the compass then shows the angle which the magnetic field makes with the vertical. What is the magnetic field of the earth at this location? Solution It is given that HE = 0. it points along the direction of the horizontal component of the earth magnetic field at the location. the compass needle would stay along the magnetic meridian of the place.11. we have cos 600 = HE BE EXAMPLE 5. So what happens if we take our compass to the magnetic pole? At the poles.9 In the ma gn etic meridia n of a certa in place. When the compass is held level. Thus.9 BE = HE cos 600 188 = 0. If the needle is only capable of moving in a horizontal plane. Example 5 . 5. the horizontal component of the earth dip angle is 60”. What one needs in such a case is a dip needle which is a compass pivoted to move in a vertical plane containing the magnetic field of the earth. the magnetic field lines are converging or diverging vertically so that the horizontal component is negligible. In some places on the earth there are deposits of magnetic minerals which cause the compass needle to deviate from the magnetic meridian. Knowing the magnetic declination at a place allows us to correct the compass to determine the direction of true north.26 = 0.Physics WHAT HAPPENS TO MY COMPASS NEEDLES AT THE POLES? A compass needle consists of a magnetic needle which floats on a pivotal point. it can point along any direction. From Fig.52 G (1/ 2) .26 G. The variation of earth term variations taking place over centuries and long term variations taking place over a period of a million years. In the present section. One would like to classify the magnetic properties of these substances.5”.5 MAGNETISATION AND MAGNETIC INTENSITY The earth abounds with a bewildering variety of elements and compounds. and basalt is emitted during volcanic activity. suggesting that the magnetic poles inside the earth change position with time. which has the fastest rotation rate among planets. while Jupiter. we have been synthesising new alloys. has no magnetic field. We have seen that a circulating electron in an atom has a magnetic moment. we define and explain certain terms which will help us to carry out this exercise. over which records are available. Geological studies of basalt containing such pieces of magnetised region have provided evidence for the change of direction of earth 5. The little iron magnets inside it align themselves parallel to the magnetic field at that place as the basalt cools and solidifies. Charged particles emitted by the sun flow towards the earth and beyond. has a fairly strong magnetic field. The variation of the earth study. Consider a long solenoid of n turns per unit length and carrying a current I. In a bulk material.11) M is a vector with dimensions L A and is measured in a units of A m . and a weaker magnetic field. compounds and even elements. On the scale of a million years. Their motion is affected by the earth affect the pattern of the earth quite different from that in other regions of the earth. In addition. the magnetic declination at London has been found to change by 3. We define magnetisation M of a sample to be equal to its net magnetic moment per unit volume: M= mnet V (5. In a span of 240 years from 1580 to 1820 AD. in a stream called the solar wind. the earth contains iron. the precise mode of these circulating currents and the energy needed to sustain them are not very well understood.Magnetism and Matter EARTH S MAGNETIC FIELD It must not be assumed that there is a giant bar magnet deep inside the earth which is causing the earth it is highly unlikely that a large solid block of iron stretches from the magnetic north pole to the magnetic south pole. these moments add up vectorially and they can give a net mag netic moment which is non. However. These are several open questions which form an important area of continuing research.zero. The earth nickel are responsible for earth which has no molten core. Venus has a slower rate of rotation. The magnetic field in the interior of the solenoid was shown to be given by 189 . .17) we obtain.18) m = m0mr = m0 (1+c). (5. is appropriately called the magnetic susceptibility.15) where H has the same dimensions as M and is measured in units of A m . It turns out that this additional field Bm is proportional to the magnetisation M of the material and is expressed as Bm = m0 M (5. Thus.16) We repeat our defining procedure. The magnetic per meability of the substance is m and it has the same dimensions and units as m0 . This influence is mathematically expressed as M = cH (5. namely M. which are called paramagnetic. the field inside the solenoid will be greater than B0 . which are termed diamagnetic.17) where c . due to external factors such as the current in the solenoid. The latter quantity can be influenced by external factors. the other two may be easily determined. is a dimensionless quantity called the relative magnetic per meability of the substance.12) If the interior of the solenoid is filled with a material with non-zero magnetisation. (5. It is small and positive for materials.16) and (5. In the latter case M and H are opposite in direction. called the magnetic intensity. The other is due to the specific nature of the magnetic material. which is defined by H= B m0 M (5. We have partitioned the contribution to the total magnetic field inside the sample into two parts: one . It is a measure of how a magnetic material responds to an external field. a dimensionless quantity. This is represented by H. mr and m are interrelated and only one of them is independent.2 lists c for some elements. B = m 0 (1 + c )H = m0 mr H = mH (5. It is small and negative for materials. From Eqs.Physics B0 = m0 nI (5.14) where m0 is the same constant (permittivity of vacuum) that appears in Biot-Savart It is convenient to introduce another vector field H. the total magnetic field B is written as B = m0 (H + M) (5. 190 The three quantities c. Table 5.13) where Bm is the field contributed by the material core. Given one.19) where mr= 1 + c. The net B field in the interior of the solenoid may be expressed as B = B0 + Bm (5. It is the analog of the dielectric constant in electrostatics. 2 MAGNETIC Diamagnetic substance Bismuth Copper Diamond Gold Lead Mercury Nitrogen (STP) Silver Silicon SUSCEPTIBILITY OF SOME ELEMENTS AT 300 K c 2.7 2.Magnetism and Matter TABLE 5. a material is diamagnetic if c is negative. The windings of the solenoid are insulated from the core and carry a current of 2A. calculate (a) H. para.9 6. Thus B = mr n0 (I + IM).10 5.3 g ives one a better feeling for these materials. (b) The magnetic field B is given by B = mr m0 H = 400 4p 10 (N/A2) 2 103 (A/m) = 1. we describe these materials in some detail.2 2. Using I = 2A.8 10 10 10 10 10 10 10 10 10 c Paramagnetic substance Aluminium Calcium Chromium Lithium Magnesium Niobium Oxygen (STP) Platinum Tungsten Exa mple 5. B = 1 T. If the number of turns is 1000 per metre.3 1. Here e is a small positive number introduced to quantify paramagnetic materials.1 0 A solen oid has a core of a material with relative permeability 400.1 1.0 T (c) Magnetisation is given by M = (B m0 H )/ m0 = (mr m0 H m0 H )/m0 = (mr H = 399 H @ 8 105 A/m (d) The magnetising current IM is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. EXAMPLE 5.1 2.if c is positive and small. A g lance at Table 5. Next. 191 . (c) B and (d) the magnetising current Im. and ferro.9 2.0 = 2 103 A/m. In terms of the susceptibility c . Solution (a) The field H is dependent of the material of the core. paramagnetic or ferromagnetic.if c is large and positive.6 2. and is H = nI = 1000 2. (b) M. we get IM = 794 A.6 MAGNETIC PROPERTIES OF MATERIALS The discussion in the previous section helps us to classify materials as diamagnetic. However. In other words.6.3 Diamagnetic £c<0 0 £ mr < 1 Paramagnetic 0<c< e 1< mr < 1+ e Ferromagnetic c >> 1 mr >> 1 m >> m0 m < m0 m > m0 5.1 Diamagnetism Diamagnetic substances are those which have tendency to move from stronger to the weaker part of the external magnetic field. 192 . being one part in 105 . unlike the way a magnet attracts metals like iron. Some diamagnetic materials are bismu th. Diamagnetism is present in all the substances. When magnetic field is applied.12 Behaviour of magnetic field lines near a (a) diamagnetic. ferromagnetism. A superconductor repels a magnet and (by Newton diamagnetism in superconductors is called the Meissner effect. In most cases. as is evident from Table 5. This happens due to induced current in accordance with Lenz you will study in Chapter 6.Physics TABLE 5. cooled to very low temperatures which exhibits both perfect conductivity and perfect diamagnetism.6. Thus. silicon. Figure 5. When placed in a non-uniform magnetic field. Electrons in an atom orbiting around nucleus possess orbital angular momentum. the effect is so weak in most cases that it gets shifted by other effects like paramagnetism. nitrogen (at STP). for running magnetically levitated superfast trains. they get weakly attracted to a magnet. the bar will tend to move from high to low field. i. They have tendency to move from a region of weak magnetic field to strong magnetic field. for example. copper. These are metals.2.e. the substance develops a net magnetic moment in direction opposite to that of the applied field and hence repulsion. after the name of its discoverer. The simplest explanation for diamagnetism is as follows.. lead.12(a) shows a bar of diamagnetic material placed in an external magnetic field. These orbiting electrons are equivalent to current-carrying loop and thus possess orbital magnetic moment. water and sodium chloride. those electrons having orbital magnetic moment in the same direction slow down and those in the opposite direction speed up. The field lines are repelled or expelled and the field inside the material is redu ced. FIGURE 5. etc. it wou ld repel a diamagnetic substance.2 Paramagnetism Paramagnetic substances are those which get weakly magnetised when placed in an external magnetic field. Diamagnetic substances are the ones in which resultant magnetic moment in an atom is zero. 5. Superconducting mag nets can be gainfully exploited in variety of situations. The most exotic diamagnetic materials are superconductors . (b) paramagnetic substance. Here the field lines are completely expelled! c = mr = 0. this reduction is slight. and at low temperatures.. oxygen (at STP) and copper chloride. In the presence of an external field B0 . This is shown in Fig. no net magnetisation is seen. using Eqs. Figure 5.org/EducationResources/CommunityCollege/ MagParticle/Physics/MagneticMatls.13 (a) Randomly oriented domains. 193 .13(a). Beyond this. being one part in 105 . for a paramagnetic material both c and mr depend not only on the material.12(b) shows a bar of paramagnetic material placed in an external field. However. i. sodium. the bar will tend to move from weak field to strong. In the first instant. at which point all the dipoles are perfectly aligned with the field.e. this enhancement is slight. (b) Aligned domains.2.ndt-ed. On account of the ceaseless random thermal motion of the atoms. c =C m0 5.17) M =C http://www. They have strong tendency to move from a region of weak magnetic field to strong magnetic field. When placed in a non-uniform magnetic field.: [5. The field lines gets concentrated inside the material. The constant C is called Curie . The explanation of this cooperative effect requ ires quantum mechanics and is beyond the scope of this textbook. Some paramag netic materials are aluminiu m. (5. the individual atomic dipole moment can be made to align and point in the same direction as B0 . as is evident from Table 5. Experimentally. When we apply an external magnetic field B0 . 5. Each domain has a net magnetisation. bu t also (in a simple fashion) on the sample temperatu re.Magnetism and Matter The individual atoms (or ions or molecules) of a paramagnetic material possess a permanent magnetic dipole moment of their own. the magnetisation varies randomly from domain to domain and there is no bulk magnetisation. one finds that the magnetisation of a paramagnetic material is inversely proportional to the absolute temperature T . Curie longer valid. As the field is increased or the temperature is lowered. they interact with one another in such a way that they spontaneously align themselves in a common direction over a macroscopic volume called domain. One may observe this under a microscope after sprinkling a liqu id su spension of powdered FIGURE 5.htm Magnetic materials.3 Ferromagnetism Ferromagnetic substances are those which gets strongly magnetised when placed in an external magnetic field. the domains orient themselves in the direction of B0 and simultaneously the domain oriented in the direction of B0 grow in size. Thus. domain. after its discoverer Pieree Curie (18591906). the magnetisation increases until it reaches the saturation value Ms. and the field inside is enhanced. In most cases.20(a)] [5. This existence of domains and their motion in B0 are not speculations.6. B0 T or equivalently. which is strong enough. Typical domain size is 1mm and the domain contains about 1011 atoms. The individual atoms (or ions or molecules) in a ferromagnetic material possess a dipole moment as in a paramagnetic material.12) and (5.20(b)] T This is known as Curie . they get strongly attracted to a magnet. calciu m. etc. This motion of suspension can be observed. nickel. The temperature of transition from ferromagnetic to paramag netism is called the Curie tempe ratur e Tc . cobalt. an alloy of iron.9 g/cm3. Assume that each iron atom has a dipole moment of 9. etc.html Hysterisis in magnetic materials: c= C T . Figure 5. http://hyperphysics. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain.edu/hbase/solids/hyst. The susceptibility above the Cu rie temperatu re. gadolinium.Physics ferromagnetic substance of samples. At high enough temperature. In non-uniform magnetic field. Soft iron is one such material. The domain structure disintegrates with temperature. In some ferromagnetic materials the magnetisation persists. The molecular mass of iron is 55 g/mole and its density is 7. The naturally occurring lodestone is another. in a ferromagnetic material the field lines are highly concentrated. On the other hand. in the paramagnetic phase is described by. aluminium.gsu. . This disappearance of magnetisation with temperature is gradual.4 CURIE Material Cobalt Iron Fe2 O3 Nickel Gadolinium EXAMPLE 5. nickel. The relative magnetic permeability is >1000! The ferromagnetic property depends on temperature.11 A domain in ferromagnetic iron is in the form of a cube of side length 1mm. cobalt and copper.12(b) shows the situation when the domains have aligned and amalgamated to form a single Thu s. such materials are called soft ferromagnetic materials ..e. i. Alnico. is one such material.Tc ( T > Tc ) (5. We may wonder as to what happens when the external field is removed.4 lists the Cu rie temperature of certain ferromagnets. there is a class of ferromagnetic materials in which the magnetisation disappears on removal of the external field.27 10 A m2. It is a phase transition reminding us of the melting of a solid crystal. the sample tends to move towards the region of high field.phy-astr. which are ferromagnetic: iron. Such materials form permanent magnets to be used among other things as a compass needle.21) TABLE 5. Appropriately enough. Such materials are called hard magnetic materials or hard ferromagnets . a ferromagnet becomes a paramagnet. Table 5. There are a number of elements.11 TEMPERATURE TC OF SOME FERROMAGNETIC MATERIALS Tc (K) 1394 1043 893 631 317 194 Example 5. 023 ´ 1023 55 10 = 8.Magnetism and Matter Solution The volume of the cubic domain is V = (10 m)3 = 10 m3 = 10 cm3 Its mass is volume density = 7. mmax = (8.0 = 8. Note that the curve Oa does not retrace itself as H is reduced. Certain domains are flipped until the net field inside stands nullified.11 5.9 10 g It is given that Avagadro number (6. At H = 0. The curve cd depicts this. For a given value of H. In Fig.9 ´ 10. The word hysterisis means lagging behind (and not EXAMPLE 5.0 10 A m2 The consequent magnetisation is Mmax = mmax/Domain volume = 8. The domains are not completely randomised even though the external driving field has been removed.9 g cm 10 cm3= 7.65 10 atoms Th e maximum possible dipole moment mma x is achieved for the (unrealistic) case when all the atomic moments are perfectly aligned. Permanent .14 The magnetic the reversed current is increased in magnitude. This phenomenon is called hysterisis .7 PERMANENT MAGNETS AND ELECTROMAGNETS 195 Substances which at room temperature retain their ferromagnetic property for a long period of time are called per manent magne ts . B is not unique but depends on previous history of the sample. This behaviour represents the alignment and merger of domains until no further enhancement is possible. It is often not linear and it depends on the magnetic history of the sample. saturated magnetic field Bs ~ 1. where the su bscript R denotes retentivity . the current in the solenoid is reversed and slowly increased. The value of B at H = 0 is called retentivity or remanence . Let the material be unmagnetised initially. Hence. Next. the number of atoms in the domain is N = 7.5 T. This is represented by the curve bc.27 10 ) = 8. The cycle repeats itself. As FIGURE 5. Next.2 T.14. 5. The ferromagnetic materials. Thus. Figure 5.14 Hc ~ . The value of H at c is called coercivity. BR ~ 1.14 depicts the behaviour of the material as we take it through one cycle of magnetisation. In Fig. we decrease H and reduce it to zero.0 10 Am2/10 m3 105 Am The relation between B and H in ferromagnetic materials is complex. the current is reduced (curve de) and reversed (curve ea). we once hysteresis loop is the B-H curve for again obtain saturation.12 ´ 6.023 1023) of iron atoms have a mass of 55 g.65 1010) (9. 5. Next. It is pointless to increase the current (and hence the magnetic intensity H ) beyond this. This is represented by the curve ab. B ¹ 0. The magnetic field B in the material rises and saturates as depicted in the curve Oa. We place it in a solenoid and increase the current through the solenoid. The energy dissipated and the heating will consequently be small. are alnico. alway s in the same sense to make a permanent magnet. The magnetic field of the solenoid magnetises the rod.15 A blacksmith The hysteresis curve (Fig. The illustration is from a 400 y ear old book to emphasise that the making of permanent magnets is an old art. When we switch off the solenoid current. Electromagnets are used in electric bells.14) allows us to select suitable forging a permanent magnet by materials for permanent magnets. the magnetism is effectively switched off since the soft iron core has a low retentivity. Steel is work published in 1600 and one-favoured choice. One can hold an iron rod in the north-south direction and hammer it repeatedly. the material goes through an ac cycle of magnetisation for a long period.res.iigm. loudspeakers and telephone diaphragms. direction with a hammer. cobalt steel and ticonal. Other suitable materials for permanent magnets Elizabeth of England. the material should have a high permeability. .in India FIGURE 5. 5. One can also hold a steel rod and stroke it with one end of a bar magnet a large number of times. The method is illustrated in Fig. we increase the mag netism of the solenoid by a thousand fold. soft iron but this is outweighed by the much smaller coercivity the court physician to Queen of soft iron. FIGURE 5. The sketch is recreated from an temperatu re flu ctu ations or minor mechanical damage. The arrangement is shown in Fig. Soft iron is a suitable material for electromagnets. It has a slightly smaller retentivity than authored by William Gilbert. The hysteresis curve of such materials must be narrow.15. Giant electromagnets are used in cranes to lift machinery. 196 In certain applications. An efficient way to make a permanent magnet is to place a ferromag netic rod in a solenoid and pass a current.Physics magnets can be made in a variety of ways. and bulk quantities of iron and steel. 5. This is the case in transformer cores and telephone diaphragms. The material must have a high resistivity to lower eddy current losses. Core of electromagnets are made of ferromagnetic materials which have high permeability and low retentivity. a Further. On placing a soft iron rod in a solenoid and passing a current. The material should have striking a red-hot rod of iron high retentivity so that the magnet is strong and high coercivity kept in the north-south so that the magnetisation is not erased by stray magnetic fields.16. 5. http://iigs. illustration in De Magnete. You will study about eddy currents in Chapter 6.16 A soft iron core in solenoid acts as an electromagnet. The IIG monitors (via its nation-wide observatories).B . Its services are used by the Oil and Natural Gas Corporation Ltd. all ar ea el ements D S å Bg D S = 0 The earth dipole located at the centre of the earth. (a) the force on it is zero. In India over a dozen observatories exist. and under the ocean and in space. These observatories work under the aegis of the Indian Institute of Geomagnetism (IIG). Now India has a permanent station called Gangotri. (b) the torque on it is m B . in Colaba. where r >>l. the magnetic field B due to this bar is.Magnetism and Matter MAPPING INDIA S MAGNETIC FIELD Because of its practical application in prospecting. the geomagnetic fields and fluctuations on land. and cutting a bar magnet in two leads to two smaller magnets. Similarly. 2. m0 m 2p r3 m 0m 4 p r3 (along axis) (along equator) Gauss any closed surface is zero fB = 5. (ONGC). communication. The IIG grew out of the Colaba and Alibag observatories and was formally established in 1971. Magnetic poles cannot be isolated. It is a part of the world-wide network which ceaselessly updates the geomagnetic data. the magnetic field of the earth is mapped by most nations with an accuracy comparable to geographical mapping. like magnetic poles repel and unlike ones attract. This dipole is aligned making a small angle with the rotation axis of the earth. the National Institu te of Oceanography (NIO) and the Indian Space Research Organisation (ISRO). extending from Trivandrum (now Thrivuvananthapuram) in the south to Gulmarg in the north. It has been known since ancient times that magnetic materials tend to point in the north-south direction. the pole near the geographic south pole is called the south magnetic pole. where we choose the zero of energy at (c) its potential energy is the orientation when m is perpendicular to B . 3. Consider a bar magnet of size l and magnetic moment m. The science of magnetism is old. The pole near the geographic north pole of the earth is called the north magnetic pole. . at a distance r from its mid-point. When a bar magnet of dipole moment m is placed in a uniform magnetic field B . B= = 4. and navigation. The magnitude of the field on the earth » 4 10 T. 197 . Mumbai. SUMMARY 1. and ferromagnetic. The magnetic field B in the material is. B = mH 198 . B = m0 ( H + M) For a linear material M = c H. and the magnetic permeability m are related as follows: m = m0 mr m0 mr = 1+ c 9. paramagnetic. Ferromagnetic materials have large c and are characterised by non-linear relation between B and H. retain their ferromagnetic property for a long period of time are called permanent magnets. Physical quantity Permeability of free space Magnetic field. H= B0 8.Physics 6. Substances. The magnetic intensity is defined as. Three quantities are needed to specify the magnetic field of the earth on its surface and the magnetic dip. The three quantities. Magnetic induction. which at room temperature. For diamagnetic materials c is negative and small and for paramagnetic materials it is positive and small. The magnetisation M of the material is its dipole moment per unit volume. Magnetic flux density Magnetic moment Magnetic flux Magnetisation Magnetic intensity Magnetic field strength Magnetic susceptibility Relative magnetic permeability Magnetic permeability Symbol Nature Scalar Vector Dimensions [MLT A ] [MT A ] Units TmA T (tesla) Remarks m0 B m0/4p = 10 104 G (gauss) = 1 T m Vector Scalar Vector Vector [L A] A m2 W (weber) Am Am W = T m2 Magnetic moment Volume fB M H [ML2T A ] [L [L A] A] B = m0 (H + M) c mr m Scalar Scalar Scalar [MLT A ] TmA NA M = cH B = m0 m. the relative magnetic permeability mr . Consider a material placed in an ex ternal magnet ic field B 0. H m = m0 m. c. They show the property of hysteresis. So that B = m H and c is called the magnetic susceptibility of the material. 10. Magnetic materials are broadly classified as: diamagnetic. 7. These are known as the elements of the earth magnetic field. for example. All other charges are integral multiples of this smallest unit charge. If you slice a magnet in half. isolated positive and negative charges exist. anti-ferromagnetic. 9. There exists a smallest unit of charge. with properties which are exotic and mysterious. the electrostatic lines of force begin on a positive charge and terminate on the negative charge (or fade out at infinity). The earth earth are responsible for the earth sustains this current.or ferromagnetic. The BCS theory was proposed in1957 and was eventually recognised by a Nobel Prize in physics in 1970. This is a metal at very low temperatures. spin glass. There exists a perfect diamagnet. scientific understanding is not a necessary condition for engineering applications. charge is quantised. The stress. Thus. Magnetic monopoles do not exist. mr = 0. yields radical ly different behavio ur: diamagnet ic versus paramagnet ic. 3. The external magnetic field is totally expelled. A similar interpretation can be gi ven to the B -H magnetic hysteresis curve. 6. science and engineering go hand-in-hand. Cooper. However. In contrast. We have cl assified mat eri al s as di amagneti c. and why the earth million years or so. m = 0. here H and B (o r M) are no t l inearly rel at ed. 8. etc. you get two smaller magnets.6 10 C. Strai n may no t be pro po rt io nal to stress. A satisfactory understanding of magnetic phenomenon in terms of moving charges/currents was arrived at aft er 1800 AD . A miniscule difference in the value of c. 7. For diamagnetic materials c = 10 whereas c = +10 fo r paramagnet ic materials. We do not know why magnetic monopoles do not exist or why electric charge is quantised. A quant um-mechanical theory by Bardeen. However. paramagneti c. a superconductor. this material is also a perfect conductor. The pheno menon o f magnetic hysteresis is remi ni scent o f simil ar behavi our concerni ng the elast i c pro perti es of materi als.st rai n curve exhibi ts hysteresis and area enclosed by it represents the energy dissipated per unit volume. 4. In this case c = 1. there exist additional types of magnetic material such as ferrimagnetic. 5. the magnetic susceptibility. On the other hand. Ideally. there exists no classical theory which ties these two properti es together. 199 . Interestingly. namely.Magnetism and Matter POINTS TO PONDER 1. But i t is weak and hard to detect if the substance is para. Di amagneti sm i s universal. 2. the electronic charge with value |e| = 1. and ferromagnetic. A consequence of the fact that magnetic monopoles do not exist is that the magnetic field lines are continuous and form closed loops. and Schrieffer (BCS theory) explains these effects. But t echno logical exploitation of the directional properties of magnets predates this scientific understanding by two thousand years. we do not know. one leading and assisting the other in tandem. It is present i n al l materi al s. In other words. on what time scale does it change appreciably? (b) The earth regard this as a source of the earth (c) The charged currents in the outer conducting regions of the earth What might be the these currents? (d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years.15 T.2 5.] A short bar magnet placed with its axis at 30” with a uniform external magnetic field of 0. Check the order of magnitude of this number in some way.2 is meant mainly to arouse your curiosity. and (b) unstable equilibrium? What is the potential energy of the magnet in each case? 5.1 Answer the following questions regarding earth (a) A vector needs three quantities for its specification.000 km). Would you expect a greater or smaller dip angle in Britain? (c) If you ma de a map of magn etic field lines at Melbourne in Australia. Name the three independent quantities conventionally used to specify the earth (b) The angle of dip at a location in southern India is about 18”. [Note: Exercise 5. Ca n such a wea k field be of any significant consequence? Explain. if located right on the geomagnetic north or south pole? (e) The earth due to a dipole of magnetic moment 8 1022 J T located at its centre. If the bar is free to rotate in the plane of the field.4 200 . which orientation would correspond to its (a) stable.32 JT is placed in a uniform magnetic field of 0. (f ) Geologists claim that besides the main magnetic N-S poles. For details. there are several local poles on the earth directions. How is such a thing possible at all? Answer the following questions: (a) The earth Does it also change with time? If so. How can geologists know about the earth (e) The earth large distances (greater than about 30. What is the magnitude of magnetic moment of the magnet? A short bar magnet of magnetic moment m = 0.25 T experiences a torque of magnitude equal to 4.Physics EXERCISES 5.5 10 J. What agencies may be responsible for this distortion? (f ) Interstellar space has an extremely weak magnetic field of the order of 10 T. you should consult a good text on geomagnetism.3 5. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. would the lines seem to go into the ground or come out of the ground? (d) In which direction would a compass free to move in the vertical plane point to. (b) the equatorial lines (normal bisector) of the magnet.48 J T .13 5. When the coil is turned slightly an d released. A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. where will the new null points be located? 5.22 T.6 5. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null centre of the magnet? (At null points.0 A. it oscillates a bout its stable equilibrium with a frequency of 2. (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)? A closely wound solenoid of 2000 turns and area of cross-section 1. (a) What is the magnetic moment associated with the solenoid? (b) Wh at is th e force and torq ue on th e solenoid if a uniform horizontal magnetic field of 7.11 5.25 T is applied. Explain the sense in which the solenoid acts like a bar magnet. A short bar magnet has a magnetic moment of 0. The earth zero.5 J T lies aligned with the direction of a uniform magnetic field of 0. What is its associated magnetic moment? If the solenoid in Exercise 5.0 10 T.16 G.35 G.14 201 .0 s .7 5. The horizontal component of the earth at the place is known to be 0. What is the moment of inertia of the coil about its axis of rotation? A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22” with the horizontal.75 A rests with its plane normal to an external field of magnitude 5. what is the magnitude of torque on the solenoid when its axis makes an angle of 30 with the direction of applied field? A bar magnet of magnetic moment 1. is suspended through its centre allowing it to turn in a horizontal plane. a compass points 12” west of the geographic north.0 A. The coil is free to turn about an axis in its plane perpendicular to the field direction. The north tip of the magnetic needle of a dip circle pla ced in th e pla ne of magnetic meridian points 60” above the horizontal. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. Determine the magnitude of the earth At a certain location in Africa.13 is turned around by 180”.8 5.5 A closely wound solenoid of 800 turns and area of cross section 2.Magnetism and Matter 5.10 5.6 10 m2. carrying a current of 4.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0. field due to a magnet is equal and opposite to the horizontal component of earth If the bar magnet in exercise 5.5 10 m2 carries a current of 3.9 5. The horizontal component of the earth to be 0. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction.5 10 T is set up at an angle of 30” with the axis of the solenoid? A circular coil of 16 turns and radius 10 cm carrying a current of 0. Specify the direction and magnitude of the earth at the location.12 5. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis. ) A telephone cable at a place has four long straight horizontal wires carrying a current of 1.19 . Suggest a method. Locate the line of neutral points (ignore the thickness of the cable)? (At neutral points. If the material is to go through repeated cycles of magnetisation.16 Answer the following questions: (a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled? (b) Wh y is dia ma gn etism.17 5. (d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player. The 5. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point. The earth is zero. or for building modern computer? (e) A certain region of space is to be shielded from magnetic fields.25 10 J T is placed with its axis perpendicular to the earth distance from the centre of the magnet.Physics 5. the resultant field is inclined at 45” with earth Magnitude of the earth Ignore the len gth of the magnet in compa rison to the distances involved. is it more for lower or higher fields? (e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. will the field in the core be (slightly) greater or (slightly) less than when the core is empty? (d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not.18 202 5. A long straight horizontal cable carries a current of 2. which piece will dissipate greater heat energy? (c) a device for storing memory? statement.) Why? (f ) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet? Answer the following questions: (a) Expla in q ua litatively on th e basis of doma in picture th e irreversibility in the magnetisation curve of a ferromagnet. (b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth magnetic field. in contrast.5 A in the direction 10” south of west to 10” north of east. ADDITIONAL EXERCISES 5.15 A short bar magnet of magnetic movement 5. The magnetic meridian of the place happens to be 10” west of the geographic meridian. a lmost in dependen t of temperature? (c) If a toroid uses bismuth for its core.0 A in the same direction east to west. of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by: m s = e/m) S . (a) Determine the horizontal component of the earth at the location.8 K? (Assume Curie A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the total dipole moment of the sample for a magnetic field of 0. Take the magnetic declination at the places to be zero. The degree of magnetic saturation achieved is equal to 15%.23 5. The coil is in a vertical plane making an angle of 45” with the magnetic meridian. Predict the direction of the needle.64 T.21 5. 5.35 A.04 G normal to the initial direction. and the coil is rotated about its vertical axis by an angle of 90” in the anticlockwise sense looking from above. respectively.0 1024 atomic dipoles each of dipole moment 1. (b) The current in the coil is reversed. and cooled to a temperature of 4. and one of the fields has a magnitude of 1.Magnetism and Matter earth 35”.24 5.11 10 31 kg). What is the magnetic field B in the core for a magnetising current of 1. The sample is placed under a homogeneous magnetic field of 0. [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth from the electron gun to the screen in a TV set. the needle points west to east.] A sample of paramagnetic salt contains 2. If the dipole comes to stable equilibrium at an angle of 15” with this field.0 cm below the cable? A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm.22 5. ml = e/2m)l Which of these relations is in accordance with the result expected classically ? Outline the derivation of the classical result.2 K.2 10 T. what is the magnitude of the other field? A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0. The angle between the field directions is 60”. The magnetic declination is nearly zero.98 T and a temperature of 2. When the current in the coil is 0.25 203 .5 10 J T .20 5. A magnetic dipole is under the influence of two magnetic fields. Estimate the up or down deflection of the beam over a distance of 30 cm (me = 9.2 A? The magnetic moment vectors m s and m l associated with the intrinsic spin an gular momentum S an d orbital a ngular momentum l . What are the resultant magnetic fields at points 4. In this chapter.Physics Chapter Six ELECTROMAGNETIC INDUCTION 6. he was asked. This naturally raises the questions like: Is the converse effect possible? Can moving magnets produce electric currents? Does the nature permit such a relation between electricity and magnetism? The answer is resounding yes! The experiments of Michael Faraday in England and Joseph Henry in USA. conducted around 1830. The phenomenon in which electric current is generated by varying magnetic fields is appropriately called electromagnetic induction. “What is the use of it?” His reply was: “What is the use of a new born baby?” The phenomenon of electromagnetic induction 204 .1 INTRODUCTION Electricity and magnetism were considered separate and unrelated phenomena for a long time. demonstrated conclusively that electric currents were induced in closed coils when subjected to changing magnetic fields. They found that moving electric charges produce magnetic fields. When Faraday first made public his discovery that relative motion between a bar magnet and a wire loop produced a small current in the latter. For example. we will study the phenomena associated with changing magnetic fields and understand the underlying principles. In the early decades of the nineteenth century. experiments on electric current by Oersted. an electric current deflects a magnetic compass needle placed in its vicinity. Ampere and a few others established the fact that electricity and magnetism are inter-related. efficient telegraph. Experiment 6. He made important improvements in electromagnets by winding coils of insulated wire around iron pole pieces and invented an electromagnetic motor and a new. Imagine a world where there is no electricity – no electric lights. Josheph Henry [1797 – 1878] American experimental physicist professor at Princeton University and first director of the Smithsonian Institution. Instead. Moreover.Electromagnetic Induction is not merely of theoretical or academic interest but also of practical utility.1 Figure 6. Experiment 6. which indicates reversal of the current’s direction. The pioneering experiments of Faraday and Henry have led directly to the development of modern day generators and transformers. The galvanometer does not show any deflection when the magnet is held stationary. He discoverd self-induction and investigated how currents in one circuit induce currents in another. the same effects are observed. when the South-pole of the bar magnet is moved towards or away from the coil. indicating the presence of electric current in the coil. When the North-pole of a bar magnet is pushed towards the coil. no trains. the pointer in the galvanometer G deflects. when the bar magnet is held fixed and the coil C1 is moved towards or away from the magnet. the pointer in the galvanometer deflects. The deflection lasts as long as the bar magnet is in motion. 6. 205 .2 THE EXPERIMENTS HENRY OF FARADAY AND JOSEPH HENRY (1797 – 1878) The discovery and understanding of electromagnetic induction are based on a long series of experiments carried out by Faraday and Henry. the galvanometer shows deflection in the opposite direction. As coil C2 is FIGURE 6. It shows that it is the relative motion between the magnet and the coil that is responsible for generation (induction) of electric current in the coil. When the magnet is pulled away from the coil. no telephones and no personal computers. Further.1 When the bar magnet is pushed towards the coil. We shall now describe some of these experiments. * Wherever the term ‘coil or ‘loop’ is used. 6.1 shows a coil C1* connected to a galvanometer G. the deflections in the galvanometer are opposite to that observed with the North-pole for similar movements.2 the bar magnet is replaced by a second coil C2 connected to a battery. it is assumed that they are made up of conducting material and are prepared using wires which are coated with insulating material. the deflection (and hence current) is found to be larger when the magnet is pushed towards or pulled away from the coil faster. Today’s civilisation owes its progress to a great extent to the discovery of electromagnetic induction. The steady current in the coil C2 produces a steady magnetic field.2 In Fig. a momentory deflection is observed again. Magnetic flux through 206 . 6.3 MAGNETIC FLUX Faraday’s great insight lay in discovering a simple mathematical relation to explain the series of experiments he carried out on electromagnetic induction. When C2 is moved away. respectively. we must get familiar with the notion of magnetic flux.magnet. the galvanometer shows a deflection. Interactive animation on Faraday’s experiments and Lenz’s law: http://micro.3 The above two experiments involved relative motion between a magnet and a coil and between two coils. The pointer in the galvanometer returns to zero immediately. It is observed that the galvanometer shows a momentary deflection when the tapping key K is pressed. The deflection lasts as long as coil C2 is in motion.2 Current is induced in coil C1 due to motion of the current carrying coil C2.3 Experimental set-up for Experiment 6. before we state and appreciate his laws.3 shows two coils C1 and C2 held stationary. there is no deflection in the galvanometer. When the coil C2 is held fixed and C1 is moved. Coil C1 is connected to galvanometer G while the second coil C2 is connected to a battery through a tapping key K. It is also observed that the deflection increases dramatically when an iron rod is inserted into the coils along their axis. it is the relative motion between the coils that induces the electric current. but in the opposite direction. FIGURE 6. Experiment 6. However. If the key is held pressed continuously. the same effects are observed.Physics moved towards the coil C 1. Faraday showed that this relative motion is not an absolute requirement. This indicates that electric current is induced in coil C1. Through another experiment. When the key is released.edu/electromag/java/faraday FIGURE 6.fsu. Figure 6. Magnetic flux is defined in the same way as electric flux is defined in Chapter 1.3. Again. but this time in the opposite direction. Φ B. the galvanometer shows a deflection again. . there is no change in the magnetic flux through coil C1 and the current in coil C1 drops to zero. It is the change in magnetic flux through coil C1 that produces an induced emf in coil C1. change the magnetic flux associated with coil C1.Electromagnetic Induction a plane of area A placed in a uniform magnetic field B (Fig. then the magnetic flux through the surface is given by Φ = B i dA + B i dA + . where ‘all’ stands for summation over all the area elements dAi comprising the surface and Bi is the magnetic field at the area element dAi. the magnetic flux through the neighbouring coil C1 also increases. = B 1 1 2 2 ∑ Bi i dA i all (6.2) FIGURE 6. The motion of a magnet towards or away from coil C1 in Experiment 6.4 A plane of surface area A placed in a uniform magnetic field B. This results in a decrease in magnetic flux through coil C1 and hence again induces an electric current in coil C1*. current in coil C2 is constant. Equation (6.1) where θ is angle between B and A. 6. It was this induced emf which caused electric FIGURE 6. 207 . When the key is held pressed.5 Magnetic field Bi current to flow in coil C1 and through the galvanometer. The law is stated below. Experimental observations discussed in Section 6. The common point in all these observations is that the time rate of change of magnetic flux through a circuit induces emf in it.3 is represents area vector of the as follows: When the tapping key K is pressed.4) can be written as Φ B = B .2 can be explained using this concept. 6.. The notion of the area as a vector has been discussed earlier in Chapter 1. the current in C2 and the resulting magnetic field decreases from the maximum value to zero in a short time. Faraday arrived at a conclusion that an emf is induced in a coil when magnetic flux through the coil changes with time. The change in magnetic flux induces emf in coil C1. coil C2 (and the resulting magnetic field) rises from zero to a maximum value in a short time.5. If the magnetic field has different magnitudes and directions at various parts of a surface as shown in Fig. dAi plausible explanation for the observations of Experiment 6. the current in i th area element. Therefore.1 and moving a current-carrying coil C2 towards or away from coil C1 in Experiment 6.1) can be extended to curved surfaces and nonuniform fields. The SI unit of magnetic flux is weber (Wb) or tesla meter squared (T m2). * Note that sensitive electrical instruments in the vicinity of an electromagnet can be damaged due to the induced emfs (and the resulting currents) when the electromagnet is turned on or off. A at the i th area element. Faraday stated experimental observations in the form of a law called Faraday’s law of electromagnetic induction. 6. Magnetic flux is a scalar quantity. A = BA cos θ (6.2.4 FARADAY’S LAW OF INDUCTION From the experimental observations. When the key is released. Consequently. 208 EXAMPLE 6. and the fact that the plane of polarisation is rotated in an electric field. Determine the magnitudes of induced emf and current during this time-interval. The relative motion between the two coils will cause the bulb to glow and thus demonstrate the presence of an induced current.1) and (6.2 EXAMPLE 6. the kind one finds in a small torch light. (ii) Connect the coil to a powerful battery. viz. (a) What would you do to obtain a large deflection of the galvanometer? (b) How would you demonstrate the presence of an induced current in the absence of a galvanometer? Solution (a) To obtain a large deflection. Example 6. the expression for the total induced emf is given by ε = –N dΦB dt (6.1 Consider Experiment 6. A uniform magnetic field of 0.2 in Section 6. an emf is induced in the respective coils. the laws of electrolysis. He is also credited with the invention of the electric motor.2 A square loop of side 10 cm and resistance 0. benzene. From Eqs.2).1 . Therefore. In Experiments 6. Mathematically. Michael Faraday who is ranked as one of the best experimentalists ever. was legendary for his innovative skills. we see that the flux can be varied by changing any one or more of the terms B. In experimental physics one must learn to innovate. change of flux associated with each turn. The negative sign indicates the direction of ε and hence the direction of current in a closed loop.4) The induced emf can be increased by increasing the number of turns N of a closed coil. The magnetic field is decreased to zero in 0. the discovery of electromagnetic induction. is the same. the flux is changed by varying B.3) MICHAEL FARADAY (1791–1867) Michael Faraday [1791– 1867] Faraday made numerous contributions to science. (6. This will be discussed in detail in the next section. the induced emf is given by ε=– dΦB dt (6.Physics The magnitude of the induced emf in a circuit is equal to the time rate of change of magnetic flux through the circuit. the electric generator and the transformer. In the case of a closely wound coil of N turns.2. and (iii) Move the arrangement rapidly towards the test coil C1. In these cases too.1 and 6.10 T is set up across the plane in the north-east direction. or rotating a coil in a magnetic field such that the angle θ between B and A changes.. A and θ. (b) Replace the galvanometer by a small bulb.2.70 s at a steady rate. Example 6. He is widely regarded as the greatest experimental scientist of the nineteenth century.5 Ω is placed vertically in the east-west plane. by shrinking it or stretching it) in a magnetic field. The flux can also be altered by changing the shape of a coil (that is. one or more of the following steps can be taken: (i) Use a rod made of soft iron inside the coil C2. Horizontal component of the earth’s magnetic field at the place is 3. Φmin = 0 The change in flux is brought about in 0. ε=N ∆Φ ∆t = 500 × (6π × 10–7)/0. But it is a steady field (which does not change within the time span of the experiment) and hence does not induce any emf. the initial magnetic flux is Φ = BA cos θ = 0. ΦB (final) = 3. 500 turns and resistance 2 Ω is placed with its plane perpendicular to the horizontal component of the earth’s magnetic field. From Eq.2 ε = Example 6.1).Electromagnetic Induction Solution The angle θ made by the area vector of the coil with the magnetic field is 45°. EXAMPLE 6. It is rotated about its vertical diameter through 180° in 0.0 × 10–5 × (π ×10–2) × cos 0º = 3π × 10–7 Wb Final flux after the rotation. ΦB (initial) = BA cos θ = 3.70 s.3 A circular coil of radius 10 cm.1 × 10 –2 2 Wb Final flux.3 209 .0 × 10–5 T.25 = 3. (6. estimated value of the induced emf is.3). From Eq.0 × 10–5 × (π ×10–2) × cos 180° = –3π × 10–7 Wb Therefore.7 = 1.8 × 10–3 V I = ε/R = 1. I = EXAMPLE 6.0 mV And the magnitude of the current is 10 –3 V = 2 mA R 0.25 s. Their instantaneous values are different and depend upon the speed of rotation at the particular instant. the magnitude of the induced emf is given by ε= ∆ΦB ∆t = (Φ – 0 ) ∆t = 10 –3 2 × 0.5Ω Note that the earth’s magnetic field also produces a flux through the loop. Estimate the magnitudes of the emf and current induced in the coil. Solution Initial flux through the coil. (6.9 × 10–3 A Note that the magnitudes of ε and I are the estimated values. 1. the induced current in the coil flows in clockwise direction and its Southpole faces the receding North-pole of the bar magnet. Where does the energy spent by the person go? This energy is dissipated by Joule heating produced by the induced current. 6. FIGURE 6. In this situation.1 in Section 6.Physics 6. Hence current is induced in the coil in such a direction that it opposes the increase in flux. They provide an easier way to understand the direction of induced currents. known as Lenz’s law which gives the polarity of the induced emf in a clear and concise fashion. 6. Similarly. Now consider the correct case shown in Fig. 210 .2. To counter this decrease in magnetic flux. the South-pole due to the induced current will face the approaching North-pole of the magnet. if the Northpole of the magnet is being withdrawn from the coil. 6. the bar magnet experiences a repulsive force due to the induced current. Consider Figs. the magnetic flux through the coil will decrease. an emf is induced across the open ends of the circuit. we see that the North-pole of a bar magnet is being pushed towards the closed coil.1. 6. The statement of the law is: The polarity of induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produced it. The direction of the induced emf can be found using Lenz’s law. Suppose that the induced current was in the direction opposite to the one depicted in Fig. Note that the direction shown by and indicate the directions of the induced currents. the magnetic flux through the coil increases.6 (a) and (b). This violates the law of conservation of energy and hence can not happen.5 LENZ’S LAW AND CONSERVATION OF ENERGY In 1834. one could construct a perpetual-motion machine by a suitable arrangement. This is possible only if the current in the coil is in a counter-clockwise direction with respect to an observer situated on the side of the magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in flux. We can understand Lenz’s law by examining Experiment 6.6(a). The bar magnet will then be attracted towards the coil at an ever increasing acceleration. In that case. a person has to do work in moving the magnet. In Fig. German physicist Heinrich Friedrich Lenz (1804-1865) deduced a rule. If this can happen. What will happen if an open circuit is used in place of the closed loop in the above example? In this case too. The negative sign shown in Eq. Therefore.6 Illustration of Lenz’s law. Note that magnetic moment associated with this current has North polarity towards the North-pole of the approaching magnet. A gentle push on the magnet will initiate the process and its velocity and kinetic energy will continuously increase without expending any energy. A little reflection on this matter should convince us on the correctness of Lenz’s law. (6.3) represents this effect.6(a). As the North-pole of the bar magnet moves towards the coil. (ii) Due to the outward motion.8) to a field-free region with a constant velocity v.4 Figure 6. Note that there are no induced current as long as the loops are completely inside or outside the region of the magnetic field. (iii) As the magnetic flux decreases due to motion of the irregular shaped loop abcd out of the region of magnetic field. EXAMPLE 6. Determine the direction of induced current in each loop using Lenz’s law.5 211 . so as to oppose the change in flux.Electromagnetic Induction Example 6. 6. In which loop do you expect the induced emf to be constant during the passage out of the field region? The field is normal to the loops.4 EXAMPLE 6. Can we hope to generate current in the loop by using very strong magnets? (b) A closed loop moves normal to the constant electric field between the plates of a large capacitor. Example 6. Is a current induced in the loop (i) when it is wholly inside the region between the capacitor plates (ii) when it is partially outside the plates of the capacitor? The electric field is normal to the plane of the loop.7 shows planar loops of different shapes moving out of or into a region of a magnetic field which is directed normal to the plane of the loop away from the reader. FIGURE 6. the induced current flows along cdabc. due to the motion of the loop into the region of magnetic field.7 Solution (i) The magnetic flux through the rectangular loop abcd increases. magnetic flux through the triangular loop abc decreases due to which the induced current flows along bacb. (c) A rectangular loop and a circular loop are moving out of a uniform magnetic field region (Fig.5 (a) A closed loop is held stationary in the magnetic field between the north and south poles of two permanent magnets held fixed. so as to oppose change in flux. The induced current must flow along the path bcdab so that it opposes the increasing flux. 5 6. PQRS forms a closed circuit enclosing an area that changes as PQ moves. the rate of change of area of the loop during its passage out of the field region is not constant.9. Current can not be induced by changing the electric flux. (d) The polarity of plate ‘A’ will be positive with respect to plate ‘B’ in the capacitor. 6. thus decreasing the area of the rectangular loop. However strong the magnet may be.8 (d) Predict the polarity of the capacitor in the situation described by Fig. This movement induces a current I as shown. the magnetic flux ΦB enclosed by the loop PQRS will be FIGURE 6. Assume that there is no loss of energy due to friction. (c) The induced emf is expected to be constant only in the case of the rectangular loop. It is placed in a uniform magnetic field B which is perpendicular to the plane of this system.10 The arm PQ is moved to the left side. current can be induced only by changing the magnetic flux through the loop. (b) No current is induced in either case. the rate of change of flux ΦB will induce an emf given by: ε= – dΦB d = – ( Blx ) dt dt dx = Blv dt 212 = – Bl (6. The rod PQ is moved towards the left with a constant velocity v as shown in the figure. In the case of circular loop. EXAMPLE 6.10 shows a rectangular conductor PQRS in which the conductor PQ is free to move.Physics FIGURE 6. FIGURE 6.6 MOTIONAL ELECTROMOTIVE FORCE Let us consider a straight conductor moving in a uniform and timeindependent magnetic field. Figure 6. hence induced emf will vary accordingly. ΦB = Blx Since x is changing with time.9 Solution (a) No.5) . If the length RQ = x and RS = l. This is the fundamental significance of the Faraday’s discovery. the force on its charges is given by F = q (E + v × B) = qE (6. Therefore.6 A metallic rod of 1 m length is rotated with a frequency of 50 rev/s. We stress that our presentation is not wholly rigorous.davidson. When the rod moves with speed v.edu/physlet_resources/bu_semester2/index. A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. W = qvBl Since emf is the work done per unit charge.5).6) since v = 0. we are able to produce induced emf by moving a conductor instead of varying the magnetic field. Consider any arbitrary charge q in the conductor PQ. In the case of a stationary conductor.netfirms.11). Thus. But it does help us to understand the basis of Faraday’s law when the conductor is moving in a uniform and time-independent magnetic field. that is. On the other hand. (6.5) by invoking the Lorentz force acting on the free charge carriers of conductor PQ.htm http://webphysics. All charges experience the same force. Conversely. a changing magnetic field) can exert a force on the stationary charge. it is not obvious how an emf is induced when a conductor is stationary and the magnetic field is changing – a fact which Faraday verified by numerous experiments. by changing the magnetic flux enclosed by the circuit. we learnt that charges in motion (current) can exert force/torque on a stationary magnet. a bar magnet in motion (or more generally. to explain the existence of induced emf or induced current. any force on the charge must arise from the electric field term E alone. In Chapter 4. Thus. and its direction is towards Q.6 213 . The Lorentz force on this charge is qvB in magnitude. we hasten to add that electric fields produced by static electric charges have properties different from those produced by time-varying magnetic fields. Electricity and magnetism are related. with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m. However. It is also possible to explain the motional emf expression in Eq. (6.com/englishhtm/Induction. Example 6.Electromagnetic Induction where we have used dx/dt = –v which is the speed of the conductor PQ. 6. in magnitude and direction. the charge will also be moving with speed v in the magnetic field B. http://ngsir. irrespective of their position in the rod PQ. The induced emf Blv is called motional emf.html Interactive animation on motional emf: ε = W q = Blv This equation gives emf induced across the rod PQ and is identical to Eq. we must assume that a time-varying magnetic field generates an electric field. about an axis passing through the centre and perpendicular to the plane of the ring (Fig. The work done in moving the charge from P to Q is. What is the emf between the centre and the metallic ring? EXAMPLE 6. If θ is the angle between the rod and the radius of the circle at P at time t. Hence. Thus. free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. we can imagine a closed loop OPQ in which point O and P are connected with a resistor R and OQ is the rotating rod. Hence. This gives 1 × 1. B ωR 2 2 ε = ∫ dε = ∫ Bv dr = ∫ B ωr dr = 0 0 R R Note that we have used v = ω r.11 Solution Method I As the rod is rotated.Physics FIGURE 6.0 × 2 π × 50 × (12 ) 2 = 157 V ε = Method II To calculate the emf. the resulting separation of charges produces an emf across the ends of the rod.5). The potential difference across the resistor is then equal to the induced emf and equals B × (rate of change of area of loop).6 ε =B × [Note: d 1 2  1 dθ B ω R 2 R θ  = BR 2 =  dt  2 2 dt 2  214 dθ = ω = 2πν ] dt This expression is identical to the expression obtained by Method I and we get the same value of ε. Using Eq. the area of the sector OPQ is given by θ 1 = R 2θ 2π 2 where R is the radius of the circle. At a certain value of emf. the induced emf is π R2 × EXAMPLE 6. the magnitude of the emf generated across a length dr of the rod as it moves at right angles to the magnetic field is given by dε = Bv dr . . there is no more flow of electrons and a steady state is reached. (6. Electromagnetic Induction Example 6.7). We assume that the remaining arms QR. 6. (6. If HE = 0.7 ENERGY CONSIDERATION: A QUANTITATIVE STUDY In Section 6. ε I = r Bl v = (6. Note that this force arises due to drift velocity of charges (responsible for current) along the rod and the consequent Lorentz force acting on them.4 × 10–4 × (0. RS and SP have negligible resistances compared to r. we discussed qualitatively that Lenz’s law is consistent with the law of conservation of energy. there will be a force on the arm PQ. P = Fv F = I lB = B 2l 2v 2 (6.4 G at the place. PJ = I 2r =   2 215 . is directed outwards in the direction opposite to the velocity of the rod.10. The current I in the loop is. B 2l 2v r where we have used Eq.7 A wheel with 10 metallic spokes each 0.5. Alternatively. the arm PQ is being pushed with a constant speed v. 6.5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of earth’s magnetic field HE at a place. Where does this mechanical energy go? The answer is: it is dissipated as Joule heat. and is given by = B 2l 2v 2  Blv  r =  r  r which is identical to Eq. Solution Induced emf = (1/2) ω B R2 EXAMPLE 6.28 × 10–5 V The number of spokes is immaterial because the emf’s across the spokes are in parallel. the power required to do this is. the overall resistance of the rectangular loop is r and this does not change as PQ is moved. The magnitude of this force is.7) r On account of the presence of the magnetic field.8) r The agent that does this work is mechanical. what is the induced emf between the axle and the rim of the wheel? Note that 1 G = 10–4 T.7 = (1/2) × 4π × 0. This force I ( l × B). Let r be the resistance of movable arm PQ of the rectangular conductor shown in Fig. Now we shall explore this aspect further with a concrete example.8).5)2 = 6. Thus. (6. Only the arm PQ possesses substantial resistance r. Obtain expressions for the flux. the force necessary to pull the arm and the power dissipated as Joule heat. outwards.8 0≤ x <b b ≤ x < 2b = Blb The induced emf is. There is an interesting relationship between the charge flow through the circuit and the change in the magnetic flux. (a) FIGURE 6. Sketch the variation of these quantities with distance.Physics Thus. mechanical energy which was needed to move the arm PQ is converted into electrical energy (the induced emf) and then to thermal energy. 6. ∆ΦB ∆t However. and is then moved back to x = 0 with constant speed v. The arm PQ of the rectangular conductor is moved from x = 0.8 Refer to Fig. we have learnt that the magnitude of the induced emf is. ∆Q = ∆Q r ∆t ∆ΦB r Example 6. From Faraday’s law.12 Solution Let us first consider the forward motion from x = 0 to x = 2b The flux ΦB linked with the circuit SPQR is ΦB = B l x EXAMPLE 6.12(a). ε = ε = Ir = Thus. the induced emf. Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b. The uniform magnetic field is perpendicular to the plane and extends from x = 0 to x = b and is zero for x > b. dΦ ε=− B dt = − Blv 0≤ x <b =0 b ≤ x < 2b 216 . the current I is (in magnitude) I = Bl v r (b) FIGURE 6.12(b).Electromagnetic Induction When the induced emf is non-zero. 6.12 The force required to keep the arm PQ in constant motion is I l B.8 One obtains similar expressions for the inward motion from x = 2b to x = 0. 217 . Its direction is to the left. In magnitude B 2l 2v 0≤ x <b r =0 b ≤ x < 2b The Joule heating loss is F = PJ = I 2r B 2l 2v 2 r =0 = 0≤ x <b b ≤ x < 2b EXAMPLE 6. One can appreciate the whole process by examining the sketch of various quantities displayed in Fig. 6. Eddy currents are minimised by using laminations of metal to make a metal core. the braking effect is smooth. Even when bulk pieces of conductors are subjected to changing magnetic flux. This fact is helpful in reducing eddy currents in the metallic cores of transformers. When the coil oscillates. heat loss is substantially reduced. their flow patterns resemble swirling eddies in water. and leaving the region of Note that magnetic moments of the induced currents (which magnetic field.Physics 6. (ii) Electromagnetic damping: Certain galvanometers have a fixed core made of nonmagnetic metallic material. This effect was discovered by physicist Foucault (1819-1868) and these currents are called eddy currents . Eddy currents are undesirable since they heat up the core and dissipate electrical energy in the form of heat. A copper plate is allowed to swing like a simple pendulum between the pole pieces of a strong magnet. induced currents are produced in them. oppose the motion) depend upon the area enclosed by the currents (recall equation m = I A in Chapter 4). Since the dissipation of electrical energy into heat depends on the square of the strength of electric current.14 Cutting slots in the copper plate reduces the effect of eddy currents. It is found that the motion is damped and in a little while the plate comes to a halt in the magnetic field. The laminations are separated by an insulating material like lacquer. 6. 218 . the eddy currents induced in the rails oppose the motion of the train. area available to the flow of eddy currents is less. while entering electromagnetic damping and the plate swings more freely.14. We can explain this phenomenon on the basis of electromagnetic induction. electric motors and other such devices in which a coil is to be wound over metallic core. Directions of eddy currents are opposite when the plate swings into the region between the poles and when it swings out of the region. Thus. As there are no mechanical linkages. The plane of the laminations must be arranged parallel to the magnetic field. When the electromagnets are activated. FIGURE 6. This arrangement reduces the strength of the eddy currents.13.13 Eddy currents are in Fig. If rectangular slots are made in the copper plate as shown FIGURE 6. Consider the apparatus shown in Fig. Magnetic flux associated with the plate keeps on changing as the plate moves in and out of the region between magnetic poles. The flux change induces eddy currents in the plate. the pendulum plate with holes or slots reduces generated in the copper plate. so that they cut across the eddy current paths.8 EDDY CURRENTS So far we have studied the electric currents induced in well defined paths in conductors like circular loops. However. Eddy currents are used to advantage in certain applications like: (i) Magnetic braking in trains: Strong electromagnets are situated above the rails in some electrically powered trains. the eddy currents generated in the core oppose the motion and bring the coil to rest quickly. . respectively.9 INDUCTANCE An electric current can be induced in a coil by flux change produced by another coil in its vicinity or flux change produced by the same coil. These two situations are described separately in the next two sub-sections. dΦB dI ∝ dt dt For a closely wound coil of N turns. the flux through a coil is proportional to the current. You will observe that the magnet dropped through the PVC pipe takes the same time to come out of the pipe as it would take when dropped through the same height without the pipe. if the geometry of the coil does not vary with time then. Electric currents are induced in the disc by magnetic fields produced by sinusoidally varying currents in a coil. Note the time it takes to come out of the pipe in each case. the motion of the magnet. Note that eddy currents are not generated in PVC pipe as its material is an insulator whereas aluminium is a conductor. You will see that the magnet takes much longer time in the case of aluminium pipe. Take a small cylinderical magnet having diameter slightly smaller than the inner diameter of the pipes and drop it through each pipe in such a way that the magnet does not touch the sides of the pipes during its fall. ELECTROMAGNETIC DAMPING Take two hollow thin cylindrical pipes of equal internal diameters made of aluminium and PVC. We shall see that inductance depends only on the geometry of the coil 219 . Such phenomena are referred to as electromagnetic damping. a term called flux linkage is used which is equal to NΦB for a closely wound coil and in such a case NΦB ∝ I The constant of proportionality. Why is it so? It is due to the eddy currents that are generated in the aluminium pipe which oppose the change in magnetic flux. i. 6. A high frequency alternating current is passed through a coil which surrounds the metals to be melted. ΦB α I. by melting the constituent metals. Fix them vertically with clamps on retort stands. Further. When the flux ΦB through the coil changes.e. However. each turn contributes to the induced emf. The retarding force due to the eddy currents inhibits the motion of the magnet. in this relation. Therefore. That is. is called inductance. You can observe the rotating shiny disc in the power meter of your house. (iv) Electric power meters: The shiny metal disc in the electric power meter (analogue type) rotates due to the eddy currents.Electromagnetic Induction (iii) Induction furnace: Induction furnace can be used to produce high temperatures and can be utilised to prepare alloys. the same magnetic flux is linked with all the turns. in both the cases. The eddy currents generated in the metals produce high temperatures sufficient to melt it. it in turn sets up a magnetic flux through S1.Physics and intrinsic material properties. The corresponding quantities for the outer solenoid S2 are r2 and n2. respectively. It is also referred to as the coefficient of mutual induction. The SI unit of inductance is henry and is denoted by H. implying l >> r2. (6.10) where n1l is the total number of turns in solenoid S1. Thus. independently of Faraday in England. The flux due to the current I1 in S1 can be assumed to be confined solely inside S1 since the solenoids are very long. A current I1 is passed through the solenoid S1 and the flux linkage with coil S2 is. respectively. The resulting flux linkage with coil S1 is. Let us denote it by Φ1. This aspect is akin to capacitance which for a parallel plate capacitor depends on the plate area and plate separation (geometry) and the dielectric constant K of the intervening medium (intrinsic material property). For these simple co-axial solenoids it is possible to calculate M12. It is named in honour of Joseph Henry who discovered electromagnetic induction in USA. It has the dimensions of [M L2 T –2 A–2] given by the dimensions of flux divided by the dimensions of current. The magnetic field due to the current I2 in S2 is µ0n2I2. We now consider the reverse case. We denote the radius of the inner solenoid S1 by r1 and the number of turns per unit length by n1. (6. FIGURE 6. = µ0n1n 2 πr12l I 2 2 M12 = µ0n1n2πr 1 l (6.1 Mutual inductance Consider Fig.15 which shows two long co-axial solenoids each of length l. Let N1 and N2 be the total number of turns of coils S1 and S2.9) M12 is called the mutual inductance of solenoid S1 with respect to solenoid S2.10). flux linkage with solenoid S2 is 220 N 2Φ2 = (n 2l ) πr12 ( ) (µ n I ) 0 1 1 . N 1Φ1 = (n1l ) πr12 ( ) (µ n I ) 0 2 2 (6. When a current I2 is set up through S2. The corresponding flux linkage with solenoid S1 is N1 Φ1 = M 12 I 2 (6. N2Φ2 = M21 I1 (6.15 Two long co-axial solenoids of same length l. 6. Thus.9.12) M21 is called the mutual inductance of solenoid S2 with respect to solenoid S1. This is a good approximation keeping in mind that the solenoid is long.11) Note that we neglected the edge effects and considered the magnetic field µ0n2I2 to be uniform throughout the length and width of the solenoid S2. Inductance is a scalar quantity. 6. from Eq.9) and Eq. etc. depends on their separation as well as their relative orientation. Note that if the inner solenoid was much shorter than (and placed well inside) the outer solenoid. Therefore. 221 .12). However. Instead.14) M12 = M 21 = µ0 π r12 2 r2 EXAMPLE 6. such that r1 << r2. solenoids. Solution Let a current I2 flow through the outer circular coil. Since the other co-axially placed coil has a very small radius.9 Two concentric circular coils.14) We have demonstrated this equality for long co-axial solenoids. (6. Example 6. 2 M21 = µ0n1n2πr 1 l (6. 2 Φ1 = πr 1 B2 = µ0 πr12 2r2 I2 = M12 I2 Thus. In this case. are placed co-axially with centres coinciding. The field at the centre of the coil is B2 = µ 0I 2 / 2 r2. the relation is far more general. we get M12 = M21= M (say) (6. M12 = µ0 πr12 2r2 From Eq. we can accept this value because r1 << r2. it would be extremely difficult to calculate the flux linkage with the outer solenoid as the magnetic field due to the inner solenoid would vary across the length as well as cross section of the outer solenoid. one of small radius r1 and the other of large radius r2. B2 may be considered constant over its cross-sectional area. the calculation of M12 would be easy. The equality M12=M21 is very useful in such situations. the mutual inductance would be 2 M =µr µ0 n1n2π r1 l It is also important to know that the mutual inductance of a pair of coils. Hence. (6. We explained the above example with air as the medium within the solenoids.11) and Eq. the calculation of M21 would also be extremely difficult in this case.Electromagnetic Induction where n2l is the total number of turns of S2. Obtain the mutual inductance of the arrangement. (6.9 Note that we calculated M12 from an approximate value of Φ1.12). (6. From Eq. assuming 2 the magnetic field B2 to be uniform over the area π r1 . then we could still have calculated the flux linkage N1Φ1 because the inner solenoid is effectively immersed in a uniform magnetic field due to the outer solenoid.13) Using Eq. However.. However. if a medium of relative permeability µr had been present. ε1 = – M 6. d ( N 1Φ1 ) dt dt Since induced emf in coil C1 is given by = d ( MI 2 ) ε1 = – We get. It is also called the coefficient of self-induction of the coil.9). When the current is varied. d ( N 1Φ1 ) dt dI 2 dt It shows that varying current in a coil can induce emf in a neighbouring coil.2. having n turns per unit length. The total flux linked with the solenoid is ε = –L 222 N ΦB = (nl ) ( µ0n I ) ( A ) . from Eq. the induced emf is given by ε=– d ( N ΦB ) dt dI (6. emf is induced in coil C1 wherever there was any change in current through coil C2.15). The magnitude of the induced emf depends upon the rate of change of current and mutual inductance of the two coils. Using Eq. The magnetic field due to a current I flowing in the solenoid is B = µ0 n I (neglecting edge effects. we considered the flux in one solenoid due to the current in the other. It is also possible that emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil. the flux linked with the coil also changes and an emf is induced in the coil. In that experiment. the self-induced emf always opposes any change (increase or decrease) of current in the coil.2 Self-inductance In the previous sub-section. let us recollect Experiment 6. Let us calculate the self-inductance of a long solenoid of crosssectional area A and length l.15) where constant of proportionality L is called self-inductance of the coil.3 in Section 6. we have N1Φ1 = MI2 For currents varrying with time. flux linkage through a coil of N turns is proportional to the current through the coil and is expressed as N ΦB ∝ I N ΦB = L I (6. In this case. as before). This phenomenon is called self-induction.16) dt Thus. It is possible to calculate the self-inductance for circuits with simple geometries. Let Φ1 be the flux through coil C1 (say of N1 turns) when current in coil C2 is I2.Physics Now. Then.9. (6. (6. then. using Faraday’s law. It is the electromagnetic analogue of mass in mechanics. Equation (6. For the current I at an instant in a circuit. (6. Physically. The self-induced emf is also called the back emf as it opposes any change in the current in a circuit. then using Eq.Electromagnetic Induction = µ 0n 2 Al I where nl is the total number of turns.e. Thus. the self-inductance plays the role of inertia. the self-inductance is. the energy required to build up the current I is..16). the rate of work done is dW = ε I dt If we ignore the resistive losses and consider only inductive effect. L = µr µ0 n 2 Al (6. 1 2 (6.19) LI 2 This expression reminds us of mv 2/2 for the (mechanical) kinetic energy of a particle of mass m. Therefore. work needs to be done against the back emf (ε ) in establishing the current. which has a high value of relative permiability). So. L is electrical inertia and opposes growth and decay of current in the circuit).17) If we fill the inside of the solenoid with a material of relative permeability µr (for example soft iron. dW dI =L I dt dt Total amount of work done in establishing the current I is W = ∫ d W = ∫ L I dI 0 I Thus.9) would be modified into W = N1 Φ1 = M11 I1 + M12 I 2 where M11 represents inductance due to the same coil. dI 1 dI − M 12 2 dt dt ε1 = − M 11 223 . This work done is stored as magnetic potential energy. The flux linked with one coil will be the sum of two fluxes which exist independently. L = ΝΦΒ I = µ0n 2 Al (6. Consider the general case of currents flowing simultaneously in two nearby coils.18) The self-inductance of the coil depends on its geometry and on the permeability of the medium. and shows that L is analogus to m (i. (6.19). Therefore. respectively.20) 2 µ0 We have already obtained the relation for the electrostatic energy stored per unit volume in a parallel plate capacitor (refer to Chapter 2.77) have been derived for special cases: a solenoid and a parallel plate capacitor. 1 ε0E 2 (2.10 AC GENERATOR The phenomenon of electromagnetic induction has been technologically exploited in many ways.magnet.20) and (2. we shall describe the basic principles behind this machine. (b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor? Solution (a) From Eq.Physics M11 is the self-inductance and is written as L1. An exceptionally important application is the generation of alternating currents (ac). But they are general and valid for any region of space in which a magnetic field or/and an electric field exist. area A and length l of the solenoid. As was pointed out in Section 6. one method to induce an emf 224 . ε1 = − L1 dI 1 dI − M12 2 dt dt Example 6.for a solenoid) =  B  1 ( µ0n 2 Al )  2  µ0n   [from Eq. the magnetic energy is http://micro.77) 2 In both the cases energy is proportional to the square of the field strength.10 6.10 (a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B. Equations (6. The Yugoslav inventor Nicola Tesla is credited with the development of the machine. Eq.edu/electromag/java/generator/ac. (6. 2.html UB = 1 2 LI 2 = 1  B  L 2   µ0 n   2 ( since B 2 = µ0nI .fsu.17)] Interactive animation on ac generator: 1 B 2 Al 2 µ0 (b) The magnetic energy per unit volume is. uΕ = EXAMPLE 6. = uB = = UB V (where V is volume that contains flux) UB Al B2 = (6. The modern ac generator with a typical output capacity of 100 MW is a highly evolved machine.3. In this section.77). where θ is the angle between A and B. The ends of the coil are connected to an external circuit by means FIGURE 6. As a result. then ε=–N ε = ε0 sin ωt (6. Eq (6.23) give the instantaneous value of the emf and ε varies between +ε0 and –ε0 periodically.16.1). Since ω = 2πν. If we denote NBAω as ε0. As the coil rotates in a magnetic field B. The direction of the current changes periodically and therefore the current is called alternating current (ac). dΦB d = – NBA (cos ω t ) dt dt Thus. and from Eq. the sign. which occurs when sin ωt = ±1. the effective area of the coil exposed to the magnetic field lines changes with time. 6. or polarity of the emf changes with time. so an emf is induced in the coil. The coil (called armature) is mechanically rotated in the uniform magnetic field by some external means. 225 . We shall learn how to determine the time-averaged value for the alternating voltage and current in the next chapter. When the coil is rotated with a constant angular speed ω. the instantaneous value of the emf is ε = NBA ω sin ωt (6.23) where ν is the frequency of revolution of the generator’s coil. (6. It consists of a coil mounted on a rotor shaft.16 AC Generator of slip rings and brushes. the flux at any time t is ΦB = BA cos θ = BA cos ωt From Faraday’s law. 6. This method of producing a flux change is the principle of operation of a simple ac generator. Note from Fig. An ac generator converts mechanical energy into electrical energy. the induced emf for the rotating coil of N turns is then.22) Since the value of the sine fuction varies between +1 and –1.22) can be written as ε = ε0sin 2π ν t (6. The rotation of the coil causes the magnetic flux through it to change.21) where NBAω is the maximum value of the emf. The axis of rotation of the coil is perpendicular to the direction of the magnetic field. The basic elements of an ac generator are shown in Fig. as the change of flux is greatest at these points.17 that the emf has its extremum value when θ = 90º or θ = 270º.Electromagnetic Induction or current in a loop is through a change in the loop’s orientation or a change in its effective area.22) and (6. the effective area of the loop (the face perpendicular to the field) is A cos θ. (6. Note that Eq. the angle θ between the magnetic field vector B and the area vector A of the coil at any instant t is θ = ωt (assuming θ = 0º at t = 0). 5 Hz. The frequency of rotation is 50 Hz in India. The steam at high pressure produces the rotation of the armature. Alternatively. Employing Eq.1 m2 and B = 0. A = 0. if a nuclear fuel is used.11 Kamla peddles a stationary bicycle the pedals of the bicycle are attached to a 100 turn coil of area 0.11 = 100 × 0. i.21) ε0 = NBA (2 π ν) EXAMPLE 6. (6.17 An alternating emf is generated by a loop of wire rotating in a magnetic field.14 × 0. These are called thermal generators. it is 60 Hz. from dams.314 V The maximum voltage is 0.314 V. What is the maximum voltage generated in the coil? Solution Here f = 0. Modern day generators produce electric power as high as 500 MW. Instead of coal. We urge you to explore such alternative possibilities for power generation. Example 6.5 = 0. In certain countries such as USA. N =100.01 T perpendicular to the axis of rotation of the coil. we get nuclear power generators. the mechanical energy required for rotation of the armature is provided by water falling from a height. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T.e. for example. the coils are held stationary and it is the electromagnets which are rotated.1 × 2 × 3. one can light up 5 million 100 W bulbs! In most generators.Physics FIGURE 6. 226 . In commercial generators. These are called hydro-electric generators..01 × 0. water is heated to produce steam using coal or other sources.10 m2. They dissipate electrical energy as heat. 6. The magnetic flux through a surface of area A placed in a uniform magnetic field B is defined as. dΦB dt Here ΦΒ is the flux linked with one turn of the coil. all three are mutually perpendicular. Such currents are called eddy currents. Lenz’s law states that the polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces it. There has been a suggestion that electromagnetic induction may provide a clue to these migratory patterns. the velocity of the bird v. the induced emf (called motional emf ) across its ends is ε = −N ε = Bl v 5. Changing magnetic fields can set up current loops in nearby metal (any conductor) bodies. As far as we know birds contain no ferromagnetic material.5). 4. So electromagnetic induction seems to be the only reasonable mechanism to determine direction. Thus. Consider the optimal case where the magnetic field B. and v = 10 m/s. ε = Blv Taking B = 4 × 10–5 T. Eq. In birds no such cells have been identified. When a metal rod of length l is placed normal to a uniform magnetic field B and moved with a velocity v perpendicular to the field. It is equal to NΦ/I. However. l = 2 cm wide. special cells have been identified which detect small voltage differences. Faraday’s laws of induction imply that the emf induced in a coil of N turns is directly related to the rate of change of flux through it. Certain kinds of fish are able to detect small potential differences. SUMMARY 1. the migration patterns of birds continues to remain a mystery. ΦB = Bi A = BA cos θ where θ is the angle between B and A. every winter birds from Siberia fly unerringly to water spots in the Indian subcontinent. and indeed all of science. 227 . From the formula for motional emf.Electromagnetic Induction MIGRATION OF BIRDS The migratory pattern of birds is one of the mysteries in the field of biology. Inductance is the ratio of the flux-linkage to current. a current I = ε/R is set up in it. 3. in these fish. 2. The negative sign in the expression for Faraday’s law indicates this fact. (6. If the circuit is closed. It would be of great benefit to migratory birds to use this field to determine the direction. we obtain ε = 4 × 10–5 × 2 × 10–2 × 10 V = 8 × 10–6 V = 8 µV This extremely small potential difference suggests that our hypothesis is of doubtful validity. where R is the resistance of the circuit. and two relevant points of its anatomy separated by a distance l. The earth’s magnetic field has existed throughout evolutionary history. For example. A changing current in a coil (coil 2) can induce an emf in a nearby coil (coil 1). 3. M12 = M21 ε1 = − M 12 8. mechanical energy is converted to electrical energy by virtue of electromagnetic induction. the experiments of Oersted. In the early part of the nineteenth century. 2. electric currents are induced so as to oppose the changing magnetic flux. the experiments of Faraday and Henry demonstrated that a moving magnet can induce electric current. Ampere and others established that moving charges (currents) produce a magnetic field. around 1830. There exists a general equality. The self-inductance of a long solenoid. 10. then the motional emf produced is ε = NBA ( 2πν) sin (2π νt) where we have assumed that at time t = 0 s. the core of which consists of a magnetic material of permeability µr. One can similarly define M21. When a current in a coil changes. Electricity and magnetism are intimately related. the coil is perpendicular to the field. It is as per the law of conservation of energy.Physics 7. If coil of N turn and area A is rotated at ν revolutions per second in a uniform magnetic field B. The self-induced emf is given by. In a closed circuit. How is it related to the flux change? The motional emf discussed in Section 6. Somewhat later. This relation is given by. It is a measure of the inertia of the coil against the change of current through it. l its length and n the number of turns per unit length. However. an emf is induced across its ends.5 can be argued independently from Faraday’s law using the Lorentz force on moving charges. Quantity Magnetic Flux EMF Mutual Inductance Self Inductance Symbol Units Wb (weber) V (volt) H (henry) H (henry) Dimensions [M L2 T –2 A–1] [M L2 T –3 A–1] [M L2 T –2 A–2] [M L2 T –2 A–2] Equations ΦB ε M L ΦB = B i A ε = − d( N ΦB )/ dt ε1 = −M12 ( dI 2 /dt ) ε = − L ( dI / dt ) POINTS TO PONDER 1. it induces a back emf in the same coil. is given by L = µr µ0 n2 A l ε = −L where A is the area of cross-section of the solenoid. in case of an open circuit. In an ac generator. dI 2 dt The quantity M12 is called mutual inductance of coil 1 with respect to coil 2. 228 . However. dI dt L is the self-inductance of the coil. 9. FIGURE 6. The motion of a copper plate is damped when it is allowed to oscillate between the magnetic pole-pieces. moving charges in static field and static charges in a time-varying field seem to be symmetric situation for Faraday’s law.Electromagnetic Induction even if the charges are stationary [and the q (v × B) term of the Lorentz force is not operative]. EXERCISES 6.1 Predict the direction of induced current in the situations described by the following Figs. 6. How is the damping force. Thus. an emf is nevertheless induced in the presence of a time-varying magnetic field.18(a) to (f ).18 229 . produced by the eddy currents? 4. This gives a tantalising hint on the relevance of the principle of relativity for Faraday’s law. Calculate the average power loss due to Joule heating.Physics 6. Calculate the emf developed between the centre and the ring.5 T parallel to the axis exists everywhere. (b) shorter side of the loop? For how long does the induced voltage last in each case? A 1.1 s.0 A to 4.0 m s–1.0 × 10–2 T. FIGURE 6.0 A in 0. A pair of adjacent coils has a mutual inductance of 1.10 230 .0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s–1 in a uniform horizontal magnetic field of magnitude 3.5 s.5 6.7 6.0 m long metallic rod is rotated with an angular frequency of 400 rad s–1 about an axis normal to the rod passing through its one end. If the current carried by the solenoid changes steadily from 2.0 A to 0. If the coil forms a closed loop of resistance 10 Ω. Where does this power come from? A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5. what is the induced emf in the loop while the current is changing? 6. calculate the maximum value of current in the coil. A constant and uniform magnetic field of 0.30 × 10–4 Wb m–2. A circular coil of radius 8.19 6.19: (a) A wire of irregular shape turning into a circular shape.6 6. give an estimate of the self-inductance of the circuit.9 6. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side.3 A long solenoid with 15 turns per cm has a small loop of area 2.8 6.0 A in 0. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential? Current in a circuit falls from 5.1 s. 6. 0. Obtain the maximum and average emf induced in the coil. The other end of the rod is in contact with a circular metallic ring.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0. What is the voltage difference developed between the ends of the wing 6. what is the change of flux linkage with the other coil? A jet plane is travelling towards west at a speed of 1800 km/h.2 Use Lenz’s law to determine the direction of induced current in the situations described by Fig.5 H. If the current in one coil changes from 0 to 20 A in 0.0 cm2 placed inside the solenoid normal to its axis. (b) A circular loop being deformed into a narrow straight wire.3 T directed normal to the loop. If an average emf of 200 V induced. at right angles to the horizontal component of the earth’s magnetic field. It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet.50 Ω. Figure 6. It has a gradient of 10 – 3 T cm–1 along the negative x-direction (that is it increases by 10 – 3 T cm – 1 as one moves in the negative x-direction). (a) Suppose K is open and the rod is moved with a speed of 12 cm s–1 in the direction shown.3 T at the rate of 0. how much power is dissipated by the loop as heat? What is the source of this power? A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s –1 in the positive x. Assume the field to be uniform.5 mC. The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7. resistance of the closed loop containing the rod = 9.6 Ω. if the Earth’s magnetic field at the location has a magnitude of 5 × 10–4 T and the dip angle is 30°. and the magnetic field are in three mutual perpendicular directions.14 FIGURE 6. The combined resistance of the coil and the galvanometer is 0.20 shows a metal rod PQ resting on the smooth rails AB and positioned between the poles of a permanent magnet. the rod.Electromagnetic Induction having a span of 25 m. B = 0. The rails. A galvanometer G connects the rails through a switch K.02 T s–1. one can give it a quick 90° turn to bring its plane parallel to the field direction). ADDITIONAL EXERCISES 6. 6. and it is decreasing in time at the rate of 10 – 3 T s–1. The field is neither uniform in space nor constant in time.50 T.20 (b) Is there an excess charge built up at the ends of the rods when K is open? What if K is closed? (c) With K open and the rod moving uniformly. Equivalently. and then quickly snatched out of the field region. A small flat search coil of area 2 cm2 with 25 closely wound turns. is positioned normal to the field direction. there is no net force on the electrons in the rod PQ even though they do 231 . Give the polarity and magnitude of the induced emf.11 Suppose the loop in Exercise 6.13 6.0 mΩ. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.4 is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0. Length of the rod = 15 cm. Estimate the field strength of magnet. If the cut is joined and the loop has a resistance of 1.direction in an environment containing a magnetic field in the positive z-direction.50 mΩ.12 6. carries a current of 2. It is given by.5 A.21. B = – B0 k (r ≤ a.1 m and assume that the loop has a large resistance. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.22). Take a = 0. 6. (b) Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity. Calculate the induced emf in the loop at the instant when x = 0. 6.15 6. The current is suddenly switched off in a brief time of 10–3 s. Explain.Physics experience magnetic force due to the motion of the rod. A uniform magnetic field extends over a circular region within the rim. area of cross-section 25 cm2 and number of turns 500. a < R) =0 (otherwise) What is the angular velocity of the wheel after the field is suddenly switched off ? 232 FIGURE 6. (d) What is the retarding force on the rod when K is closed? (e) How much power is required (by an external agent) to keep the rod moving at the same speed (=12 cm s–1) when K is closed? How much power is required when K is open? (f ) How much power is dissipated as heat in the closed circuit? What is the source of this power? (g) What is the induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular? An air-cored solenoid with length 30 cm. v = 10 m/s.2 m.17 A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. 6.22 . The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. (a) Obtain an expression for the mutual inductance between a long straight wire and a square loop of side a as shown in Fig.16 FIGURE 6.21 6. . AC circuits exhibit characteristics which are exploited in many devices of daily use. * The phrases ac voltage and ac current are contradictory and redundant. Further. The main reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and efficiently converted from one voltage to the other by means of transformers. The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. This is mainly because most of the electrical energy sold by power companies is transmitted and distributed as alternating current. Still. Further. For example. we are taking advantage of a special property of ac circuits – one of many that you will study in this chapter. But voltages and currents that vary with time are very common. most of the electrical devices we use require ac voltage. Such a voltage is called alternating voltage (ac voltage) and the current driven by it in a circuit is called the alternating current (ac current)*. Today. electrical energy can also be transmitted economically over long distances. respectively. whenever we tune our radio to a favourite station. These currents do not change direction with time. alternating current voltage and alternating current current.Chapter Seven ALTERNATING CURRENT 7. the abbreviation ac to designate an electrical quantity displaying simple harmonic time dependance has become so universally accepted that we follow others in its use. voltage – another phrase commonly used means potential difference between two points. since they mean. literally.1 INTRODUCTION We have so far considered direct current (dc) sources and circuits with dc sources. like the applied voltage. The minima. we can write this equation as ∑ ε (t ) = 0 .1 AC voltage applied to a resistor. the sum of the instantaneous current values over one complete cycle is zero. and which helped usher in the age of electric power. The voltage across a pure resistor and the current through it.1 to get vm sin ω t = i R vm sin ω t R Since R is a constant. we apply Kirchhoff’s loop rule shown in Fig. and the average current is zero. and the high frequency induction coil (the Tesla coil) used in radio and television sets and other electronic equipment. We see that.3) is just Ohm’s law which for resistors works equally well for both ac and dc voltages. be given by v = vm sin ω t (7.2 AC VOLTAGE APPLIED TO A RESISTOR Figure 7. Clearly. does . also called ac voltage. He conceived the idea of the rotating magnetic field. in particular that both v and i reach zero. The SI unit of magnetic field is named in his honour.2 In a pure resistor. however. NICOLA TESLA (1836 – 1943) FIGURE 7.3) R Equation (7. the voltage and current are in phase with each other. Note. minimum and maximum values at the same time.1 shows a resistor connected to a source ε of ac voltage. which is the basis of practically all alternating current machinery. Thus. The fact that the average current is zero. 7. He also invented among other things the induction motor.2) FIGURE 7. 7. 234 vm (7.2. zero and maxima occur at the same respective times.Physics 7.2) are plotted as a function of time in Fig. To find the value of current through the resistor. (7. We consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference. the current varies sinusoidally and has corresponding positive and negative values during each cycle. to the circuit or i = i = im sin ω t where the current amplitude im is given by im = (7. Nicola Tesla (1836 – 1943) Yugoslov scientist. given by Eqs.1) where vm is the amplitude of the oscillating potential difference and ω is its angular frequency. The symbol for an ac source in a circuit diagram is ~ .1) and (7. inventor and genius. the polyphase system of ac power. the voltage and current are in phase. The instantaneous power dissipated in the resistor is 2 p = i 2 R = im R sin2 ω t The average value of p over a cycle is* (7. Since. * The average value of a function F (t ) over a period T is given by F (t ) = 1 1  sin 2ω t  T 1 = [ sin 2ω T − 0] = 0 2ω   0 2ω T 1 F (t ) dt T∫ 0 T T ** < cos 2ω t > = T ∫ cos 2ω t dt = T   0 235 . i m and R are constants. As you know. a special value of current is defined and used.4) 2 [7.5(c)] 2 To express ac power in the same form as dc power (P = I2R).3) and is denoted by Irms or I. an advocate of direct curr ent. we have. p= George Westinghouse (1846 – 1914) A leading proponent of the use of alternating current over direct current.. p) denotes its average value and <. Westinghouse was convinced that the technology of alternating current was the key to the electrical future.5(a)] p = < i 2R > = < im R sin2 ω t > where the bar over a letter(here..3 The rms current I is related to the peak current im by I = i m / 2 = 0.. we have < sin2 ωt > = (1/2) (1– < cos 2ωt >) and since < cos2ωt > = 0**. < sin 2 ω t > = 1 2 Thus. 7.707 im. 1 2 im R [7. It is called. he came into conflict with Thomas Alva Edison. Joule heating is given by i 2R and depends on i 2 (which is always positive whether i is positive or negative) and not on i . there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor.. Thus. sin ω t = 1/2 (1– cos 2ωt ). GEORGE WESTINGHOUSE (1846 – 1914) FIGURE 7. Thus.5(b)] p = im R < sin2 ωt > 2 Using the trigonometric identity.> denotes taking average of the quantity 2 inside the bracket. He founded the famous Company named after him and enlisted the services of Nicola Tesla and other inventors in the development of alternating current motors and apparatus for the transmission of high tension current. root mean square (rms) or effective current (Fig. 2 [7. pioneering in large scale lighting.Alternating Current not mean that the average power consumed is zero and that there is no dissipation of electrical energy.. V = IR (7. we have V= v m = i mR vm (7. denoted by P is 1 2 im R = I 2 R 2 Similarly.707 im (7.450A 220 V V 236 . the average power. vm = 2 V = (1.3). the household line voltage of 220 V is an rms value with a peak voltage of or. (7.7) can also be written as P = V2 / R = I V (since V = I R ) Example 7.1 A light bulb is rated at 100W for a 220 V supply. Solution (a) We are given P = 100 W and V = 220 V. The resistance of the bulb is R= V 2 ( 220 V ) = = 484Ω 100 W P 2 vm = im (b) The peak voltage of the source is EXAMPLE 7.414)(220 V) = 311 V In fact.6) In terms of I. and (c) the rms current through the bulb. This shows the advantage of introducing the concept of rms values. Equation (7.7) = 0.7)] and relation between current and voltage in ac circuits are essentially the same as those for the dc case.1 vm = 2V = 311 V (c) Since. (7. In terms of rms values. the I or rms current is the equivalent dc current that would produce the same average power loss as the alternating current. (b) the peak voltage of the source.9) gives the relation between ac current and ac voltage and is similar to that in the dc case.Physics It is defined by I = i2 = i 1 2 im = m 2 2 = 0. P = I V I = P 100 W = = 0.707 vm 2 From Eq. Find (a) the resistance of the bulb. we define the rms voltage or effective voltage by P = p= (7. For example. It is customary to measure and specify rms values for ac quantities. the equation for power [Eq.9) Equation (7.8) R 2 2 or. we use the notion of phasors.e. But this is not so in the case of an inductor. curves in Fig. Thus. corresponding to the circuit shown in Fig. di =0 (7. Figure 7. Usually. From Fig. a capacitor or a combination of these circuit elements.3 REPRESENTATION OF AC CURRENT AND VOLTAGE BY ROTATING VECTORS — PHASORS In the previous section.4 (a) A phasor diagram for the circuit in Fig 7. The projection of voltage and current phasors on vertical axis.5 shows an ac source connected to an inductor.10) dt where the second term is the self-induced Faraday emf in the inductor. 7. In order to show phase relationship between voltage and current in an ac circuit. respectively represent the value of voltage and current at that instant. 7. The vertical components of phasors V and I represent the sinusoidally varying quantities v and i. but we shall assume that this inductor has negligible resistance. and L is the self-inductance of v −L FIGURE 7.4 AC VOLTAGE APPLIED TO AN INDUCTOR Figure 7. The magnitudes of phasors V and I represent the amplitudes or the peak values vm and im of these oscillating quantities. It so happens that the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions.. vm sinωt and im sinωt. A phasor* is a vector which rotates about the origin with angular speed ω.4(a) shows the FIGURE 7. 7. The rotating vectors that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know. the circuit is a purely inductive ac circuit.5 An ac source connected to an inductor.4(a) we see that phasors V and I for the case of a resistor are in the same direction.. inductors have appreciable resistance in their windings. The analysis of an ac circuit is facilitated by the use of a phasor diagram. (b) Graph of v and voltage and current phasors and their i versus ωt. relationship at time t1 for the case of an ac source connected to a resistor i.e. This means that the phase angle between the voltage and the current is zero.1. as shown in Fig. 7. Let the voltage across the source be v = vm sinωt.Alternating Current 7.4(b) are generated. 237 . Using the Kirchhoff’s loop rule. ∑ ε (t ) = 0 . * Though voltage and current in ac circuit are represented by phasors – rotating vectors. This is so for all times.1. i.4. we learnt that the current through a resistor is in phase with the ac voltage. As they rotate with frequency ω. 7. and since there is no resistor in the circuit. They are scalar quantities. they are not vectors themselves. animations. denoted by XL: XL = ω L The amplitude of the current is. (7.10).12). The inductive reactance limits the current in a purely inductive circuit in the same way as the resistance limits the current in a purely resistive circuit.6(b). L. Combining Eqs. they generate the voltage and current given by Eqs. . The current phasor I is π/2 behind the voltage phasor V. R. Using π  − cos(ω t ) = sin  ω t −  .1) and (7. The inductive reactance is directly proportional to the inductance and to the frequency of the current. i =− di vm L ∫ sin(ωt )dt http://www.unsw. the current it sustains also oscillates symmetrically about zero.html vm cos( ωt ) + constant ωL The integration constant has the dimension of current and is timeindependent. Figure 7. the integration constant is zero.6 (a) shows the voltage and the current phasors in the present case at instant t1. To obtain the current.11) implies that the equation for i(t). respectively and as shown in Fig.Physics the inductor.11) L dt L Equation (7. 7. (7. (7.12) vm is the amplitude of the current.12) for the source voltage and the current in an inductor shows that the current lags the voltage by π/2 or one-quarter (1/4) cycle.1) and (7.13) (7. The negative sign follows from Lenz’s law (Chapter 6).physics. with the same phase as the source voltage and an amplitude given by vm/L. Therefore. so that no constant or time-independent component of the current exists. The quantity ω L is ωL analogous to the resistance and is called inductive reactance.1) and (7. we have  2 π  i = i m sin  ωt −   2 where im = (7. must be such that its slope di/dt is a sinusoidally varying quantity. then im = vm XL (7. Since the source has an emf which oscillates symmetrically about zero.au//jw/AC. A comparison of Eqs. we integrate di/dt with respect to time: Interactive animation on Phasor diagrams of ac circuits containing. When rotated with frequency ω counterclockwise.14) 238 The dimension of inductive reactance is the same as that of resistance and its SI unit is ohm (Ω). C and RLC series circuits: ∫ dt dt = and get. the current as a function of time. we have d i v vm = = sin ω t (7.edu. Alternating Current FIGURE 7.2 239 . Figure 7. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz. You have seen that an voltage by one-fourth of a period  = ω  4  inductor has reactance that limits current similar to resistance in a dc circuit.5. Solution The inductive reactance.85 Ω EXAMPLE 7. We see that the current reaches its maximum value later than the π /2  T . (b) Graph of v and i versus ωt. the average power supplied to an inductor over one complete cycle is zero.6 (a) A Phasor diagram for the circuit in Fig. 7. the average power over a complete cycle is =− PL = − =− i m vm sin ( 2ω t ) 2 i m vm sin ( 2ω t ) = 0. The instantaneous power supplied to the inductor is π  p L = i v = im sin  ω t −  ×vm sin ( ωt )  2 = −im vm cos ( ωt ) sin ( ωt ) i m vm sin ( 2ωt ) 2 So.0 mH is connected to a source of 220 V.2 A pure inductor of 25. Thus. 2 since the average of sin (2ωt) over a complete cycle is zero. X L = 2 π ν L = 2 × 3. Does it also consume power like a resistance? Let us try to find out.14 × 50 × 25 × 10 –3 W = 7.85Ω The rms current in the circuit is I = 220 V V = = 28A X L 7. Example 7.7 explains it in detail. The power is. ENERGY IS ABSORBED FROM THE SOURCE. Since the direction of current has changed.. So their product p is positive. The current and voltage are both negative.Physics 0-1 Current i through the coil entering at A increase from zero to a maximum value. negative. 2-3 Current i becomes negative i. ENERGY IS ABSORBED.e. 1-2 Current in the coil is still positive but is decreasing. therefore. Note that the current lags the voltage. the core gets magnetised. One complete cycle of voltage/current. .. 3-4 Current i decreases and reaches its zero value at 4 when core is demagnetised and flux is zero. The product of voltage and current is negative. 240 FIGURE 7.7 Magnetisation and demagnetisation of an inductor. it enters at B and comes out of A. So their product p is positive. Flux lines are set up i. The voltage v is negative (since di/dt is negative).e. The core gets demagnetised and the net flux becomes zero at the end of a half cycle. the polarity of the magnet changes. ENERGY ABSORBED DURING THE 1/4 CYCLE 2-3 IS RETURNED TO THE SOURCE. and ENERGY IS BEING RETURNED TO SOURCE. With the polarity shown voltage and current are both positive. The voltage is positive but the current is negative. v= vm sin ω t = q C To find the current. When the capacitor is connected to an ac source.Alternating Current 7.17) 241 . 7.15) C From the Kirchhoff’s loop rule. im = Xc= 1/ωC so that the amplitude of the current is im = vm XC (7.8 An ac source does not completely prevent the flow of charge. capacitor is alternately charged and discharged as the current reverses each half cycle. we use the relation i = i = d (vm C sin ω t ) = ω C vm cos(ω t ) dt dq dt π Using the relation. as in Fig. a purely capacitive ac circuit. opposing the current. but FIGURE 7. The instantaneous voltage v across the capacitor is q (7. the voltage across them increases. the current in the circuit falls to zero. Let q be the charge on the capacitor at any time t. We can rewrite it as vm (1/ ω C ) Comparing it to im= vm/R for a purely resistive circuit. The connected to a capacitor.16)  2 where the amplitude of the oscillating current is im = ωCvm.18) (7. cos(ω t ) = sin  ω t +  . a capacitor in a dc circuit will limit or oppose the current as it charges. That is.5 AC VOLTAGE APPLIED TO A CAPACITOR Figure 7.8 shows an ac source ε generating ac voltage v = vm sin ωt connected to a capacitor only. As charge accumulates on the capacitor plates. current will flow for the short time required to charge the capacitor. When a capacitor is connected to a voltage source in a dc circuit. the voltage across the source and the capacitor are equal. When the capacitor is fully charged. It is called capacitive reactance and is denoted by Xc. we have   2 π  i = im sin  ω t +  (7. we find that (1/ωC) plays the role of resistance. it limits or regulates the current.8. If the frequency is doubled.4 EXAMPLE 7.1) shows that in Fig.9(a) shows the phasor diagram at an instant t1. Figure 7. Thus. Example 7. Figure 7. PC = Example 7. What happens in each case if the capacitance of the capacitor is reduced? Solution When a dc source is connected to a capacitor. the current leads the voltage by π/2. the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow.19) i m vm i v sin(2ωt ) = m m sin(2ωt ) = 0 2 2 since <sin (2ωt)> = 0 over a complete cycle. The instantaneous power supplied to the capacitor is pc = i v = im cos(ωt)vm sin(ωt) = imvm cos(ωt) sin(ωt) i m vm sin(2ωt ) 2 So. the current lags the voltage by π/2 and in the case of a capacitor.0 × 10 −6 F) 242 The rms current is . (7.9(b) shows the variation of voltage and current with time.0 µF capacitor is connected to a 220 V. the lamp will shine. A comparison of Eq. Figure 7. (b) Graph of v and i versus ωt. Reducing C will increase reactance and the lamp will shine less brightly than before.3 Solution The capacitive reactance is XC = 1 1 = = 212 Ω 2 π ν C 2π (50Hz)(15. we see that in the case of an inductor. With ac source. the average power = (7. the current is π /2 ahead of voltage. the capacitor offers capacitative reactance (1/ω C ) and the current flows in the circuit. Eq. 50 Hz source.9 (a) A Phasor diagram for the circuit equation of source voltage. The capacitive reactance limits the amplitude of the current in a purely capacitive circuit in the same way as the resistance limits the current in a purely resistive circuit. as in the case of an inductor.Physics The dimension of capacitive reactance is the same as that of resistance and its SI unit is ohm (Ω).10 explains it in detail. 7.16) with the FIGURE 7. (7.3 A lamp is connected in series with a capacitor. what happens to the capacitive reactance and the current? EXAMPLE 7. Find the capacitive reactance and the current (rms and peak) in the circuit. There will be no change even if C is reduced. Here the current phasor I is π /2 ahead of the voltage phasor V as they rotate counterclockwise. Consequently.8. But it is inversely proportional to the frequency and the capacitance. Predict your observations for dc and ac connections. We see that the current reaches its maximum value earlier than the voltage by one-fourth of a period.4 A 15. THE ENERGY ABSORBED DURING THE 1/4 CYCLE 0-1 IS RETURNED DURING THIS QUARTER. The power is negative. The voltage vc = q/C is in phase with q and reaches maximum value at 1. vc becomes zero at 4 when the capacitor is fully discharged. The plate A is charged to positive polarity while negative charge q builds up in B reaching a maximum at 1 until the current becomes zero. Their product. 243 . ENERGY IS ABSORBED FROM THE SOURCE DURING THIS QUAR TER CYCLE AS THE CAPACITOR IS CHARGED.Alternating Current 0-1 The current i flows as shown and from the maximum at 0. The capacitor ABSORBS ENERGY during this 1/4 cycle. the capacitor is discharged during this quarter cycle. The accumulated charge is depleted i. Current and voltage are both positive. the capacitor is charged to reversed polarity i. So p = vci is positive. reaches a zero value at 1.e.. the power is negative.10 Charging and discharging of a capacitor. 2-3 As i continues to flow from A to B. One complete cycle of voltage/current. The accumulated charge is depleted and the magnitude of the voltage vc is reduced. Note that the current leads the voltage. 1-2 The current i reverses its direction. Their product p is positive.ENERGY ABSORBED DURING 2-3 IS RETURNED TO THE SOURCE.. NET ENERGY ABSORBED IS ZERO. The current is negative. the plate B acquires positive and A acquires negative charge.The voltage gets reduced but is still positive. Both the current and the voltage are negative. FIGURE 7. 3-4 The current i reverses its direction at 3 and flows from B to A.e. the current is doubled. the capacitive reactance is halved and consequently. If q is the charge on the capacitor and i the current.04 A ) = 1. a larger fraction of the applied ac voltage appears across the inductor. First. 244 .20) analytically to obtain the time– dependence of i .41)(1. at time t.47 A. the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Solution As the iron rod is inserted.12 A series LCR circuit connected to an ac source.04 A X C 212 Ω The peak current is EXAMPLE 7.47A and –1. FIGURE 7. we have.5 A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig.11.12 shows a series LCR circuit connected to an ac source ε. (7.20) dt C We want to determine the instantaneous current i and its phase relationship to the applied alternating voltage v. we take the voltage of the source to be v = vm sin ωt. Consequently. 7. As usual. we use the technique of phasors and in the second method.5 7. leaving less voltage across the bulb. the inductive reactance of the coil increases. an iron rod is inserted into the interior of the inductor. L FIGURE 7. from Kirchhoff’s loop rule: di q +iR + =v (7.Physics I = 220 V V = = 1.47 A This current oscillates between +1.11 The switch is closed and after sometime.6 AC VOLTAGE APPLIED TO A SERIES LCR CIRCUIT Figure 7. If the frequency is doubled. we solve Eq. As a result. the glow of the light bulb decreases. (c) is unchanged. Give your answer with reasons. the inductance of the coil increases. Hence. EXAMPLE 7.4 i m = 2I = (1. We shall solve this problem by two methods. Therefore. (b) decreases. Example 7. as the iron rod is inserted. The glow of the light bulb (a) increases. and is ahead of the voltage by π/2. we see that the resistor. 7. capacitor and the source.22) The voltage Equation (7. VR.13 (a) Relation between the phasors VL. The phasor relation whose vertical component gives the above equation is VL + VR + VC = V (7. we have 2 vm = (im R )2 + (i m X C − im X L )2 2 2 2  = im R + ( X C − X L )   or. VC and I are shown in Fig.25(b)] (7. 7. VR. Therefore.26) where Z = R 2 + ( X C − X L )2 245 .21). inductor and capacitor are in series. 7.11. we introduce the impedance Z in an ac circuit: im = vm Z [7. resistor.23) FIGURE 7. Let I be the phasor representing the current in the circuit as given by Eq.13(b).22) into the above equation.20) for the circuit can be written as vL + vR + vC = v (7. respectively. the ac current in each element is the same at any time. VR.21) where φ is the phase difference between the voltage across the source and the current in the circuit. (b) Relation between the phasors VL. vCm. VC and VL are: vRm = im R. Since VC and VL are always along the same line and in opposite directions. From previous section. let VL. vLm = im XL (7. they can be combined into a single phasor (VC + VL) which has a magnitude vCm – vLm. 7.13(a) with apppropriate phaserelations. Since V is represented as the hypotenuse of a right-traingle whose sides are VR and (VC + VL).24) This relation is represented in Fig. we know that VR is parallel to I. VC. i m = vm R + ( X C − X L )2 2 [7. having the same amplitude and phase. VC is π/2 behind I and VL is π/2 ahead of I. and I. The length of these phasors or the amplitude of VR.12. VR. Let it be i = im sin(ωt+φ ) (7. (7. Further.25(a)] By analogy to the resistance in a circuit.1 Phasor-diagram solution From the circuit shown in Fig. VC.6. the pythagorean theorem gives: 2 2 vm = v Rm + (vCm − v Lm ) 2 Substituting the values of vRm. VL. vCm = im XC. and V represent the voltage across the inductor.Alternating Current 7. we shall construct a phasor diagram for the present case. (7. and (VL + VC) for the circuit in Fig. and vLm from Eq. On the basis of what we have learnt in the previous sections. in terms of q. (7. the phasor diagram say nothing about the initial condition. Eq. Thus. solution.21) is completely specified. (7. (7. the current in the circuit lags the source voltage. the effects of the transient solution die out and the behaviour of the circuit is described by the steady-state solution. as done throughout this chapter) and draw different phasors which show the relative angle between different phasors. the current in the circuit leads the source voltage. we have XC − X L (7. Figure 7. Consequently. v = 0. If XC > XL. φ is negative and the circuit is predominantly inductive.27) are graphically shown in Fig. φ is positive and the circuit is predominantly capacitive.6. After a sufficiently long time. Additionally. Consequently. di/dt = d2q/dt2. One can take any arbitrary value of t (say.22). But this method of analysing ac circuits suffers from certain disadvantages.26) and (7. The general solution is the sum of the transient solution and the steady-state solution. we do have a (b) Graphs of v and i versus ω t for a series LCR transient solution which exists even for circuit where XC > XL.27) gives the phase angle. Z as its hypotenuse. t1.14: tan φ = vCm − v Lm v Rm Using Eq. tan φ = 7.15 (a) Phasor diagram of V and I. the phase angle φ is the angle between VR and V and can be determined from Fig. Therefore. Thus. First. 7.15 shows the phasor diagram and variation of v and i with ω t for the case XC > XL.2 Analytical solution The voltage equation for the circuit is L di q + Ri + = v C dt = vm sin ωt 246 We know that i = dq/dt. we have obtained the amplitude and phase of current for an LCR series circuit using the technique of phasors.27) R Equations (7.Physics Since phasor I is always parallel to phasor VR. (7. FIGURE 7. With these. If XC < X L.14 Impedance This is called Impedance diagram which is a right-triangle with diagram. The solution so obtained is called the steady-state solution. the voltage equation becomes . Equation 7.14).25(a) gives the amplitude of the current and Eq. This is not a general FIGURE 7. Multiplying and (7.37(b)} in Class XI Physics Textbook]. damped oscillator.29(a)] [7.34(a)] = im cos(ωt + θ ) or i = imsin(ωt + φ ) where im = vm = Z vm 2 −1 and φ = tan R + ( X C − X L )2 XC − X L R 247 . Let us assume a solution L q = qm sin (ω t + θ ) so that and dq = qm ω cos(ω t + θ ) dt [7.28).31) and simplifying.31) Now. the current in the circuit is [7.33(b)] and θ − φ = − i = dq = qm ω cos(ω t + θ ) dt (7.28) 2 dt C dt This is like the equation for a forced. we see that vm = qm ω Z = im Z where im = qm ω π π or θ = − + φ 2 2 Therefore. we have (X − X L ) R  qm ω Z  cos(ω t + θ ) + C sin(ω t + θ ) = vm sin ω t Z Z   2 (7. [see Eq. we get: φ = tan −1 (7.29(c)] d2 q = −qm ω 2 sin(ω t + θ ) 2 dt Substituting these values in Eq. we get qm ω [ R cos(ω t + θ ) + ( X C − X L ) sin(ω t + θ )] = vm sin ωt dividing Eq. (7.30) where we have used the relation Xc= 1/ωC.34) [7.30) by Z = R 2 + ( X c − X L ) . (7.33) qm ω Z cos(ω t + θ − φ ) = vm sin ω t Comparing the two sides of this equation.33(a)] [7.Alternating Current d2 q dq q +R + = vm sin ω t (7.29(b)] [7. (7.32) (7. XL = ω L. {14. let and so that R = cos φ Z (XC − X L ) = sin φ Z XC − X L R Substituting this in Eq. for example.Physics Thus.00×10 rad/s. Xc = XL. mechanism of a radio or a TV set. the current amplitude is maximum. 7. This frequency is called the system’s natural frequency. Since im = vm / R at resonance. For an RLC circuit driven with voltage of amplitude vm and frequency ω.16 Variation of im with ω for two applications. The phenomenon of resonance is common among systems that have a tendency to oscillate at a particular frequency. then at a particular frequency 2 2 ω0. im = vm/R. the current amplitude for case (i) is twice to that for case (ii).3 Resonance An interesting characteristic of the series RLC circuit is the phenomenon of resonance. the amplitude of oscillation is found to be large. We see that the current amplitude is maximum at the resonant frequency. So if ω is varied. C = 1. If the child pulls on the rope at regular intervals and the frequency of the pulls is almost the same as the frequency of swinging. A familiar example of this phenomenon is a child on a swing. the amplitude of the swinging will be large (Chapter 14. Figure 7. ω0 for this case is   LC   = 1. (ii) R = 200 Ω. we found that the current amplitude is given by im = vm = Z vm R + ( X C − X L )2 2 with Xc = 1/ωC and XL = ω L .00 mH. in the tuning cases: (i) R = 100 Ω.16 shows the variation of im with ω in a RLC series circuit with L = 1. Class XI). . The signals picked up in the antenna acts as a source in the 248 tuning circuit of the radio. For the source applied vm =  1  6 100 V.00 nF for two values of R : (i) R = 100 Ω and (ii) R = 200 Ω. The antenna of L = 1. a radio accepts signals from many broadcasting stations. Resonant circuits have a variety of FIGURE 7.35) At resonant frequency. If such a system is driven by an energy source at a frequency that is near the natural frequency.6. and the impedance is minimum Z = R + 0 = R . This ( ) frequency is called the resonant frequency: X c = X L or 1 = ω0 L ω0 C or ω 0 = 1 LC (7. the analytical solution for the amplitude and phase of the current in the circuit agrees with that obtained by the technique of phasors. The the swing has a natural frequency for swinging back and forth like a pendulum.00 mH. so the circuit can be driven at many frequencies. At this value.16). This means that we cannot have resonance in a RL or RC circuit. max m for ω1 = ω0 + ∆ω. To get an expression for ∆ω. Suppose we choose a value of ω for which the current amplitude is 1/ 2 times its maximum value. The maximum value is max im = vm / R . From the curve in Fig. Only then do the voltages across L and C cancel each other (both being out of phase) and the current amplitude is vm/R. the amplitude of the current is less than the maximum value. we see that there are two such values of ω. Sharpness of resonance The amplitude of the current in the series LCR circuit is given by im = vm  1  R + ω L − ωC   2 2 and is maximum when ω = ω 0 = 1/ L C . ω1 and ω2. one greater and the other smaller than ω0 and symmetrical about ω0. the power dissipated by the circuit becomes half. we tune the radio. Therefore. (7.Alternating Current But to hear one particular radio station. we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. We may write ω1 = ω0 + ∆ω ω2 = ω0 – ∆ω The difference ω1 – ω2 = 2∆ω is often called the bandwidth of the circuit. It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. When this happens. we note that the current amplitude im is (1/ 2 ) i at ω1 . the amplitude of the current with the frequency of the signal of the particular radio station in the circuit is maximum. say. the sharper or narrower is the resonance. the total source voltage appearing across R. For values of ω other than ω0. In tuning. The quantity (ω0 / 2∆ω) is regarded as a measure of the sharpness of resonance. vm  1  R +  ω 1L − ω 1C    2 max im im = 2 = 2 = vm R 2 249 . The smaller the ∆ω. Therefore. Q of the circuit. larger the .36 (c)].36(a)] R 2L The sharpness of resonance is given by. [7. ∆ω = ω0 ω L = 0 R 2 ∆ω The ratio [7.36 (b)] and [7. [7. we see that 2 ∆ω = ω0 Q .Physics or  1  R +  ω 1L − ω1 C    2 2 2 =R 2 or  1  = 2R 2 R 2 +  ω1 L − ω1 C    ω 1L − 1 =R ω 1C 1 =R (ω 0 + ∆ω )C which may be written as. we get LC ω 0 L 1 +   ω0 L ∆ω  − =R  ω0   ∆ω  1+  ω0    −1  ∆ω  We can approximate 1 + ω0     ∆ω ∆ω  as 1 − since ω <<1. (ω 0 + ∆ω ) L − ω 0 L 1 +   ∆ω  − ω0   1  ∆ω  ω 0C 1 + ω0    =R 2 Using ω 0 = 1 in the second term on the left hand side.36(b)] ω0 L R R is also called the quality factor. So.36(c)] Q= ω0L 250 From Eqs.  ω  0 0  ω 0 L 1 +    ∆ω  ∆ω  − ω 0 L 1 − =R  ω0  ω0    or ω0 L 2 ∆ω ω0 =R [7. Example 7. resolve the paradox. The two voltages are out of phase by ninety degrees. 50 Hz ac source.0 µF are connected in series to a 220 V.15. Therefore. if the phase difference between two voltages is properly taken into account. From Eq.5 Ω (b) Since the current is the same throughout the circuit.14 × 50 × 10 −6 F)−2 = (200 Ω)2 + (212 Ω)2 = 291.. We see from Fig.755 A)(212. Is the algebraic sum of these voltages more than the source voltage? If yes. ν = 50 Hz (a) In order to calculate the current. VR and VC is 311. the current in the circuit is I = 220 V V = = 0. they cannot be added like ordinary numbers. Eq. [7.5 Ω Therefore.6 2 2 VR +C = VR + VC = 220 V Thus. the circuit is more selective. that if the resonance is less sharp. (b) Calculate the voltage (rms) across the resistor and the capacitor. R is low or L is large. the circuit is close to resonance for a larger range ∆ω of frequencies and the tuning of the circuit will not be good. we need the impedance of the circuit.3 V The algebraic sum of the two voltages. not only is the maximum current less. the two voltages are not in the same phase. less is the selectivity of the circuit or vice versa. the total voltage across the resistor and the capacitor is equal to the voltage of the source. So. It is 2 Z = R2 + X C = R 2 + (2π ν C )−2 = (200 Ω)2 + (2 × 3.36(c)] can be equivalently expressed as Q = 1/ω0CR.0 × 10−6 F V = 220 V. Solution Given R = 200 Ω. the smaller is the value of 2∆ω or the bandwidth and sharper 2 is the resonance.0 µF = 15. (a) Calculate the current in the circuit.Alternating Current value of Q.755 A Z 291.e.36). Using ω 0 = 1/ L C . Therefore. less sharp the resonance. 251 .6 A resistor of 200 Ω and a capacitor of 15. C = 15. the total of these voltages must be obtained using the Pythagorean theorem: EXAMPLE 7. (7. i. we have VR = I R = (0.3 Ω) = 160. How to resolve this paradox? As you have learnt in the text.3 V which is more than the source voltage of 220 V.755 A)(200 Ω) = 151 V VC = I X C = (0. 7. we see that if quality factor is large. cosφ = 1 and P = I 2Z = I 2 R. the instantaneous power p supplied by the source is p = v i = (vm sin ω t ) × [im sin(ω t + φ )] vm i m [ cos φ − cos(2ω t + φ )] (7. Even in such cases. P = I 2 Z cos φ [7. Let us discuss the following cases: Case (i) Resistive circuit: If the circuit contains only pure R. In that case φ = 0.37). Therefore. There is maximum power dissipation. φ may be non-zero in a RL or RC or RCL circuit.38(a)] = V I cos φ This can also be written as.38) where φ = tan–1 (Xc – XL )/ R. we know that the phase difference between voltage and current is π/2. The quantity cosφ is called the power factor. Therefore.37) 2 The average power over a cycle is given by the average of the two terms in R. (7. (b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit.7 (a) For circuits used for transporting electric power. power is dissipated only in the resistor. Explain. of Eq.S. 252 . It is only the second term which is time-dependent.7 POWER vm Z IN AC CIRCUIT: THE POWER FACTOR We have seen that a voltage v = vm sinωt applied to a series RLC circuit drives a current in the circuit given by i = im sin(ωt + φ) where im =  X − XL  and φ = tan −1  C    R Therefore. (7. = P = v i vm i m cos φ = m m cos φ 2 2 2 [7. Case (iv) Power dissipated at resonance in LCR circuit: At resonance Xc – XL= 0. maximum power is dissipated in a circuit (through R) at resonance. it is called resistive. Explain. Case (ii) Purely inductive or capacitive circuit: If the circuit contains only an inductor or capacitor. Case (iii) LCR series circuit: In an LCR series circuit.38(b)] So. This current is sometimes referred to as wattless current. cos φ = 1. cos φ = 0. and no power is dissipated even though a current is flowing in the circuit. the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle φ between them. That is. Its average is zero (the positive half of the cosine cancels the negative half).Physics 7. power dissipated is given by Eq. and φ = 0.7 Example7. So. Therefore. a low power factor implies large power loss in transmission.H. EXAMPLE 7. We can improve the power factor (tending to 1) by making Z tend to R.17 the applied voltage V and Iq perpendicular to the applied voltage.7. (b) the phase difference between the voltage across the source and the current. with the help of a phasor diagram (Fig. (b)Suppose in a circuit.48 mH.17) how this can be achieved.7 EXAMPLE 7. This can be done by connecting a capacitor of appropriate value in parallel so that Iq and I′ q cancel each other and P is effectively Ip V. if cosφ is small. But this will lead to large power loss (I2R) in transmission. (c) the power dissipated in the circuit. current I lags the voltage by an angle φ. there is no power loss. Find (a) the impedance of the circuit. Iq as you have learnt in Section 7. Solution (a) To find the impedance of the circuit.8 253 . we must completely neutralize the lagging wattless current Iq by an equal leading wattless current I′q. we first calculate XL and XC. Then power factor cosφ =R/Z. we have to increase current accordingly. To supply a given power at a given voltage. Let us resolve I into two components. XL = 2 πνL = 2 × 3. IP is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit. L = 25. and C = 796 µF. Let us understand.8 A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω. and (d) the power factor. Example 7. 7.Alternating Current Solution (a) We know that P = I V cosφ where cosφ is the power factor. It’s clear from this analysis that if we want to improve power factor. Ip along FIGURE 7. is called the wattless component since corresponding to this component of current.48 × 10–3 Ω = 8 Ω 1 XC = 2 πν C EXAMPLE 7.14 × 50 × 25. 35 kW You can see that in the present case. the current. 254 .14 × 50 × 796 × 10 −6 Therefore. I = im 2 = 1  283    = 40A 2 5  Therefore.8.9 The power dissipated at resonance is P = I 2 × R = (66.48 × 10 −3 ω0 = × 796 × 10 −6 = 222.1° = 0.Physics 1 = 4Ω 2 × 3. power dissipated at resonance is more than the power dissipated in Example 7.1rad/s νr = ω0 2π = 221.8 Now. Solution (a) The frequency at which the resonance occurs is 1 = LC 1 25. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance.1°  3   XC − X L R Since φ is negative.4Hz 2 × 3. the current in the circuit lags the voltage across the source.14 (b) The impedance Z at resonant condition is equal to the resistance: Z = R = 3Ω The rms current at resonance is = V V  283  1 = = = 66. φ = tan–1  4 − 8 = tan −1  = −53.1 Hz = 35. (c) The power dissipated in the circuit is P = I 2R EXAMPLE 7. = Z = R 2 + ( X L − X C )2 = 32 + (8 − 4)2 =5Ω (b) Phase difference.6 Example 7.7A  2 3 Z R  EXAMPLE 7. P = (40A )2 × 3 Ω = 4800 W (d) Power factor = cos φ = cos 53.9 Suppose the frequency of the source in the previous example can be varied. and the power dissipated at the resonant condition.7)2 × 3 = 13. in fact.40) d2 x 2 + ω0 x = 0 for a simple harmonic 2 dt oscillator. walking through a coil of many turns. The charge. When a capacitor (initially charged) is connected to an inductor.39) C dt i = – (dq/dt ) in the present case (as q decreases. in the present case. the charge on the capacitor and the current in the circuit exhibit the phenomenon of electrical oscillations similar to oscillations in mechanical systems (Chapter 14.e. we have cos φ =1 or φ = 0. the metal detector emits a sound.10 At an airport. the charge on the capacitor starts decreasing. Since q = qm at t = 0. therefore. On what principle does this detector work? Solution The metal detector works on the principle of resonance in ac circuits. i. Let a capacitor be charged qm (at t = 0) and connected to an inductor as shown in Fig. Therefore. Therefore. respectively. 255 . Eq. 7. When you walk through with metal in your pocket.18 At the instant shown. giving rise to current in the circuit. (7.41) (7. When you walk through a metal detector. for security reasons.10 7. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm.8 LC OSCILLATIONS We know that a capacitor and an inductor can store electrical and magnetic energy. The coil is connected to a capacitor tuned so that the circuit is in resonance. the impedance of the circuit changes – resulting in significant change in current in the circuit.. the induced emf in L will have polarity as shown. i increases). the current is increasing so the polarity of induced emf in the inductor is as shown. d2 q 1 + q=0 2 LC dt This equation has the form (7. If she/he is carrying anything made of metal. Let q and i be the charge and current in the circuit at time t. oscillates with a natural frequency LC and varies sinusoidally with time as q = qm cos ( ω 0 t + φ ) ω0 = 1 (7. Since di/dt is positive.18.Alternating Current Example 7.42) where qm is the maximum value of q and φ is a phase constant. The moment the circuit is completed. a person is made to walk through the doorway of a metal detector. you are. q di −L =0 (7. EXAMPLE 7. According to Kirchhoff’s loop rule. Class XI).39) becomes: FIGURE 7. vb < va. As the current reaches its maximum value im. there is no current in the circuit.19(e)]. it sets up a magnetic field in the inductor and thereby. (7. (at t = T/4) as in Fig. The current now starts charging the capacitor. The whole process just described will now repeat itself till the system reverts to its original state.Physics q = qm cos(ω 0t ) dq  The current i  =−   is given by  dt  i = im sin(ω 0t ) (7. .19(b)].43) UE = U = UE = 2 1 qm 2 C FIGURE 7. But it is charged with a polarity opposite to its initial state in Fig. This process continues till the capacitor is fully charged (at t = T/2) [Fig. Figure 7. the total energy of LC circuit is. 7. the switch is closed and the capacitor starts to discharge [Fig.19(c). 7. the energy in the system oscillates between the capacitor and the inductor. Thus. all the energy is stored in the magnetic field: UB = (1/2) Li2 . 7. as in Fig. You can easily check that the m maximum electrical energy equals the maximum magnetic energy. The capacitor now has no charge and hence no energy.19(d). energy in the inductor 2 C is zero. As the current increases.44) where im = ω 0qm Let us now try to visualise how this oscillation takes place in the circuit.19(a). The figure depicts one-half of a cycle.19(a) shows a capacitor with initial charge qm connected to an ideal inductor. 256 At t = 0.19 The oscillations in an LC circuit are analogous to the oscillation of a block at the end of a spring. 7. 7. Thus. some energy gets stored in the inductor in the form of magnetic energy: UB = (1/2) Li2. Since. The electrical energy stored in the charged capacitor is 2 1 qm . the equation is d2 x + ω2 x =0 2 0 dt Here. So. In fact.Alternating Current The LC oscillation is similar to the mechanical oscillation of a block attached to a spring. for a block of a mass m oscillating with frequency ω0. 7. ω 0 = k / m . we see that L is analogous to mass m: L is a measure of resistance to change in current. 1/C is analogous to k. The effect of this resistance is to introduce a damping effect on the charge and current in the circuit and the oscillations finally die away. ω 0 = k / m . In case of a mechanical system F = ma = m (dv/dt) = m (d2x/dt2 ). Comparing these two equations. 257 . (ii) Even if the resistance were zero. In case of LC circuit. For an electrical system. radio and TV transmitters depend on this radiation. The constant k (= F/x ) tells us the (external) force required to produce a unit displacement whereas 1/C (=V/q ) tells us the potential difference required to store a unit charge. ε = –L (di/dt ) = –L (d2q/dt 2 ). Table 7. The lower part of each figure in Fig. TABLE 7.19 depicts the corresponding stage of a mechanical system (a block attached to a spring). As noted earlier. So. and k is the spring constant. It is radiated away from the system in the form of electromagnetic waves (discussed in the next chapter). x corresponds to q.1 ANALOGIES BETWEEN MECHANICAL AND ELECTRICAL QUANTITIES Mechanical system Mass m Force constant k Displacement x Velocity v = dx/dt Mechanical energy E = 1 1 k x 2 + m v2 2 2 Slectrical system Inductance L Reciprocal capacitance 1/C Charge q Current i = dq/dt Electromagnetic energy U = 1 q2 1 2 + Li 2 C 2 Note that the above discussion of LC oscillations is not realistic for two reasons: (i) Every inductor has some resistance. the total energy of the system would not remain constant.1 gives the analogy between mechanical and electrical quantities. ω 0 = 1/ LC and for mass on a spring. 34). qm. of course. although they use different symbols and parameters.33) and (7. 258 . b Spring constant. which we reproduce here for convenience: A= F0 1/ 2 {m 2 2 2 2 2 (ω 2 − ω d ) + ωd b } Replace each parameter in the above equation by the corresponding electrical quantity. Equation [14. R Inverse capacitance. ω . you will see that there is a perfect match. be aware of the differences too. q Time. x Time. ω Amplitude of forced oscillations. F0 Charge on capacitor. L Resistance. ωd Natural frequency of oscillations. We suggest that you may try to find more such parallel situations from different areas of physics. with that of an LCR circuit when an ac voltage is applied in it. C. using XL= ωL. (7. 1/C Driving frequency.Physics TWO DIFFERENT PHENOMENA. and you come across another situation. qm Amplitude of applied voltage. ω Natural frequency of LCR circuit. Eliminate L.28) here. t Self inductance.37(b)] of Class XI Textbook is exactly similar to Eq. vm You must note that since x corresponds to q.10 of Class XI physics textbook.39 (a)] of Class XI gives the amplitude of oscillations in terms of other parameters. and ω . Let us therefore list the equivalence between different quantities in the two situations: Forced oscillations Driven LCR circuit 2 m d x + b dx + kx = F cos ω d t dt dt 2 L d2q dq q +R + = vm sin ω t 2 dt C dt Displacement. t Mass. and 0 2 ω0 = 1/ LC . and see what happens. One must. You will come across numerous such situations in physics where diverse physical phenomena are represented by the same mathematical equation. A Amplitude of driving force. When you use Eqs. ω0 Maximum charge stored. (7. We have already remarked that Eq. If you have dealt with one of them. you may simply replace the corresponding quantities and interpret the result in the new context. m Damping constant. SAME MATHEMATICAL TREATMENT You may like to compare the treatment of a forced damped oscillator discussed in Section 14. XC = 1/ωC. k Driving frequency. [14. the amplitude A (maximum displacement) will correspond to the maximum charge stored. Note that it is equal to the initial energy of the capacitor. the sum of energies stored in the capacitor and the inductor is constant in time. 259 .11 Show that in the free oscillations of an LC circuit. Let the charged capacitor be connected to an inductor of inductance L. They are wound on a soft-iron core. this LC circuit will sustain an oscillation with frquency  ω  = 2π ν =  1   LC  At an instant t. insulated from each other. One of the coils called the primary coil has Np turns.9 TRANSFORMERS For many purposes. A transformer consists of two sets of coils.20(a) or on separate limbs of the core as in Fig. either one on top of the other as in Fig. charge q on the capacitor and the current i are given by: q (t) = q0 cos ωt i (t) = – q0 ω sin ωt Energy stored in the capacitor at time t is UE = 2 1 1 q 2 q0 C V2 = = cos 2 (ωt ) 2 2 C 2C Energy stored in the inductor at time t is UM = = 1 L i2 2 1 2 L q0 ω 2 sin 2 (ωt ) 2 2 q0 sin 2 (ωt ) 2C = (∵ω 2 = 1/ LC ) Sum of energies U E +U M = 2 q0 cos2 ωt + sin 2 ωt   2C  EXAMPLE 7. As you have studied in Section 7. The other coil is called the secondary coil.11 = 2 q0 2C This sum is constant in time as qo and C. Solution Let q0 be the initial charge on a capacitor. Often the primary coil is the input coil and the secondary coil is the output coil of the transformer.20(b). 7.8. This is done with a device called transformer using the principle of mutual induction. Why it is so? Think! 7. it is necessary to change (or transform) an alternating voltage from one to another of greater or smaller value.Alternating Current Example 7. 7. both are time-independent. it has Ns turns. the primary current would be infinite since the primary has zero resistance(as assumed).Physics FIGURE 7. (b) two coils on separate limbs of the core. If this were not so. Let φ be the flux in each turn in the core at time t due to current in the primary when a voltage vp is applied to it.45) and (7. If the secondary is an open circuit or the current taken from it is small. This is εs = − N s dφ (7.45(a)] [7.46) can be written as ε p = −N p vs = − N s v p = −N p dφ dt [7. the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it.46 (a)].47) . When an alternating voltage is applied to the primary.46) dt But εp = vp. then to a good approximation εs = vs where vs is the voltage across the secondary. We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings. [7.45) dt The alternating flux φ also induces an emf. (7.45 (a)] and [7. we have 260 vs N = s vp N p (7. Eqs. in the secondary with Ns turns is dφ (7. called back emf in the primary. Therefore.46(a)] dφ dt From Eqs.20 Two arrangements for winding of primary and secondary coil in a transformer: (a) two coils on top of each other. The value of this emf depends on the number of turns in the secondary. Then the induced emf or voltage εs. Combining Eqs. (7. However. 261 . But in actual transformers. we can see how a transformer affects the voltage and current.48). the voltage is stepped down. and (iii) the secondary current is small. Now. ipvp = isvs (7. The effect is reduced by having a laminated core. That is. in this arrangement. we have i p vs N = = s (7. (ii) the same flux links both the primary and the secondary as very little flux escapes from the core. we have a step-down transformer. a 220V input at 10A will step-up to 440 V output at 5. or reduced. The resulting expenditure of energy in the core appears as heat and is kept to a minimum by using a magnetic material which has a low hysteresis loss. (iv) Hysteresis: The magnetisation of the core is repeatedly reversed by the alternating magnetic field. and the current is increased. there is less current in the secondary than in the primary (Np/Ns < 1 and Is < Ip). For example. (7. that is. Ns/Np = 2 and Np/Ns=1/2. This type of arrangement is called a step-up transformer. If the secondary coil has less turns than the primary(Ns < Np). these are minimised by using thick wire.47) and (7. (iii) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core and causes heating. In this case.50)  Ns    Np  That is. small energy losses do occur due to the following reasons: (i) Flux Leakage: There is always some flux leakage. We have: N   Np  Vs =  s  V p and I s =  Ip (7. this is a good approximation. Eq. If the transformer is assumed to be 100% efficient (no energy losses). not all of the flux due to primary passes through the secondary due to poor design of the core or the air gaps in the core. It can be reduced by winding the primary and secondary coils one over the other. The equations obtained above apply to ideal transformers (without any energy losses).48) Although some energy is always lost.Alternating Current Note that the above relation has been obtained using three assumptions: (i) the primary resistance and current are small. if the primary coil of a transformer has 100 turns and the secondary has 200 turns. (ii) Resistance of the windings: The wire used for the windings has some resistance and so. In high current. Vs < Vp and Is > Ip. since a well designed transformer may have an efficiency of more than 95%.49) also gives the ratio of the amplitudes or rms values of corresponding quantities. and since p = i v.49) i v N s p p Since i and v both oscillate with the same frequency as the ac source. the power input is equal to the power output.0 A. low voltage windings. the voltage is stepped up(Vs > Vp). Thus. energy is lost due to heat produced in the wire (I 2R). if the secondary coil has a greater number of turns than the primary (Ns > Np). For an alternating current i = im sinωt passing through a resistor R. the average power supplied to a capacitor over one complete cycle is zero. SUMMARY 1. The average power supplied to an inductor over one complete cycle is zero. To express it in the same form as the dc power (P = I 2R). An ac voltage v = vm sinωt applied to a capacitor drives a current in the capacitor: i = im sin (ωt + π/2). The voltage output of the generator is stepped-up (so that current is reduced and consequently. For a series RLC circuit driven by voltage v = vm sinωt. XL = ωL is called inductive reactance. drives a current in the inductor i = im sin (ωt – π/2). XC = XC ωC is called capacitive reactance.Physics The large scale transmission and distribution of electrical energy over long distances is done with the use of transformers. It is then transmitted over long distances to an area sub-station near the consumers. im = vm 1 . An alternating voltage v = vm sin ω t applied to a resistor R drives a current i = im sinωt in the resistor. As in the case of inductor. = 0. The current is in phase with R the applied voltage. the current is given by i = im sin (ωt + φ ) where im = vm R + ( XC − X L ) XC − X L R 2 2 2 and φ = tan −1 262 Z = R2 + ( X C − X L ) is called the impedance of the circuit. Here.707 im Similarly. vm . i m = 2. There the voltage is stepped down. a m special value of current is used. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes. It is called root mean square (rms) current and is donoted by I: I = im 2 vm 2 = 0. the I 2R loss is cut down). The current in the inductor lags the voltage by π/2. 5.707 vm 4. . The current through the capacitor is π/2 ahead of the applied voltage. the rms voltage is defined by V = 3. We have P = IV = I 2R An ac voltage v = vm sinωt applied to a pure inductor L. where im = vm/XL. the average power loss P (averaged over a cycle) due to joule heating is ( 1/2 )i 2 R. we have a step-down transformer. i. An interesting characteristic of a series RLC circuit is the phenomenon of resonance . If the primary coil is connected to an ac source.Alternating Current The average power loss over a complete cycle is given by P = V I cosφ The term cosφ is called the power factor. the amplitude of the current is maximum at the resonant frequency. The LC energy in the system oscillates between the capacitor and the inductor but their sum or the total energy is constant in time. 8. If the secondary coil has turns less than the primary. 1 . Q= ω0 L R = 1 ω 0CR 9. The charge q of the capacitor satisfies the equation of simple harmonic motion: d2 q 1 q=0 + dt 2 LC and therefore. The phase relationship between current and voltage in an ac circuit can be shown conveniently by representing voltage and current by rotating vectors called phasors . A phasor is a vector which rotates about the origin with angular speed ω. ω 0 = 7. the primary and secondary voltages are related by N  Vs =  s  V p  Np  and the currents are related by  Np Is =   Ns  Ip   If the secondary coil has a greater number of turns than the primary. The quality factor Q defined by LC is an indicator of the sharpness of the resonance. A transformer consists of an iron core on which are bound a primary coil of N p turns and a secondary coil of N s turns. The circuit exhibits resonance.. The analysis of an ac circuit is facilitated by the use of a phasor diagram.e. In such cases. In a purely inductive or capacitive circuit. A circuit containing an inductor L and a capacitor C (initially charged) with no ac source and no resistors exhibits free oscillations . 263 . the frequency ω of free oscillation is ω 0 = 1 . The magnitude of a phasor represents the amplitude or peak value of the quantity (voltage or current) represented by the phasor. the voltage is stepped-up (Vs > Vp). current is referred to as a wattless current. the higher value of Q indicating sharper peak in the current. 6. cosφ = 0 and no power is dissipated even though a current is flowing in the circuit. This type of arrangement is called a stepup transformer. 10. vm is the amplitude of the ac voltage. The power consumed in an circuit is never negative. im is the amplitude of the ac current. The voltage across the terminals of an outlet in your room is normally 240 V. it is ordinarily the rms value. Resonant frequency ωr or ω0 [T ] –1 Hz ω0 = 1 LC for a series RLC circuit Quality factor Q Dimensionless Q= ω0 L R = 1 ω0 C R for a series RLC circuit. Power factor Dimensionless = cos φ . Both alternating current and direct current are measured in amperes. When a value is given for ac voltage or current. rms current I [ A] A I= im 2 . But how is the ampere defined for an alternating current? It cannot be derived from the mutual attraction of two parallel wires carrying ac currents. Reactance: Inductive Capacitive Impedance XL XC Z [M L T A ] [M L T A ] [M L2 T –3 A–2] 2 –3 –2 2 –3 –2 Ω Ω Ω XL = ω L XC = 1/ ω C Depends on elements present in the circuit. The power rating of an element used in ac circuits refers to its average power rating.Physics Physical quantity Symbol Dimensions Unit Remarks rms voltage V [M L2 T –3 A–1] V V = vm 2 . 4. This refers to the rms value of the voltage. 3. φ is the phase difference between voltage applied and current in the circuit. The amplitude of this voltage is vm = 2V = 2(240) = 340 V 2. POINTS TO PONDER 1. as the dc ampere is derived. An ac current changes direction 264 . For example. The only element that dissipates energy in an ac circuit is the resistive element. and there is one ampere of rms value of alternating current in a circuit if the current produces the same average heating effect as one ampere of dc current would produce under the same conditions. the roles of input and output are reversed. 9. these quantities are not really vectors themselves. In a generator. Though in a phasor diagram. the presence of both L and C LC elements in the circuit is a must. resonance phenomenon occur when X L = X C or 2 2 VR + VC and not 7. then the total voltage across RC combination is VRC = VR + VC since VC is π/2 out of phase of VR. Thus. With only one of these ( L or C ) elements. while adding voltages across different elements. 5. 11. A transformer (step-up) changes a low-voltage into a high-voltage. Both devices simply transform energy from one form to another. one should take care of their phases properly. In a RLC circuit. mechanical energy is the input and electric energy is the output. In a motor. if VR and VC are voltages across R and C. The ‘rotating vectors’ that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know as the law of vector addition. voltage and current are represented by vectors. the ac ampere must be defined in terms of some property that is independent of the direction of the current. They are scalar quantities. For resonance to occur. It so happens that the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions. 8. electric energy is the input and mechanical energy is the output. The choice of whether the description of an oscillatory motion is by means of sines or cosines or by their linear combinations is unimportant. 12.Alternating Current with the source frequency and the attractive force would average to zero. respectively in an RC circuit. ω0 = 1 . In generators and motors. The current is reduced by the same proportion. In an ac circuit. 6. no resonance is possible. There are no power losses associated with pure capacitances and pure inductances in an ac circuit. This does not violate the law of conservation of energy. 10. The power factor in a RLC circuit is a measure of how close the circuit is to expending the maximum power. there is no possibility of voltage cancellation and hence. 265 . Joule heating is such a property. since changing the zero-time position transforms the one to the other. 6 7.5 7. Determine the rms value of the current in the circuit. what is the average power transferred to the circuit in one complete cycle? A radio can tune over the frequency range of a portion of MW broadcast band: (800 kHz to 1200 kHz). What is the Q-value of this circuit? A charged 30 µF capacitor is connected to a 27 mH inductor. What is the peak current? A 44 mH inductor is connected to 220 V. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.] Figure 7.10 7.7 7.1 A 100 Ω resistor is connected to a 220 V.Physics EXERCISES 7. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. (c) Determine the rms potential drops across the three elements of the circuit. Determine the rms value of the current in the circuit.11 FIGURE 7.4 7. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle? (a) The peak voltage of an ac supply is 300 V. Explain your answer. C = 32 µF and R = 10 Ω. 50 Hz ac supply.3 and 7.0H.3 7. A 60 µF capacitor is connected to a 110 V.21 shows a series LCR circuit connected to a variable frequency 230 V source. What is the total energy stored in the circuit initially ? What is the total energy at later time? A series LCR circuit with R = 20 Ω. 50 Hz ac supply.9 7..21 266 (a) Determine the source frequency which drives the circuit in resonance. .2 7. L = 5. L = 1. 7. the frequency of free oscillations of the LC circuit should be equal to the frequency of the radiowave.e.4.5 H and C = 35 µF is connected to a variable-frequency 200 V ac supply. the natural frequency i. what is the net power absorbed by each circuit over a complete cycle. C = 80µF. what must be the range of its variable capacitor ? [Hint: For tuning. When the frequency of the supply equals the natural frequency of the circuit. In Exercises 7.7 is 6 mC. Show that the potential drop across the LC combination is zero at the resonating frequency.0 H. If its LC circuit has an effective inductance of 200 µ H. What is the angular frequency of free oscillations of the circuit? Suppose the initial charge on the capacitor in Exercise 7. 60 Hz ac supply.8 7. R = 40 Ω. Obtain the resonant frequency ωr of a series LCR circuit with L = 2. . stored in the inductor)? (d) At what times is the total energy shared equally between the inductor and the capacitor? (e) If a resistor is inserted in the circuit. stored in the capacitor)? (ii) completely magnetic (i.13 if the circuit is connected to a high frequency supply (240 V.e. Obtain the current rms value in each branch of the circuit for the elements and source specified in Exercise 7. Compare this behaviour with that of a capacitor in a dc circuit after the steady state.14 7. Keeping the source frequency equal to the resonating frequency of the series LCR circuit.. and the total power absorbed.13 7. 50 Hz supply. The resistance of the circuit is negligible.18 7. C and R are arranged in parallel.18 has a resistance of 15 Ω. how much energy is eventually dissipated as heat? A coil of inductance 0. Hence. (a) What is the total energy stored initially? Is it conserved during LC oscillations? (b) What is the natural frequency of the circuit? (c) At what time is the energy stored (i) completely electrical (i. 60 Hz supply. A circuit containing a 80 mH inductor and a 60 µF capacitor in series is connected to a 230 V. L. (a) Obtain the current amplitude and rms values.12 An LC circuit contains a 20 mH inductor and a 50 µF capacitor with an initial charge of 10 mC. Let the instant the circuit is closed be t = 0. show that the total current in the parallel LCR circuit is minimum at this frequency. 50 Hz ac supply.] Suppose the circuit in Exercise 7.15 7. 7. if the three elements.e. explain the statement that a capacitor is a conductor at very high frequencies.Alternating Current ADDITIONAL EXERCISES 7.17 7. The resistance of the circuit is negligible.11 for this frequency. explain the statement that at very high frequency.50 H and resistance 100 Ω is connected to a 240 V.15 if the circuit is connected to a 110 V.16 7. (a) What is the maximum current in the coil? (b) What is the time lag between the voltage maximum and the current maximum? Obtain the answers (a) to (b) in Exercise 7. 12 kHz supply? Hence. an inductor in a circuit nearly amounts to an open circuit. (a) What is the maximum current in the circuit? (b) What is the time lag between the current maximum and the voltage maximum? Obtain the answers to (a) and (b) in Exercise 7. (b) Obtain the rms values of potential drops across each element.19 267 . (c) What is the average power transferred to the inductor? (d) What is the average power transferred to the capacitor? (e) What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’. How does an inductor behave in a dc circuit after the steady state? A 100 µF capacitor in series with a 40 Ω resistance is connected to a 110 V. Obtain the average power transferred to each element of the circuit. 10 kHz). 25 7.24 7. explain why high voltage transmission is preferred? 7. The resistance of the two wire line carrying power is 0. the water pressure head is at a height of 300 m and the water flow available is 100 m3s–1. (a) What is the source frequency for which current amplitude is maximum.23 7. estimate the electric power available from the plant (g = 9. What should be the number of turns in the secondary in order to get output power at 230 V? At a hydroelectric power plant.21 7.8 ms–2 ).12 H. (e) Why is choke coil needed in the use of fluorescent tubes with ac mains? Why can we not use an ordinary resistor instead of the choke coil? A power transmission line feeds input power at 2300 V to a stepdown transformer with its primary windings having 4000 turns. R = 23 Ω is connected to a 230 V variable frequency supply.22 7.26 268 . The town gets power from the line through a 4000-220 V step-down transformer at a sub-station in the town. assuming there is negligible power loss due to leakage? (c) Characterise the step up transformer at the plant.0 H. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Obtain the value of this maximum power.Physics 7. Show that the dc signal will appear across C and the ac signal across L. The lamp is seen to shine brightly. Do the same exercise as above with the replacement of the earlier transformer by a 40. C = 480 nF. C = 27 µF.20 A series LCR circuit with L = 0. Predict the corresponding observations if the connection is to an ac line.000-220 V step-down transformer (Neglect. (b) How much power must the plant supply. as before. leakage losses though this may not be a good assumption any longer because of the very high voltage transmission involved). Obtain this maximum value. (d) A choke coil in series with a lamp is connected to a dc line. Suggest a suitable way. is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage? (b) A capacitor is used in the primary circuit of an induction coil. Answer the following questions: (a) In any ac circuit. (b) What is the source frequency for which average power absorbed by the circuit is maximum. The circuit consists of an inductor and a capacitor in series. If the turbine generator efficiency is 60%.4 Ω.5 Ω per km. It is desired to improve the sharpness of the resonance of the circuit by reducing its ‘full width at half maximum’ by a factor of 2. (a) Estimate the line power loss in the form of heat. (c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies? (d) What is the Q-factor of the given circuit? Obtain the resonant frequency and Q-factor of a series LCR circuit with L = 3. and R = 7. (c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. A small town with a demand of 800 kW of electric power at 220 V is situated 15 km away from an electric plant generating power at 440 V. Hence. Is the converse also true? Does an electric field changing with time give rise to a magnetic field? James Clerk Maxwell (1831-1879). Maxwell formulated a set of equations involving electric and magnetic fields. Maxwell noticed an inconsistency in the Ampere’s circuital law. The speed of the waves. argued that this was indeed the case – not only an electric current but also a time-varying electric field generates magnetic field. He suggested the existence of an additional current. we learnt that an electric current produces magnetic field and that two current-carrying wires exert a magnetic force on each other. While applying the Ampere’s circuital law to find magnetic field at a point outside a capacitor connected to a time-varying current. we have seen that a magnetic field changing with time gives rise to an electric field.Chapter Eight ELECTROMAGNETIC d WAVES e o n h s T i l R b E u C p N re © e b o t t 8. The most important prediction to emerge from Maxwell’s equations is the existence of electromagnetic waves. These equations are known as Maxwell’s equations. they mathematically express all the basic laws of electromagnetism. called by him. which are (coupled) timevarying electric and magnetic fields that propagate in space. the displacement current to remove this inconsistency. the charge and current densities. in Chapter 6. and their sources.1 INTRODUCTION In Chapter 4. Further. according to these equations. Together with the Lorentz force formula (Chapter 4). turned out to be very close to . So we have B (2πr) = µ0i (t ) (8 . This effect is of great importance because it explains the existence of radio waves.2 DISPLACEMENT CURRENT We have seen in Chapter 4 that an electrical current produces a magnetic field around it. gamma rays and visible light. the left side of Eq.1(a) shows a parallel plate capacitor C which is a part of circuit through which a time-dependent current i (t ) flows .Physics the speed of light( 3 ×108 m/s). Then we present a descriptive account of electromagnetic waves. He derived the thermal velocity distribution of molecules in a gas and was among the first to obtain reliable estimates of molecular parameters from measurable quantities like viscosity. experimentally demonstrated the existence of electromagnetic waves. Maxwell’s work thus unified the domain of electricity.2) 270 . o n James Clerk Maxwell (1831 – 1879) Born in Edinburgh.1(a)]. obtained from optical measurements. Maxwell’s greatest acheivement was the unification of the laws of electricity and magnetism (discovered by Coulomb. Scotland. To see how a changing electric field gives rise to a magnetic field. we consider a plane circular loop of radius r whose plane is perpendicular to the direction of the current-carrying wire. was among the greatest physicists of the nineteenth century. From symmetry. magnetism and light. a changing electric field must also produce a magnetic field. as well as all other forms of electromagnetic waves.dl = µ0 i (t ) (8. 8.1) is B (2π r). Maxwell did not agree with the idea (strongly suggested by the Faraday’s laws of electrolysis) that electricity was particulate in nature. in a region outside the parallel plate capacitor. The broad spectrum of electromagnetic waves. let us consider the process of charging of a capacitor and apply Ampere’s circuital law given by (Chapter 4) “B. How the electromagnetic waves are sent and received for communication is discussed in Chapter 15. and which is centred symmetrically with respect to the wire [Fig. Hertz. For this.1) to find magnetic field at a point outside the capacitor. Maxwell showed that for logical consistency. stretching from γ rays (wavelength ~10–12 m) to long radio waves (wavelength ~106 m) is described. Oersted. in 1885. JAMES CLERK MAXWELL (1831–1879) d e h s T i l R b E u C p N re © e b o t t 8. the magnetic field is directed along the circumference of the circular loop and is the same in magnitude at all points on the loop so that if B is the magnitude of the field. we first discuss the need for displacement current and its consequences. In this chapter. Its technological use by Marconi and others led in due course to the revolution in communication that we are witnessing today. From these he arrived at the most important conclusion that light is an electromagnetic wave. This led to the remarkable conclusion that light is an electromagnetic wave. etc. Ampere and Faraday) into a consistent set of equations now called Maxwell’s equations. Figure 8. Let us find the magnetic field at a point such as P. (8. Interestingly. 2. it is ΦE = E A = 1 Q Q A= ε0 A ε0 Now if the charge Q on the capacitor plates changes with time. (8.3). called displacement current or Maxwell’s displacement current. given by Eq.2 shows the electric and magnetic fields inside the parallel plate capacitor discussed above. since no current passes through the surface of Fig. the magnitude of the electric field E between the plates is (Q/A)/ε0 (see Eq. It is. no matter what surface is used. It has the same magnitude over the area A of the capacitor plates. we find that the left hand side of Eq. So we have a contradiction. The source of a magnetic field is not just the conduction electric current due to flowing ε0 ⎜ o n d e h s T i l R b E u C p N re © e b o t t (8. The current.1) has not changed but the right hand side is zero and not µ0 i. 8. consider a different surface. The generalisation made by Maxwell then is the following. B at a point P outside the plates [Fig.4) ⎝ dt ⎟ ⎠ This is the missing term in Ampere’s circuital law.Electromagnetic Waves Now. If we generalise this law by adding to the total current carried by conductors through the surface. to determine magnetic field at a point P on the loop.1(c). there is no contradiction in the value of B obtained anywhere using the generalised Ampere’s law. 8. 8. This is a pot like surface [Fig. is a new term.41). its mouth is the circular loop mentioned above.1(b) and (c).1(b)] which nowhere touches the current. and is due to changing electric field (or electric displacement. The arrows show uniform electric field between the capacitor plates. the total has the same value of current i for all surfaces.4). Figure 8. another term which is ε0 times the rate of change of electric flux through the same surface. as part of a circuit through which a time dependent current i (t) flows. there is a current i = (dQ/dt). (a) a loop of radius r. as it should be. Another such surface is shaped like a tiffin box (without the lid) [Fig. ⎛ dΦE ⎞ =i (8.3) FIGURE 8.1(c)]. Is there anything passing through the surface S between the plates of the capacitor? Yes. but has its bottom between the capacitor plates. 271 . Since the contradiction arises from our use of Ampere’s circuital law. so that using Eq.1 A parallel plate capacitor C. The missing term must be such that one gets the same magnetic field at point P.1(a)] is the same as at a point M just inside. calculated one way. So what is the electric flux ΦE through the surface S ? Using Gauss’s law. which has the same boundary.1(c). (8. 8. we have dΦE d ⎛ Q ⎞ 1 dQ = = dt dt ⎜ ⎝ ε0 ⎟ ⎠ ε 0 dt This implies that for consistency. If this is done. (8. and vanishes outside it. of course. this law must be missing something. therefore. 8. 8. calculated another way. On applying Ampere’s circuital law to such surfaces with the same perimeter. The current carried by conductors due to flow of charges is called conduction current. B at the point P is non-zero no matter which surface is used for calculating it. the magnetic field at P is zero. (c) a tiffinshaped surface with the circular loop as its rim and a flat circular bottom S between the capacitor plates. the electric field! If the plates of the capacitor have an area A. The field is perpendicular to the surface S of Fig. (b) a pot-shaped surface passing through the interior between the capacitor plates with the loop shown in (a) as its rim. an old term still used sometimes). We can actually guess the missing term by looking carefully at Fig. there is a magnetic field at a point P. and a total charge Q. steady electric fields in a conducting wire. More precisely.2(a) can be measured and is seen to be the same as that just outside (at P). and the displacement current denoted by id (= ε0 (dΦE/ dt)). but also the time rate of change of electric field. (b) A cross insulating medium.e. id = 0. i. On the other hand. The generalised law is i = ie + id = ic + ε0 dΦE (8. (8. since the emf between two points 1 and 2 is the work done per unit charge in taking it from 1 to 2. the fact that an electric field changing with time gives rise to a magnetic field. In all respects. gives rise to an electric field. 272 . the existence of an emf implies the existence of an electric field. 8. The generalised (and correct) Ampere’s circuital law has the same form as Eq. One thing we immediately notice is that the laws of electricity and magnetism are now more symmetrical*. we can rephrase Faraday’s law of electromagnetic induction by saying that a magnetic field. is the symmetrical counterpart..5) dt In explicit terms.e. The displacement current has (literally) far reaching consequences. but there is only a displacement current due to time-varying electric fields. a magnetic field (say at point M) between the plates of the capacitor in Fig.Physics charges. of space where there is no conduction current. ic = 0. In other FIGURE 8. this means that outside the capacitor plates. with one difference: “the total current passing through any surface of which the closed loop is the perimeter” is the sum of the conduction current and the displacement current. In most of the cases. there is no conduction current. So. as there exist no perfectly conducting or perfectly the point M. In such a region. at region of space. For example. though there is no (conduction) current source nearby! The prediction of such a displacement current can be verified experimentally. so that id = i. we expect a magnetic field. Most interestingly. changing with time. the total current i is the sum of the conduction current denoted by ic. So we have dΦE (8. the displacement current may be zero since the electric field E does not change with time.2 (a) The cases.1). Now. Faraday’s law of induction states that there is an induced emf equal to the rate of change of magnetic flux. the displacement current has the same physical effects as the conduction current. inside the capacitor. (a). and no displacement current. they both may be present in the same the capacitor plates. both conduction electric and magnetic and displacement currents may be present in different regions of fields E and B between space. there may be large regions sectional view of Fig. we have only conduction current ic = i.6) dt and is known as Ampere-Maxwell law. and there is only displacement current.. there are no known sources of magnetic field (magnetic monopoles) analogous to electric charges which are sources of electric field. for example. the charging capacitor above. and is o n d e h s T i l R b E u C p N re © e b o t t ∫ Bidl = μ0 i c + μ0 ε0 * They are still not perfectly symmetrical. In some cases. Then. for example. i. with the current being the total current.1 A parallel plate capacitor with circular plates of radius 1 m has a capacitance of 1 nF. At t = 0.1 Consider now a circular loop of radius (1/2) m parallel to the plates passing through P. 2. as in Eq.3 E XAMPLE 8.5).) Solution The time constant of the CR circuit is τ = CR = 10–3 s. it is connected for charging in series with a resistor R = 1 M Ω across a 2V battery (Fig. o n d e h s T i l R b E u C p N re © e b o t t MAXWELL’S EQUATIONS (Gauss’s Law for electricity) (Gauss’s Law for magnetism) (Faraday’s Law) ε0 dΦE dt (Ampere – Maxwell Law) FIGURE 8. where the time constant τ is equal to CR. time. ∫ E idA = Q / ε0 ∫ BidA = 0 ∫ E idl = –dΦB dt ∫ Bidl = μ0 i c + μ0 Example 8.3). 4. 8. One very important consequence of this symmetry is the existence of electromagnetic waves. 3. halfway between the centre and the periphery of the plates. The magnetic field B at all points on the loop is 273 . A = π (1) m = area of the plates.Electromagnetic Waves a consequence of the displacement current being a source of a magnetic field. Calculate the magnetic field at a point P. 1. which we discuss qualitatively in the next section. (The charge on the capacitor at time t is q (t) = CV [1 – exp (–t/τ )]. Thus. (8. after t = 10–3 s.dependent electric and magnetic fields give rise to each other! Faraday’s law of electromagnetic induction and Ampere-Maxwell law give a quantitative expression of this statement. we have q(t) = CV [1 – exp (–t/τ)] = 2 × 10–9 [1– exp (–t/10–3)] The electric field in between the plates at time t is E= q (t ) q 2 2 = ε 0 A πε 0 . Then. 74 × 10–13 T 8. as in the Hertz’s experiment (1887). The frequency of the electromagnetic wave naturally equals the frequency of oscillation of the charge. working at Calcutta (now Kolkata).Physics along the loop and of the same value. that is not possible. The energy associated with the propagating wave comes at the expense of the energy of the source – the accelerated charge. while the frequency that we get even with modern electronic circuits is hardly about 1011 Hz. (An oscillating charge is an example of accelerating charge. This is why the experimental demonstration of electromagnetic wave had to come in the low frequency region (the radio wave region). yellow light. Hertz’s successful experimental test of Maxwell’s theory created a sensation and sparked off other important works in this field. The oscillating electric and magnetic fields thus regenerate each other. is a source of oscillating electric field. Consider a charge oscillating with some frequency. while the latter produces magnetic fields that. as the wave propagates through the space.3 ELECTROMAGNETIC WAVES 8. The former produces only electrostatic fields.1 Sources of electromagnetic waves How are electromagnetic waves produced? Neither stationary charges nor charges in uniform motion (steady currents) can be sources of electromagnetic waves. say.1 id = ε0 dΦE 1 dq = = 0. and so on. which produces an oscillating magnetic field. qualitative reasoning. Now. Two important achievements in this connection deserve mention. however. so to speak. The frequency of yellow light is about 6 × 1014 Hz. It is an important result of Maxwell’s theory that accelerated charges radiate electromagnetic waves.3. which in turn.) This produces an oscillating electric field in space. The proof of this basic result is beyond the scope of this book. it might appear easy to test the prediction that light is an electromagnetic wave. We might think that all we needed to do was to set up an ac circuit in which the current oscillate at the frequency of visible light. Jagdish Chandra Bose. . we get ⎛ 1⎞ B × 2π × ⎜ ⎟ = μ0 (i c + i d ) = μ0 ( 0 + i d ) = 0. From the preceding discussion. The flux ΦE through this loop is ΦE = E × area of the loop πE q ⎛ 1⎞ = = = E ×π×⎜ ⎝ 2⎟ ⎠ 4 4ε 0 2 The displacement current o n 274 d e h s T i l R b E u C p N re © e b o t t EXAMPLE 8. But. B = 0. applying Ampere-Maxwell law to the loop.5×10–6 µ exp(–1) 0 ⎝ 2⎠ or. alas. Seven years after Hertz. do not vary with time.5 × 10 –6 exp ( –1) 4 dt dt at t = 10–3s. but we can accept it on the basis of rough. His experiment.7(b)] By= B0 sin (kz–ωt ) Here k is related to the wave length λ of the wave by the usual equation (8. 8. and measured their wavelength and speed. reflection and refraction was the same as that of light and heat waves. at a given time t). [8. 8. It appears reasonable. establishing their identity for the first time.4. FIGURE 8. He also pioneered research on discharge of electricity through gases. So B and E are perpendicular in this case. The speed of propagation of the wave is ( ω/ k ). one finds that k= 2π o n d e h s T i l R b E u C p N re © e b o t t HEINRICH RUDOLF HERTZ (1857–1894) Heinrich Rudolf Hertz (1857 – 1894) German physicist who was the first to broadcast and receive radio waves. At around the same time. The electric field inside the plates of the capacitor is directed perpendicular to the plates.4 A linearly polarised electromagnetic wave. and discovered the photoelectric effect. Consider Fig.2 Nature of electromagnetic waves It can be shown from Maxwell’s equations that electric and magnetic fields in an electromagnetic wave are perpendicular to each other. Guglielmo Marconi in Italy followed Hertz’s work and succeeded in transmitting electromagnetic waves over distances of many kilometres. we show a typical example of a plane electromagnetic wave propagating along the z direction (the fields are shown as a function of the z coordinate. at a given time.3. The electric and magnetic fields Ex and By are perpendicular to each other. sent them through space. The magnetic field this gives rise to via the displacement current is along the perimeter of a circle parallel to the capacitor plates. k is the magnitude of the wave vector (or propagation vector) k and its direction describes the direction of propagation of the wave.8) λ and ω is the angular frequency. like that of Hertz’s.Electromagnetic Waves succeeded in producing and observing electromagnetic waves of much shorter wavelength (25 mm to 5 mm). and to the direction of propagation. propagating in the z-direction with the oscillating electric field E along the x-direction and the oscillating magnetic field B along the y-direction. 275 .7(a)] [8. The magnetic field By is along the y-axis. and to the direction z of propagation. say from our discussion of the displacement current. 8. He showed that the nature of their vibration.7(a) and (b)] for Ex and By and Maxwell’s equations. This is a general feature. Using Eqs.2. and again varies sinusoidally with z. Marconi’s experiment marks the beginning of the field of communication using electromagnetic waves. was confined to the laboratory. We can write Ex and By as follows: Ex= E0 sin (kz–ωt ) [8. In Fig. He produced electromagnetic waves. and varies sinusoidally with z. The electric field Ex is along the x-axis. 4 of class XI Physics textbook).hawaii. This relation is often written in terms of frequency. the velocity of light becomes. λ (=2π/k) as ⎛ 2π ⎞ 2πν = c ⎜ ⎟ or ⎝ λ ⎠ νλ = c [8.html (ii) http://www. These replace ε0 and µ0 in the description to electric and magnetic fields in Maxwell’s equations with the result that in a material medium of permittivity ε and magnetic permeability µ.Physics ω = ck. Section 15. The velocity of electromagnetic waves in free space or vacuum is an important fundamental constant.com/waves. They differ from all the other waves we have studied so far.11) με Thus. They called this medium ether. where. the velocity of light depends on electric and magnetic properties of the medium. Electric and magnetic fields.edu/~teb/java/ntnujava/emWave/emWave. that there is even a novel called The Poison Belt by Sir Arthur Conan Doyle (the creator of the famous detective Sherlock Holmes) where the solar system is supposed to pass through a poisonous region of ether! We now accept that no such physical medium is needed. We have seen earlier that the total electric and magnetic fields inside a medium are described in terms of a permittivity ε and a magnetic permeability µ (these describe the factors by which the total fields differ from the external fields). It has been shown by experiments on electromagnetic waves of different wavelengths that this velocity is the 276 Simulate propagation of electromagnetic waves (i) http://www. The famous experiment of Michelson and Morley in 1887 demolished conclusively the hypothesis of ether.html The relation ω = ck is the standard one for waves (see for example. 1 v= (8. for example.9(a)] o n d e h s T i l R b E u C p N re © e b o t t B0 = (E0/c) (8. an electromagnetic wave.9(b)] It is also seen from Maxwell’s equations that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as . can sustain each other in vacuum. They were so convinced of the reality of this medium.phys. Transverse waves on the surface of water consist of water moving up and down as the wave spreads horizontally and radially onwards. or vacuum. which is rigid and that resists shear. ν (=ω/2π) and wavelength. does propagate through glass. Transverse elastic (sound) waves can also propagate in a solid. oscillating in space and time. c = 1/ μ0 ε 0 [8. They are self-sustaining oscillations of electric and magnetic fields in free space.10) We here make remarks on some features of electromagnetic waves.amanogawa. But what if a material medium is actually there? We know that light. which responds to electric and magnetic fields just as any elastic medium does. in respect that no material medium is involved in the vibrations of the electric and magnetic fields. We shall see in the next chapter that the refractive index of one medium with respect to the other is equal to the ratio of velocities of light in the two media. Sound waves in air are longitudinal waves of compression and rarefaction. Scientists in the nineteenth century were so much used to this mechanical picture that they thought that there must be some medium pervading all space and all matter. i. the metre is now defined as the distance travelled by light in vacuum in a time (1/c) seconds = (2. there is an energy density (ε0E2/2). 8. it responds to the electric part of the signal.. it can be shown that the magnitude of the total momentum delivered to this surface (for complete absorption) is. The fact that electromagnetic waves are polarised can be easily seen in the response of a portable AM radio to a broadcasting station.e. Since it carries momentum. o n d e h s T i l R b E u C p N re © e b o t t (8.Electromagnetic Waves same (independent of wavelength) to within a few metres per second. Namely. response also depends on the orientation of the radio with respect to the station.99792458 × 108)–1 seconds. Thus. Similarly. which are sensitive to the magnetic component of the electromagnetic wave. etc. Since the frequency of the wave was known (being equal to the frequency of the oscillator).9979246 × 108 m/s. unit of length can be defined in terms of c and the unit of time! Hertz not only showed the existence of electromagnetic waves. as seen in Chapter 6. he obtained the speed of the wave using the formula v = νλ and found that the waves travelled with the same speed as the speed of light. refracted and polarised. the signal will be greatly diminished. The basic unit of time can be defined very accurately in terms of some atomic frequency. This has come about for the following reason. Since c is such a strongly fixed number. When the antenna is turned horizontal. there is a non-zero energy density associated with it. In such cases. called radiation pressure. on this plane. If the total energy transferred to a surface in time t is U. Now consider a plane perpendicular to the direction of propagation of the electromagnetic wave (Fig. Further. Do electromagnetic waves carry energy and momentum like other waves? Yes. they do. As electromagnetic wave contains both electric and magnetic fields.) converged to a value of about 2. he conclusively established the wave nature of the radiation. The basic unit of length is harder to define as accurately in a direct way. which had wavelength ten million times that of the light waves. he produced stationary electromagnetic waves and determined their wavelength by measuring the distance between two successive nodes. The constancy of the velocity of em waves in vacuum is so strongly supported by experiments and the actual value is so well known now that this is used to define a standard of length. Earlier measurement of c using earlier units of length (metre rods. electric charges. out of a value of 3×108 m/s.4). but also demonstrated that the waves. If there are. an electromagnetic wave also exerts pressure. could be diffracted. If an AM radio has a telescopic antenna.12) p= U c 277 . We have seen in chapter 2 that in a region of free space with electric field E. This just illustrates the fact that an electromagnetic wave (like other waves) carries energy and momentum. Such a radio must remain horizontal in order to receive the signal. frequency of light emitted by an atom in a particular process. associated with a magnetic field B is a magnetic energy density (B2/2µ0). The charges thus acquire energy and momentum from the waves. they will be set and sustained in motion by the electric and magnetic fields of the electromagnetic wave. Some portable radios have horizontal antenna (usually inside the case of radio). The great technological importance of electromagnetic waves stems from their capability to carry energy from one place to another. E = 6. the magnitude of B is o n 278 d e h s T i l R b E u C p N re © e b o t t B= E XAMPLE 8. on a surface of area 10 cm2. Using vector algebra. Thus. the force due to radiation is only about 7 × 10–9 N. B = 2. the American scientists Nicols and Hull succeeded in measuring radiation pressure of visible light and verified Eq. Therefore.5 × 103x + 1. At a particular j V/m.10).5 × 103 1 = ν = 1.1 × 10–8 k Example 8.5 × 1011 t ) V/m E XAMPLE 8. Since.3 V/m = = 2. What is B at this point? point in space and time. λ = and 2π m = 1. The radio and TV signals from broadcasting stations carry energy.2 E c 6.3 ˆ Solution Using Eq. In 1903. thus making life possible on the earth. 0. the electric field component along the z-axis is obtained as Ez = 60 sin (0.3 The magnetic field in a plane electromagnetic wave is given by By = 2 × 10–7 sin (0. we note that E is along y-direction and the wave propagates along x-axis.3 ( ) . you feel the energy being absorbed from the electromagnetic waves (your hands get warm).9 GHz T (b) E0 = B0c = 2×10–7 T × 3 × 108 m/s = 6 × 101 V/m The electric field component is perpendicular to the direction of propagation and the direction of magnetic field. (+ ˆ ˆT Thus. B should be in a direction perpendicular to both x. It was found to be of the order of 7 × 10–6 N/m2.12).and y-axes. (8. E × B should ˆ)= ˆ j ) × (+ k i . be along x-direction. Light carries energy from the sun to the earth. Therefore. Solution (a) Comparing the given equation with ⎡ ⎛ x t ⎞⎤ + ⎟⎥ By = B0 sin ⎢2π ⎜ ⎣ ⎝ λ T ⎠⎦ We get.Physics When the sun shines on your hand. Example 8.2 A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. the amount of momentum transferred is extremely small and you do not feel the pressure.26 cm. (8. (a) What is the wavelength and frequency of the wave? (b) Write an expression for the electric field.5×1011t ) T.5 × 1011 /2π = 23.5×103x+1. Electromagnetic waves also transfer momentum to your hand but because c is very large. B is along the z-direction.1 × 10 –8 T 3 × 108 m/s To find the direction. Since the electric field in a light beam is sinusoidal. Solution The bulb.16 × 10 −3 = = 1.5 E0 = 279 . find the average force exerted on the surface during a 30 minute time span. the total momentum delivered (for complete absorption) is U 6.16 × 10–3 kg m/s 3 × 108 m/s The average force exerted on the surface is p 2.07 V/m Thus. At a distance of 3 m. which is of the order of a few microvolts per metre. Compare it with electric field strength of TV or FM waves.4 Light with an energy flux of 18 W/cm2 falls on a nonreflecting surface at normal incidence. you see that the electric field strength of the light that you use for reading is fairly large. as a point source.18 × 104 How will your result be modified if the surface is a perfect reflector? F= Example 8. Assume that the efficiency of the bulb is 2.9 V/m The value of E found above is the root mean square value of the electric field. If the surface has an area of 20 cm2.48 × 105 J Therefore. radiates light in all directions uniformly.48 × 105 J p= c = = 2. E0 is 2E rms = 2 × 2.Electromagnetic Waves Example 8.2 × 10 −6 N t 0.022 W/m2 Half of this intensity is provided by the electric field and half by the magnetic field.5 Calculate the electric and magnetic fields produced by the radiation coming from a 100 W bulb at a distance of 3 m.5% and it is a point source. Solution The total energy falling on the surface is U = (18 W/cm2) × (20 cm2) × (30 × 60) = 6.9 V/m = 4.022 V/m EXAMPLE 8. the peak electric field.85 × 10 )(3 × 10 ) −12 8 d e h s T i l R b E u C p N re © e b o t t EXAMPLE 8.4 ) 0.022 W/m 2 = 2 Erms = ( ) = 2. 1 1 2 I = ε 0 Erms c 2 2 1 0. the surface area of the surrounding sphere is A = 4 π r 2 = 4π (3)2 = 113 m2 The intensity at this distance is I = Power 100 W × 2. o n ( (8.5 % = Area 113 m 2 = 0. It is o n 280 d e h s T i l R b E u C p N re © e b o t t 8. The existence of ultraviolet and infrared waves was barely established.4 ELECTROMAGNETIC SPECTRUM At the time Maxwell predicted the existence of electromagnetic waves.5 Erms 2. the magnetic field strength is evidently very weak. Electromagnetic spectrum http://www. The various regions do not have sharply defined boundaries. microwaves.fnal. X-rays. FIGURE 8.gov/docs/science/ EXAMPLE 8.gsfc. The classification is based roughly on how the waves are produced and/or detected. X-rays and gamma rays had also been discovered. radio waves. the only familiar electromagnetic waves were the visible light waves.6 × 10–9 T c 3 × 108 m s −1 Again.5 The electromagnetic spectrum.5). The classification of em waves according to frequency is the electromagnetic spectrum (Fig. let us calculate the strength of the magnetic field. 8.Physics Now. We now know that.gov/pub/inquiring/more/light http://imagine. with common names for various part of it. gamma rays. electromagnetic waves include visible light waves.4 × 10–8 T.9 V m −1 = = 9. the peak magnetic field is B0 = 2 Brms = 1. By the end of the nineteenth century.nasa. Note that although the energy in the magnetic field is equal to the energy in the electric field. Brms = . since the field in the light beam is sinusoidal. There is no sharp division between one kind of wave and the next. ultraviolet and infrared waves. they are suitable for the radar systems used in aircraft navigation.4. on the other hand. the energy of the random motion of atoms and molecules increases and the molecules travel or vibrate or rotate with higher energies. Due to their short wavelengths. In the microwave oven. One should use porcelain vessels and not metal containers in a microwave oven because of the danger of getting a shock from accumulated electric charges. Cellular phones use radio waves to transmit voice communication in the ultrahigh frequency (UHF) band. they do not get heated up. Thus. and automobiles. Higher frequencies upto 54 MHz are used for short wave bands. in order of decreasing wavelengths. All food items such as fruit. Now. heating up the food.Electromagnetic Waves We briefly describe these different types of electromagnetic waves. its molecules absorb this radiation. the vessel on the burner gets heated first. Radar also provides the basis for the speed guns used to time fast balls. This way energy is not wasted in heating up the vessel.1 Radio waves Radio waves are produced by the accelerated motion of charges in conducting wires. 8. which is 3 gigahertz (GHz). This raises the temperature of any food containing water. What is the principle of a microwave oven and how does it work? Our objective is to cook food or warm it up.2 Microwaves Microwaves (short-wavelength radio waves). These molecules share this energy with neighbouring food molecules. energy is directly delivered to water molecules which is shared by the entire food. vegetables. The spectrum of electromagnetic radiation contains a part known as microwaves. The AM (amplitude modulated) band is from 530 kHz to 1710 kHz. tennisserves. and then the food inside gets heated because of transfer of energy from the vessel. o n d e h s T i l R b E u C p N re © e b o t t MICROWAVE OVEN 281 . They are used in radio and television communication systems. contain water as a constituent. cereals. In the conventional heating method.. They are generally in the frequency range from 500 kHz to about 1000 MHz. The FM (frequency modulated) radio band extends from 88 MHz to 108 MHz. How these waves are transmitted and received is described in Chapter 15. are produced by special vacuum tubes (called klystrons. In such ovens. The frequency of rotation of water molecules is about 300 crore hertz. etc. magnetrons and Gunn diodes). the frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves is transferred efficiently to the kinetic energy of the molecules. what does it mean when we say that a certain object has become warmer? When the temperature of a body rises. and thus cannot absorb microwaves. the basic principle of a microwave oven is to generate microwave radiation of appropriate frequency in the working space of the oven where we keep food. Hence.4. meat. Metals may also melt from heating. The porcelain container remains unaffected and cool. These waves have frequency and energy smaller than visible light and wavelength larger than it. because its large molecules vibrate and rotate with much smaller frequencies. which is equivalent to heating up water. If water receives microwaves of this frequency. 8. TV waves range from 54 MHz to 890 MHz. Microwave ovens are an interesting domestic application of these waves. with frequencies in the gigahertz (GHz) range. This radiation is trapped by greenhouse gases such as carbon dioxide and water vapour. Ultraviolet (UV) radiation is produced by special lamps and very hot bodies. For example. Ozone layer in the atmosphere plays a protective role. Different animals are sensitive to different range of wavelengths. UV lamps are used to kill germs in water purifiers. 8. Our eyes are sensitive to this range of wavelengths. But fortunately. This is because water molecules present in most materials readily absorb infrared waves (many other molecules. Hence. This band lies adjacent to the low-frequency or long-wave length end of the visible spectrum. for example.4. Due to its shorter wavelengths. also absorb infrared waves). UV light in large quantities has harmful effects on humans. that is. one cannot get tanned or sunburn through glass windows. Visible light emitted or reflected from objects around us provides us information about the world. Infrared waves are sometimes referred to as heat waves.3 Infrared waves Infrared waves are produced by hot bodies and molecules. It is the part of the spectrum that is detected by the human eye. NH3.4. Exposure to UV radiation induces the production of more melanin. snakes can detect infrared waves. After absorption. Welders wear special glass goggles or face masks with glass windows to protect their eyes from large amount of UV produced by welding arcs. Infrared detectors are used in Earth satellites. CO2. and the ‘visible’ range of many insects extends well into the utraviolet. they heat up and heat their surroundings. UV radiation is absorbed by ordinary glass.4. most of it is absorbed in the ozone layer in the atmosphere at an altitude of about 40 – 50 km. Infrared radiation also plays an important role in maintaining the earth’s warmth or average temperature through the greenhouse effect. o n 282 d e h s T i l R b E u C p N re © e b o t t 8. The sun is an important source of ultraviolet light. video recorders and hi-fi systems.4 Visible rays It is the most familiar form of electromagnetic waves.Physics 8. UV radiations can be focussed into very narrow beams for high precision applications such as LASIK (Laserassisted in situ keratomileusis) eye surgery. their thermal motion increases. causing tanning of the skin. both for military purposes and to observe growth of crops. Electronic devices (for example semiconductor light emitting diodes) also emit infrared and are widely used in the remote switches of household electronic systems such as TV sets. . and hence its depletion by chlorofluorocarbons (CFCs) gas (such as freon) is a matter of international concern. Incoming visible light (which passes relatively easily through the atmosphere) is absorbed by the earth’s surface and reradiated as infrared (longer wavelength) radiations.6 nm). It runs from about 4 × 1014 Hz to about 7 × 1014 Hz or a wavelength range of about 700 – 400 nm. Infrared lamps are used in physical therapy.5 Ultraviolet rays It covers wavelengths ranging from about 4 × 10–7 m (400 nm) down to 6 × 10–10m (0. the demarcation between different region is not sharp and there are over laps. Infrared photographic film The eye Photocells Photographic film 700 nm to 400 nm Electrons in atoms emit light when they move from one energy level to a lower energy level 400 nm to 1nm Inner shell electrons in atoms moving from one energy level to a lower level X-ray tubes or inner shell electrons Radioactive decay of the nucleus Photocells Photographic film 1nm to 10–3 nm Photographic film Geiger tubes Ionisation chamber -do<10–3 nm Gamma rays 283 . They are used in medicine to destroy cancer cells. As mentioned earlier. It covers wavelengths from about 10 –8 m (10 nm) down to 10 –13 m (10–4 nm). X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer. We are familiar with X-rays because of its medical applications.1 summarises different types of electromagnetic waves.4. Type Radio Microwave Infra-red Light Ultraviolet X-rays o n d e h s T i l R b E u C p N re © e b o t t TABLE 8. One common way to generate X-rays is to bombard a metal target by high energy electrons.1 m Rapid acceleration and decelerations of electrons in aerials Klystron valve or magnetron valve Receiver’s aerials 0.1m to 1 mm Point contact diodes 1mm to 700 nm Vibration of atoms and molecules Thermopiles Bolometer.1 DIFFERENT TYPES OF ELECTROMAGNETIC WAVES Production Wavelength range Detection > 0.7 Gamma rays They lie in the upper frequency range of the electromagnetic spectrum and have wavelengths of from about 10–10m to less than 10–14m. Table 8.4. This high frequency radiation is produced in nuclear reactions and also emitted by radioactive nuclei. 8.6 X-rays Beyond the UV region of the electromagnetic spectrum lies the X-ray region. Because X-rays damage or destroy living tissues and organisms. care must be taken to avoid unnecessary or over exposure. their production and detections.Electromagnetic Waves 8. Electric and magnetic fields oscillate sinusoidally in space and time in an electromagnetic wave. total momentum delivered to this surface is p = U/c. we have E = Ex (t) = E0 sin (kz – ω t ) id = ε0 dΦΕ dt ⎡ ⎛z ⎡ ⎛ z t ⎞⎤ ⎞⎤ − νt ⎟ ⎥ = E 0 sin ⎢2π ⎜ − ⎟ ⎥ ⎢ 2π ⎜ ⎝ ⎠ λ ⎣ ⎦ ⎣ ⎝ λ T ⎠⎦ B = By(t) = B0 sin (kz – ω t) = E0 sin = B0 sin ⎢2π ⎜ ⎝ ⎡ ⎣ ⎛z ⎡ ⎛ z t ⎞⎤ ⎞⎤ − νt ⎟ ⎥ = B0 sin ⎢2π ⎜ − ⎟ ⎥ ⎠ λ ⎦ ⎣ ⎝ λ T ⎠⎦ 5. The spectrum of electromagnetic waves stretches. An electric charge oscillating harmonically with frequency ν. 3. He thus verified a basic prediction of Maxwell’s equations. Different regions are known by different . in principle. 4. produces electromagnetic waves of the same frequency ν . An accelerating charge produces electromagnetic waves. where µ is the permeability of the medium and ε its permittivity. The speed c of electromagnetic wave in vacuum is related to µ0 and ε0 (the free space permeability and permittivity constants) as follows: c = 1/ μ 0 ε 0 . Light is an electromagnetic wave. and to the direction of propagation of the electromagnetic wave. For a wave of frequency ν. The value of c equals the speed of light obtained from optical measurements. Maxwell found an inconsistency in the Ampere’s law and suggested the existence of an additional current. Electromagnetic waves transport momentum as well. called displacement current. When these waves strike a surface. They are related by E0/B0 = c. E and B are perpendicular to each other. This displacement current is due to time-varying electric field and is given by o n 284 d e h s T i l R b E u C p N re © e b o t t 2. Electromagnetic waves other than light also have the same velocity c in free space. a pressure is exerted on the surface. and acts as a source of magnetic field in exactly the same way as conduction current. over an infinite range of wavelengths. 7. or of electromagnetic waves in a material medium is given by v = 1/ μ ε 6. therefore. An electric dipole is a basic source of electromagnetic waves. Electromagnetic waves carry energy as they travel through space and this energy is shared equally by the electric and magnetic fields. to remove this inconsistency. also the speed of light. If total energy transferred to a surface in time t is U. The speed of light.Physics SUMMARY 1. c is. propagating along z-direction. The oscillating electric and magnetic fields. Electromagnetic waves with wavelength of the order of a few metres were first produced and detected in the laboratory by Hertz in 1887. wavelength λ. This is why infrared waves are often called heat waves.. Visible radiation emitted by atoms is. the waves differ considerably in their mode of interaction with matter. The charging current is constant and equal to 0. Therefore. o n d e h s T i l R b E u C p N re © e b o t t POINTS TO PONDER EXERCISES 8. gamma radiation. (a) Calculate the capacitance and the rate of charge of potential difference between the plates. having wavelength of 10–14 m to 10–15 m. with frequencies lower than those of visible light. however. Infrared waves. The same basic principle is utilised in practically all modern receiving devices.Electromagnetic Waves names. vibrate not only the electrons. 1. microwaves and radio waves in order of increasing wavelength from 10–2 Å or 10–12 m to 106 m. It is because humans have evolved with visions most sensitive to the strongest wavelengths from the sun. 5.6 shows a capacitor made of two circular plates each of radius 12 cm.15A. 2. much longer in wavelength than atomic size. They interact with matter via their electric and magnetic fields which set in oscillation charges present in all matter. typically originate from an atomic nucleus. High frequency electromagnetic waves are detected by other means based on the physical effects they produce on interacting with matter. Consequently. X-rays. infrared rays. ultraviolet rays. scattering. The oscillating fields of an electromagnetic wave can accelerate charges and can produce oscillating currents. Radio waves are produced by accelerating electrons in a circuit. visible rays. Hertz original ‘receiver’ worked in exactly this way. The basic difference between various types of electromagnetic waves lies in their wavelengths or frequencies since all of them travel through vacuum with the same speed. The capacitor is being charged by an external source (not shown in the figure). an apparatus designed to detect electromagnetic waves is based on this fact. the temperature of the substance. but entire atoms or molecules of a substance.1 Figure 8. 3. The centre of sensitivity of our eyes coincides with the centre of the wavelength distribution of the sun. Thus. The detailed interaction and so the mechanism of absorption. The wavelength of the electromagnetic wave is often correlated with the characteristic size of the system that radiates. depend on the wavelength of the electromagnetic wave. X-rays are emitted from heavy atoms. and the nature of the atoms and molecules in the medium. This vibration increases the internal energy and consequently. 4. etc. and separated by 5. 285 . γ-rays. A transmitting antenna can most efficiently radiate waves having a wavelength of about the same size as the antenna. Accelerated charged particles radiate electromagnetic waves.0 cm. 0 cm from the axis between the plates.0 MHz.10 What physical quantity is the same for X-rays of wavelength 10–10 m.5 MHz to 12 MHz band.Physics (b) Obtain the displacement current across the plates. The terminology of different parts of the electromagnetic spectrum is given in the text. and λ. (a) Determine.7 ) made of circular plates each of radius R = 6. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz. What is the frequency of the electromagnetic waves produced by the oscillator? The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B 0 = 510 nT. red light of wavelength 6800 Å and radiowaves of wavelength 500m? A plane electromagnetic wave travels in vacuum along z-direction.6 8.6 8.4 8. FIGURE 8.9 8. o n 286 d e h s T i l R b E u C p N re © e b o t t FIGURE 8. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? In a plane electromagnetic wave. (b) Find expressions for E and B. (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain. (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3. what is its wavelength? A radio can tune in to any station in the 7.ω. Use the formula E = hν (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum.3 8.0 × 1010 Hz and amplitude 48 V m–1. What is the amplitude of the electric field part of the wave? Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is ν = 50.7 8. k.0 cm has a capacitance C = 100 pF.2 A parallel plate capacitor (Fig. What is the corresponding wavelength band? A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz.5 8. B0. the electric field oscillates sinusoidally at a frequency of 2. . 8.7 8. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s–1.8 8. 1 N/C) cos [(1.15 o n d e h s T i l R b E u C p N re © e b o t t 287 . Use the formula λ m T = 0.12 8. About 5% of the power of a 100 W light bulb is converted to visible radiation. Why? (d) The small ozone layer on top of the stratosphere is crucial for human survival. What is the average intensity of visible radiation (a) at a distance of 1m from the bulb? (b) at a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection. Why? (b) It is necessary to use satellites for long distance TV transmission. (d) 5890 Å . (a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).Electromagnetic Waves (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. known as Lamb shift).4 keV [energy of a particular transition in 57 Fe nucleus associated with a famous high resolution spectroscopic method (Mössbauer spectroscopy)].29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. State the part of the electromagnetic spectrum to which each belongs.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe]. Why? (c) Optical and radiotelescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth.5896 Å [double lines of sodium] (e) 14. (b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen. (c) 2.14 8. Why? (e) If the earth did not have an atmosphere.13 8. What do the numbers that you obtain tell you? Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics.4 × 106 rad/s)t]} ˆ i.11 Suppose that the electric field part of an electromagnetic wave in vacuum is E = {(3. [c = 3 × 108 m s–1. What might be the basis of this prediction? 8. Answer the following questions: (a) Long distance radio broadcasts use short-wave bands. (a) What is the direction of propagation? (b) What is the wavelength λ ? (c) What is the frequency ν ? (d) What is the amplitude of the magnetic field part of the wave? (e) Write an expression for the magnetic field part of the wave. would its average surface temperature be higher or lower than what it is now? (f ) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth.] ADDITIONAL EXERCISES 8.8 rad/m) y + (5. describe the basic aspects of observational skills. Sensitivity to Diversity : Individual and Cultural Differences (Box 9. and explain the nature and process of counselling.1) Observational Skills Specific Skills Communication Skills Characteristics of Communication (Box 9.3) Psychological Testing Skills Essentials of Psychological Assessment Skills (Box 9. you would be able to: understand the need to develop skills among psychologists. understand the importance of psychological testing skills in individual assessment.4) Interviewing Skills Types of Interview Questions (Box 9.DEVELOPING PSYCHOLOGICAL SKILLS After reading this chapter.2) Some Tips to Improve Your Listening Skills (Box 9.5) Key Terms Counselling Skills Summary Review Questions Project Ideas Weblinks Pedagogical Hints T i l R b E u C p N re © e b o t t h s d e . CONTENTS o n Introduction Developing as an Effective Psychologist General Skills Intellectual and Personal Skills. know the significance of developing communication skills. etc. assessment of her/his personality and also on other important . It is generally assumed that psychologists have interest in people. Over the past few years. psychologists have found. Almost all areas of life are calling for help from psychologists.Introduction One aspect common to all the applied areas within psychology is a universal agreement on the basic assumptions about human nature and the role of a psychologist in different settings. Various sections of society have developed a keen interest towards integrating the knowledge of psychology within their core activities. in management of health-related problems and more recently even in sports. what kind of personality make-up is required to become a successful human being? Or what would be the right vocation for a student studying in Class XII ? Similarly. For instance. These are not the only areas. A psychologist from any 178 Psychology o n As you have already studied in the previous class. It would be important to learn that there are certain factors which contribute in making psychology application-oriented. and certain intervention models are developed to assist the youth experiencing such problems early in their lives.. As problems encountered by individuals as well as societies have become more evident and acute. employee behaviour. where in some states it has become mandatory for the schools to have a trained counsellor. as a scientific method-driven research discipline. Both of these are interrelated and inseparable. rapid increase in depressive tendencies and suicidal rates among adolescents is one such area. First. both in the past and present. Psychology has two kinds of application-related images: first. there are other questions and curiosities about human beings which psychologists are asked to answer. their abilities and temperaments. groups. where psychological knowledge dedicated to the issues of adolescent development is utilised for getting a clearer understanding of the occurrence of these phenomena. psychologists have responded with concrete solutions. that solutions to many problems faced by the individuals. as a service-oriented discipline and second. workplace stress. Psychological testing stands as a prime example of psychology in action. her/his socio-cultural environment. in the field of management. For example. Psychology touches on many questions related to our lives. One can find active involvement of a psychologist in obtaining the client’s history. T i l R b E u C p N re © e b o t t h s d e field is required to have interest in other people and exhibits a willingness to provide help by using her/his knowledge of the discipline. psychology began as an application-oriented discipline. The other reason is that expertise of psychologists has been highly valued in the marketplace. there is a growing acceptability and demand of psychology as a profession. which seeks help of psychologists to combat problems related to recruitment and assessment. organisations and societies require an understanding of psychological principles. It can be seen particularly in the army or in the field of education. sometimes even proves disastrous. Recognition and application of these skills and competencies have strengthened the foundation and practices of applied psychology in a positive way. There still remains a question o n T i l R b E u C p N re © e b o t t EFFECTIVE of how to differentiate between a pseudopsychologist from a real psychologist. another distinction should be made. institutional affiliation. a client may refer to an individual/group/organisation who on her/his own seeks help. These are: assessment of individual differences . S/he tries to find the answer in terms of mathematical probability. However. Generally people pick up such terms from popular writings and media. Some regularity in human behaviour is frequently observed by us to warrant generalisation. and broader community. We. They cover a range of knowledge that a psychologist should possess when entering the profession after completing their education and training. An answer can be constructed by asking such questions like professional training. We need both types of psychologists. This kind of everyday amateur psychology often misfires. and her/his experience in providing service. How can one develop into a professional psychologist? DEVELOPING AS PSYCHOLOGIST AN Most people think that they are some kind of psychologists. These apply to practitioners. their methods and theories are required to develop psychological expertise. It is important to know about the conditions and competencies that are necessary besides research skills for a psychologist. They take their problem to the laboratory or study it in field settings to answer various problems. The Webster dictionary defines it as “possession of the qualities required to do something or get something done”. There are a lot of common sense notions about human behaviour that one develops in the course of their lives. a professional psychologist addresses the problem at the scientific level. behaviour modification skills. In psychology. Some psychologists carry out r esearch to propound or investigate theoretical formulations while others are concerned with our daily life activities and behaviour. academicians. American Psychological Association (1973) in their task force constituted with the objective to identify skills essential for professional psychologists recommended at least three sets of skills. and counselling and guidance skills . For example. educational background. industry. mental blocks. at times. talk about intelligence.dimensions. There are conditions and competencies for psychologists which have come to be recognised internationally. It is now recognised that the knowledge of tools used by psychologists. It is recognised that it is difficult to develop. implement and h s d e Chapter 9 • Developing Psychological Skills 179 . identity crisis. One might think that client is a special term. facility or dexterity that is acquired or developed through training and experience. attitude. communication barriers and so many other terms. guidance or intervention from a psychologist with respect to any problem faced by her/him. what is critical is training as a researcher and internalisation of certain professional values. business. inferiority complex. Here. We need some scientists to develop theories and others to find solutions to human problems. The term ‘skill ’ may be defined as proficiency. stress. Only then does s/he arrive at psychological principles or laws that can be depended upon. which is mostly used in a clinical or counselling set-up. and researchers whose roles involve consulting with students. (b) Observational Skills. (a) General Skills. Ability to promote diversity in cultural beliefs and respecting it to promote positive life outcomes. The basic skills or competencies which psychologists have identified for becoming an effective psychologist fall into three broad sections. tolerance of ambiguity and uncertainty. behaviours and ability to be sensitive to one’s own behaviour or others. It is expected that it will not be proper to provide any form of professional training (in clinical or organisational fields) to students who do not possess these skills. feelings and infor mation in verbal.1 o n • • • • • • 180 1. 3. Personal Skills: personal or ganisation. Once a student has these skills. They use all the senses. to develop respect for/interest in others’ cultures. and consultation. Sensitivity to Diversity : Individual and Cultural Differences Knowledge of self (one’s own attitudes. time management. noticing h s d e Psychology . etc. 2. openness to new ideas. or are acting as consultants. points of view. These skills include personal as well as intellectual skills. Expressive Skills: ability to communicate one’s ideas. or in environmental settings. Cognitive Skills: ability to solve problems. values. personal hygiene. social. and related str engths/limitations) as one operates in the professional settings with diverse others. Personality/Attitude: desire to help others. attitudes. Ability to work ef fectively with diverse backgrounds in assessment. These are discussed in detail here. and appropriate dress.1. honesty/integrity/ value ethical behaviour. goals and desires. 6. 7. treatment. Knowledge about the nature and impact of individual and cultural diversity in different situations. whether they are working in the field of clinical and health psychology.measure competencies required in a subject like psychology as the criteria for specification. Ability to respect and appreciate different cultural nor ms and beliefs. Af fective Skills: emotional control and balance. Being sensitive to one’s preferences and also to one’s prefer ence for own group. These skills are expressed verbally and/or non-verbally. fears. T i l R b E u C p N re © e b o t t Intellectual and Personal Skills O BSERVATIONAL SKILLS A great deal of what psychologists as researchers and practitioners do in the field is to pay attention. tolerance/understanding of interpersonal conflict. namely. experiences. Box 9. engage in critical thinking and organised reasoning. Reflective Skills: ability to examine and consider one’s own motives. Interpersonal Skills: ability to listen and be empathic. identification and evaluation are not yet fully agreed upon. GENERAL SKILLS These skills are generic in nature and are needed by all psychologists irrespective of their field of specialisation. industrial/ organisational. 5. openness to receive feedback. These skills are essential for all professional psychologists. non-verbal. personal courage. 4. subsequent training in her/his area of specialisation would only r efine and further hone these skills required by a professional within her/his field of specialisation. and having intellectual curiosity and flexibility. watch and listen carefully. and (c) Specific Skills. Some examples of such skills are given in Box 9. values. educational. and written forms. Participant Observation is the variation of the method of naturalistic observation. focus more on developing observational skills this year. an observer may take a part-time job in a shopping mall showroom to become an insider in order to observe variations in the behaviour of customers. Naturalistic Observation is one of the primary ways of learning about the way people behave in a given setting. For this. This technique is widely used by anthropologists whose objective is to gain a firsthand perspective of a system from within which otherwise may not be readily available to an outsider. To begin with. and motivations. viz. etc. One disadvantage of it is that events being observed are subject to bias due 181 Chapter 9 • Developing Psychological Skills . Here the observer is actively involved in the process of observing by becoming an active member of the setting where the observation takes place. Let us now consider developing skills about them. can be trained to use it. gender. for the problem mentioned above. • Pay close attention to your physical surroundings — who. These small. For instance. etc. You are already familiar with two major approaches to observation. naturalistic observation and participant observation. or touched. A psychologist. S/he might look at the colour of the floor/ceiling. you want to learn how people behave in response to a heavy discount provided by a company while visiting a shopping mall. • Observe with an optimistic curiosity. and how. A psychologist records such details because something of significance may be revealed in the process of observation. etc. tasted. when. and irrelevant looking signals influence human behaviour. norms of people you are observing. is like an instrument that absorbs all sources of information from the environment. you have to respect privacy.). smelt. h s d e Advantages and Disadvantages of Observation Its major advantage is that it allows behaviour to be seen and studied in its natural setting. take o n T i l R b E u C p N re © e b o t t • • • care not to disclose any information to anyone. In addition to physical surroundings. which is why a psychologist notes such signals in the surroundings. what. • Be yourself. size of the window/doors. People from outside. ways of dealing and relating with others. We will. if asked. you could visit the shopping mall where the discounted items are showcased and systematically observe what people do and say before and after the purchases have been made. give information about yourself. race. a psychologist may begin with carefully scrutinising the physical setting in order to capture its “atmosphere”. emotions. subtle. Suppose. a psychologist actively engages in observing people and their actions. or those already working in a setting. type of lighting. stature. The following points are taken into consideration while making an observation: • Observe patiently. heard . A psychologist engages in observing various facets of surroundings including people and varying events. • Ask questions that can be answered while observing. You have already studied about observation in Class XI. thus. artefacts/paintings/sculptures. pattern of behaviours in the presence of others. therefore. This may include the demographic features (age. • Be aware of people’s reactions. and • Be ethical. where. Making comparison of this kind may provide you with useful insights into what is going on.what is seen. For this. Another potential pitfall is that the actual behaviour and responses of others may get influenced by the p resence of the observer. psychologists working in clinical settings need to be trained in various techniques of therapeutic interventions. behavioural skills to bring about individual. Have each member evaluated by other members in the group as best as you can on the dimensions given above. organisational psychologists working in the organisational context need to have 182 Psychology o n T i l R b E u C p N re © e b o t t Communication Skills skills in assessment. SPECIFIC SKILLS These skills are core/basic to the field of psychological service. and counselling. Activity 9. Let us understand the basics of communication process and see what role it plays in fostering relationships and personal effectiveness. Similarly. Though.• • to the feelings of the people involved as well as of the observers. Relevant specific skills and competencies can be classified as follows: (a) Communication Skills • Speaking • Active listening • Body language or non-verbal skills (b) Psychological Testing Skills (c) Interviewing Skills (d) Counselling Skills • Empathy • Positive regard • Authenticity h s d e The skills we are going to discuss may appear abstract. It is one of the most important skills you will need to succeed in life. They are not exclusive to an area. Learning how to be an effective communicator is not just an academic exercise. You will. distinguish facts from opinions. however. thus. etc.1 Evaluating Your Interpersonal Skills Write down on a piece of paper how you would describe yourself on the following aspects : (a) Friendliness (b) Mood (c) Sense of humour (d) Career motivation (e) Ability to work in teams and groups (f) Independence (g) Desire to be accepted by others (h) Discipline Now form groups of three to five members. group. nonetheless. For example. Each member concludes by analysing the similarities and dif ferences between their selfperceptions on these dimensions and how they ar e per ceived by other members of the gr oups. understand them better when you engage in exercises related to them. specific skills and competencies are required for a very specialised professional functioning. For example. facilitation and consultation. Generally day-to-day activities in a given setting are fairly routine. You will be required to have good presentation . to do well you should be able to ask and answer questions. you will also need listening skills in order to comprehend the information presented in class and what others say verbally or non-verbally. all skill sets do overlap quite a bit. which can go unnoticed by the observer. summarise opinions. defeating the very purpose of observation. Your success in this class may well depend on your ability to communicate. psychological assessment. and interact fruitfully with your peers and teachers. team and organisational development besides research skills. It can be said that communication is a conscious or unconscious.e. we cannot erase it. received. and focusing on self. from the wide range of repertory of words. For example. and we react to our own speech and actions. we communicate selectively. The process of communication can be accidental (having no intent). o n T i l R b E u C p N re © e b o t t interview and small group discussions. we encode (i. intentional or unintentional process in which feelings and ideas are expressed as verbal and/or non-verbal messages that are sent. Communication is interactive because we are constantly in contact with other people and with ourselves. take ideas. Communication is irreversible because once we send a message we cannot take it back. Others react to our speech and actions.Characteristics of Communication Communication is dynamic because the process is constantly in a state of change. or engaged in an emotional outburst. or indirect. The message is sent to someone who receives it using her or his primary signal system. available to us. Components of Human Communication h s d e When we communicate. and send the idea through a channel. smelling. As the expectations. Box 9. interpersonal. These are examples of verbal communication which express how you 183 Chapter 9 • Developing Psychological Skills .e. Our apologies or denials can make it light but cannot stamp out what was communicated. expressive (resulting from the emotional state of the person). When we communicate. So. It encompasses such activities as thought processes. attitudes.e. Forms of interpersonal communication include faceto-face or mediated conversations. hearing.2.. we choose that which we believe is best suited for the idea we wish to express. and then react to those r eactions. such as face-to-face messages delivered by the speaker to an audience. It is composed of our primary signal system based on our senses (i. etc. and public levels. Once we have made a slip of tongue. Our brain r emains active. Intrapersonal communication involves communicating with yourself. and touching). a cycle of action and reaction is the basis of our communication. feelings. Thus. what do we mean by communication process. given a meaningful glance. The characteristics of communication are outlined in Box 9. and comprehended . Human communication occurs on the intrapersonal. Public communication is characterised by a speaker sending a message to an audience. such as message relayed over radio or television. and emotions of the persons who ar e communicating change.. give them meaning and put them into message forms). Interpersonal communication refers to the communication that takes place between two or more persons who establish a communicative relationship. the nature of their communication also changes. S/he decodes (i. seeing. translates message into understandable forms). you may say that you heard a bell or an object feels soft. whether we are asleep or awake we are always processing ideas or thoughts.. Communication is continuous because it never stops. It may be direct. or rhetorical (resulting from the specific goal of the communicator). actions. tasting. That is.2 skills to give briefings or to present reports on projects that are part of classroom assignments. personal decision making. Your friend received the message using her sensory agent (ears) and decoded it (understood that you want a pen). You have just participated in an effective communication transaction. your fingers pull away quickly.1 shows the process of communication involving different stages. You (communicator A) encoded a message (you need a pen) and used a channel for transmitting it (vocal chords producing sound waves) to your friend (communicator o n 184 T i l R b E u C p N re © e b o t t Speaking Fig. To be effective. “Sorry. the degree to which the communication is effective depends on the communicators’ mutual understanding of the signals or codes being used in transmitting a message and in receiving. The communication would have been still effective if she had said.1 : Basic Communication Process B). a communicator must know how to use language appropriately. As you can see that in any communication process. Suppose you are about to take an examination and suddenly realise that you have not brought your pen to class. You can also respond at a nonverbal level. You touch a hot stove. the utterance of sound. Communication is broader. Speech is only a biological act. because I am carrying only one pen. You ask your friend. Language involves use of symbols which package meaning within them.” You may remember that the act of speech itself is not communication. I cannot. possibly the use of language. h s d e One important component of communication is speaking with the use of language.9. “Do you have an extra pen that you can spare for me?” She says “yes” and gives you the pen. and your eyes well up with tears. it involves a relationship among two or more people in which they attempt to share meaning so that the intent of message received is the same as the intent of the message sent. The pulling away of fingers and welling up of eyes with tears will communicate to an onlooker the pain suffered by you. Your friend’s feedback (the word “yes” and appropriate behaviour of giving a pen to you) indicated that the message was successfully received and decoded.understand the signals your senses have received. The model given in Figure 9. Because language is Psychology . Hearing is a biological activity that involves reception of a message through sensory channels. Try to paraphrase what you heard and if you cannot do so. you should seek immediate clarification. Communication takes place within a context. attention. It allows you to understand how much s/he understood of what was communicated. and euphemism can sometimes become obstacles in good communication. or both. o n T i l R b E u C p N re © e b o t t Paraphrasing h s d e How would you know that someone has been listening? Ask her/him to restate what you had said. If not. then s/he probably did not get the whole message or did not understand it. Remember that slang expressions. if possible. This is called ‘paraphrasing ’. and listener’s response to the message presented.e. and personal happiness. A message could be auditory and/or visual. So one needs to consider the other’s frame of reference.symbolic. and appearance. is received. If someone cannot repeat or write down a summary of what was said. words unique to a culture or region. But this image of passivity is far from true. They observe a person’s facial expressions. the word or visual. a process that involves reception. to a large extent. Attention Once the stimulus. Divided attention makes it difficult for you to receive signals or messages. it reaches the attention stage of the human processing system. At first. You are also worried about the forthcoming examination. Listening posture. We can keep this in mind when we are listening to our teacher in the class or to others. depend upon your ability to listen effectively. Reception The initial step in the listening process is the reception of a stimulus or message. In this phase. Your academic success. So your attention is being pulled in different directions. Consider. the other stimuli recede so that we can concentrate on specific words or visual symbols. it is necessary to be as clear and precise as possible when using words. S/he should be patient. The person in front of you is constantly whispering to her/his friend. 185 Chapter 9 • Developing Psychological Skills . The hearing process is based on a complex set of physical interactions that take place involving the ear and the brain. it is important to adjust your vocabulary level and choice of words to fit the level of the listener. you are watching a movie. i. There is a buzz in the sound system. S/he makes a summary of the ideas just r eceived and provides you with a restatement of what s/he understands. non-judgmental and yet have the capacity to analyse and respond. people listen through their visual system. Hearing and listening are not the same. and what is going on in your mind. and what is happening around you. It is only a part of listening. that is. Normally your attention is divided between what you are attempting to listen to. In addition to using the hearing mechanism. Listening is an important skill that we use daily. Also whether s/he shares your interpretation. the context used by the sender to say something. movement. employment achievement. The person in doing this does not repeat your exact words. listening may appear to you as a passive behaviour. as it involves silence. Listening requires a person to be attentive. assignment of meaning. which provide important cues that may not be obvious merely by listening to the verbal part of the message. We all know that it is possible to communicate a great deal even without using verbal language. Asian cultures. T i l R b E u C p N re © e b o t t Some Tips to Improve Your Listening Skills Do you believe that when you communicate with another person. Control distractions. Assignment of Meaning The process of putting the stimulus we have received into some predetermined category develops as we acquire language. Body language is composed of all those messages that people exchange besides h s d e Psychology . has a notion called ‘mindfulness’. our categorising system for the word ‘cheese’ may include such factors as a dairy product. That is. For instance. Be careful to the emotional responses which certain words are likely to bring about. If in doubt. Visualise what is being said. We develop mental categories for interpreting the message we receive. Refrain from forming an early judgment about information that is being communicated. Repeat to the speaker what you think she or he had just said in order to check whether you both received and understood the same thing. Be aware that your posture af fects your listening. We are aware that non-verbal acts are symbolic and closely connected to any talk in progress. the culture in which we have been brought up also influences our listening and learning abilities. for instance. emphasise on Box 9. Box 9. all of which help us to relate the word ‘cheese’ to the sense in which it is used. Avoid ego speak. then you are mistaken. Be a patient listener. in many cultures. However. Also check with the sender whether s/he has been correctly understood by you. such as India. try to translate the message in the form of a concrete action.3 • • o n • • • • • • • 186 Recognise that both the sender as well as the receiver have equal responsibility in making ef fective communication.3 gives some tips to improve your listening skills. That is. Be open to all ideas. do not talk only about what you want to talk about. try to paraphrase.Try verbally paraphrasing the next time you are engaged in a demanding conversation. Buddhism. Role of Culture in Listening listening by being a silent communicator when receiving messages from seniors or elders. your words communicate the complete meaning of the message? If your answer is yes. Give consideration also to others and to what they say. T raining in ‘mindfulness’ which starts in childhood can help to develop longer attention spans and therefore. lead not only to better listening but also to sympathetic listening. such listening enhancing concepts are not present. its peculiar taste and colour. You will be surprised how many times conflicts result from miscommunication. Body Language Like the brain. Such non-verbal acts are part of what is called ‘body language’. such as when you are receiving directions or when you are in a conflict situation with a friend. Do not be in a hurry to respond. Some cultures focus on controlling attention. This means devoting your complete attention to whatever you are doing. Do you find any similarities between your non-verbal behaviour or body language and theirs? Share it with the class. crossed arms accompanied by an erect posture. a set clenched jaw. you would not have noticed the change. scientific orientation. Psychologists study these differences based on factors such as occupation. as well as harmony between verbal and non-verbal communication. Factors such as gestures. etc. For example. eye contact. age. organisation. in business and industry. They involve psychological assessment. Much of the use of body language occurs in conversing with others without conscious realisation. For instance. and the community. and standardised interpretation must be kept in mind. clothing style. While reading body language. we must remember that a single non-verbal signal does not carry complete meaning. Also. vocational fitness. In other words. “you do not look well today”. and narrowing of the eyes are likely to communicate anger. social attitudes. educational achievement. it should be found that the performance on the job Activity 9. tightened body muscles. Psychological tests have also been used for studying a variety of psychological studies on groups besides making an assessment of a particular individual. When you say to your friend. evaluation and problem solving with individuals and groups. While using psychological tests an attitude of objectivity. personality. postures. you are basing your statement on an evaluation of the person’s appearance today and comparing it with how s/he looked in the past. But. and you see that difference. the personnel department wants to know whether a certain psychological test can help it to identify potentially best stenographers. For example. and various nonintellectual characteristics. Let us recall how much we use body language to encourage or discourage conversation. and body movement — all of them have to be considered together. Note their non-verbal behaviour or body language when they are talking to someone.words. Suppose. it is essential to use actual performance records or ratings as a criterion for establishing validity of a test.2 Observing Non-verbal Behaviour in Communication Carefully observe members of your family. in verbal communication. in a cluster. Psychologists have always been interested in understanding individual differences from the time of Galton in the late 19th century. we consciously wave at waiters or friends to catch their attention. where specialised tests are used to select individuals for specific jobs. The consistency between current and past patterns of behaviour. culture. is termed as congruency. something has changed. If you did not have experience to draw on. Then focus on your own non-verbal behaviour in a similar way. Psychological tests have been devised and are primarily used for the determination and analysis of individual differences in general intelligence. in organisational and personnel work. that is. o n T i l R b E u C p N re © e b o t t h s d e Chapter 9 • Developing Psychological Skills 187 . A person’s background and past patterns of behaviour are also considered when we analyse body language. non-verbal signs can have many different meanings. In addition. differential aptitudes. crossing arms over the chest may suggest that a person likes to keep aloof. Psychological Testing Skills The next set of competencies which psychologists require is concerned with the knowledge base of the discipline of psychology. gender. education. it must be established that the test differentiates among employees of several performance levels. how much error is possible in scores? • The standardisation sample. age groups. and research interview.? • Influence of any special factors. e. who wer e tested. Box 9.4 presents the essentials of psychological assessment skills.e. radio-television interview. who. Anyone who uses a psychological test has to be a pr ofessionally qualified psychologist trained in psychological testing. Some other formats ar e information gathering interview. literate/semi-literate. Once the objectives of the interview are established. psychological testing. The skills needed for psychological assessment are : • Ability to select and implement multiple methods and means of evaluation in ways that are responsive to. stress situations.e. and gr oups. That is. i. the interviewer prepares an . • Ability to for mulate and apply diagnoses. • The r eliability indices. and respectful of diverse individuals. There are many kinds of interviews.? • T ime taken in the administration of test. what is to be scored and what method is to be used? • Norms. the areas in which it has been found working. Indians or Americans.g. etc. INTERVIEWING SKILLS An interview is a purposeful conversation between two or more people that follows a basic question and answer format. interrogatory interview. • Capacity for ef fective use of supervision to implement and enhance skills.e. • Knowledge of issues r elated to integration of different data sources. Interviewing is more formal than most other conversations because it has a pre188 Psychology o n T i l R b E u C p N re © e b o t t Interview Format h s d e set purpose and uses a focused structure. etc. The employment interview is one which most of you are likely to face. i. That is. e. i. of a newly employed worker selected on the basis of a test indeed matches with her/his test scores. and evaluation of the outcomes of psychological services. what it has to be used for? • T arget population for which it can be used. on what basis it can be said that it measur es what it claims to measur e? • The exter nal criteria of validation. i.Box 9. It includes knowledge of comprehensive and integrated assessment of persons based on interviewing. The facts r equired for this purpose are as follows: • Purpose of the test. what kinds are available? What is the appropriate group which is to be used for interpretation of scores.e.g. • Ability to integrate assessment data fr om different sources for diagnostic purposes. male/female. when the test was constructed. • Knowledge of psychometric issues and bases of assessment methods. i.g. pr esence of others. Psychological tests ar e administered strictly based on information given in the test manual. i.4 Essentials of Psychological Assessment Skills Psychological assessment is a basic competency r equired by psychologists. counselling interview. and what it cannot assess? Conditions in which it may not give good assessment. rural/urban. etc. families. to understand strengths and limitations of current diagnostic appr oaches. e. • Type of validation done on the test.e. • Scoring patterns. • Ability to utilise systematic approaches to gather data r equired for taking decisions. couples.e. • Limitations of the test. 5. is divided into three stages. In this stage. or categories s/he wants to cover. etc.e. the interviewer prepares a set of questions. also called a schedule. • Answer one question at a time. For example. ask for clarification. satisfaction with the past job. the interviewer starts the conversation and does most of the talking at the outset. For example. the interviewer must first decide on the domains/categories under which information is to be generated. views on product. These categories and the questions within them are framed ranging from easy-to-answer to difficult-to-answer. “Wouldn’t you say you are in favour of having officer’s union in the company?” Mirror Question: They are intended to get a person to reflect on what she or he had said and expand on it. the interviewer selected several categories such as nature of the organisation last worked for. and the closing. “Would you like to work for the company?” Leading Question: It encourages a r esponse in favour of a specific answer. “Where did you last work?” Open-ended Question: They are less direct and specify only the topic. and gives the interviewee time to become comfortable with the situation and the interviewer. • Restate the question in your answer. For example. For example. opening. Answering Interview Questions • If you do not understand the question. Generally. For example. To do this. For example.5 Chapter 9 • Developing Psychological Skills 189 .” Please explain as to why this happens. “How happy were you with your job on the whole?” Close-ended Question: They provide response alternatives. The basic format. in the questions used in job interview given in Box 9. narrowing the response variations. The body of the interview is the heart of the process. for different domains. i. This serves two functions.interview for mat. o n T i l R b E u C p N re © e b o t t Types of Interview Questions h s d e Box 9. Opening of the Interview The opening of interview involves establishing rapport between two communicators. Questions are also formulated to assess facts as well as subjective assessment. • Try to tur n negative questions into positive ones. For example. The purpose is to make the interviewee comfortable. Body of the Interview asks questions in an attempt to generate information and data that are required for the purpose. regardless of the interview’s purpose. it establishes the goal of interview. It r equires a yes or no response. “Do you think knowledge of a product or communication skill is mor e important for a salesperson?” Bipolar Question: It is a form of close-ended question. We would now discuss these three stages briefly. namely. the body. Sequence of Questions To accomplish the purpose of an interview. you said “I work so hard but I am unable to get success. the interviewer Direct Question: They are explicit and require specific information. The focus is more on the person and how s/he defines the problem. A counsellor is most often interested in building an understanding of the clients problem by focusing on what understanding the client has of her/his problem and how s/he feels about it. concepts. psychologists must undergo proper training and education under guided supervision.Closing the Interview While closing the interview. satisfaction in doing the job well. Meaning and Nature of Counselling COUNSELLING SKILLS Another prerequisite for developing as a psychologist is the competence in the domain of counselling and guidance. Contrary to this. h s d e Psychology . the interviewer should give a chance to the interviewee to ask questions or offer comments. s/he can develop serious problems of adjustments. The consequences of getting into a wrong vocation are pretty serious. The actual or objective facts of the problems are considered less important. In order to develop these competencies. develop negative emotions. If a person enters a job for which s/he does not have requisite aptitude. the interviewer should summarise what s/he has been able to gather. The positive feeling thus generated would have tremendous impact on overall life adjustment. analysing the counsellor’s and client’s behaviour. if anyone who takes a vocation for which s/he is well adapted. we will discuss skills. One should end with a discussion of the next step to be taken. Counselling is also one such domain where a person entering the field is required to engage in self-introspection in order to assess her/his inclination and basic skill set for being effective in her/his vocation. These difficulties may then come to be projected onto someone else.2 : Pre-requisites of Counselling Process Counselling provides a system for planning the interview. When the interview is ending. In this section. and determining the developmental impact on the client. and methods that are designed to help develop concrete competencies. suffer from inferiority complex.9. there will be considerable o n 190 T i l R b E u C p N re © e b o t t Fig. etc. and it is considered more important to work on the feelings and their acknowledgement by the clients. Counselling is not influencing attitudes. B may also share her/ his observations with A and C about A’s communication. B will share with A some problems that s/he has been facing in life lately. three students are needed. The feedback session may be of 10 minutes. Therefore. Counsellors and clients both transmit and receive verbal and non-verbal messages during the process. It takes place when a client approaches a counsellor. developing new insight. o n T i l R b E u C p N re © e b o t t h s d e Listening and Paraphrasing Activity 9. who is capable of or trained to help in a setting that permits help to be given and received (see Fig. but they are trained in developing a more healthy and helpful relationship than others. awareness and sensitivity to the nature of the message is an important prerequisite for a counsellor’s effectiveness. After the interaction is over. B. Counselling is not selection and placement of individuals onto jobs or for courses. 3. A and B will interact for about 10 minutes. Her/his role will be to repeat to the client in different words what s/he listened. that includes someone seeking help. or compelling. 5. threatening. and C. 191 Chapter 9 • Developing Psychological Skills . it is essential for clients to start developing more positive r elationships with other persons. A will act as a counsellor .3 For this activity. counselling usually has an allinclusive outcome for the clients. learning to engage in different behaviours. Like all of us. It may show up in the for m of a client taking greater responsibility. Counselling involves a basic acceptance of the client’s perceptions and feelings. and someone willing to give help. Counselling involves responding to the feelings. C will share her/his observations. without using any evaluative standards. Counselling is not the same as interviewing though interviewing may be involved. A will listen not only to what was said but also how it was said (body language) and the feelings behind it. admonishing. beliefs and behaviour by persuading. In brief.9. thoughts. Physical facilities that preserve this quality are important. who will practice listening. summarise what you have learned. 4.Counselling involves helping relationship. A counsellor never uses any kind of coercion for obtaining information. Counselling is voluntary. effective or satisfying relationships are almost non-existent or infrequent. Confidentiality and privacy constitute essential ingredients in the counselling setting. switch roles so that all three may get a chance to play the three roles. Counselling is not giving advice. and actions of the clients. counsellors too are not perfect. The counselling relationship is the initial vehicle through which this begins. The following elements about counselling are common to the major theoretical approaches to counselling : 1. C will act as an observer and take notes on how good a listener A is. After it is over. Breaking the Myths of Counselling • • • Developing Effective Relationships For most people who seek help from a counsellor. At the end of the activity. Since change in behaviour is often created and supported by a network of social support. • • Counselling is not merely giving information. A. Effective behavioural change that takes place in the client is multifaceted.2). 2. and they also sometimes communicate this image to you. the following guidelines may be kept in mind: (1) When you are speaking. To a large extent. words. (5) Do not label either yourself or the other person (e.e. Those who come into close contact with you also build their own image of you for themselves. we experience feelings of uncertainty and anxiety. (ii) Positive regard for others . The selfperceived “I” is revealed through ideas.g. These collective judgments by people you respect. any or all of which can enhance or diminish the helping process. Do not interrupt or cut in. we will discuss four qualities that are associated with effective counsellors. actions. also called ‘significant others’. in the context of the verbal exchange taking place. (3) Give the other person the freedom to share feelings or anything s/he wants to say. personal qualities and behaviour of a counsellor. You are also evaluated by persons you respect. (4) Do not assume that the other person knows what you are thinking. and (iv) Paraphrasing. clothing. the success of a counselling process depends on the skill.). (ii) Positive Regard for Others: In a counselling-counsellor relationship. The degree to which you are aware of these perceptions of others as well as of your own perception of your self indicates that you are self-aware. For example. In order to show positive regard to others. Your teachers and parents praise and/ or criticise you. friends tell you what they like and dislike about you. Let us understand these qualities briefly. “I understand” rather than “you should not”. This perception may be the same as or different from your own self-perception of ‘I’. (iii) Empathy: This is one of the most critical competencies that a counsellor needs to have. These include: (i) Authenticity . develop into a ‘me’. (2) Respond to what the other person has said. In this section. All of these communicate your “I” to others. the counsellor has the responsibility for ensuring that her/his client is benefited from counselling and its therapeutic effects are achieved. attitude. An example of this would be. This other perceived ‘me’ is the person that others perceive you to be. “you are an introvert”. We should remember that when we form a new relationship. There is a difference between sympathy and h s d e 192 Psychology . etc.. Authenticity means that your behavioural expressions are consistent o n T i l R b E u C p N re © e b o t t with what you value and the way you feel and relate to your inner self-image. (iii) Ability to empathise . after checking with her/him. It is like stepping into someone else’s shoes and trying to understand the pain and troubled feelings of the other person. Characteristics of Effective Helper Being a trained helper. Such feelings get minimised when a counsellor extends a positive regard to the client by accepting that it is all right to feel the way the client is feeling. (i) Authenticity : Your image or perception of yourself makes up your “I”. i. get into the habit of using “I” messages rather than “you” messages. knowledge. You have already read in Chapter 5 that empathy is the ability of a counsellor to understand the feelings of another person from her/his perspective. It reflects acceptance of the idea that the feelings of both are important. and your life-style. however. Express yourself according to the frame of reference. a good relationship allows freedom of expression.and making an effort to develop more effective relationships. standards. knowledge of statutes. Awareness of the ethical standards and codes is extremely important. the possible reasons for a particular fear to exist. After everyone has finished writing their fears. Now. The same process is repeated until all the slips in that basket have been read. • Practice appropriate professional assertiveness related to ethical issues. all the slips are collected and thrown in a big basket. Sensitivity. and case law relevant to the practice of psychology. • What elements other than fear would have also contributed in the process of self-awareness? • How did you feel when someone was reading out your fears? empathy. The American Psychological Association (APA) has developed a code of ethical conduct for behaviour and decision-making in actual clinical settings. Problem solving skills. A critical criterion for any professional group is the development and implementation of appropriate ethical standards. • Seek appropriate information and consultation when faced with ethical issues. marriage counsellors. Some of the APA practice guidelines are: • Knowledge of ethical/professional codes. etc. Ethics of Counselling In recent years. Intervention and consultation skills. counsellors have taken important steps to develop their Key Terms o n T i l R b E u C p N re © e b o t t professional identity. Not following the ethical standards may have legal implications. Reflective skills.Activity 9. In sympathy. it is important for you to know that the client-counsellor relationship is built on ethical practice. because counselling is a part of the service sector. Ethical observation. and psychologists — all have their ethical codes. The practical knowledge of these ethical domains can guide the practice of counselling in achieving its desired purpose. Objectivity. At the end of this activity. you play the saviour. Everyone in the group writes her/his worst fears/doubts/ anxieties on a piece of paper without disclosing her/his name. h s d e Applied psychology. the relevant agencies who are in some form related to instilling that fear . You will recall that this involves the ability of a counsellor to reflect on what the client says and feels using different words. Trustworthiness. the potential ways to overcome it. Open mindedness. everyone in the group picks up one slip from the basket and reads out aloud the fears mentioned on that list. Self-awareness. Chapter 9 • Developing Psychological Skills 193 . Intrapersonal awareness. Assessment skills. Psychological test.4 Confronting One’s Fear This activity can be carried out in a group situation. Social workers. • Recognise and understand the ethical dimensions/features of her/his own attitudes and practice in the clinical setting. You may think that someone deserves your kindness. Competence. rules. Then the group engages in a discussion about the nature of the fear. Psychological assessment. regulations. (iv) Paraphrasing: This skill has already been discussed in the section on communication earlier. you should check: • If this activity helped you in understanding something more about your self. and guidelines. While learning about the competencies of a counsellor. Counselling. family therapists. • Recognise and analyse ethical and legal issues across the range of professional activities in the clinical setting. Cognitive skills. The qualities that are associated with effective counsellors are (i) Authenticity. 8. becomes pertinent for a psychologist to equip herself/himself with these indispensable competencies. What competencies are required for becoming an effective psychologist? 2. psychological assessment. hand movements. Define communication. etc. Before entering any professional arena. etc. Psychology T i l R b E u C p N re © e b o t t The general and specific skills form the core competencies essential for a psychologist to act in a more responsive and ethical manner . What is the typical for mat of a counselling interview? 6. Proper training is required for administration. These skills are essential for all professional psychologists. Observational skills are basic skills and are used by psychologists as a starting point for providing insights into behaviour . h s d e . Communication is a process that helps in transmitting meaning from one person to another . integrity. Language is important for communication. and (iv) Paraphrasing. whether they are working in the field of clinical and health psychology. 7. What do you understand by the ter m counselling? Explain the characteristics of an ef fective counsellor. professional and scientific responsibility. therefore. What are the generic skills needed by all psychologists? 3. it is mandatory that s/he undergoes professional training. Non-verbal cues such as gestures. psychologists working in clinical settings need to be trained in various techniques of therapeutic interventions. educational. (ii) Positive regard for others. Developing the skills of psychological testing is important since tests are important tools used for the assessment of individuals for various purposes. Counselling involves helping relationship. scoring and interpretation of tests. To be an effective counsellor. and the closing stage. What ar e the ethical considerations in client-counsellor relationships? 9. postures. and someone willing to give help. For example. Interviewing is a process of face-to-face communication. it. The two major appr oaches to observation are naturalistic observation and participant observation. Its use should be done according to the characteristics of audience. the question and answer (the body). one needs to have certain characteristics such as competence. (iii) Ability to empathise. social. Do you agr ee with this statement? Give reasons in support of your arguments. respect for people’s rights and dignity. As a student of psychology.• • • • • • • • • • • Review Questions o n 194 1. In order to become an effective psychologist. that includes someone seeking help. Specific skills are core/basic to the field of psychological service. tackling environmental noise. Which component of the communication pr ocess is most important? Justify your answer with relevant examples. Describe the set of competencies that must be kept in mind while administering a psychological test. think of specific ways in which you can devise a programme to help your friend modify or solve her/his problem. Speaking and listening ar e central to interpersonal communication. 5. It proceeds through three stages which include the warm up (opening stage). industrial/organisational. Identify an aspect of your friend’s personal life that s/he wants to change. and counselling. General skills include personal as well as intellectual skills. are also used to communicate ideas. or in environmental settings or are acting as consultants. and providing feedback are ways of reducing distortions and making effective communication. 4. Creating a proper message. On the basis of the obtained information suggest some steps to enhance that skill.com www. Weblinks www.allpsych. would be particularly helpful.edu. 3. Students can also be asked to brainstor m on the possible skills and competencies needed by psychologists working in differ ent areas. Students could be asked to share their views on the incr easing applications of psychology in different ar eas of life. 2. you can choose a clinical psychologist. paraphrasing. 2. etc. a counsellor. h s d e Chapter 9 • Developing Psychological Skills 195 .unisa. Gather information about the theoretical and practical aspects of that particular skill. You can either develop a questionnaire or conduct personal interviews with all of them to identify the competencies related to the kind of work that these psychologists undertake.asp o n T i l R b E u C p N re © e b o t t Pedagogical Hints 1.library. Obtain information about the type of work they do and the skills that are used by these psychologists in their unique setting. For instance. and an educational psychologist. Make a presentation in the class. Choose any one skill from the list of competencies for a psychologist. ef fective listening. Prepare a report and discuss in the class.Project Ideas 1.au/resources/subject/counsel. Identify 3 –4 separate fields of psychology. Use of innovative methods such as narration of case vignette and roleplay to demonstrate communication skills. apprehension. poor work record. aggr essiveness. thoughts or ideas regar ding a topic. Anorexia nervosa: Disorder involving severe loss of body weight. Alienation: The feeling of not being part of society or a group. Alar m reaction: The first stage of the general adaptation syndrome characterised by an emergency r eaction involving the mobilisation of energy through adr enal and sympathetic activity. 200 o n T i l R b E u C p N re © e b o t t Attitudes: States of the mind. Cardinal trait: According to Allport. including deficits in language. accompanied by an intense fear of gaining weight or becoming “fat”. promiscuity. Archetypes: Jung’s term for the contents of the collective unconscious. Attribution: Explaining our own or others’ behaviour by pointing out the cause(s). or logically consistent with each other. Anxiety: A state of psychic distr ess characterised by fear. Antisocial personality: A behavioural disorder characterised by truancy. delinquency. managerial) to give orders and to expect the orders to be obeyed. images or symbols expressing the inherited patterns for the organisation of experience. Aptitude: A combination of characteristics indicative of individual’s potential to acquire some specific skills with training. irrationality. Autism: Pervasive developmental disor der beginning in infancy and involving a wide range of abnor malities. and physiological arousal.. h s d e Psychology . Arousal: The tension experienced at the thought of others being present. and behaviour as a result of theoretical and experimental psychology. fighting. Air pollution: Degraded quality of air is air pollution. theft. which occurs during the child’s second year. Pleasure is focused on the anus and on retention and expulsion of faeces. Beliefs: The cognitive component of the thoughts or ideas regarding a topic. The disor der is characterised by feelings of vulnerability. per ceptual. apprehension. Attitude object: The target of an attitude. either physically or verbally. containing a cognitive. impulsiveness. reckless behaviour. and inability to plan ahead. and motor development. Adaptation: Structural or functional change that enhances the organism’s survival value. and for the same experience or behaviour in the case of another person (observer). vandalism. brain. Aptitude tests: T ests meant to measur e individual’s potential to predict futur e performance. and/or per formance being evaluated. The particular pattern of behaviour varies from individual to individual. a single trait that dominates an individual’s entir e personality. affective and behavioural component.g. and between the other individual (O) and the attitude object (X) are in the same direction. Anxiety disorders: Disorders in which anxiety is a central symptom. Balance: The state of an attitude system in which the attitudes between a person (P) and another individual (O). defective reality testing. Anal stage: The second of Freud’s psychosexual stages. and social withdrawal. the person (P) and the attitude object (X). Behaviour therapy: Therapy based on the principles of behaviouristic learning theories in order to change the maladaptive behaviour. Aggression: An overt behaviour intended to hurt someone. or fear. violation of common social rules. Authority: The rights inherent in a position (e.GLOSSARY Actor-observer ef fect: The tendency to make different attributions for one’s own experience or behaviour (actor). Applied psychology : The practical application of what is known about the mind. and other higher mental processes. needs or goals. drives. Componential intelligence: In Ster nberg’s triarchic theory. Central traits: The major trait consider ed in forming an impression of others. Cognitive consistency: A state in which thoughts or ideas are logically in line with each other. as postulated by Carl Jung. Contextual intelligence: In Ster nberg’s triarchic theory. language. it is the practical intelligence used in solving everyday problems. the administration and interpretation of tests. Client-centred (Rogerian) therapy: Th e therapeutic appr oach developed by Carl Rogers in which therapist helps clients to clarify their true feelings and come to value who they are. Competition: Mutual striving between two individuals or groups for the same objective. The unconscious shared by all human beings. and vocational assistance. it r efers to ability to think critically and analytically. to compete with many others for even basic resources. perception of crampedness. Cohesiveness: All forces (factors) that cause group members to remain in the group. 201 Glossary . such as a cr owded residence. and problem solutions that are novel and appropriate. not holding authority. objects. decisionmaking. Conformity: A type of social influence in which individuals change their attitudes or behaviour in order to adher e to existing social norms. Creativity: The ability to produce ideas. Competition tolerance: The ability to put up with a situation in which individuals would have o n T i l R b E u C p N re © e b o t t Congruent attitude change: Attitude change in the same direction as that of the existing attitude.Case study: An intensive study of an individual or a situation to develop general principles about behaviour. Crowding: A psychological feeling of too little space. Centrality of attitude: The extent to which a specific attitude af fects the entire attitude system. vocational choice. or from the envir onment to man or animal. Cognition: The process of knowing. Conflict: A state of disturbance or tension resulting from opposing motives. Culture-fair test: A test that does not discriminate examinees on the basis of their cultural experiences. Cognitive therapies: For ms of therapy focused on changing distorted and maladaptive patterns of thought. Cognitive assessment system: A battery of tests designed to measure the four basic PASS (Planning-Attention-SimultaneousSuccessive) processes. Collective unconscious: Inherited portion of the unconscious. The mental activities associated with thought. Communicable disease: An illness due to specific infectious agent capable of being directly or indirectly transmitted fr om man to man. Compliance: A for m of social influence in which one or more persons. Coping: The process of trying to manage demands that are appraised as taxing or exceeding one’s resources. accepts direct requests from one or more others. therapeutic discussion. h s d e Counselling: A broad name for a wide variety of procedures for helping individuals achieve adjustment. Crowding tolerance: The ability to mentally deal with a high density or cr owded environment. animal to animal. etc. including physical space. Cognitive dissonance: The state of an attitude system in which two cognitive elements are logically contradictory. or inconsistent. Counselling interview: An interview whose purpose is counselling or providing guidance in the area of personality. such as the giving of advice. Coaction: A situation in which many people are performing the same task individually in the presence of others. being able to motivate oneself and r estrain one’s impulses. a defence mechanism. Environment: Totality. Depersonalisation disorder: Dissociative disorder in which there is a loss of the sense of self. Empathy: Reacting to another’s feelings with an emotional response that is similar to the other’s feelings. it is the ability to use past experiences creatively to solve novel problems. one person alone will not be held responsible for doing. Psychology . something. Diffusion of responsibility: The thought that when others are present. projection. an electric current through the brain. It identifies items that are homogeneous or inter nally consistent and independent of others. Discrimination: Behaviour that shows a distinction being made between two or more persons. Factor analysis : Mathematical pr ocedure . Exhaustion: State in which energy resources have been used up and responsiveness is reduced to a minimum. and behaviour operate independently from others. sublimation. Ecology: That branch of biology which deals with the r elations of or ganisms to their environment. Diathesis-stress model: A view that the interaction of factors such as biological predisposition combined with life stress may cause a specific disorder. etc. Disaster: A disaster is an unforeseen and often sudden event that disrupts the nor mal conditions within a society and causes widespread damage.Defence mechanisms: According to Freud. ways in which the ego unconsciously tries to cope with unacceptable id impulses. as in repr ession. often on the basis of the person’s (or persons’) membership of a particular group. for sorting trait terms or test responses into clusters or factors. destruction. A biological treatment for unipolar depression in which electrodes attached to a patient’s head send 202 o n T i l R b E u C p N re © e b o t t Environmental psychology: The branch of psychology that concentrates on the interaction between the physical world and human behaviour. or any aspect of physical and social set-up that surround and affect an individual organism. Delusions: Irrational beliefs that are held despite overwhelming evidence to the contrary. Emotional intelligence: A cluster of traits or abilities relating to the emotional side of life — abilities such as recognising and managing one’s own emotions. rationalisation. other members are also responsible and will ther efore do the task. and human suf fering. Evaluation apprehension: The fear of being evaluated negatively by others who are present (an audience). Deinstitutionalisation: The transfer of for mer mental patients from institutions into the community. used in the development of tests designed to discover basic personality traits. feelings. causing a convulsion. Experiential intelligence: In Sternberg’s triarchic theory. Extremeness of attitude: Refers to how far an attitude is from the neutral point. a key concept in psychoanalytic theory. or not doing. Electroconvulsive therapy (ECT): Commonly called “shock tr eatment”. and handling interpersonal relationships in an ef fective manner. Dissociation: A split in consciousness whereby certain thoughts. It is expressed in the form of an emotional quotient (EQ) score. Ego: The part of the personality that provides a buf fer between the id and the outside world. Exor cism: Religiously inspired tr eatment procedure designed to drive out evil spirits or forces from a “possessed” person. recognising and managing others’ emotions. It is effective in the treatment of cases of several depr ession that fail to respond to drug therapy. reaction for mation. Displacement: Redirecting an impulse towards a less threatening or safer target. h s d e Extraversion: One of the dimensions of personality in which interests are directed outwards to nature and other people rather than inwards to the thoughts and feelings of self (introvert). involving correlations. In the contemporary context. and challenge. can make choices about their lives.e. or image that comes to mind. feelings. are interdependent. Humanistic approach: The theory that people are basically good and tend to grow to higher levels of functioning. Group: Two or more persons who interact with one another. h s d e Identity: The distinguishing character of the individual: who each of us is. Halo effect: The tendency to link positive qualities with other positive qualities about which information is not available. typically by a specially trained person. the impulsive and unconscious part of the psyche that operates through the pleasure principle toward the gratification of instinctual drives. The id is conceived as the true unconscious. and consider themselves as members of group. Humanistic therapy: A therapy in which the underlying assumption is that people have contr ol over their behaviour. Group test: A test designed to be administered to mor e than one individual at the same time. The Binet and Wechsler intelligence tests ar e examples of individual tests. It is an abnor mal phenomenon. g-factor: General intelligence factor referring to a basic intellectual capacity underlying all manifestations of intelligence. Individual differ ences: Distinctiveness and unique variations among people’s characteristics and behaviour patterns. Homeostasis: A state of physiological balance within the body. Free association: A psychodynamic technique in which the patient describes verbally any thought. a resistance phase during which the organism makes ef forts to cope with the threat. Identification: The process of feeling one with another person. and how they interact. Hallucination: A false perception which has a compulsive sense of the reality of objects although relevant and adequate stimuli for such perception is lacking. control. Also called ego-ideal/idealised self-image. o n T i l R b E u C p N re © e b o t t Hypochondriasis: A psychological disorder in which the individual is dominated by preoccupation with bodily processes and fear of presumed diseases despite r eassurance from doctors that no physical illness exists. and are essentially responsible for solving their own pr oblems. Id: According to Fr eud. It has three characteristics. and an exhaustion phase which occurs if the organism fails to overcome the t h reat and depletes its physiological resources. usually resulting fr om liking or extr eme regard for the other person. commitment. an example of group polarisation. or the deepest part of the psyche. in contrast to individual test. Incongruent attitude change: Attitude change in a direction opposite to that of the existing attitude. i. even if it seems unimportant. and solve problems. Fundamental attribution error: The tendency to attribute internal causes more than external causes for behaviour. Genetics: The study of how the qualities of living things are passed on in their genes. Gestalt therapy: An approach to therapy that attempts to integrate a client’s thoughts. the world. Individual test: A test that must be given to a single individual at a time. r eason abstractly. what our roles are. Ideal self: The kind of person we would like to be.Fluid intelligence: Ability to perceive complex relationships. Groupthink: A mode of thinking in which the desire to reach unanimous agreement over rides the wish to adopt proper. the emphasis has Glossary 203 . General adaptation syndrome (GAS): It consists of thr ee phases : an alar m phase which promotes sympathetic nervous system activity. and behaviour into a unified whole. feeling. Industrial/organisational psychology: A subfield of psychology that focuses on relationship between people and work. Hardiness: It is a set of beliefs about oneself. decision-making procedures. have shar ed goals. and what we are capable of. rational. Observational method: A method in which researcher observes a phenomenon that occurs naturally without being able to manipulate any of the factors. selfesteem. Neurotransmitter: Chemicals that carry messages acr oss the synapse to the dendrite (and sometimes the cell body) of a receiver neur on. Interest: An individual’s pr eference for one or more specific activities. when they were small and limited in their knowledge about the world. to think rationally. Mental age (MA): A measure of intellectual functioning expressed in ter ms of age. Metaneeds: In the hierarchy of needs. Libido: Freud introduced this ter m. Introversion: One of the dimensions of personality in which interests are directed inwards rather than outwards (extravert). Ker nel of truth: The small element of truth that may be perceived in overgeneralised clusters of beliefs about groups (stereotypes). Obsessive-compulsive disor der: A disorder characterised by obsessions or compulsions. Instrumental perspective: The appr oach that suggests that the physical environment exists mainly for use by human beings for their comfort and well-being. Mood disorder: Disorder affecting one’s emotional state. The other groups are outgroups. one that brings about a negative affective response. such as self-actualisation. Intelligence tests: Tests designed to measure person’s level of intelligence. Meditation : A technique of tur ning one’s concentration inward and achieving an altered state of consciousness. Latency period: In Freud’s theory of psychosexual stages. and fewer and fewer occur towards the extr emes. h s d e Normal probability curve: A symmetrical. the period between the phallic stage and the mature genital stage (period from age 4 or 5 to about 12) during which interest in sex is sublimated. including depression and bipolar disorder. Life skills: Abilities fo r adaptive and positive behaviour that enable individuals to deal effectively with the environment. Intelligence: The capacity to understand the world. . which includes industrial and all other organisations. Obedience: Confirming behaviour in reaction to the commands of others. Mental retardation: Sub-average intellectual functioning combined with varying degrees of deficits in adaptive behaviour. Noise: An unwanted sound. Most scores are found near the middle. Norms: Standards of test per formance that per mit the comparison of one person’s score on the test to the scores of others who have taken the same test. Inferiority complex: Accor ding to Adler. the overall pattern of decisions and behaviours that determine health and quality of life. aesthetic needs. The group with which one identifies. Ingroup: The social gr oup to which an individual perceives herself or himself as belonging (“us”). In Fr eud’s treatment. Many psychological characteristics are distributed in this manner. a complex developed by adults who have not been able to overcome the feelings of inferiority they developed as children. and to use resources ef fectively when faced with challenges. which can only be satisfied when lower order needs are satisfied. libido was quite simply a direct or indirect sexual expression. Intelligence quotient (IQ): An index derived from standardised intelligence tests indicating a ratio of mental age to chronological age. Lifestyle: In the context of health psychology. those at the top.shifted fr om industrial psy chology to organisational psychology. bellshaped frequency distribution. and the like. Intellectual giftedness: Exceptional general intellectual efficiency shown in superior performance in a wide range of tasks. 204 Psychology o n T i l R b E u C p N re © e b o t t Modelling: A process of lear ning in which an individual acquires responses by observing and imitating others. Interview: Verbal interaction between a respondent and a researcher to gather information about the respondent. flora and fauna. lack of public amenities. persistent. a sense of purpose in life. Prejudice: A prejudgment. Pro-social behaviour: Behaviour that does good to another person. Peace: It is the absence of hostility and an expr ession of harmony with fellow human beings and the envir onment. distinct being. Phobia: A strong. ambiguous. mal. is done without any pr essure from outside. o n T i l R b E u C p N re © e b o t t h s d e Projection: A defence mechanism. Primacy effect: The stronger r ole of information that comes first. and resilience to stress. or subjective processes to others. and monitoring of goal-orientation. Planning: In Das’s PASS model of intelligence. and incr eased susceptibility to diseases. Optimism: The tendency to seek out. and without any expectation of a reward or return. which lie between an animal and its goal. Poverty alleviation: Measures/programmes taken up to reduce poverty. Phallic stage: Third of Freud’s psychosexual stages (at about age five) when pleasure is focused on the genitals and both males and females experience the “Oedipus complex”. air. Psychodynamic therapy: First suggested by Freud. and in which the members. Positive health: It includes a healthy body. meet face-to-face. attitudes. Psychodynamic approach: Appr oach that strives for explanation of behaviour in terms of motives. Poverty: Poverty is the economic deprivation. Pro-environmental behaviour: Willingness and activities of human beings to pr otect the envir onment ar e pr o-envir onmental behaviour. and is often towards a group. Personal space: The small ar ea around an individual considered belonging to her/him whose invasion is experienced as threatening or unpleasant. Post-traumatic stress disor der: Patter ns of symptoms involving anxiety r eactions. therapy based on the premise that the primary sources of abnor mal behaviour are unresolved past conflicts and the possibility 205 Glossary . tensions. r emember. Outgroup: Any group of which an individual is not a member. illiteracy. unstructured stimulus objects or situations in order to elicit the individual’s characteristic modes of perceiving her/his world or of behaving in it. physical or conceptual. Problem solving behaviour: The activity and mental processes involved in overcoming the obstacles. overcrowding. temperature. Primary group: Group in which each member is personally known to each of the other member. It is associated with low income. hunger. strategy selection. Persuasibility: The degree to which people can be made to change their attitudes. and expect pleasurable experiences. the task requiring overt motor responses other than verbal. Per formance test: A test in which the role of language is minimised. water. poor housing. usually a negative attitude that is unverified. the process of unwittingly attributing one’s own traits. or drives. Proximity: The principle of Gestalt psychology that stimuli close together tend to be perceived as a group. low caste and class status. and depr ession following a disaster such as an earthquake or a flood. nightmar es. trauma and change.and under -nutrition. Personal identity: Awareness of oneself as a separate. good interpersonal r elationships. Physical envir onment: It is the natur e that includes climate. it involves goal setting. Projective techniques: The utilisation of vague.Oedipus complex: The Freudian concept in which the young child develops an intense desire to replace the parent of the same sex and enjoy that af fection of the opposite sex parent. and irrational fear of some specific object or situation that presents little or no actual danger to a person. at least on occasion. Prototype: A schema in the form of a category representing all the possible qualities of an object or a person. one’s attitude toward oneself along a positive-negative dimension. Scapegoating: Placing the blame on a group for something that has gone wrong. potential and limitations. Resilience: The maintenance of positive adjustment under challenging life conditions. Recency effect: The stronger role of infor mation that comes last. etc. Psychoneuroimmunology: Interactions among behavioural. neuroendocrine. in which with the help of correlation prediction is made. or as satisfactory a state as possible. remember. Psychotherapy: The use of any psychological technique in the tr eatment of mental/ psychological disorder or maladjustment. Self-actualisation: A state of self-fulfilment in which people realise their highest potential in their own unique way. the expectation that one can master a situation and produce positive outcomes. withdrawal from reality. Self-awareness: Insight into one’s own motives. Rational emotive therapy (RET): A therapeutic system developed by Albert Ellis. 206 o n T i l R b E u C p N re © e b o t t h s d e Self-esteem: The individual’s personal judgment of her or his own worth. Regression: A defence mechanism that involves a retur n to behaviours characteristic of an earlier stage in life. Schizophrenia: A group of psychotic reactions characterised by the breakdown of integrated personality functioning. Sensitivity: Tendency to respond to very low levels of physical stimulation. Roles: An important concept in social psychology which refers to the behaviour expected of an individual in accordance with the position s/he holds in a particular society. Situationism: A principle which states that situations and circumstances outside oneself have the power to influence behaviour. emotional blunting and distortion. Relaxation training: A procedure in which clients are taught to release all the tension in their bodies. It helps in making sense of other people and ourselves. Rehabilitation: Restoring an individual to normal. Psychology . Self-efficacy: Bandura’s term for the individual’s beliefs about her or his own effectiveness. and later use social information. attempts by the patient to block treatment. interpret. It seeks to replace irrational. because the blamed group cannot defend itself. or contains many sub-attitudes (multiplex). Psychological test: An objective and standardised instrument for measuring an individual’s mental and behavioural traits. Social cognition: The pr ocesses through which we notice. Reaction formation: A defence mechanism in which a person denies a disapproved motive through giving str ong expression to its opposite. Self-fulfilling prophecy: Behaving in a way that confirms the prediction others make. pr oblem-pr ovoking outlooks with more realistic ones. Rationalisation: A defence mechanism that occurs when one attempts to explain failure or shortcomings by attributing them to mor e acceptable causes. following an illness. Self-regulation: Refers to our ability to organise and monitor our own behaviour. used by psychologists to help people make decisions about their lives and understand more about themselves. The term is also used in statistics. Resistance: In psychoanalysis.that unacceptable unconscious impulses will enter consciousness. and disturbances in thought and behaviour. and immunological pr ocesses of adaptation. Schema: A mental structure that guides social (and other) cognition. anxiety-provoking thoughts and impulses into the unconscious to avoid confronting them directly. criminal episode. Simultaneous processing: Cognitive processing in the PASS model that involves integrating elements of the stimulus situation into composite and meaningful patterns. Simplicity or complexity (multiplexity) of attitude: Whether the whole attitude consists of a single or very few sub-attitudes (simple). Repression: A defence mechanism by which people push unacceptable. Somatoform disorders: Conditions involving physical complaints or disabilities occurring in the absence of any identifiable organic cause. Unconditional positive regard: An attitude of acceptance and respect on the part of an observer. Cattell’s term for clusters of observable trait elements (responses) that seem to go together. Superego: According to Freud. it r epresents society’s standards of right and wrong as handed down by person’s parents. Substance abuse: The use of any drug or chemical to modify mood or behaviour that results in impair ment. which is a pr ocess of a relatively brief duration. Factor analysis of the correlations reveals source traits.B.Social facilitation: The tenden cy for people’s performance to improve in the presence of others.g. Trait : A r elatively persistent and consistent behaviour patter n manifested in a wide range of circumstances. Surface traits: R. Glossary 207 . Stressors: Events or situations in our environment that cause stress. the final personality structure to develop. Social identity: A person’s definition of who she or he is. Social influence: The process by which the actions of an individual or group affect the behaviour of others. Stress: Our response to events that disrupt or t h reaten to disrupt our physical and psychological functioning. Successive processing: Cognitive processing in the P ASS model where elements of the stimulus situation ar e responded to sequentially. Syndrome: Group or pattern of symptoms that occur together in a disorder and represent the typical pictur e of the disorder. Social inhibition: Social restraint on conduct. Status: Social rank within a group. Contrast with function. e. o n T i l R b E u C p N re © e b o t t Systematic desensitisation: A for m of behavioural therapy in which phobic client learns to induce a relaxed state and then exposed to stimuli that elicit fear or phobia. arising out of structure. Spiritual perspective: The perspective that specifies to do activities what are desirable in accordance with the scriptures. Social loafing: In a gr oup. and other important figures. h s d e Transference: Strong positive or negative feelings toward the therapist on the part of individual undergoing psychoanalysis. contractual nature of the relationship and limited duration of the therapy ar e its two major components. esteemed and valued. Therapeutic alliance: The special relationship between the client and the therapist. Structure: The enduring for m and composition of a complex system or phenomenon. Typology: Ways of categorising individuals into discr ete categories or types. no matter what the other person says or does. Human beings af fect the environment and in turn are affected by the environment. includes personal attributes (selfconcept) along with membership in various groups. and part of a network of communication and mutual obligation. Transactional approach: It includes interactions between people and environment. each additional individual puts in less ef fort. teachers. when they behave in ways the hospital staff consider to be desirable. Type-A personality. or an audience. Social support: Information from other people that one is loved and cared for. Stereotype: An overgeneralised and unverified prototype about a particular group. Trait approach: An approach to personality that seeks to identify the basic traits necessary to describe personality. It pleads for a harmony between man and nature. thinking that others will be putting in their effort. Token economy: Forms of behaviour therapy based on operant conditioning in which hospitalised patients earn tokens they can exchange for valued rewards. Values: Enduring beliefs about ideal modes of behaviour or end-state of existence. Verbal test: Test in which a subject’s ability to understand and use words and concepts is important in making the required responses. Valence of attitude: Whether an attitude is positive or negative. characterising any activity or mental structure which a person is not aware of. o n 208 T i l R b E u C p N re © e b o t t h s d e Psychology . attitudes that have a strong evaluative and ‘ought’ aspect.Unconscious : In psychoanalytic theory. & Weber. S.. 1997. 1997. Elsevier. R. John Wiley & Sons..R. 1998. H. • McMahon.. Psychology. Kaplan & Sadock’s Comprehensive Textbook of Psychiatry (8th ed.. Psychology.H.SUGGESTED READINGS For developing further understanding on the topics. Norton & Company.B. • Davis. Psychology : Themes and Variations. J. • Malim. Addison Wesley Longman. & Romano. J. R. 1995. Wadsworth. J. 1998. • Weiten. • Davison... • Gleitman. A. Introductory Psychology. T. (Eds. & Hersen.W. Allyn & Bacon. Vol.. 1997. W..C. Abnormal Psychology and Modern Life. A.). • Bellack. Comprehensive Clinical Psychology. Butcher. 2001/Indian reprint 2002. Inc. London. J. P. Delhi.N.A.F. Inc. G. 1996. Longman. & Palladino. Basic Psychology. 2001. 1998. A. • Gerow. II). you may like to read the following books • Baron. • Lahey.B.W. V.S. B..G. 2004. Psychology : An Introduction. M. West Publishing Company. & Birch. Inc. Thomson Learning. Macmillan Press Ltd. Psychology (5th ed. Pearson Education. Tata McGraw-Hill. F. Abnormal Psychology. W. Psychology and You. Prentice-Hall.) 2004. & Sadock. Lippincott Williams & Wilkins. • Carson. McMahon. o n T i l R b E u C p N re © e b o t t h s d e . S. 1998.L. Inc.C.. T.J.A. Psychology : An Introduction. • Zimbardo. & Mineka. • Sadock. New York. B. NOTES o n 210 T i l R b E u C p N re © e b o t t h s d e Psychology . 11 1.9 1. (c) 1.8 1.2 1. (b) 15 N m2/C Zero.18 1.8 nC – 6.1 × 10–3 N along OA Total charge is zero.6 1.4 × 1039.Physics o n 288 d e h s T i l R b E u C p N re © e b o t t ANSWERS CHAPTER 1 1. (a) 1.19 1.24 N 5.7 × 10–3 N Charges 1 and 2 are negative.3 1.14 1. only that the net charge inside is zero.07 µC (b) No. (b) Zero. Charge is not created or destroyed. 2.5 × 10–8 C m along z-axis.22 1.24 6 × 10–3 N (repulsive) (a) 12 cm (b) 0. charge 3 is positive.20 1.1 1. (a) 30 N m2/C. Dipole moment = 7. Zero N (a) 5.45 × 10–3 C (b) 1. because the charge enclosed is the same in the two cases. (b) Yes.67 nC (a) 1.16 1.4 × 106 N C–1 along OB (b) 8.21 1. Particle 3 has the highest charge to mass ratio. (b) –8.10 1.6 × 108 Nm2/C 10 µC/m (a) Zero.5 1.2 × 105 N m2/C 1.23 1. 10–4 N m (a) 2 × 1012.9 N/C .9 × 105 N m2/C (a) –103 N m2/C. from wool to polythene. The number of lines entering the cube is the same as the number of lines leaving the cube.2 N (attractive) 2.5 × 10–2 N (b) 0.12 1. It is merely transferred from one body to another.17 1. but of a negligible amount ( = 2 × 10–18 kg in the example). (a) 0. This is the ratio of electric force to the gravitational force (at the same distance) between an electron and a proton.15 1.13 1. Only (c) is right. The force is 10–2 N in the negative z-direction. (a) is wrong because field lines must be normal to a conductor. (b) The mid-point of the line joining the two charges is a null-point. n 2.34 CHAPTER 2 2. (d) is wrong because field lines cannot intersect each other.25 1.udd. the rest cannot represent electrostatic field lines. Hint: Consider the conductor with the hole filled up. But by Gauss’s law. in the direction of decreasing electric field. You will see that the net force takes it away from the null-point. stability of equilibrium needs restoring force in all directions. Displace a test charge from the null-point slightly along the line. 40 cm away from the positive charge on the side of the negative charge. Then the field ˆ and is zero inside. not enclosing any charge. That is. There is a restoring force.31 1. That is. normal to the line.1 2.2 2.Answers 1. say. (a) Hint: Prove it by contradiction. 1. Hence. Remember. torque is zero. n.6 cm 1.uud. (c) Enclose the instrument fully by a metallic surface. all field lines near the null point should be directed inwards towards the null-point.27 1. these fields are equal and opposite. Hence. You can check that this is also the direction of decreasing potential energy of the dipole.81 × 10–4 mm. that is. Suppose the equilibrium is stable. Inside the conductor. there is a net inward flux of electric field through a closed surface around the null-point. (b) is wrong because field lines cannot start from a negative charge. the equilibrium cannot be stable. Outside they are equal both in magnitude and direction. the field due to the rest of the conductor is 1. View this field as a just outside is ( σ/ ε 0) n superposition of the field due to the filled up hole plus the field due to the rest of the charged conductor.4 10 cm. (e) is wrong because electrostatic field lines cannot form closed loops. But displace it. (a) Zero 289 . 2. (b) Normal to the plane in the direction AB.26 9. then the test charge displaced slightly in any direction will experience a restoring force towards the null-point.28 1.7 × 106 V (a) The plane normal to AB and passing through its mid-point has zero potential everywhere. the flux of electric field through a surface. must be zero. (a) Hint: Choose a Gaussian surface lying wholly within the conductor and enclosing the cavity.29 p.3 o n d e h s T i l R b E u C p N re © e b o t t σ 2 ε0 ˆ.32 1. (b) Gauss’s law on the same surface as in (a) shows that q must induce –q on the inner surface of the conductor. 4 × 105 V. take a closed loop. Hence. there are no field lines inside the cavity (i. part of which is inside the cavity along a field line and the rest inside the conductor. the + sign when the point is closer to q and the – sign when it is closer to – q. the net charge on the inner surface enclosing the cavity (not having any charge) must be zero. A flat portion may be equated to a spherical surface of large radius. 0).6 V 1.6 × 105 Vm–1 in the direction that makes an angle of about 69° to the line joining charge 2.5 × 10–8 J 6 × 10–6 J 1. To dispose of this possibility.11 2.6 eV (c) 2. o n 290 d e h s T i l R b E u C p N re © e b o t t 2.15 2 − q /(4 π r12 ).5 2.8 2. and no charge on the inner surface of the conductor. as expected by symmetry. No.2 eV (b) 13. 3 × 10–10 C. 2. C = 108 pF.08 × 10–8 C (b) Q = 1. (b) 2. 4. perpendicular to the axis. (Q + q ) / (4 π r2 ) (b) By Gauss’s law. (a) 2.2 eV. 6. and a pointed portion to one of small radius. V = 16. where r is the distance of the point from the common axis of the cylinders.e.7 2. The ratio of electric field of the first to the second is (b/a ). the zero of potential energy is taken to be at infinity. The cavity may have positive and negative charges with total charge zero.6 eV. 4 × 10–10 C 18 pF.17 λ/(2 π ε0 r ). potential is (± 1/4 π ε0 ) p/(x 2 – a2) where p = 2qa is the magnitude of the dipole moment. field is zero. 1.13 2.6 eV. at points (x.Physics 2.19 2.5 µC. 2. We know this is impossible for an electrostatic field.12 2. because work done by electrostatic field between two points is independent of the path connecting the two points. 13. the point R is irrelevant to the answer.9 (b) 105 N C–1 (c) 4. (b) The dependence on r is 1/r 2 type. 2. (a) –27.4 × 104 N C–1 96 pF (a) 3 pF (b) 40 V (a) 9 pF (b) 2 × 10–10 C. (a) 2. C = 108 pF.10 2.8 × 10–9 C. Q = 1. this gives a net work done by the field in carrying a test charge over a closed loop. Since field inside the conductor is zero. potential is zero.8 × 10–9 C (a) V = 100 V.18 –13.20 –19.5 µC. whatever be its shape. For a cavity of arbitrary shape.0 × 105 V. Normal to the axis. y.5 µC to 1.0 × 105 Vm–1 from charge 2. The field is radial.2 J.14 Potential = 4q/( 3 π ε0 b ).5 µC to 1. Note in the latter choice the total energy of the hydrogen atom is zero. (a) On the axis of the dipole.. (c) Zero. no field).21 .6 2. this is not enough to claim that the electric field inside must be zero. potential is continuous. 30 cm Hint: By Gauss’s law. this potential difference is always positive. (b) Same as in (a). (True only if the field line is a straight line. So.27 2.67 × 10 –2 J Hint: Suppose we increase the separation of the plates by ∆x. (e) No. because charge distributions on the spheres will not be uniform. using u = (1/2) ε0 E 2.113 J m –3. (c) Concentric spheres centred at the origin.28 2. (b) No. each row consisting of 3 capacitors in series. except that planes differing by a fixed potential get closer as field increases. 1130 km2 Equivalent capacitance = (200/3) pF.36 2.26 2. field is E.32 2.30 2.5 × 102 V (c) 1. This goes to increase the potential energy of the capacitor by u a ∆x where u is energy density. (c) Not necessarily.55 × 10 –6 J (b) u = 0. (a) Our body and the ground form an equipotential surface.Answers 2. not that of velocity. potential difference between the sphere and the shell is independent of q2. However. Work done (by external agency) = F ∆x.34 o n d e h s T i l R b E u C p N re © e b o t t 291 2. V1 = 100 V . u = (½) ε0 E 2 2. (f ) A single conductor is a capacitor with one of the ‘plates’ at infinity. and inside it is zero. 2.) The field line gives the direction of acceleration. the average value E/2 contributes to the force. in general. F = u a which is easily seen to be (1/2) QE. (d) Zero.55 × 10 –8 C. V4 = 200 V (a) 2.33 2. If q1 is positive.5 × 10–9 F (b) 4. quadrupole potential goes like 1/r 3. Therefore. 1. The physical origin of the factor 1/2 in the force formula lies in the fact that just outside the conductor.35 2. dipole potential goes like 1/r 2. (a) 5.23 2.3 × 10–11 F (a) No. Hence. As we step out into the open. (d) A periodically varying shape near the grid which gradually reaches the shape of planes parallel to the grid at far distances. field between the sphere and the shell is determined by q1 alone.2 × 10–10 F.25 For large r. Q1 = 10 –8 C. no matter what the shape of the complete orbit. the original equipotential surfaces of 2. Eighteen 1 µF capacitors arranged in 6 parallel rows. detailed explanation of the value of dielectric constant requires microscopic theory and is beyond the scope of the book.9 × 104 V 19 cm2 (a) Planes parallel to x-y plane.22 2. monopole potential goes like 1/r.37 . (g) A water molecule has permanent dipole moment.24 2. Q2 = Q3 = 10 –8 C V2 = V3 = 50 V Q4 = 2.31 2. 7) ≅ 2.14 3. Divide it by current to obtain time = 283 s. Why? (a) 1. (d) Light energy involved in lightning. In its absence. The steady discharging current in the atmosphere charges up the aluminium sheet gradually and raises its voltage to an extent depending on the capacitance of the capacitor (formed by the sheet. the series resistor limits the current drawn from the external source.5 V.17 30 A 17 Ω. (b) Yes.63) × (8.4 A. in AD = (6/17) A.13 3. in equilibrium. in BC = (6/17) A.2 3. 5 A.6 3. 2. (c) The galvanometer will show no current. in BD. 4A. the resistivity of the alloy manganin is nearly independent of temperature. it is preferred for long suspensions of cables. = (–2/17) A. 11.5 cm from A. heat and sound energy in the accompanying thunder. o n 292 d e h s T i l R b E u C p N re © e b o t t CHAPTER 3 3. Still this method gives you only an estimate. total current = (10/17) A. the current will be dangerously high.2 Ω. (b) 60.25 V 2.0 × 10–7 Ω m 0. Ohm’s law is valid to a high accuracy. slab and the ground). to minimise resistance of the connection which are not accounted for in the bridge formula. The mass (or weight) ratio of copper to aluminium wire is (1.005 A. keeping our head and the ground at the same potential. 4 V. The two opposing currents are. impossible because a starter motor requires large current ( ~ 100 A) for a few seconds.4 3. (c) The atmosphere is continually being charged by thunderstorms and lightning all over the globe and discharged through regions of ordinary weather.9 V (b) 0.1 3.9 3.37 × 106 m and obtain total charge of the globe.10 3.8 3. Since aluminium is lighter. in CD = (–4/17) A.72/2. 6 V (a) (20/19) Ω (b) 10A.5 V (a) 6 Ω (b) 2 V.12 3.9/2. it is not strictly correct.16 3.15 3.0039 ºC–1 867 ºC Current in branch AB = (4/17) A. 11.7 × 104 s (7. (a) X = 8.7 3.5 3. .3 3. on an average.5 h) Take the radius of the earth = 6. 19A 1027 ºC 2.2. 8.11 3.Physics open air change. (c) nearly independent of. (d) 1022. (ii) all in parallel. (b) (i) Join 1 Ω.21 3. if the circuit is shorted (accidentally). therefore the current is 3. towards south 0. (b) lower.18 (a) Only current (because it is given to be steady!).20 3. as it is.5 × 10–5 T 4 × 10–6 T.75 Ω or 11. the current drawn will exceed safety limits. (a) greater. (ii) parallel combination of 2 Ω and 3 Ω in series with 1 Ω. 1. (e) No.7 Ω 3. there will be no balance point on the wire AB. examples of non-ohmic elements: vacuum diode.24 CHAPTER 4 4. the balance point will be at larger length of the wire and the percentage error will be much smaller. if internal resistance is not large. (a) (i) in series.6 N m–1 8. The obvious thing to do is to reduce current in the outside circuit (and hence potential drops across R and X ) suitably by putting a series resistor. (f ) The circuit. (d) Because.8 Ω. Clearly.25 V. would be unsuitable.3 4. (b) No. (iv) all in parallel. If ε is greater than the emf of the driver cell of the potentiometer.1 × 10–4 T 3.5 4. (ii) 5 R. (c) (i) (16/3) Ω. because the balance point (for ε of the order of a few mV) will be very close to the end A and the percentage error in measurement will be very large. 2 Ω in parallel and the combination in series with 3Ω. If there is no balance point.2 4. it means potential drops across R or X are greater than the potential drop across the potentiometer wire AB.7 π × 10–4 T o n d e h s T i l R b E u C p N re © e b o t t 293 3.2 × 10–5 T.Answers 3. X = 11.23 3. (b) To reduce current through the galvanometer when the movable contact is far from the balance point. (c) Because the maximum current drawn from a source = ε/r. The circuit is modified by putting a suitable resistor R in series with the wire AB so that potential drop across AB is only slightly greater than the emf to be measured. vertical up 1. n2. (d) No.19 3. Hint: Let X be the equivalent resistance of the infinite network.4 4. attractive force normal to A towards B .1 4. direction of force given by Fleming’s left-hand rule 2 × 10–5 N. 2 + X/(X +1) = X which gives X = (1 + 3 ) Ω. The rest depends on the area of cross-section inversely.22 3. (iii) all in series. semiconductor diode. Then.1 × 10–2 N.7 A. (a) ε = 1. (c) No.6 4. the answer is not unique because only the ratio of charge to mass is determined.20 Deuterium ions or deuterons.18 4.0 × 10–2 T. number of turns about 400. Answer (b) corresponds to the mean radius r = 25. Note. B= 4. (b) Yes. (a) Zero.0 mm normal to B. The terms R linear in d/R cancel giving a uniform field B in a small region: B= 4 5 3/2 µ0 IN R − 0. where n in this case is given by n = 2 π r . .10 4.72 µ0 IN R 4.2 cm 18 MHz (a) 3. terms containing d2/R 2 and higher powers of d/R are neglected since d << 1 . The field is non-zero only inside the core surrounded by the windings. (c) zero.9 4.12 4.19 (a) Circular trajectory of radius 1.96 N m (a) 1. (b) Helical trajectory of radius 0. because magnetic force can change the direction of v. current about 10 A.6 × 10–3 T towards west.16 µ0 IR 2 N 2 × R +d 2 2 −3 / 2 +R 2 + R −d 2 2 −3 / 2 +R 2 − µ0 IR 2 N 2 × 5R 2 4 −3 / 2 × 1+ 4d 5R −3 / 2 + 1− 4d 5R −3 / 2 − µ0 IR 2 N 2R 3 × 4 5 3/2 × 1− 6d 6d +1+ 5R 5R where in the second and third steps above. These particulars are not unique. etc. 2. the field varies slightly across the cross-section of the toroid as r varies from the inner to outer radius.17 Hint: B for a toroid is given by the same formula as for a solenoid: N B = µ0 nI.5 × 10–2 T o n 294 d e h s T i l R b E u C p N re © e b o t t 4. 4.1 Nm. 5π × 10–4 T = 1. 4.Physics 4. (b) In a small region of length 2d about the mid-point between the coils. (b) 1 4. (b) No. Li+++ . (c) B should be in a vertically downward direction. (a) Initial v is either parallel or anti-parallel to B.5 mm with velocity component 2.15 Length about 50 cm.13 4.3 × 107 m s–1 along B.11 4. not its magnitude.14 8π × 10–3 T 0. the answer is unchanged because the formula τ = N I A × B is true for a planar loop of any shape.4.8 4. Some adjustment with limits is possible. (b) 3.5 cm. Other possible answers are He++. radius about 4 cm. one gets B = 0. because Britain is closer to the magnetic north pole.2 N m –1 . (c) Field lines of B due to the earth’s magnetism would seem to come out of the ground. 1.22 4.28 (a) Zero. r = 6.24 µ0 I1 I 2 for force per unit length is strictly valid for infinitely 2π r long conductors. magnetised mineral deposits. (b) Greater in Britain (it is about 70º). (d) A compass is free to move in a horizontal plane.68 N vertically downwards Use τ = IA × B and F = I1 × B (a) 1.Answers 4. Note: Answer (c) denotes only the magnetic force.1 (a) Magnetic declination.21 (a) A horizontal magnetic field of magnitude 0. 4. equal to 20 cm) (c) 1.3 G.1 N vertically downwards (b) 2. (c) force on each electron is evB = IB/(nA) = 5 × 10–25 N. for instance.26 T normal to the conductor in such a direction that Fleming’s left-hand rule gives a magnetic force upward.23 4. obtaining total force on the wire as 1. So the compass can point in any direction there. (b) 1.8 × 10–2 N m along –y direction (b) same as in (a) (c) 1. Local N-S poles may arise due to.8 × 10–2 N m along –x direction (d) 1.26 4. repulsive.4 × 106 m. while the earth’s field is exactly vertical at the magnetic poles. is only approximately correct because the formula F = 4. angle of dip.2 × 0. (a) 2.8 × 10–2 N m at an angle of 240º with the +x direction (e) zero (f ) zero Force is zero in each case. (b) zero. (f) Why not? The earth’s field is only approximately a dipole field. which checks with the order of magnitude of the observed field on the earth.1 N vertically downwards (true for any angle between current direction and B since l sin θ remains fixed. Case (e) corresponds to stable.27 4.84 N.7 = 0. horizontal component of earth’s magnetic field. Note.176 N. (e) Use the formula for field B on the normal bisector of a dipole of magnetic moment m. and case (f ) corresponds to unstable equilibrium. 108 A Resistance in series = 5988 Ω Shunt resistance = 10 mΩ CHAPTER 5 5. µ m BE = − 0 3 4π r Take m = 8 × 1022 J T –1.25 4. o n d e h s T i l R b E u C p N re © e b o t t 295 . 10 5. (2µ0m )/(4πr13 ) = 0. (a) (b) ( µ0m )/(4πr 3 ) = 0.35 sec 22º 0. (ii) Zero. Over a small distance. 7.4 (a) m parallel to B. 0. r1 = 21/ 3 r = 6.13 5. (b) 0.33 J in a direction that tends to align the magnitude moment vector along B. the deflection due to the circular orbit of such large R may not be noticeable. its variations are not completely negligible. m = NIA to get I = 1.36 J T –1 5.66 J 5. (a) 1. cosmic rays. (b) m anti-parallel to B.. (f ) From the relation R = mv . You should consult a good modern text on geomagnetism for a proper view of the question. Time scale for appreciable change is roughly a few hundred years.42 × 10−4 which gives r = 5.048 Nm in a direction that tends to align the axis of the solenoid (i.33 J (ii) 0. 5.8 × 10–2 J. U = +mB = +4. the deflection can significantly affect the passage of charged particles. the field gets modified due to the field of ions in motion (in the earth’s ionosphere).7 (b) (i) Torque of magnitude 0. (e) At large distances. At 14 × 2–1/3 = 11. But even on a much smaller scale of a few years. But nobody really knows. its magnitude moment vector) along B.48 G along N-S direction.2 (a) Yes. The earth’s lies in the vertical plane 12º west of the geographic meridian making an angle of 60º (upwards) with the horizontal (magnetic south to magnetic north) direction.9 5.8 (b) Force is zero in uniform field.e.e.3 0.96 G along S-N direction. the solar wind.14 5. Use I = mB /(4π 2ν 2 ). an extremely minute field bends eB charged particles in a circle of very large radius. it does change with time. U = –mB = – 4.42 × 10−4 i.28 A m2 along the axis in the direction related to the sense of current via the right-handed screw rule.1 cm on the normal bisector.0 cm.54 G in the direction of earth’s field. ..Physics 5. (b) Because molten iron (which is the phase of the iron at the high temperatures of the core) is not ferromagnetic. (a) 0.38 G. o n 296 d e h s T i l R b E u C p N re © e b o t t (d) Earth’s magnetic field gets weakly ‘recorded’ in certain rocks during solidification. 5. The latter is sensitive to extra-terrestrial disturbances such as.32 G.60 J T –1 along the axis of the solenoid determined by the sense of flow of the current. but over the gigantic interstellar distances.2 × 10−4 kg m 2 . (c) One possibility is the radioactivity in the interior of the earth. Magnitude = 0. 5.5 ×10–2 J (a) (i) 0.3 cm.15 B = 0.11 5. Analysis of this rock magnetism offers clues to geomagnetic history.8 × 10–2 J: stable. unstable. 5. torque = 0.6 5.5 0. for example.12 5. 57 G. because heat lost per cycle is proportional to the area of hysteresis loop. saturation requires impractically high magnetising fields. the field lines meet this medium nearly normally. (d) No. (When one of the media has µ >> 1. (b) The induced dipole moment in a diamagnetic sample is always opposite to the magnetising field.16 (a) The tendency to disrupt the alignment of dipoles (with the magnetising field) arising from random thermal motion is reduced at lower temperatures.2 = 0. (f ) Yes.39 cos35º – 0. (c) Slightly less. In other words. But of course.).e. (e) Surround the region by soft iron rings.39 sin35º = 0.39 cos350 + 0.224 G R = 0. how many cycles of magnetisation it has gone through.18 5. Apart from minor differences in strength of the individual atomic dipoles of two different materials.52 G Rv = 0. Its value for a particular field depends both on the field and also on history of magnetisation (i.19 R = R h 2 + Rv 2 = 0. it is clear that m is greater for lower fields.12 G Rv = 0. a paramagnetic sample with saturated magnetisation will have the same order of magnetisation. (d) Ceramics (specially treated barium iron oxides) also called ferrites. If information bits can be made to correspond to these cycles. (b) Carbon steel piece. Below the cable: Rh = 0. Magnetic field lines will be drawn into the rings.) Details are beyond the scope of this book.Answers 5.25 G θ = tan −1 o n d e h s T i l R b E u C p N re © e b o t t 23º Rv = 62º Rh Above the cable: Rh = 0. no matter what the internal motion of the atoms is. the system displaying such a hysteresis loop can act as a device for storing information. the value of magnetisation is a record or memory of its cycles of magnetisation. θ 297 . since bismuth is diamagnetic.17 5. as it evident from the magnetisation curve. From the slope of magnetisation curve.2 = 0.5 cm. (e) Proof of this important fact (of much practical use) is based on boundary conditions of magnetic fields (B and H) at the interface of two media.. and the enclosed space will be free of magnetic field. But this shielding is only approximate. (c) Magnetisation of a ferromagnet is not a single-valued function of the magnetising field.22 G 5. unlike the perfect electric shielding of a cavity in a conductor placed in an external electric field. etc. Parallel to and above the cable at a distance at 1. 1 (a) Along qrpq (b) Along prq.e.20 (a) Bh = ( µ0 IN /2r ) cos 45o = 0.23 5.9 J T –1 Use the formula B = 5. it is easily seen that µ and l are antiparallel. Of the two.. Since charge of the electron is negative (= – e). 5.e.5 × 10-23 × 2. Clearly. twice the classically expected value.4 × 10−3 T 5.25 B = 4. total dipole moment = 0.3. We get 4 mm.8) = 7. This latter result (verified experimentally) is an outstanding consequence of modern quantum theory and cannot be obtained classically. µ l = − (e /2m ) l..48 T.2/2.15 × 1.3/11.24 µr µo NI where µr (relative permeability) to get 2π R 5.39 G (b) East to west (i.5 × (0. i. deflection Initially. along yzx . the needle will reverse its original direction).3 m Up or down deflection = R (1-cosθ ) where sinθ = 0.84) × (4. It follows easily from the definitions of µl and l: µl = IA = (e / T )πr 2 2πr 2 T where r is the radius of the circular orbit which the electron of mass l = mvr = m m and charge (–e) completes in time T. the relation µl = – (e/2m) l is in accordance with classical physics. µl / l = e / 2m . Therefore. CHAPTER 6 6.2 × 10 −2 × sin15o sin 45o o n 298 d e h s T i l R b E u C p N re © e b o t t = 4.Physics 5.22 R= meV eB = 2m e × kinetic energy eB = 11. both normal to the plane of the orbit.5 J T –1 Use Curie’s Law m ∝ B/T to get the final dipole moment = 4.21 Magnitude of the other field = 1. Note µs/S in contrast to µl / l is e/m.0 ×1024 = 4.98/0. (a) 1.1 V (using significant figures).5 6. lasting 2 s (b) 0. (b) West to East. the source of the power dissipated as heat in the coil is the external rotor.2 (c) (d) (e) (f ) (a) 6.02 = 3.4 6.44 × 10 –5 Wb s–1 Rate of change of flux due to motion of the loop in a non-uniform B = 144 × 10 –4 m2 × 10–3 T cm–1 × 8 cm s–1 = 11.12 The induced current causes a torque opposing the rotation of the coil.9 6.6 6.3 6. 4H 30 Wb Vertical component of B = 5. Rate of change of flux due to explicit time variation in B = 144 × 10 – 4 m2 × 10–3 T s–1 = 1.2 × 10–5 V Induced current = 2 × 10 –5 A Power loss = 6.4 × 10 –10 W Source of this power is the external agent responsible for changing the magnetic field with time.8 6. Thus.018 W 2 6.0 × 10 – 4 sin 30° = 2. lasting 8 s 100 V Flux through each turn of the loop = π r 2B cos(ωt) ε = –N ω π r 2B sin(ωt) εmax = –N ω π r 2B = 20 × 50 × π × 64 × 10– 4 × 3.0 × 10 –2 = 0.7 6. Induced emf = 8 × 2 × 10 – 4 × 0.125 V The emf induced is 3.10 6.603 V εavg is zero over a cycle Imax = 0. (c) Eastern end.52 × 10–5 Wb s–1 o n d e h s T i l R b E u C p N re © e b o t t 299 .0603 A Paverage = 1 ε max I max = 0.5 × 10 – 4 × 25 × 500 = 3. The direction of the wing is immaterial (as long as it is horizontal) for this answer.11 6. (b) Along a′d′c′b′ (flux decreases during the process) 7.6 × 10–4 V.Answers Along yzx Along zyx Along xry No induced current since field lines lie in the plane of the loop.5 × 10–3 V. An external agent (rotor) must supply torque (and do work) to counter this torque in order to keep the coil rotating uniformly.5 × 10–6 V (a) 2.5 × 10 – 4 T ε = Blv ε = 2. so induced current produces opposing flux). Along adcd (flux through the surface increases during shape change.4 × 10–4 V. 13 Q= Idt ti 1 = R tf ε dt Φf ti =− = N R dΦ Φi N (Φi − Φf ) R for N = 25.t )dx a a dΦ dB ( x . t ) = a dx dt dt 0 d B ∂B ∂ B dx = + dt ∂t ∂x dt using. Q = 7. induced current = 2. R = 0.75 T . Φi = 1. the current will be seen to be anti-clockwise. The direction of induced current is such as to increase the flux through the loop along positive z-direction.88 × 10–2 A. t ) ∂B ( x . A = 2. dΦ ∂B ( x . tf 6.5 × 10–3 C Φf = 0. you will still appreciate that flux change can occur both due to the motion of the loop as well as time variations in the magnetic field. induced emf = 12.96 × 10–5 V. If for the observer the loop moves to the right. Even if you do not understand this formal proof (which requires good familiarity with calculus).50 Ω. ∂t ∂B ∂x and v are given to be constants in the problem. = ∂B ∂B +v ∂t ∂x we get. A proper proof of the procedure above is as follows: o n 300 d e h s T i l R b E u C p N re © e b o t t 0 Φ (t ) = aB ( x .0 × 10–4 m2. Therefore.Physics The two effects add up since both cause a decrease in flux along the positive z-direction. t ) = a dx +v dt ∂ t ∂x 0 a =A ∂B ∂B +v ∂t ∂x where A = a2 The last step follows because ∂B .5 × 10–4 Wb B = Φi/A = 0. the excess charge is maintained by the continuous flow of current.50 × 0.15 m 9 mΩ (e) (f ) (g) = 75 × 10–3 N Power expended by an external agent against the above retarding force to keep the rod moving uniformly at 12 cm s–1 = 75 × 10 –3 × 12 × 10 –2 = 9. (b) Yes.14 (a) ! = vBl = 0. I2R = 1 × 1 × 9 × 10 –3 = 9.Answers 6. P positive end and Q negative end.0 × 10 –3 W When K is open.7 × 10–5 V 6. (c) Magnetic force is cancelled by the electric force set-up due to the excess charge of opposite signs at the ends of the rod. motion of the rod does not cut across the field lines. When K is closed.0 mV.15 = 9.0 × 10 –3 W The source of this power is the power provided by the external agent as calculated above. Zero.17 − B πa2 λ ˆ k MR 301 . [Note: length of PQ has been considered above to be equal to the spacing between the rails. no power is expended.16 M = µ0 a 2π ln 1 + a x ε = 1.3 × 10 −3 6.15 l (Inside the solenoid away from the ends) A l Total flux linkage " N# B= µ0 NI Φ= µ0 NI (Ignoring end variations in B) ε = ε av = ε av o n = 6. (d) Retarding force = I B l = 9 mV × 0.] 6.5 T × 0.12 × 0.5 0.5 V d (N Φ ) dt total change in flux total time d e h s T i l R b E u C p N re © e b o t t = µ0 N 2 A l I 4 π × 10 −7 × 25 × 10 −4 = × (500)2 × 2. .1 × 103 s–1 7.. 125 s–1... ν = 159 Hz q = q0 cos ω t (c) (i) Energy stored is completely electrical at t = 0.20 A (b) 484 W 7. ν = 800 kHz.e.. . C = 2 2 LC 4π ν L 7. (a) 50 rad s–1 (b) 40 Ω.5 7.. ν 4 4 4 π q0 ωT T 3T 5T = q 0 cos = ..4 7.0 J) is eventually dissipated as heat. C = 197.8 pF. V Rrms = 230 V V LCrms = I rms ω0L − 1 ω0 C =0 7. .. where T = = 6.11 For L = 200 µH.9 A 2. R damps out the LC oscillations eventually.0 J. .10 ν= 1 2π 1 1 .000 W 7. T. VCrms = 1437. electrical energy is zero) at t = (d) At t = 1 T 3T 5T ..1 V o n 302 d e h s T i l R b E u C p N re © e b o t t (b) 10 2 = 14.9 pF.6 Zero in each case. . C = 87.49 A 7. electrical energy = (e) energy.5 V...2 (a) 300 2 = 212. T 3T .9 0.e.. 25 1. . i.. (b) ω = 103 rads−1 . .1 A (c) VLrms = 1437. ν = 1200 kHz.3 ms . The variable capacitor should have a range of about 88 pF to 198 pF.. For L = 200 µH.1 A 15..8 7.7 7.6 J. .5 V .. 2. same at later times. The whole of the initial energy (= 1.. Yes. 8.1 (a) 2..12 (a) 1.. 2 2 (ii) Energy stored is completely magnetic (i. because q = q 0 cos 2 8 8 8 4 8 1q  =  0  which is half the total  2C 2   2C  q 2 2 Therefore. sum of the energies stored in L and C is conserved if R = 0.Physics CHAPTER 7 7. .3 7.. . 14 (a) I 0 = 1. I Lrms = 0.5° sin ( ω t + φ ) where tan φ = 1 ωC R = 1.75 A .82 A I = 2 V0 (b) V is maximum at t = 0. at high frequency.13) showing thereby that at high frequencies.2 ms 180 2π × 50 2 2πν L = 1.15 For a RC circuit.Answers 7. I is maximum at t = (φ / ω ) . Thus.13 For an LR circuit. 7. tan φ = Therefore.2 and is nearly zero at high frequency. In a dc circuit (after steady state) ω = 0. since the currents in L and C branch are 180° out of phase and add to zero at every instant of the cycle.1 × 10−2 A (b) tan φ = 100 π.5° Time lag = 7. if V = V0 sin ω t I = V0 2 R + (1 / ω C ) (a) I0 = 3.75 A In L branch.17 φ ≈ 0. C acts like a conductor.571 φ ≈ 57. I Rrms = 5. if V = V0 sin ω t sin (ω t − φ ) . so here L acts like a pure conductor.18 303 . IC rms = 0.5 π 1 × = 3. after steady state. time lag = 7.92 A In C branch. |Z| is maximum at ω = ω0 . where tan φ = (ω L / R ) . L nearly amounts to an open circuit. φ is close to π/2. ω = 0 and C amounts to an open circuit. In R branch.16 φ ω (a) I0 = 3.55 ms 2 1 LC 7.23 A (b) φ = 33. 2 2 R +ω L (a) I0 = 1. and the total current amplitude is minimum. For a dc circuit.88 A (b) 7.92 A Note: total current I rms = 5. Effective impedance of the parallel LCR circuit is given by 1 = Z which is minimum at ω = ω0 = Therefore. Now. (a) For V = V0 sin ω t o n 1 1 + ωC − 2 ωL R d e h s T i l R b E u C p N re © e b o t t or R 57. I0 is much smaller than the low frequency case (Exercise 7. Physics I = V0 ωL −  π sin ωt +  . 0 i.. A resistor would waste power as heat. The voltage across L and C gets subtracted because they are 180° out of phase. It will dim further when the iron core is inserted which increases the choke’s impedance. For high frequency ac. 1  2 if R = 0 ωC where – sign appears if ωL > 1/ω C. impedance of L is high and that of C is low.22 At these frequencies.20 Average power to L = Average power to C = 0 Total power absorbed = 791 W (a) ω0= 4167 rad s–1. answer to Exercise 7. Power absorbed is half the peak power at ν = 648 Hz and 678 Hz. thus avoiding sparks.21 7.18. See. Therefore. V C rms = 437 V 7. Irms = 8. average power consumed by C is zero.19 (Note: 437 V – 207 V = 230 V is equal to the applied rms voltage as should be the case. average power consumed by L is zero.e. and + sign appears if ωL < 1/ω C. current amplitude(at half the peak power points) is 10 A. reduce R to 3. (e) Total average power absorbed is zero. L has no effect. (d) For C. (e) A choke coil reduces voltage across the tube without wasting power. The same is not true for rms voltage. For ac. current amplitude is (1 / 2 ) times I max . even if it is increased by an iron core.26 A 2 Average power to R = I rms R = 791 W (b) VL rms = 207 V. (c) At ω = ω0 ± ∆ω [Approximation good if (R/2L ) << ω0 ]. ν0 = 663 Hz I0 max = 14.7 ω0= 111 rad s–1.2 Hz. the ac signal appears across L. Irms = 7. So.) (c) Whatever be the current I in L. Again. impedance of L is negligible and of C very high (infinite). when the circuit is broken. is used to charge the capacitor.1 A (b) 2 P = (1 / 2) I 0 R which is maximum at the same frequency (663 Hz) for which I0 is maximum Pmax = (1 / 2)(I max )2 R = 2300 W. so the dc signal appears across C . actual voltage leads current by π/2. for example. the lamp will shine dimly because of additional impedance of the choke. Q = 45 To double Q without changing ω0. ∆ω = R/2L = 95. ∆ν = ∆ω/2π = 15.8 rad s–1. etc. because voltages across different elements may not be in phase. 7. . voltage lags by π/2. I0 = 11.24 A o n 304 d e h s T i l R b E u C p N re © e b o t t 7.6 A. (d) For a steady state dc. (b) The high induced voltage. (c) For dc.7 Ω. (d) Q = 21. (a) Yes. 74%. (b) Power supply by the plant = 800 kW + 600 kW = 1400 kW.1 pF dQ dV =C dt dt (b) o n id = ε0 dV 0.25 (a) Line power loss = (200 A)2 × 15 Ω = 600 kW.1 × 10 –12 d ΦΕ.87 × 10 9 V s –1 dt 80.23 7. Yes. (c) (b) Power supply by the plant = 800 kW + 6 kW = 806 kW. E = (c) 8. provided by ‘current’ we mean the sum of conduction and displacement currents.5 = 15 Ω. = ε0 A dt Now.15 = = 1. this power loss is (600/1400) × 100 = 43%.8 × 100 W = 176 MW Line resistance = 30 × 0. (c) Voltage drop on the line = 200 A × 15 Ω = 3000 V. i d = ε0 A Q dΦΕ dt dE i .15 A. The step-up transformer at the plant is 440 V – 7000 V. which implies id = i = 0.1 (a) C = ε 0 A / d = 80. In Exercise 7. d e h s T i l R b E u C p N re © e b o t t 800 × 1000 W 4000 V = 200 A = 20 A Therefore. Voltage drop on the line = 20 A × 15 Ω = 300 V.Answers 7. .2 εA 0 . The step-up transformer is 440 V – 40.24 400 Hydroelectric power = h ρ g × A × v = h ρ g β where β = Av is the flow (volume of water flowing per second across a cross-section). it is only (6/806) × 100 = 0. It is clear that percentage power loss is greatly reduced by high voltage transmission. (a) Irms = Vrms ωC = 6.6 × 300 × 103 × 9.26 Current = 800 × 1000 W 40. Now across the capacitor ΦE = EA. Therefore.25. rms current in the line = 7. 300 V.000 V (a) Line power loss = (20 A)2 × (15 Ω) = 6 kW. Electric power available = 0.9µA 305 . 7. ignoring end dt corrections. CHAPTER 8 8. In this exercise. Energy of a photon that a source produces indicates the spacings of the relevant energy levels of the source. λ = 10–12 m corresponds to photon energy = 1.05 rad/m)]x – (3.12 (a) 0. This indicates that nuclear energy levels (transition between which causes γ-ray emission) are typically spaced by 1 MeV or so.8 153 N/C (a) 400 nT.6 × 10 −19 Photon energy for other wavelengths in the figure for electromagnetic spectrum can be obtained by multiplying approximate powers of ten.76 µA. ˆ (e) {(100 nT) cos[(1.10 (a) λ = (c/ν) = 1. (c) The formula B = µ0 r id 2π R 2 o n 306 d e h s T i l R b E u C p N re © e b o t t 8. 3. Since id = i. = 8.8 rad/m)y + (5. i0= 2I rms = 9.4 W/m2.14 × 10 rad/s)t ]} k Photon energy (for λ = 1 m) 8.4 × 10 rad/s)t]} k 8.5 m. we have µ r B0 = 0 2 i 0 .5 × 10–2 m (b) B0 = (E0/c) = 1. j (b) E = { (120 N/C) sin[(1. (c) 86 MHz. R = 6 cm.63 × 10 −34 × 3 × 108 eV = 1. (b) 3.05 rad/m.14 × 108 rad/s)t ]} ˆ 8 ˆ B = { (400 nT) sin[(1.7 8.05 rad/m)]x – (3. 6.5 eV. The formula shows they oscillate in phase.5 8. Wavelength band: 40 m – 25 m.11 (a) – ˆ j.6 8.6 × 10–7 T (c) Energy density in E field: uE = (1/2)ε0 E 2 Energy density in B field: uB = (1/2µ0)B 2 Using E = cB.24 × 106 eV = 1. 10 m.24 × 10 −6 eV 1.63 × 10–11 T. respectively. B0 = 1. For r = 3 cm.14 × 108 rad/s.9 6.1(b) is true even if i is oscillating in time.24 MeV. For example. 109 Hz 8. goes through even if id (and therefore B ) oscillates in time.00 m. a visible wavelength λ = 5 × 10–7 m. The derivation in Exercise 8.3 The speed in vacuum is the same for all: c = 3 × 108 m s–1.Physics (b) Yes.004 W/m2 . corresponds to photon energy = 2. (b) 0. This implies that energy levels (transition between which gives visible radiation) are typically spaced by a few eV. 8. where B0 and i0 are the amplitudes of the oscillating 2π R magnetic field and current. and c = 1 µ0 ε 0 .4 E and B in x-y plane and are mutually perpendicular. 6 8. Similarly. 1. uE = uB (d) 100 nT. 29 cm K/ T . Thus. (b) Television signals are not properly reflected by the ionosphere (see text). to obtain visible radiation. the wavelength corresponding to maximum intensity of radiation is given according to Planck’s law by the relation: λ m=0. This would cause a ‘winter’. say λ = 5 × 10–7 m. Therefore. (d) It absorbs ultraviolet radiations from the sun and prevents it from reaching the earth’s surface and causing damage to life. (e) The temperature of the earth would be lower because the Greenhouse effect of the atmosphere would be absent.14 8. o n d e h s T i l R b E u C p N re © e b o t t 307 . ( f ) The clouds produced by global nuclear war would perhaps cover substantial parts of the sky preventing solar light from reaching many parts of the globe. the source should have a temperature of about 6000 K.13 A body at temperature T produces a continuous spectrum of wavelengths. while visible and radiowaves can penetrate it. reflection is effected by satellites. (c) Atmosphere absorbs X-rays. Temperatures for other wavelengths can be found. T = 2900 K. (a) (b) (c) (d) (e) Radio (short wavelength end) Radio (short wavelength end) Microwave Visible ( Yellow ) X-rays (or soft γ-rays) region 8.15 (a) Ionosphere reflects waves in these bands. For a black body. For λ m=10 –6 m. These numbers tell us the temperature ranges required for obtaining radiations in different parts of the electromagnetic spectrum. Note: a lower temperature will also produce this wavelength but not the maximum intensity.Answers 8. PHYSICS PAR T – II © no N C tt E o R be T re pu bl is he d TEXTBOOK FOR CLASS XII FOREWORD The National Curriculum Framework (NCF), 2005 recommends that children’s life at school must be linked to their life outside the school. This principle marks a departure from the legacy of bookish learning which continues to shape our system and causes a gap between the school, home and community. The syllabi and textbooks developed on the basis of NCF signify an attempt to implement this basic idea. They also attempt to discourage rote learning and the maintenance of sharp boundaries between different subject areas. We hope these measures will take us significantly further in the direction of a child-centred system of education outlined in the National Policy on Education (NPE), 1986. The success of this effort depends on the steps that school principals and teachers will take to encourage children to reflect on their own learning and to pursue imaginative activities and questions. We must recognise that, given space, time and freedom, children generate new knowledge by engaging with the information passed on to them by adults. Treating the prescribed textbook as the sole basis of examination is one of the key reasons why other resources and sites of learning are ignored. Inculcating creativity and initiative is possible if we perceive and treat children as participants in learning, not as receivers of a fixed body of knowledge. These aims imply considerable change in school routines and mode of functioning. Flexibility in the daily time-table is as necessary as rigour in implementing the annual calendar so that the required number of teaching days are actually devoted to teaching. The methods used for teaching and evaluation will also determine how effective this textbook proves for making children’s life at school a happy experience, rather than a source of stress or boredom. Syllabus designers have tried to address the problem of curricular burden by restructuring and reorienting knowledge at different stages with greater consideration for child psychology and the time available for teaching. The textbook attempts to enhance this endeavour by giving higher priority and space to opportunities for contemplation and wondering, discussion in small groups, and activities requiring hands-on experience. © no N C tt E o R be T re pu New Delhi 20 November 2006 The National Council of Educational Research and Training (NCERT) appreciates the hard work done by the textbook development committee responsible for this book. We wish to thank the Chairperson of the advisory group in science and mathematics, Professor J.V. Narlikar and the Chief Advisor for this book, Professor A.W. Joshi for guiding the work of this committee. Several teachers contributed to the development of this textbook; we are grateful to their principals for making this possible. We are indebted to the institutions and organisations which have generously permitted us to draw upon their resources, material and personnel. We are especially grateful to the members of the National Monitoring Committee, appointed by the Department of Secondary and Higher Education, Ministry of Human Resource Development under the Chairpersonship of Professor Mrinal Miri and Professor G.P. Deshpande, for their valuable time and contribution. As an organisation committed to systemic reform and continuous improvement in the quality of its products, NCERT welcomes comments and suggestions which will enable us to undertake further revision and refinement. Director National Council of Educational Research and Training bl is he d PREFACE It gives me pleasure to place this book in the hands of the students, teachers and the public at large (whose role cannot be overlooked). It is a natural sequel to the Class XI textbook which was brought out in 2006. This book is also a trimmed version of the textbooks which existed so far. The chapter on thermal and chemical effects of current has been cut out. This topic has also been dropped from the CBSE syllabus. Similarly, the chapter on communications has been substantially curtailed. It has been rewritten in an easily comprehensible form. Although most other chapters have been based on the earlier versions, several parts and sections in them have been rewritten. The Development Team has been guided by the feedback received from innumerable teachers across the country. In producing these books, Class XI as well as Class XII, there has been a basic change of emphasis. Both the books present physics to students without assuming that they would pursue this subject beyond the higher secondary level. This new view has been prompted by the various observations and suggestions made in the National Curriculum Framework (NCF), 2005. Similarly, in today’s educational scenario where students can opt for various combinations of subjects, we cannot assume that a physics student is also studying mathematics. Therefore, physics has to be presented, so to say, in a stand-alone form. As in Class XI textbook, some interesting box items have been inserted in many chapters. They are not meant for teaching or examinations. Their purpose is to catch the attention of the reader, to show some applications in daily life or in other areas of science and technology, to suggest a simple experiment, to show connection of concepts in different areas of physics, and in general, to break the monotony and enliven the book. Features like Summary, Points to Ponder, Exercises and Additional Exercises at the end of each chapter, and Examples have been retained. Several concept-based Exercises have been transferred from end-of-chapter Exercises to Examples with Solutions in the text. It is hoped that this will make the concepts discussed in the chapter more comprehensible. Several new examples and exercises have been added. Students wishing to pursue physics further would find Points to Ponder and Additional Exercises very useful and thoughtful. To provide resources beyond the textbook and to encourage eLearning, each chapter has been provided with some relevant website addresses under the title ePhysics. These sites provide additional materials on specific topics and also provide learners the opportunites for interactive demonstrations/ experiments. The intricate concepts of physics must be understood, comprehended and appreciated. Students must learn to ask questions like ‘why’, ‘how’, ‘how do we know it’. They will find almost always that the question ‘why’ has no answer within the domain of physics and science in general. But that itself is a learning experience, is it not? On the other hand, the question ‘how’ has been reasonably well answered by physicists in the case of most natural phenomena. In fact, with the understanding of how things happen, it has been possible to make use of many phenomena to create technological applications for the use of humans. © no N C tt E o R be T re pu bl is he d © no N C tt E o R be T re pu viii For example, consider statements in a book, like ‘A negatively charged electron is attracted by the positively charged plate’, or ‘In this experiment, light (or electron) behaves like a wave’. You will realise that it is not possible to answer ‘why’. This question belongs to the domain of philosophy or metaphysics. But we can answer ‘how’, we can find the force acting, we can find the wavelength of the photon (or electron), we can determine how things behave under different conditions, and we can develop instruments which will use these phenomena to our advantage. It has been a pleasure to work for these books at the higher secondary level, along with a team of members. The Textbook Development Team, the Review Team and Editing Teams involved college and university teachers, teachers from Indian Institutes of Technology, scientists from national institutes and laboratories, as well as higher secondary teachers. The feedback and critical look provided by higher secondary teachers in the various teams are highly laudable. Most box items were generated by members of one or the other team, but three of them were generated by friends and well-wishers not part of any team. We are thankful to Dr P.N. Sen of Pune, Professor Roopmanjari Ghosh of Delhi and Dr Rajesh B Khaparde of Mumbai for allowing us to use their box items, respectively in Chapters 3, 4 (Part I) and 9 (Part II). We are very thankful to the members of the Review and Editing Workshops to discuss and refine the first draft of the textbook. We also express our gratitude to Prof. Krishna Kumar, Director, NCERT, for entrusting us with the task of presenting this textbook as a part of the national effort for improving science education. I also thank Prof. G. Ravindra, Joint Director, NCERT, for his help from time-to-time. Prof. Hukum Singh, Head, Department of Education in Science and Mathematics, NCER T, was always willing to help us in our endeavour in every possible way. We welcome suggestions and comments from our valued users, especially students and teachers. We wish our young readers a happy journey into the exciting realm of physics. A. W. JOSHI Chief Advisor Textbook Development Committee bl is he d TEXTBOOK DEVELOPMENT COMMITTEE CHAIRPERSON, A DVISORY COMMITTEE FOR T EXTBOOKS IN SCIENCE AND MATHEMATICS J.V. Narlikar, Emeritus Professor, Inter-University Centre for Astronomy and Astrophysics (IUCAA), Ganeshkhind, Pune University Campus, Pune CHIEF ADVISOR A.W. Joshi, Honorary Visiting Scientist, National Centre for Radio Astrophysics (NCRA), Pune University Campus, Pune (Formerly Professor at Department of Physics, University of Pune) MEMBERS © no N C tt E o R be T re pu Anuradha Mathur, PGT , Modern School, Vasant Vihar, New Delhi B.K. Sharma, Professor, DESM, NCERT, New Delhi Gagan Gupta, Reader, DESM, NCERT, New Delhi R. Joshi, Lecturer (S.G.), DESM, NCERT, New Delhi S.K. Dash, Reader, DESM, NCERT, New Delhi S.K. Upadhyay, PGT, Jawahar Navodaya Vidyalaya, Muzaffar Nagar V.H. Raybagkar, Reader, Nowrosjee Wadia College, Pune MEMBER-COORDINATOR V.P. Srivastava, Reader, DESM, NCERT, New Delhi A.K. Ghatak, Emeritus Professor , Department of Physics, Indian Institute of Technology, New Delhi Alika Khare, Professor, Department of Physics, Indian Institute of Technology, Guwahati Anjali Kshirsagar, Reader, Department of Physics, University of Pune, Pune Atul Mody, Lecturer (S.G.), VES College of Arts, Science and Commerce, Mumbai Chitra Goel, PGT, Rajkiya Pratibha Vikas Vidyalaya, Tyagraj Nagar, New Delhi H.C. Pradhan, Professor, Homi Bhabha Centre of Science Education (TIFR), Mumbai N. Panchapakesan, Professor (Retd.), Department of Physics and Astrophysics, University of Delhi, Delhi S. Rai Choudhary, Professor, Department of Physics and Astrophysics, University of Delhi, Delhi S.N. Prabhakara, PGT, DM School, Regional Institute of Education (NCERT), Mysore Vishwajeet Kulkarni, Teacher (Grade I ), Higher Secondary Section, Smt. Parvatibai Chowgule College, Margao, Goa bl is he d © no N C tt E o R be T re pu bl is he d ACKNOWLEDGEMENTS The National Council of Educational Research and Training acknowledges the valuable contribution of the individuals and organisations involved in the development of Physics Textbook for Class XII. The Council also acknowledges the valuable contribution of the following academics for reviewing and refining the manuscripts of this book: Anu Venugopalan, Lecturer, School of Basic and Applied Sciences, GGSIP University, Delhi; A.K. Das, PGT, St. Xavier’s Senior Secondary School, Delhi; Bharati Kukkal, PGT, Kendriya Vidyalaya, Pushp Vihar, New Delhi; D.A. Desai, Lecturer (Retd.) , Ruparel College, Mumbai; Devendra Kumar, PGT, Rajkiya Pratibha Vikas Vidyalaya, Yamuna Vihar, Delhi; I.K. Gogia, PGT, Kendriya Vidyalaya, Gole Market, New Delhi; K.C. Sharma, Reader, Regional Institute of Education (NCERT), Ajmer; M.K. Nandy, Associate Professor, Department of Physics, Indian Institute of Technology, Guwahati; M.N. Bapat, Reader, Regional Institute of Education (NCERT), Mysore; R. Bhattacharjee, Asstt. Professor, Department of Electronics and Communication Engineering, Indian Institute of Technology, Guwahati; R.S. Das, Vice-Principal (Retd.), Balwant Ray Mehta Senior Secondary School, Lajpat Nagar, New Delhi; Sangeeta D. Gadre, Reader, Kirori Mal College, Delhi; Suresh Kumar, PGT, Delhi Public School, Dwarka, New Delhi; Sushma Jaireth, Reader, Department of Women’s Studies, NCERT, New Delhi; Shyama Rath, Reader, Department of Physics and Astrophysics, University of Delhi, Delhi; Yashu Kumar, PGT, Kulachi Hans Raj Model School, Ashok Vihar, Delhi. The Council also gratefully acknowledges the valuable contribution of the following academics for the editing and finalisation of this book: B.B. Tripathi, Professor (Retd.), Department of Physics, Indian Institute of Technology, New Delhi; Dipan K. Ghosh, Professor, Department of Physics, Indian Institute of Technology, Mumbai; Dipanjan Mitra, Scientist, National Centre for Radio Astrophysics (TIFR), Pune; G.K. Mehta, Raja Ramanna Fellow, Inter -University Accelerator Centre, New Delhi; G.S. Visweswaran, Professor, Department of Electrical Engineering, Indian Institute of Technology, New Delhi; H.C. Kandpal, Head, Optical Radiation Standards, National Physical Laboratory, New Delhi; H.S. Mani, Raja Ramanna Fellow, Institute of Mathematical Sciences, Chennai; K. Thyagarajan, Professor, Department of Physics, Indian Institute of Technology, New Delhi; P.C. Vinod Kumar, Professor, Department of Physics, Sardar Patel University, Vallabh Vidyanagar, Gujarat; S. Annapoorni, Professor, Department of Physics and Astrophysics, University of Delhi, Delhi; S.C. Dutta Roy, Emeritus Professor, Department of Electrical Engineering, Indian Institute of Technology, New Delhi; S.D. Joglekar, Professor, Department of Physics, Indian Institute of Technology, Kanpur; V. Sundara Raja, Professor, Sri Venkateswara University, Tirupati. Special thanks are due to Hukum Singh, Professor and Head, DESM, NCERT for his support. The Council also acknowledges the support provided by the APC office and the administrative staff of the DESM; Deepak Kapoor, Incharge, Computer Station; Inder Kumar, DTP Operator; Mohd. Qamar Tabrez, Copy Editor ; Rishi Pal Singh, Sr. Proof Reader, NCERT and Ashima Srivastava, Proof Reader in shaping this book. The contributions of the Publication Department in bringing out this book are also duly acknowledged. © no N C tt E o R be T re pu bl is he d CHAPTER ONE ELECTRIC CHARGES AND FIELDS CHAPTER TWO ELECTROSTATIC POTENTIAL AND CAPACITANCE bl © no N C tt E o R be T re pu CHAPTER THREE CURRENT ELECTRICITY CHAPTER FOUR M OVING CHARGES AND M AGNETISM CHAPTER FIVE M AGNETISM AND MATTER CHAPTER SIX ELECTROMAGNETIC INDUCTION CHAPTER SEVEN ALTERNATING CURRENT CHAPTER EIGHT ELECTROMAGNETIC WAVES ANSWERS is 1 51 93 132 173 204 233 269 288 Contents of Physics Part I Class XII he d CONTENTS © no N C tt E o R be T re pu CHAPTER TEN WAVE OPTICS 10.1 10.2 10.3 10.4 10.5 10.6 10.7 CHAPTER ELEVEN D UAL NATURE OF R ADIATION AND MATTER 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9 Introduction Reflection of Light by Spherical Mirrors Refraction Total Internal Reflection Refraction at Spherical Surfaces and by Lenses Refraction through a Prism Dispersion by a Prism Some Natural Phenomena due to Sunlight Optical Instruments Introduction Huygens Principle Refraction and reflection of plane waves using Huygens Principle Coherent and Incoherent Addition of Waves Interference of Light Waves and Young’s Experiment Diffraction Polarisation bl is Introduction Electron Emission Photoelectric Effect Experimental Study of Photoelectric Effect Photoelectric Effect and Wave Theory of Light Einstein’s Photoelectric Equation: Energy Quantum of Radiation Particle Nature of Light: The Photon Wave Nature of Matter Davisson and Germer Experiment he 309 310 316 319 323 330 332 333 335 351 353 355 360 362 367 376 386 387 388 389 393 393 395 398 403 CHAPTER NINE R AY OPTICS AND OPTICAL INSTRUMENTS d FOREWORD PREFACE v vii CHAPTER TWELVE ATOMS 12.1 12.2 12.3 12.4 12.5 12.6 Introduction Alpha-particle Scattering and Rutherford’s Nuclear Model of Atom Atomic Spectra Bohr Model of the Hydrogen Atom The Line Spectra of the Hydrogen Atom DE Broglie’s Explanation of Bohr’s Second Postulate of Quantisation 414 415 420 422 428 430 CHAPTER THIRTEEN NUCLEI 13.1 13.2 13.3 13.4 13.5 13.6 13.7 Introduction Atomic Masses and Composition of Nucleus Size of the Nucleus Mass-Energy and Nuclear Binding Energy Nuclear Force Radioactivity Nuclear Energy © no N C tt E o R be T re pu CHAPTER FOURTEEN SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE CIRCUITS 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9 14.10 14.11 xiv Introduction Classification of Metals, Conductors and Semiconductors Intrinsic Semiconductor Extrinsic Semiconductor p-n Junction Semiconductor diode Application of Junction Diode as a Rectifier Special Purpose p-n Junction Diodes Junction Transistor Digital Electronics and Logic Gates Integrated Circuits bl is CHAPTER FIFTEEN COMMUNICATION SYSTEMS 15.1 Introduction 15.2 Elements of a Communication System 15.3 Basic Terminology Used in Electronic Communication Systems 15.4 Bandwidth of Signals 15.5 Bandwidth of Transmission Medium 15.6 Propagation of Electromagnetic Waves he 438 438 441 442 445 446 451 467 468 472 474 478 479 483 485 490 501 505 513 513 515 517 518 519 d 15.7 15.8 15.9 15.10 Modulation and its Necessity Amplitude Modulation Production of Amplitude Modulated Wave Detection of Amplitude Modulated Wave 522 524 525 526 528 ADDITIONAL INFORMATION APPENDICES ANSWERS BIBLIOGRAPHY INDEX © no N C tt E o R be T re pu xv bl is he 534 552 554 d 532 iter. bl is he d (Adapted from http://nobelprize. the Republic of Korea. The man in the bottom shows the scale. © no N C tt E o R be T re pu BACK COVER (Adapted from http://www. For details of India’s role. with abundant fuel resources.COVER DESIGN Different stages in the evolution of the universe. at Cadarache in the South of France and will provide 500 MW of fusion power.in) Cut away view of International Thermonuclear Experimental Reactor (ITER) device. Fusion is the energy source of the sun and the stars. the Russian Federation and the USA. Japan. ITER will be constructed in Europe.dae. the People’s Republic of China. ITER is a joint international research and development project that aims to demonstrate the scientific and technical feasibility of fusion power. the others being the European Union (represented by EURATOM). issue available at Department of Atomic Energy (DAE) website mentioned above. see Nuclear India .gov.org and http://www.org and the Nobel Prize in Physics 2006) . 11-12/ May-June 2006. fusion research is aimed at demonstrating that this energy source can be used to produce electricity in a safe and environmentally benign way. 39. No. to meet the needs of a growing world population. India is one of the seven full partners in the project. Vol. On earth. It is mainly through light and the sense of vision that we know and interpret the world around us. First. For many purposes. it suffices to take c = 3 × 108 m s–1. The speed of light in vacuum is the highest speed attainable in nature. a light wave can be considered to travel from one point to another. In this situation. Its presently accepted value in vacuum is c = 2. along a straight line joining bl RAY OPTICS AND OPTICAL INSTRUMENTS is he d . Electromagnetic radiation belonging to this region of the spectrum (wavelength of about 400 nm to 750 nm) is called light. The intuitive notion that light travels in a straight line seems to contradict what we have learnt in Chapter 8. It took some time for people to realise that the speed of light is finite and measurable.Chapter Nine © no N C tt E o R be T re pu 9. that light is an electromagnetic wave of wavelength belonging to the visible part of the spectrum. that it travels in a straight line.99792458 × 108 m s–1. How to reconcile the two facts? The answer is that the wavelength of light is very small compared to the size of ordinary objects that we encounter commonly (generally of the order of a few cm or larger).1 INTRODUCTION Nature has endowed the human eye (retina) with the sensitivity to detect electromagnetic waves within a small range of the electromagnetic spectrum. as you will learn in Chapter 10. There are two things that we can intuitively mention about light from common experience. that it travels with enormous speed and second. . When this is an elastic collision. The angle of reflection (i. Also that the incident ray. Such a dilemma does not occur in the wave picture of light. the colours observed due to a thin film of oil on water. we shall restrict our discussion to the special case of curved surfaces. we shall study the image formation by plane and spherical reflecting and refracting surfaces.e. With his understanding of mechanics. i. The normal in bl BY S PHERICAL MIRRORS is he d . which fall on the water. so the component of momentum in this direction also remains the same. he could come up with a simple model of reflection and refraction. which decided whether an individual corpuscle would be reflected or not. and a bundle of such rays constitutes a beam of light. It is a common observation that a ball bouncing from a smooth plane surface obeys the laws of reflection. It presumes that light energy is concentrated in tiny particles called corpuscles. The path is called a ray of light. Using the basic laws of reflection and refraction. the angle between reflected ray and the normal to the reflecting surface or the mirror) equals the angle of incidence (angle between incident ray and the normal). the corpuscles were presumed to behave as if they are identical. He himself observed many phenomena. We then go on to describe the construction and working of some important optical instruments. which were difficult to understand in terms of particle nature of light. refraction and dispersion of light. Newton argued that some of the corpuscles. In explaining other phenomena. Newton postulated that the speed of the corpuscles was greater in water or glass than in air. there is no force acting parallel to the surface. An incoming wave can be divided into two weaker waves at the boundary between air and water. however. Newton argued that smooth surfaces like mirrors reflect the corpuscles in a similar manner. In this chapter. and gravitation often blind us to his deep experimental and theoretical study of light. gets reversed in reflection. P ARTICLE MODEL OF LIGHT © no N C tt E o R be T re pu 9. get reflected and some get transmitted. spherical surfaces. was greater than Newton – the theorist. we consider the phenomena of reflection. However.. mechanics. These laws are valid at each point on any reflecting surface whether plane or curved. As the surface is smooth. Newton – the experimenter. but also sees the bottom of the pond. He made pioneering contributions in the field of optics. 9. LIGHT 310 We are familiar with the laws of reflection. including the human eye. using the ray picture of light. He further developed the corpuscular model of light proposed by Descartes.e. For example.1). He further assumed that corpuscles of light were massless elastic particles. However. In order to explain the phenomena of refraction. the magnitude of the velocity remains the same. Only the component perpendicular to the surface. the normal component of the momentum. In the field of optics. Property of partial reflection of light is yet another such example.Physics them. But what property could distinguish these two kinds of corpuscles? Newton had to postulate some kind of unpredictable. Everyone who has looked into the water in a pond sees image of the face in it. later on it was discovered that the speed of light is less in water or glass than in air. reflected ray and the normal to the reflecting surface at the point of incidence lie in the same plane (Fig.2 REFLECTION OF Newton’s fundamental contributions to mathematics. that is. chance phenomenon. 2). all distances are measured from the pole of the mirror or the optical centre of the lens. In this book.1 The incident ray. The reflected rays converge at a point F on the principal axis of a concave mirror [Fig. For a convex mirror. we shall follow the Cartesian sign convention. 9. the line joining the centre of curvature of the mirror to the point of incidence. i. making some angle with the principal axis. The line joining the pole and the centre of curvature of the spherical mirror is known as the principal axis.2 Focal length of spherical mirrors To derive the relevant formulae for reflection by spherical mirrors and refraction by spherical lenses. In the case of spherical lenses.3 shows what happens when a parallel beam of light is incident on (a) a concave mirror. This is called the focal plane of the mirror [Fig. taken as negative. the reflected rays would converge (or appear to diverge) from a point in a plane through F normal to the principal axis. reflected ray and the normal to the reflecting surface lie in the same plane.3(a)]. According to this convention.1 Sign convention © no N C tt E o R be T re pu 9. and (b) a convex mirror.2. The point F is called the principal focus of the mirror. We assume that the rays are paraxial.3(c)]. the reflected rays appear to diverge from a point F on its principal axis [Fig.3(b)]. 9. The distances measured in the same direction as the incident light are taken as positive and those measured in the direction opposite to the direction of incident light are taken as negative (Fig.2 The Cartesian Sign Convention. The heights measured upwards with respect to x -axis and normal to the principal axis (x -axis) of the mirror/ lens are taken as positive (Fig. 9. 9. the normal is along the radius.e. We have already studied that the geometric centre of a spherical mirror is called its pole while that of a spherical lens is called its optical centre. Figure 9. it turns out that a single formula for spherical mirrors and a single formula for spherical lenses can handle all different cases.2). bl is he 311 d . With a common accepted convention. the principal axis is the line joining the optical centre with its principal focus as you will see later. If the parallel paraxial beam of light were incident. they are incident at points close to the pole P of the mirror and make small angles with the principal axis.Ray Optics and Optical Instruments this case is to be taken as normal to the tangent to surface at the point of incidence. we must first adopt a sign convention for measuring distances. FIGURE 9. That is. 9.. 9. The heights measured downwards are FIGURE 9.2. 1) gives FIGURE 9.2) then gives f = R/ 2 (9. MD MD and tan 2θ = (9. Consider a ray parallel to the principal axis striking the mirror at M. Then. Then CM will be perpendicular to the mirror at M. where R is the radius of curvature of the mirror. We now show that f = R/2.1) CD FD For small θ. The image is real if the rays actually converge to the point.4. (9.2. FD = f and CD = R.3) or. it is bl is he d . FD = 9. Therefore. CD (9. tan θ ≈ θ. denoted by f . Equation (9. ∠ MCP = θ and ∠MFP = 2θ Now.3 Focus of a concave and convex mirr or. The distance between the focus F and the pole P of the mirror is called the focal length of the mirror. Therefore. that point is called the image of the first point.2) 2 Now. 9.4 Geometry of reflection of an incident ray on (a) concave spherical mirror. Eq. and MD be the perpendicular from M on the principal axis.Physics © no N C tt E o R be T re pu tan θ = MD MD =2 FD CD FIGURE 9. The geometry of reflection of an incident ray is shown in Fig. the point D is very close to the point P. for small θ . Let C be the centre of curvature of the mirror. and (b) convex spherical mirror.3 The mirror equation 312 If rays emanating from a point actually meet at another point after reflection and/or refraction. which is true for paraxial rays. tan 2θ ≈ 2θ. Let θ be the angle of incidence. real) of an object AB formed by a concave mirror. (iii) The ray passing through (or directed towards) the focus of the concave mirror or appearing to pass through (or directed towards) the focus of a convex mirror. we get B′A ′ B ′F = ( PM = AB) (9. (For paraxial rays.6) FP FP BP Equation (9. Hence this is taken as the positive direction. To reach bl is he 313 d . It does not mean that only three rays emanate from the point A. Therefore. It shows the image A′ B′ (in this case.5.4) and (9. in all directions.) Therefore. MP can be considered to be a straight line perpendicular to CP. The reflected ray follows laws of reflection.5 Ray diagram for image the focus of the mirror.6) is a relation involving magnitude of distances. we can take any two rays emanating from a point on an object. We now apply the sign convention. 9. it is convenient to choose any two of the following rays: (i) The ray from the point which is parallel to the principal axis. The reflected ray simply retraces the path. In practice. We note that light travels from the object to the mirror MPN. obtain the image of the point due to reflection at a spherical mirror.5) B′F B′ P – FP B′ P = = (9. B′A ′ B ′F = PM FP © no N C tt E o R be T re pu or B ′A ′ B ′ P = BA B P Comparing Eqs. the two right-angled triangles A′B′ F and MPF are similar.5). The reflected ray is parallel to the principal axis. An infinite number of rays emanate from any source.Ray Optics and Optical Instruments virtual if the rays do not actually meet but appear to diverge from the point when produced backwards. The reflected ray goes through FIGURE 9. (9. however. formation by a concave mirror.5 shows the ray diagram considering three rays. trace their paths. (ii) The ray passing through the centre of curvature of a concave mirror or appearing to pass through it for a convex mirror. (iv) The ray incident at any angle at the pole. (9. Thus. We now derive the mirror equation or the relation between the object distance (u ).4) BA FP ∵ Since ∠ APB = ∠ A′PB ′. In principle. the right angled triangles A′ B′P and ABP are also similar. An image is thus a point-to-point correspondence with the object established through reflection and/or refraction. image distance (v ) and the focal length ( f ). find their point of intersection and thus. point A′ is image point of A if every ray originating at point A and falling on the concave mirror after reflection passes through the point A′. From Fig. Figure 9. 7) and (9. bl is he d –v + f –v = –u – f . We define linear magnification ( m ) as the ratio of the height of the image (h′ ) to the height of the object (h ): h′ m= (9. inverted image formed by a concave mirror.Physics the object AB. and the magnification formula. The size of the image relative to the size of the object is another important quantity to consider. BP = –u Using these in Eq. (9. we have to travel opposite to the direction of incident light. image A′ B′ as well as the focus F from the pole P. (9. You should verify that Eqs. for the case of real. these are. Eq. Hence. all the three will have negative signs. Figure 9.6 Image formation by (a) a concave mirror with object between P and F. in fact.9) h u We have derived here the mirror equation.8) h h and h′ will be taken positive or negative in accordance with the accepted sign convention. Eq. In triangles A′ B′P and ABP.6). or h′ v = – (9.9) are valid for these cases as well. and (b) a convex mirror. this becomes – h′ –v = –u h so that m= v– f v = f u 1 1 1 + = (9.7) f v u This relation is known as the mirror equation. Thus.7). With the proper use of sign convention. (9. (9. FP = –f.9). we have. valid for all the cases of reflection by a spherical mirror (concave or convex) whether the image formed is real or virtual. we get © no N C tt E o R be T re pu B ′A ′ B′ P = BA BP With the sign convention.6 shows the ray diagrams for virtual image formed by a concave and convex mirror. B′ P = –v . 314 FIGURE 9. 1 © no N C tt E o R be T re pu FIGURE 9. and magnification of the image in each case. Example 9.7 Solution The ray diagram for the formation of the image of the phone is shown in Fig.5 cm 1 1 1 + = v –10 – 7. Explain why the magnification is not uniform. the for mation of its image. but taking the laws of reflection to be true for all points of the remaining part of the mirror. Then Eq.2 EXAMPLE 9. Will the distortion of image depend on the location of the phone with respect to the mirror? he d Solution You may think that the image will now show only half of the object.7) gives or 10 × 7 .Ray Optics and Optical Instruments Example 9. 9. (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. What effect will this have on the image of an object placed in front of the mirror? EXAMPLE 9. 5 The image is 30 cm from the mirror on the same side as the object.5 is covered with an opaque (non-reflective) material. half). the image will be that of the whole object. B ′C = BC. . Solution The focal length f = –15/2 cm = –7. the intensity of the image will be low (in this case. as shown in Fig. (9.1 Suppose that the lower half of the concave mirror’s reflecting surface in Fig.e. Show by suitable diagram. real and inverted. as the area of the reflecting surface has been r educed.7. Also. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. i.3 is 315 Example 9.2 A mobile phone lies along the principal axis of a concave mirr or. However. You can yourself r ealise why the image is distorted. 5 v= (i) The object distance u = –10 cm. magnification m = – v ( −30) =– =–3 u ( −10) The image is magnified. 9. It will be of the same size. nature.7.3 An object is placed at (i) 10 cm. 9..5 = – 30 cm −2. Find the position. bl EXAMPLE 9. EXAMPLE 9. m s –1 and m s–1 . and (d) 9 m away. the speed with which the image appears to move is 1 1 1 m s –1. virtual and erect. (c) 19 m. The shift in the position of image in 1 s is 39 34 1365 − 1360 5 1 − = = = m 40 35 1400 1400 280 Therefore. In case of moving vehicles.5 EXAMPLE 9.5 = 15 cm (7. It is a virtual image.5 – 5) Solution From the mirror equation. is (1/280) m s–1 Similarly. Then from Eq. we get v= © no N C tt E o R be T re pu for u = –39 m. This phenomenon can be noticed by any person sitting in a stationary car or a bus.4 9. (9. after 1 s the position of the image v (for u = –39 + 5 = –34) is (34/35 )m. –19 m and –9 m. (9.7). Magnification m = – d This image is formed at 15 cm behind the mirror. 1 1 1 + = v −5 −7.4 Suppose while sitting in a parked car. it can be seen that for u = –29 m. since R = 2 m. The direction of propagation of an obliquely incident ray of light that enters the other medium. v = fu u − f For convex mirr or. is he v 15 =– =3 u ( −5) The image is magnified. Eq. A ray of light represents a beam. If the jogger is running at a speed of 5 m s –1.7). a part of light gets reflected back into the first medium while the rest enters the other. (b) 29 m. a similar phenomenon could be observed if the vehicle in the rear is moving closer with a constant speed. changes bl Example 9. . how fast the image of the jogger appear to move when the jogger is (a) 39 m. Then (−39) × 1 39 = m −39 − 1 40 Since the jogger moves at a constant speed of 5 m s –1. 150 60 10 Although the jogger has been moving with a constant speed.3 REFRACTION 316 When a beam of light encounters another transparent medium. the average speed of the image when the jogger is between 39 m and 34 m from the mirror. the speed of his/her image appears to increase substantially as he/she moves closer to the mirr or. respectively. f = 1 m.Physics (ii) The object distance u = –5 cm. you notice a jogger approaching towards you in the side view mirror of R = 2 m.3 or v= 5 × 7 . .e. We have sini = n 21 sin r (9. then it should be clear that n 12 = 1 n 21 (9. Snell experimentally obtained the following laws of refraction: (i) The incident ray. where n 31 is the refractive index of medium 3 with respect to medium 1. but is independent of the angle of incidence.10). For example. r < i . if n 21 > 1. the refracted ray and the normal to the interface at the point of incidence. Equation (9. r > i. It is possible that mass density of an optically denser medium may be less than that of an optically rarer medium (optical density is the ratio of the speed of light in two media).10) is the well-known Snell’s law of refraction. FIGURE 9. of refraction follow immediately.9 that r2 = i1. Note: Optical density should not be confused with mass density. turpentine and water. in short) than medium 1. It also follows that if n 32 is the refractive index of medium 3 with respect to medium 2 then n 32 = n 31 × n 12. Mass density of turpentine is less than that of water but its optical density is higher. the refracted ray bends away from the normal. From Eq.. (9.8 Refraction and reflection of light.e. the emergent ray is parallel to the incident ray—there is no 317 bl is he d . 9. This phenomenon is called refraction of light. We note that n 21 is a characteristic of the pair of media (and also depends on the wavelength of light). i.9 Lateral shift of a ray refracted Some elementary results based on the laws through a parallel-sided slab. which is mass per unit volume. the refracted ray bends towards the normal. respectively. It is easily seen from Fig. In such a case medium 2 is said to be optically denser (or denser.Ray Optics and Optical Instruments at the interface of the two media. Remember that the angles of incidence (i ) and refraction (r ) are the angles that the incident and its refracted ray make with the normal. This is the case when incident ray in a denser medium refracts into a rarer medium. (ii) The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant. For a rectangular slab. i. all lie in the same plane.10) FIGURE 9. called the refractive index of the second medium with respect to the first medium. On the other hand. if n 21 <1.11) where n 21 is a constant. © no N C tt E o R be T re pu If n 21 is the refractive index of medium 2 with respect to medium 1 and n 12 the refractive index of medium 1 with respect to medium 2. refraction takes place at two interfaces (air-glass and glass-air). FIGURE 9. but it does suffer lateral displacement/ shift with respect to the incident ray. EXAMPLE 9. For example. The apparent flattening (oval shape) of the sun at sunset and sunrise is also due to the same phenomenon. The refractive index of air with respect to vacuum is 1. 318 bl is he d . For viewing near the normal direction.5).10). How much time does the sun take to shift by 1º when viewed from the earth? Solution T ime taken for 360° shift = 24 h T ime taken for 1° shift = 24/360 h = 4 min.10 Apparent depth for (a) normal. 9. The figure is highly exaggerated to show the effect.11 Advance sunrise and delayed sunset due to atmospheric refraction.11 shows the actual and apparent positions of the sun with respect to the horizon. Due to this. and (b) oblique viewing.5 The earth takes 24 h to rotate once about its axis. Another familiar observation is that the bottom of a tank filled with water appears to be raised (Fig. ( h1) is real depth (h 2) divided by the refractive index of the medium (water). The refraction of light through the atmosphere is responsible for many interesting phenomena. it can be shown that the apparent depth. the apparent shift in the direction of the sun is by about half a degree and the corresponding time difference between actual sunset and apparent sunset is about 2 minutes (see Example 9.00029. Figure 9.Physics deviation.11). By actual sunrise we mean the actual crossing of the horizon by the sun. Example 9. the sun is visible a little before the actual sunrise and until a little after the actual sunset due to refraction of light through the atmosphere (Fig. 9.5 © no N C tt E o R be T re pu FIGURE 9. the ray AO4 ). The incident ray AO1 is partially reflected (O1C) and partially transmitted (O1B) or refracted. In short.Ray Optics and Optical Instruments THE DROWNING CHILD. till for the ray AO3. refraction is not possible. see figure here. one has to differentiate it (with respect to the coordinate of X ) and find the point X when t is a minimum. Then the time taken to reach from G to C would be he 319 d . and the incident ray is totally reflected. This reflection is called the internal reflection.12.4 TOTAL INTERNAL REFLECTION When light travels from an optically denser medium to a rarer medium at the interface. Let SR be the side of the pool between G and C. A lifeguard sitting at G outside the pool notices a child drowning at a point C. whenever two mediums and two velocities are involved. Then it can be seen that t is minimum when In the case of light v 1/v2.g. LIFEGUARD AND SNELL’S LAW © no N C tt E o R be T re pu sin i v = 1 sin r v 2 To make this time minimum.. 9. On doing all this algebra (which we skip here). draw a perpendicular LM to side SR at X. the angle of refraction is π/2. Should he/she take a straight line path GAC between G and C or GBC in which the path BC in water would be the shortest. If the angle of incidence is increased still further (e. we find that the guard should enter water at a point where Snell’s law is satisfied. This is shown by the ray AO3 D in Fig. for example. 9. it is partly reflected back into the same medium and partly refracted to the second medium.12. bl t = l1 l + 2 v1 v2 is Consider a rectangular swimming pool PQSR. whether it is a wave or a particle or a human being. The guard wants to reach the child in the shortest possible time. The refracted ray is bent so much away from the normal that it grazes the surface at the interface between the two media. As the angle of incidence increases. Let ∠GXM = i and ∠CXL = r. To understand this. Suppose the guard enters water at X. so does the angle of refraction. the angle of refraction (r ) being larger than the angle of incidence (i ). the ray AO1 B in Fig. When a ray of light enters from a denser medium to a rarer medium. 9. is the refractive index n of the medium. Let GX =l1 and XC =l 2. one must follow Snell’s law if one wants to take the shortest time. the ratio of the velocity of light in vacuum to that in the medium. it bends away from the normal. or some other path GXC? The guard knows that his/her running speed v 1 on ground is higher than his/her swimming speed v2. The refractive index of denser medium 2 with respect to rarer medium 1 will be n12 = 1/sin ic . since the maximum value with a rarer medium (air).31° 24. In total internal reflection.41° © no N C tt E o R be T re pu TABLE 9.13(a)]? Now direct the laser beam from one side of the beaker such that it strikes the upper surface of water more obliquely [Fig. on the other hand. Fig.Physics This is called total internal reflection. 9. therefore. that is. Some typical critical angles are listed in Table 9.13(b)]. so that it becomes a little turbid.1 C RITICAL ANGLE OF Substance medium Water Refractive index 1. which is easily available nowadays. there is an upper limit to the value of sin i for which the law can be satisfied. Do you find that it undergoes partial reflection (which is seen as a spot on the table below) and partial refraction [which comes out in the air and is seen as a spot on the roof.42 For values of i larger than ic. SOME TRANSPARENT MEDIA Critical angle A demonstration for total internal reflection All optical phenomena can be demonstrated very easily with the use of a laser torch or pointer. howsoever smooth the reflecting surface may be.12) d . We see from Snell’s law [Eq.52 1.14° 37.62 Dense flint glass Diamond 2. 9. is called the critical angle (ic ) for the given pair of media. You will find that the path of the beam inside the water shines brightly. The angle of incidence corresponding to an angle of refraction 90º. and hence no refraction is possible. Take a glass beaker with clear water in it.1. Snell’s law of refraction cannot be satisfied. When light gets reflected by a surface. Take a laser pointer and shine its beam through the turbid water. of sin r is unity. Shine the beam from below the beaker such that it strikes at the upper water surface at the other end. is always less intense than the incident ray. no transmission of light takes place.75° 41. The reflected ray. say ∠AO3 N.33 Crown glass 1. (9.10)] FIGURE 9. normally some fraction of it gets transmitted. Adjust the direction of laser beam until you find the angle for which the refraction 320 he (9. i = ic such that sin ic = n 21 bl is 48. Stir the water a few times with a piece of soap.12 Refraction and internal reflection of rays from a point A in the denser medium that if the relative refractive index is less (water) incident at different angles at the interface than one then. and the apparent image of the tree may create 321 an illusion to the observer that the tree is near a pool of water. This is similar to what happens in optical fibres.13(c). This is total internal reflection at its simplest. 9. This phenomenon is called mirage. Some of you neglected being very might have noticed that while moving in a bus or a car during a hot thin). passes through a medium whose refractive index decreases towards the ground. beam (refraction due to glass of beaker This type of mirage is especially common in hot deserts. As a result.1 Total internal reflection in nature and its technological applications d . The observer naturally assumes Observing total that light is being reflected from the ground. Such inverted images of distant tall objects cause water with a laser an optical illusion to the observer. This is also due to mirage. To a distant observer. the light appears to be coming FIGURE 9. a ray of light from such an object successively bends away from the normal and undergoes total internal reflection. © no N C tt E o R be T re pu (i) Mirage: On hot summer days. Thus. This is shown in Fig. by a pool of water internal reflection in near the tall object. that is. (b) When the layers of air close to the ground have varying temperature with hottest layers near the ground. the air near the ground becomes hotter than the air at higher levels. FIGURE 9. Take care not to look into the laser beam directly and not to point it at anybody’s face. 9.Ray Optics and Optical Instruments above the water surface is totally absent and the beam is totally reflected back to water. as shown in Fig. Hotter air is less dense. say. especially on a highway. bl is he 9.14 (a) A tree is seen by an observer at its place when the air above the ground is at uniform temperature. If the air currents are small. Pour this water in a long test tube and shine the laser light from top. the air is still.13 from somewhere below the ground.4. and has smaller refractive index than the cooler air. Adjust the direction of the laser beam such that it is totally internally reflected every time it strikes the walls of the tube. a distant patch of road.14(b). light from a distant tree may undergo total internal reflection. But. The refractive index of air increases with its density. light from a tall object such as a tree. appears to be wet. the optical density at different layers of air increases with height. if the angle of incidence for the air near the ground exceeds the critical angle. summer day. you do not find any evidence of wetness when you reach that spot. 4°) is very small. stomach and intestines. therefore once light enters a diamond.16 Light undergoes successive total are fabricated such that light reflected at one internal reflections as it moves through an side of inner surface strikes the other at an optical fibre. Optical fibres are fabricated with high quality composite glass/quartz fibres. Thus. Optical fibres are extensively used for transmitting and receiving electrical signals which are converted to light by suitable transducers. For example. When a signal in the form of light is directed at one end of the fibre at a suitable angle. The refractive index of the material of the core is higher than that of the cladding. Their brilliance is mainly due to the total internal reflection of light inside them. By cutting the diamond suitably. In the first two cases. it undergoes repeated total internal reflections along the length of the fibre and finally comes out at the other end (Fig. Each fibre consists of a core and cladding. 9. Even if the fibre is bent. Since light undergoes total internal reflection at each stage. Optical fibres too make use of the phenomenon of total internal reflection. It is the technical skill of a diamond cutter which makes diamonds to sparkle so brilliantly. You might have seen a commonly © no N C tt E o R be T re pu bl is he d . Optical fibres FIGURE 9. angle larger than the critical angle.15(c)]. (iv) Optical fibres: Now-a-days optical fibres are extensively used for transmitting audio and video signals through long distances.Physics (ii) Diamond : Diamonds are known for their spectacular brilliance. A bundle of optical fibres can be put to several uses. the critical angle ic for the material of the prism must be less than 45º. We see from Table 9. Such a its size make use of total internal reflection. Obviously. there is no appreciable loss in the intensity of the light signal. multiple total internal reflections can be made to occur. The critical angle for diamond-air interface (≅ 24. 9.15 Prisms designed to bend rays by 90º and 180º or to invert image without changing reflection [Fig. prism is also used to invert images without changing their size [Fig. Diamonds found in nature rarely exhibit the brilliance for which they are known.15(a) and (b)]. (iii) Prism : Prisms designed to bend light by 90º or by 180º make use of total internal FIGURE 9.16). optical fibres can also be used for transmission of optical signals. an optical fibre can be used to act as an optical pipe. these are used as a ‘light pipe’ to facilitate visual examination of internal organs 322 like esophagus. 9. light can easily travel along its length.1 that this is true for both crown glass and dense flint glass. it is very likely to undergo total internal reflection inside it. An infinitesimal part of a spherical surface can be regarded as planar and the same laws of refraction can be applied at every point on the surface. Applying the formula for image formation by a single spherical surface successively at the two surfaces of a lens.) © no N C tt E o R be T re pu 9. We have.5. we take the aperture (or the lateral size) of the surface to be small compared to other distances involved. we shall obtain the lens maker’s formula and then the lens formula. NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis. FIGURE 9.Ray Optics and Optical Instruments available decorative lamp with fine plastic fibres with their free ends forming a fountain like structure. and radius of curvature R .1 Refraction at a spherical surface tan ∠NOM = tan ∠NCM = MN OM MN MC MN MI tan ∠ NIM = We have so far considered refraction at a plane interface. Just as for reflection by a spherical mirror. bl 323 is 9. for small angles.17 Refraction at a spherical surface separating two media.5 REFRACTION AT SPHERICAL SURFACES AND BY LENSES he d . This has been achieved by purification and special preparation of materials such as quartz. so that small angle approximation can be made. (Compare with what you expect for a block of ordinary window glass 1 km thick.17 shows the geometry of formation of image I of an object O on the principal axis of a spherical surface with centre of curvature C. In particular. As before. We first consider refraction by a single spherical surface and follow it by thin lenses. passes through its centre of curvature. The other end of the fibres is fixed over an electric lamp. at least one of which should be spherical. therefore. In silica glass fibres. it is possible to transmit more than 95% of the light over a fibre length of 1 km. We shall now consider refraction at a spherical interface between two transparent media. The main requirement in fabricating optical fibres is that there should be very little absorption of light as it travels for long distances inside them. the light travels from the bottom of each fibre and appears at the tip of its free end as a dot of light. Figure 9. the normal at the point of incidence is perpendicular to the tangent plane to the spherical surface at that point and. to another of refractive index n 2. The fibres in such decorative lamps are optical fibres. A thin lens is a transparent optical medium bounded by two surfaces. When the lamp is switched on. The rays are incident from a medium of refractive index n 1. is used to characterise the brightness of emitting or reflecting flat surfaces. being the stimulus of light as received by the human eye. The wavelength region in which the body emits the radiation depends on its absolute temperature. produced by a source of luminous intensity I . The sun with temperature of about 5500 K emits radiation whose energy versus wavelength graph peaks approximately at 550 nm corresponding to green light and is almost in the middle of the visible region.Physics LIGHT SOURCES AND PHOTOMETRY © no N C tt E o R be T re pu i= MN MN + OM MC Similarly. It is defined as luminous flux incident per unit area on a surface (lm/m2 or lux ). Its unit is cd/m2 (sometimes called ‘nit’ in industry) . (ii) the luminous flux or flow of light from the source. The candela is the luminous intensity.14) d .13) i. Photometry is measurement of a physiological phenomenon. The measurement of light as perceived by human eye is called photometry. of a source that emits monochromatic radiation of frequency 540 × 1012 Hz and that has a radiant intensity in that direction of 1/683 watt per steradian. r = bl is Now. A quantity named luminance (L ). is given by E = I/ r2. Therefore. A good LCD computer monitor has a brightness of about 250 nits. If a light source emits one candela of luminous intensity into a solid angle of one steradian. A standard 100 watt incadescent light bulb emits approximately 1700 lumens. The energy versus wavelength distribution graph for a given body peaks at some wavelength. and (iii) illuminance of the surface. Now. In photometry. The SI unit of luminous intensity (I ) is candela (cd). in a given direction. the only parameter. i = ∠NOM + ∠NCM (9.. the total luminous flux emitted into that solid angle is one lumen (lm). which can be measured directly is illuminance . As the temperature of the body increases radiation emitted by it is in visible region. Radiation emitted by a hot body. The illuminance E. for Δ NOC. by Snell’s law n1 sin i = n 2 sin r or for small angles 324 n1i = n 2r he (9. i is the exterior angle. for example. The main physical quantities in photometry are (i) the luminous intensity of the source. a tungsten filament lamp having temperature 2850 K are partly invisible and mostly in infrared (or heat) region. r = ∠NCM – ∠NIM MN MN − MC MI It is known that a body above absolute zero temperature emits electromagnetic radiation. where r is the normal distance of the surface from the source. which is inversely proportional to the absolute temperature of that body.e. transmitted by the optic nerves and analysed by the brain. Most light meters measure this quantity. R = + 20 cm .18(a) shows the geometry of image formation by a double convex lens. − n 2 n 1 n 2 − n1 + = DI1 DI DC2 or v = +100 cm The image is formed at a distance of 100 cm from the glass surface. (9.18(b)]. we get n1 OM + n2 MI = n 2 − n1 MC (9.18) is 325 he d .15) Here. The image I1 acts as a virtual object for the second surface that forms the image at I [Fig. Example 9. 9. (9. we get n2 n1 n − n1 − = 2 (9.18(c)]. 9. The distance of the light source from the glass surface is 100 cm. (9.16).14).2 Refraction by a lens n1 n n − n1 + 2 = 2 OB BI1 BC1 A similar procedure applied to the second interface* ADC gives.5 + = v 100 20 © no N C tt E o R be T re pu 9. we get (9. MC = +R Substituting these in Eq.5 1 0. OM. Figure 9. OM = –u. v = ?.Ray Optics and Optical Instruments Substituting i and r from Eqs.17) * Note that now the refractive index of the medium on the right side of ADC is n 1 while on its left it is n 2. Here u = – 100 cm.5. n1 = 1.13) and (9. in the direction of incident light.5. We then have 1. The image formation can be seen in terms of two steps: (i) The first refracting surface forms the image I1 of the object O [Fig.5 and radius of curvature = 20 cm). (9. and n 2 = 1.15).15) to the first interface ABC. It holds for any curved spherical surface. Further DI1 is negative as the distance is measured against the direction of incident light. bl EXAMPLE 9.6 (9. MI and MC represent magnitudes of distances. Applying the Cartesian sign convention. MI = +v .16) v u R Equation (9. Applying Eq.16) gives us a relation between object and image distance in terms of refractive index of the medium and the radius of curvature of the curved spherical surface. At what position the image is formed? Solution We use the r elation given by Eq.6 Light from a point source in air falls on a spherical glass surface (n = 1. So Eq.20). DC 2 = –R 2 © no N C tt E o R be T re pu FIGURE 9. It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature. is Equation (9. f is negative.Physics For a thin lens. R 2 positive and therefore. trace their paths using he (9. B and D are both close to the optical centre of the lens. 9. of the lens and the distance f gives its focal length. (9. Note that the formula is true for a concave lens also. Though we derived it for a real image formed by a convex lens. Applying the sign convention.18 (a) The position of object. in principle. OB → ∞ and DI = f . the formula is valid for both convex as well as concave lenses and for both real and virtual images..20) Suppose the object is at infinity.20) can be written as ⎛ 1 1 1⎞ = (n 2 1 − 1) ⎜ − f ⎝ R1 R 2 ⎟ ⎠ ⎛ n ⎞ ∵ n2 1 = 2 ⎟ ⎜ ⎝ n1 ⎠ bl The point where image of an object placed at infinity is formed is called the focus F.18). we get 1 1 1 − = v u f (9. (9.19) BC1 = + R1 .17) and (9. and the image formed by a double convex lens. BI 1 = DI1 . Adding Eqs. we get n1 n1 n + = 1 OB DI f (9. we get n1 n1 ⎛ 1 1 ⎞ + = (n 2 − n1 ) ⎜ + OB DI ⎝ BC1 DC2 ⎟ ⎠ (9. Eq.22) Again. F and F′ . It is worth mentioning that the two foci. In that case R1 is negative. (9. (b) Refraction at the first spherical surface and (c) Refraction at the second spherical surface. DI = + v.21) n1 ⎛ 1 1 ⎞ = (n 2 − n1 ) ⎜ + f ⎝ BC1 DC2 ⎟ ⎠ d (9. in the thin lens approximation. F and F′.e. i.21) is known as the lens maker’s for mula. To find the image of an object by a lens. A lens has two foci. whereas the other is called the second focal point.19) and (9. take any two rays emanating from a point on an object.19) gives .23) is the familiar thin lens formula .19). we can. on either side of it (Fig.23) 326 Equation (9. (9. From Eqs. of a double convex or concave lens are equidistant from the optical centre. The focus on the side of the (original) source of light is called the first focal point. By the sign convention. BO = – u. Proceeding in the same way as for spherical mirrors.19(a) and (b) illustrate these rules for a convex and a concave lens. Magnification (m ) produced by a lens is defined. It could be glycerine. we see that. Eq.7 A magician during a show makes a glass lens with n = 1. All these rays will pass through the same image point after refraction at the lens. as the ratio of the size of the image to that of the object. Here again it must be remembered that each point on an object gives out infinite number of rays. This gives 1/f = 0 or f → ∞. What is the refractive index of the liquid? Could the liquid be water? © no N C tt E o R be T re pu 9.5. emerges without any deviation after refraction. Clearly.7 is he 327 d .. m is positive. respectively.3 Power of a lens Solution The refractive index of the liquid must be equal to 1.24) h u When we apply the sign convention. You should practice drawing similar ray diagrams for different positions of the object with respect to the lens and also verify that the lens FIGURE 9. holds good for all cases.47 disappear in a trough of liquid.47 in order to make the lens disappear. passing through the optical centre of the lens. m= Example 9. (iii) A ray of light passing through the first principal focus (for a convex lens) or appearing to meet at it (for a concave lens) emerges parallel to the principal axis after refraction.Ray Optics and Optical Instruments the laws of refraction and find the point where the refracted rays meet (or appear to meet). This means n1 = n 2. No. while for an inverted (and real) image. (9. which a lens introduces in the light falling on it. The lens in the liquid will act like a plane sheet of glass. (ii) A ray of light. the liquid is not water. for erect (and virtual) image formed by a convex or concave lens. m is negative. In practice. it is convenient to choose any two of the following rays: (i) A ray emanating from the object parallel to the principal axis of the lens after refraction passes through the second principal focus F′ (in a convex lens) or appears to diverge (in a concave lens) from the first principal focus F. convex lens (b) concave lens. Figures 9. however. like that for a mirror. Power of a lens is a measure of the convergence or divergence. a lens of shorter focal bl EXAMPLE 9.19 Tracing rays through (a) formula. it is easily seen that for a lens h′ v = (9.23). 0 D means a concave lens of focal length – 25 cm. while converging it in case of a convex lens and diverging it in case of a concave lens. We use the lens formula of Eq.20).5 m for a glass lens. if h = 1 tan δ = or f f f for small .33. The power of a lens of focal length of 1 metre is one dioptre.8 (i) If f = 0. P= 1 f EXAMPLE 9. we have 1 1 ⎞ ⎛ 1 = (n − 1) ⎜ − ⎟ ⎝ 12 10 −15 ⎠ This gives n = 1. when an optician prescribes a corrective lens of power + 2.20 Power of a lens. n1 = 1. (9. Thus.5. 9. R 1 and R 2. R2 = –15 cm. Substituting the values.5. refractive index for air-glass = 1.4 Combination of thin lenses in contact Consider two lenses A and B of focal length f 1 and f 2 placed in contact with each other.25) (9.2 cm. What is focal length in water? (Refractive index of air-water = 1.33) ⎢ − ⎥ f ⎣ R1 R 2 ⎦ Combining these two equations. Hence. f = +20 cm.Physics length bends the incident light more. The sign convention has to be applied for f .5 D.26) value of δ. the required lens is a convex lens of focal length + 40 cm. What is the refractive index of glass? (iii) A convex lens has 20 cm focal length in air.8 © no N C tt E o R be T re pu 328 Example 9. we find f = + 78.22). is 9. d tan δ = h 1 1 δ = .33. what is the power of the lens? (ii) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Power of a lens is positive for a converging lens and negative for a diverging lens.33 1 ⎤ = (1.5.5 ⎢ − ⎥ 20 ⎣ R1 R 2 ⎦ For the same glass lens in water. n 2 = 1.5. The power P of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light falling at unit distant from the optical centre (Fig. A lens of power of – 4. FIGURE 9. Refractive index of air is taken as unity. Let the object be placed at a point O beyond the focus of he (9. (ii) Here.5. n1 = 1. n 2 = 1. bl The SI unit for power of a lens is dioptre (D): 1D = 1m–1. Thus. Therefore.) Solution (i) Power = +2 dioptre.5 −1. ⎡1 1. R 1 = +10 cm. we have f = +12 cm. Its focal length is 12 cm. the lens formula gives ⎡1 1 1 ⎤ = 0. (iii) For a glass lens in air. m 3 .. we get © no N C tt E o R be T re pu 1 1 1 1 − = + f2 v u f1 1 1 1 − = v u f so that we get 1 1 1 = + f f1 f 2 1 1 1 1 = + + +… f f1 f 2 f3 In terms of power.25) implies that the total magnification m of the combination is a product of magnification (m 1.21 Image formation by a from the first lens gets modified in accordance with the angle at which they strike the second lens. f3 . bl (9.. (9. Note that the sum in Eq. For the image formed by the first lens A.27) and (9.33) m = m1 m 2 m 3 ..31) where P is the net power of the lens combination.) of individual lenses (9. 9.. so some of the terms on the right side may be positive (for convex lenses) and some negative (for concave lenses). we get is 329 1 1 1 − = v1 u f 1 For the image formed by the second lens B. Eq. The first lens produces an image at I1.31) can be written as P = P1 + P 2 + P3 + … If the two lens-system is regarded as equivalent to a single lens of focal length f.Ray Optics and Optical Instruments the first lens A (Fig. f2 .21). the lenses are thin. Let this central point be denoted by P. Since combination of two thin lenses in contact. In fact.32) is an algebraic sum of individual powers. we assume the optical centres of the lenses to be coincident. the direction of rays emerging FIGURE 9. however. Eq.28).. we have The derivation is valid for any number of thin lenses in contact. Since the image formed by the first lens becomes the object for the second. (9..29) (9. It also enhances sharpness of the image. are in contact. the effective focal length of their combination is given by (9.32) 1 1 1 − = v v1 f 2 Adding Eqs. Combination of lenses helps to obtain diverging or converging lenses of desired magnification. producing the final image at I. be borne in mind that formation of image by the first lens is presumed only to facilitate determination of the position of the final image. If several thin lenses of focal length f1 . (9.27) he d . (9.30) (9.28) (9. Since image I1 is real. m 2.. It must. we get (9. it serves as a virtual object for the second lens B.. 9.9 or or 1 1 1 = + v3 ∞ 30 v3 = 30 cm The final image is formed 30 cm to the right of the third lens. it serves as a virtual object for the second lens. microscopes. FIGURE 9.22. or v2 = ∞ The virtual image is formed at an infinite distance to the left of the second lens.22 Solution Image formed by the first lens 1 1 1 − = v1 u1 f1 1 1 1 − = v1 −30 10 © no N C tt E o R be T re pu 1 1 1 − = v2 10 − 10 or v1 = 15 cm The image formed by the first lens serves as the object for the second. which means that the rays appear to come from it for the second lens.9 Find the position of the image formed by the lens combination given in the Fig. Example 9.6 REFRACTION THROUGH A 330 Figure 9. The angle between the emergent ray RS and the direction of the incident ray PQ is called the angle of deviation. 9. The angles of incidence and refraction at the first face AB are i and r1.Physics Such a system of combination of lenses is commonly used in designing lenses for cameras. bl PRISM is he d . This acts as an object for the third lens.23 shows the passage of light through a triangular prism ABC. δ . Though the image is real. This is at a distance of (15 – 5) cm = 10 cm to the right of the second lens. telescopes and other optical instruments. 1 1 1 − = f3 v3 u 3 EXAMPLE 9. while the angle of incidence (from glass to air) at the second face AC is r2 and the angle of refraction or emergence e. e. Equation (9. δ = i + e–A (9.Ray Optics and Optical Instruments In the quadrilateral AQNR. Therefore.35) gives Dm = 2i – A. A plot between the angle of deviation and angle of incidence is shown in Fig. 9. 9. we get (9. i. ∠A + ∠QNR = 180º From the triangle QNR.36) (9. i = e which implies r1 = r2 . the angle of deviation depends on the angle of incidence. We have δ = Dm. δ remains the same if i and e are interchanged. This. For a small angle prism. bl 331 is he d r1 + r2 + ∠QNR = 180º . You can see that. is expected from the symmetry of i and e in Eq. Dm is also very small. this is related to the fact that the path of ray in Fig. or i = (A + Dm )/2 n 2 sin [( A + Dm )/ 2] = n1 sin [ A / 2] (9.38) thus provides a method of determining refractive index of the material of the prism. (9. FIGURE 9. resulting in the same angle of deviation..24.34) r1 + r2 = A The total deviation δ is the sum of deviations at the two faces. thin prisms do not deviate light much. in general. corresponds to two values i and hence of e . a thin prism.34) gives The angles A and D m can be measured experimentally. Comparing these two equations.23 can be traced back.35). the refracted ray inside the prism becomes parallel to its base. in fact. Physically. Eq.. and we get FIGURE 9.23 A ray of light passing through a triangular glass prism. two of the angles (at the vertices Q and R) are right angles.35) Thus. except for i = e. At the minimum deviation Dm. i. the sum of the other angles of the quadrilateral is 180º. © no N C tt E o R be T re pu 2r = A or r = A 2 In the same way.37) The refractive index of the prism is n 21 = (9. (9. Equation (9. any given value of δ. ( A + Dm ) /2 A/2 Dm = ( n 21 –1)A It implies that.24 Plot of angle of deviation (δ ) versus angle of incidence ( i ) for a triangular prism.e.38) n 21 = sin [( A + Dm )/ 2] sin [ A /2] δ = (i – r 1 ) + (e – r2 ) that is. Each ray splits into component colours when it enters the glass prism. from γ -rays to radio deviation of different colours shown is highly waves. is incident on a glass prism. Thus. but in an inverted position. light is at the long wavelength end (~700 nm) while the violet light is at the short wavelength end (~ 400 nm). while the violet light bends the most (Fig. the bending © no N C tt E o R be T re pu bl is he d . does not exist.25 Dispersion of sunlight or white light magnetic spectrum over the large range on passing through a glass prism. red dispersion of white light.25). Isaac Newton settled the issue once for all. b lue. The phenomenon of splitting of light into its component colours is known as dispersion . The relative of wavelengths. It must be understood here that a ray of light. y ellow. usually called white light. For example. green. Does the prism itself create colour in some way or does it only separate the colours already present in white light? In a classic experiment known for its simplicity but great significance. but broadly. The resulting emergent beam was found to be white light. The pattern of colour components of light is called the spectrum of light. When those coloured rays come out on the other side. and let the emergent beam from the first prism fall on the second prism (Fig. of which the spectrum of light exaggerated. the emergent light is seen to be consisting of several colours. The word spectrum is now used in a much more general sense: we discussed in Chapter 8 the electroFIGURE 9. An actual ray is really a beam of many rays of light. as defined mathematically. indigo.7 DISPERSION BY A PRISM It has been known for a long time that when a narrow beam of sunlight. 9. they again produce a white beam.26 Schematic diagram of We now know that colour is associated with Newton’s classic experiment on wavelength of light. FIGURE 9. Though the reason for appearance of spectrum is now common knowledge. He put another similar prism. the different component colours that appear in sequence are: v iolet. The red light bends the least.26). orange and r ed (given by the acronym VIBGYOR). 9. In the visible spectrum. The explanation was clear— the first prism splits the white light into its component colours. while the inverted prism recombines them to give white light. white light itself consists of light of different colours. There is actually a continuous variation of colour. Dispersion takes place because the refractive index of medium 332 for different wavelengths (colours) is different. (visible spectrum) is only a small part.Physics 9. which are separated by the prism. it was a matter of much debate in the history of physics. This is a phenomenon due to combined effect of dispersion. white clouds. glass is a dispersive medium. SUNLIGHT http://www.639 1. This also follows from the fact that sunlight reaches us in the form of white light and not as its components. consider Fig.1 589.517 1.3 656.627 1.9 486. therefore.htm Formation of rainbows The interplay of light with things around us gives rise to several beautiful phenomena. the redhue at sunrise and sunset.622 d .2 REFRACTIVE INDICES FOR DIFFERENT WAVELENGTHS Colour Violet Blue Yellow Red Wavelength (nm) 396.27(a). Next.8 SOME N ATURAL PHENOMENA DUE TO The variation of refractive index with wavelength may be more pronounced in some media than the other. thick lenses show chromatic aberration due to dispersion of light. Equivalently. the speed of light is independent of wavelength. refraction and reflection of sunlight by spherical water droplets of rain. vacuum (or air approximately) is a non-dispersive medium in which all colours travel with the same speed. Sunlight is first refracted as it enters a raindrop. shells. The spectacle of colour that we see around us all the time is possible only due to sunlight. Longer wangelength of light (red) are bent the least while the shorter wavelength (violet) are bent the most. (9. the rainbow. red light travels faster than violet light in a glass prism. these component rays strike bl 333 is he Flint glass 1.ucar. are just a few of the natural wonders we are used to.eo. The blue of the sky.3 Crown glass 1.663 1.8. TABLE 9. and wings of birds. An observer can therefore see a rainbow only when his back is towards the sun. In vacuum.2 gives the refractive indices for different wavelength for crown glass and flint glass. Thick lenses could be assumed as made of many prisms.atoptics. On the other hand.Ray Optics and Optical Instruments of red component of white light is least while it is most for the violet.co.533 1.523 1. of course.515 © no N C tt E o R be T re pu 9. Table 9.uk/bows. which causes the different wavelengths (colours) of white light to separate. 9. Thus. The conditions for observing a rainbow are that the sun should be shining in one part of the sky (say near western horizon) while it is raining in the opposite part of the sky (say eastern horizon).edu/rainbows http://www. In order to understand the formation of rainbows. the brilliant colours of some pearls. We describe some of them here from the point of view of physics.1 The rainbow The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. The reflected light is refracted again as it comes out of the drop as shown in the figure.27 Rainbow: (a) The sun rays incident on a water drop get refracted twice and reflected internally by a drop. For other colours. bl is he d .Physics © no N C tt E o R be T re pu 334 FIGURE 9. the inner surface of the water drop and get internally reflected if the angle between the refracted ray and normal to the drop surface is greater then the critical angle (48º. and (c) secondary rainbow is formed by rays undergoing internal reflection twice inside the drop. It is found that the violet light emerges at an angle of 40º related to the incoming sunlight and red light emerges at an angle of 42º. (b) Enlarge view of internal reflection and refraction of a ray of light inside a drop form primary rainbow. angles lie in between these two values. in this case). a secondary rainbow is formed as shown in Fig. therefore.9 OPTICAL INSTRUMENTS As sunlight travels through the earth’s atmosphere. Further. lenses and prisms. 9. A number of optical devices and instruments have been designed utilising reflecting and refracting properties of mirrors. When light rays undergoes two internal reflections inside a raindrop. clouds FIGURE 9. microscopes are some bl is he 335 d . The violet from drop 1 and red light from drop 2 are directed at level above or below the observer. i. are generally white. The intensity of light is reduced at the second reflection and hence the secondary rainbow is fainter than the primary rainbow. This is known as Rayleigh scattering). For a << λ. Thus.Ray Optics and Optical Instruments Figure 9. 9. telescopes. and the scatterer (of typical size. But since our eyes are more sensitive to blue than violet. we see the sky blue. (The amount of scattering is inversely proportional to the fourth power of the wavelength. the sun’s rays have to pass through a larger distance in the atmosphere (Fig. refraction.28). This explains the reddish appearance of the sun and full moon near the horizon. say.28 Sunlight travels through a longer which have droplets of water with a >> λ distance in the atmosphere at sunset and sunrise. large dust or ice particles) this is not true. one has Rayleigh scattering which is proportional to 1/λ4. it gets scattered (changes its direction) by the atmospheric particles. the bluish colour predominates in a clear sky. kaleidoscope. having a shorter wavelength.27(c). raindrops.e. Thus the observer sees a rainbow with red colour on the top and violet on the bottom.2 Scattering of light © no N C tt E o R be T re pu 9. Most of the blue and other shorter wavelengths are removed by scattering. 9.27(b) explains the formation of primary rainbow. the order of the colours is reversed in it as is clear from Fig. the sun looks reddish. Periscope. violet gets scattered even more than blue. At sunset or sunrise. It is due to four-step process. Large particles like dust and water droplets present in the atmosphere behave differently. the primary rainbow is a result of three-step process.. since blue has a shorter wavelength than red and is scattered much more strongly. all wavelengths are scattered nearly equally. reflection and refraction. 9.8. Light of shorter wavelengths is scattered much more than light of longer wavelengths. large scattering objects (for example. Thus.27(c). Hence. binoculars. The least scattered light reaching our eyes. For a >> λ. that is. We see that red light from drop 1 and violet light from drop 2 reach the observers eye. a). The relevant quantity here is the relative size of the wavelength of light λ. In fact. instead of one as in the primary rainbow. and the image is blurred.5 cm). To compensate this. the image appears blurred. of course. the lens cannot curve enough to focus the image on to the retina. The closest distance for which the lens can focus light on the retina is called the least distance of distinct vision.) This distance increases with age. the light from a distant object arriving at the eye-lens may get converged at a point in front of the retina. the focal length of the eye lens becomes shorter by the action of the ciliary muscles. It is corrected by using a converging lens for reading. we then go on to describe the principles of working of the microscope and the telescope. They deserve our appreciation for their courage and conviction. For example. The shape (curvature) and therefore the focal length of the lens can be modified somewhat by the ciliary muscles. When the object is brought closer to the eye. one of the most important optical device the nature has endowed us with. For example. respectively. Yet many amongst us bravely face this challenge by effectively overcoming their limitations to lead a normal life. 9. © no N C tt E o R be T re pu 336 bl is he d . if an elderly person tries to read a book at about 25 cm from the eye. This property of the eye is called accommodation .Physics examples of optical devices and instruments that are in common use. This means that the eye is producing too much convergence in the incident beam. we interpose a concave lens between the eye and the object. in order to maintain the same image-lens distance (≅ 2. Our eye is. The retina is a film of nerve fibres covering the curved back surface of the eye. This condition (defect of the eye) is called presbyopia.29(b)]. In spite of all precautions and proactive action. the focal length is about 2. We shall restrict our discussion to some common optical defects of the eye. our eyes may develop some defects due to various reasons. This type of defect is called nearsightedness or myopia . when the muscle is relaxed. our eyes are marvellous organs that have the capability to interpret incoming electromagnetic waves as images through a complex process. (Often the near point is given the symbol D. or the near point. The retina contains rods and cones which sense light intensity and colour. and may increase to as much as 200 cm at 60 years of age. Starting with the eye.9. Imagine the world without a pair of functional eyes. 9. The standard value for normal vision is taken as 25 cm. with the diverging effect desired to get the image focussed on the retina [Fig. The near point may be as close as about 7 to 8 cm in a child ten years of age. the cornea. Light enters the eye through a curved front surface. because of the decreasing effectiveness of the ciliary muscle and the loss of flexibility of the lens. The size of the pupil can change under control of muscles. The light is further focussed by the eye lens on the retina. It passes through the pupil which is the central hole in the iris. and transmit electrical signals via the optic nerve to the brain which finally processes this information. Thus.29 (a) shows the eye. If the object is too close to the eye. These are our greatest assets and we must take proper care to protect them.5 cm and objects at infinity are in sharp focus on the retina. Thus.1 The eye Figure 9. (b) shortsighted or myopic eye and its correction. the desired focal length is given by 1 1 1 = − f v u 1 1 1 1 − = or = f –50 – 25 50 orf = + 50 cm (convex lens).29(c)].29 (a) The structure of the eye. focussing in either the vertical or the horizontal plane may not be as sharp as in the other plane. If a person with such a defect in eye-lens looks at a wire mesh or a grid of lines. (c) farsighted or hypermetropic eye and its correction. Similarly. Example 9.10 What focal length should the reading spectacles have for a person for whom the least distance of distinct vision is 50 cm? Solution The distance of normal vision is 25 cm. its image should be formed at v = –50 cm. a convergent lens is needed to compensate for the defect in vision. So if a book is at u = –25 cm. Another common defect of vision is called astigmatism.10 is 337 he d . This defect is called farsightedness or hypermetropia [Fig. This defect can occur along with myopia or hypermetropia. 9. Astigmatism results in lines in one direction being well focussed while those in a perpendicular direction may appear distorted [Fig.Ray Optics and Optical Instruments © no N C tt E o R be T re pu FIGURE 9. Therefore. This occurs when the cornea is not spherical in shape. if the eye-lens focusses the incoming light at a point behind the retina. Astigmatism can be corrected by using a cylindrical lens of desired radius of curvature with an appropriately directed axis. and (d) astigmatic eye and its correction. bl EXAMPLE 9.29(d)]. For example. 9. the cornea could have a larger curvature in the vertical plane than in the horizontal plane or vice-versa. it produces virtual image of the object) at the far point of the eye which then can be focussed by the eye-lens on the retina.. (c) The above person prefers to remove his spectacles while reading a book.12 © no N C tt E o R be T re pu EXAMPLE 9. Explain why? 338 EXAMPLE 9. the person prefers to remove the spectacles while reading. which then gets focussed on the retina.. Hence the person prefers not to use the spectacles for far objects.11 (a) The far point of a myopic person is 80 cm in front of the eye. The angular size of the book (or its image) at the greater distance is evidently less than the angular size when the book is placed at 25 cm and no spectacles are needed. but because it brings the object (i. In order to read a book with the spectacles.11 Solution (a) Solving as in the previous example. However. Example 9.25 dioptres.e.e. i. such a person must keep the book at a distance greater than 25 cm so that the image of the book by the concave lens is produced not closer than 25 cm. bl is he d . The angular size of this image is the same as that of the object.12 (a) The near point of a hypermetropic person is 75 cm from the eye. (b) The corrective lens produces a virtual image (at 75 cm) of an object at 25 cm.e. it may have enough converging power to focus parallel rays from infinity on the retina of the shortened eyeball. i.e.. reduces the size of the object. i. v = – 75 cm 1/f = 1/25 – 1/75. we find that the person should use a concave lens of focal length = – 80 cm.. What is the power of the lens required to enable the person to read clearly a book held at 25 cm from the eye? (b) In what way does the corrective lens help the above person? Does the lens magnify objects held near the eye? (c) The above person prefers to remove the spectacles while looking at the sky. The eye is able to see distant objects not because the corrective lens magnifies the object. In this sense the lens does not magnify the object but merely brings the object to the near point of the hypermetric eye..e.5 cm. (c) The myopic person may have a normal near point. Wearing spectacles of converging lenses (used for near vision) will amount to more converging power than needed for parallel rays. of power = – 1. Explain why? Solution (a) u = – 25 cm. What is the power of the lens required to enable him to see very distant objects clearly? (b) In what way does the corrective lens help the above person? Does the lens magnify very distant objects? Explain carefully. Hence. in fact. but the angle subtended by the distant object at the eye is the same as the angle subtended by the image (at the far point) at the eye.67 dioptres. (b) No. about 25 cm (or even less). The corrective lens needs to have a converging power of +2. The concave lens.Physics Example 9. the angular size is greater than that of the same object at the near point (75 cm) viewed without the spectacles. (c) A hypermetropic eye may have normal far point i. f = 37. and the eye is positioned close to the lens on the other side. magnified and virtual image of the object at a distance so that it can be viewed comfortably. We will now find the magnification when the image is at infinity.e.. Suppose the object has a height h. to have a magnification of six.39) f ⎠ ⎝ Since D is about 25 cm. 9. one needs a convex lens of focal length. The angle subtended is then given by (9. it causes some strain on the eye. (b) the angle object and h ′ the size of the image. magnifying lens is located such that the Note that m = h ′/ h where h is the size of the image is at the near point.. the image is at infinity. ) What a single-lens simple magnifier achieves is that it allows the object to be brought closer to the eye than D.40) bl 339 is he d .e. by a simple microscope can be obtained by using the relation m= v ⎛1 1 ⎞ ⎛ v⎞ = v ⎜ – ⎟ = ⎜1 – ⎟ f⎠ u ⎝v f ⎠ ⎝ © no N C tt E o R be T re pu Now according to our sign convention. Therefore. i. the image is virtual and closer than infinity. if placed at D for near the focal point of the lens. However.Ray Optics and Optical Instruments 9. if the object is at a distance slightly less than the focal length of the lens. In order to use such a lens as a microscope. the lens is held near the object. the magnification is ⎛h ⎞ tan θ o = ⎜ ⎟ ≈ θ o ⎝D⎠ D⎞ ⎛ m = ⎜1 + ⎟ (9. 9. (a) the f = 5 cm. which is h/u. is when it is at the near point. and the second in Fig. The idea is to get an erect. the image formed at infinity is often considered most suitable for viewing by the relaxed eye. This is also the subtanded by the object.9. v is negative. the image comfortable viewing. is the same as ratio of the angle subtended by the image that at the near point. one focal length away or less.30). The linear magnification m .30 A simple microscope.2 The microscope A simple magnifier or microscope is a converging lens of small focal length (Fig. The maximum angle it can subtend. In this case we will have to obtained the angular magnification. and is equal in magnitude to D. at 25 cm or more. for the image formed at the near point D. If the object is at a distance f .30(b) and (c). 9. FIGURE 9. We show both cases. and be clearly visible (without a lens). and (c) the object to that subtended by the object. actually subtended by the object at the eye. Although the closest comfortable distance for viewing the image is when it is at the near point (distance D ≅ 25 cm). i. the first in Fig.30(a). a distance D. Thus. (Note that this is not the angle is far off but closer than infinity. namely h ′/ h. Eq. produces the final image. equals (9. 9. tanθ i = θi = ⎜ ⎟ ⎝f⎠ ⎛θ ⎞ D m =⎜ i⎟ = f ⎝ θo ⎠ ⎛h⎞ as is clear from Fig. due to the objective. inverted. 9. the final image is inverted with respect to the original object. which is enlarged and virtual. In subsequent discussions of optical instruments (microscope and telescope) we shall assume the image to be at infinity. A simple microscope has a limited maximum magnification (≤ 9) for realistic focal lengths. the eyepiece .43) 340 he (9. The lens nearest the object.31. therefore is bl © no N C tt E o R be T re pu h′ L = fo h where we have used the result mO = ⎛ h ⎞ ⎛ h ′⎞ tan β = ⎜ ⎟ = ⎜ ⎟ ⎝ fo ⎠ ⎝ L ⎠ This is one less than the magnification when the image is at the near point. The first inverted image is thus near (at or within) the focal plane of the eyepiece. 9. The angular magnification is. magnified image of the object. Clearly. at a distance appropriate for final image formation at infinity.29(c). forms a real. or a little closer for image formation at the near point.Physics We now find the angle subtended at the eye by the image when the object is at u. (9. For much larger magnifications. This serves as the object for the second lens. one uses two lenses.41) (9. which functions essentially like a simple microscope or magnifier. A schematic diagram of a compound microscope is shown in Fig. when it = ⋅ = −v −v u −u is at u = –f. but the viewing is more comfortable and the difference in magnification is usually small. called the objective.31 shows that the (linear) magnification microscope. The ray diagram of formation of image by a compound Fig. The angle subtended by the object.39). one compounding the effect of the other. We now obtain the magnification due to a FIGURE 9. This is known as a compound microscope. From the relations h′ v =m = h u we have the angle subtended by the image h′ h v h ≈θ .42) d .31 Ray diagram for the compound microscope. contribute to the quality and visibility of the image. i.33)].46) α fe h fe In this case.39)]. As the first inverted image is near the focal point of the eyepiece.0 cm. is http://astro. (9..45) ⎝ fo ⎠ ⎝ fe ⎠ Clearly. The eyepiece magnifies this image producing a final inverted image.42)] is me = (D/f e ) [9. For example. (9. the objective and eyepiece should have small focal lengths. the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length fe ) is called the tube length of the compound microscope. the objective has a large focal length and a much larger aperture than the eyepiece. to achieve a large magnification of a small object (hence the name microscope).32). The distance L.44(b)] bl Thus.0 cm. the length of the telescope tube is fo + fe .44(a)] fe ⎠ ⎝ When the final image is formed at infinity.Ray Optics and Optical Instruments Here h ′ is the size of the first image. the total magnification [(according to Eq. is ⎛ D⎞ m e = ⎜1 + ⎟ [9.e. is ⎛ L⎞ ⎛D⎞ m = mome = ⎜ ⎟ ⎜ ⎟ (9. The first image is formed near the focal point of the eyepiece. 9. The magnifying power m is the ratio of the angle β subtended at the eye by the final image to the angle α which the object subtends at the lens or the eye. Also large lenses are required to make L large. = o m ≈ ≈ (9.3 Telescope 20 25 × = 250 1 2 Various other factors such as illumination of the object. the magnification is ⎛ L⎞ m = mome = ⎜ ⎟ ⎝ fo ⎠ = ⎛D⎞ ⎜ ⎝ fe ⎟ ⎠ © no N C tt E o R be T re pu 9.9. with an objective with fo = 1. But here. the object size being h and f o being the focal length of the objective. multicomponent lenses are used for both the objective and the eyepiece to improve image quality by minimising various optical aberrations (defects) in lenses. (9. it is difficult to make the focal length much smaller than 1 cm. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. In modern microscopes. when the final image is formed at the near point. when the image is formed at infinity. and a tube length of 20 cm.html The world’s largest optical telescopes he 341 d . we use the result from the discussion above for the simple microscope to obtain the (angular) magnification m e due to it [Eq.nineplanets. Hence β h fo f . and an eyepiece with focal length fe = 2. the angular magnification due to the eyepiece [Eq. In practice. It also has an objective and an eyepiece. The telescope is used to provide angular magnification of distant objects (Fig.org/bigeyes. if a parabolic reflecting surface is chosen. 342 which now passes through a hole in the objective primary mirror. This is what is done in the very large 200 inch (~5. Second. Mt. The magnifying power of this telescope is m = 100/1 = 100. The viewer sits near the focal point of the FIGURE 9. Refracting telescopes can be used both for terrestrial and astronomical observations. not just over its rim. consider a telescope whose objective has a focal length of 100 cm and the eyepiece a focal length of 1 cm. spherical aberration is also removed. is shown © no N C tt E o R be T re pu bl is he d . modern telescopes use a concave mirror rather than a lens for the objective. or the ability to observe two objects distinctly. The main considerations with an astronomical telescope are its light gathering power and its resolution or resolving power. in addition. The stars appear as though they FIGURE 9. Let us consider a pair of stars of actual separation 1′ (one minute of arc). Further. also depends on the diameter of the objective. For example. Another solution to telescope (Cassegrain). The resolving power. difficult to make and support by their edges. So. With larger diameters. obstructing some light (depending on the size of the observer cage). a pair of inverting lenses to make the final image erect.67º. Palomar telescope. it is rather difficult and expensive to make such large sized lenses which form images that are free from any kind of chromatic aberration and distortions. the problem is to deflect the light being focussed by another mirror. First. Such big lenses tend to be very heavy and therefore. in a small cage. California. fainter objects can be observed. The former clearly depends on the area of the objective.32 A refracting telescope.02 m).33 Schematic diagram of a reflecting mirror. One such arrangement using a convex secondary mirror to focus the incident light. which are in very nearly the same direction. One must have an eyepiece and the observer right there.08 m) diameters. They have several advantages. It is at the Yerkes Observatory in Wisconsin. USA. there is no chromatic aberration in a mirror. For these reasons. are separated by an angle of 100 × 1 ′ = 100 ′ =1. Telescopes with mirror objectives are called reflecting telescopes. One obvious problem with a reflecting telescope is that the objective mirror focusses light inside the telescope tube. and can be supported over its entire back surface. The largest lens objective in use has a diameter of 40 inch (~1. Mechanical support is much less of a problem since a mirror weighs much less than a lens of equivalent optical quality. the desirable aim in optical telescopes is to make them with objective of large diameter.Physics Terrestrial telescopes have. refracted ray and normal lie in the same plane. f is positive for a convex mirror. after multiple internal reflections. total internal reflection occurs. are some examples of total internal reflection. of refractive index n 2 placed in a medium of refractive index n 1. respectively) n2 n1 n 2 − n 1 − = v u R Thin lens formula 1 1 1 − = f v u bl 343 is he d . reflection and refraction are i. where the incident ray. This is known as a Cassegrain telescope. n 21 = n 2 sin ⎡ ⎣ ( A + D m ) /2⎤ ⎦ = n1 sin ( A / 2) 6. The largest telescope in India is in Kavalur.34 m diameter reflecting telescope (Cassegrain). Light incident at an angle at one end comes out at the other. SUMMARY 1. © no N C tt E o R be T re pu 3.Ray Optics and Optical Instruments in Fig. is that angle for which the angle of refraction is 90°. even if the fibre is bent. For i > i c. For a prism of the angle A. All distances are measured from the pole/optic centre of the mirror/lens on the principal axis. Reflection is governed by the equation ∠i = ∠r ′ and refraction by the Snell’s law. reflected ray. Tamil Nadu. Optical fibres consist of glass fibres coated with a thin layer of material of lower refractive index.33. USA. where Dm is the angle of minimum deviation. r ′ and r. f is negative for concave mirror. those measured in the opposite direction are negative. The heights measured downwards are taken as negative. polished. set up. Multiple internal reflections in diamond (ic ≅ 24. 2. It has the advantages of a large focal length in a short telescope. Angles of incidence. 1 1 1 + = f v u where u and v are object and image distances. It was ground. 9. and is being used by the Indian Institute of Astrophysics. The heights measured upwards above x-axis and normal to the principal axis of the mirror/ lens are taken as positive. Cartesian sign convention: Distances measured in the same direction as the incident light are positive. It is a 2. Mirror equation: 5. The critical angle of incidence ic for a ray incident from a denser to rarer medium. f is (approximately) half the radius of curvature R.4°). The largest reflecting telescopes in the world are the pair of Keck telescopes in Hawaii. 4. Bangalore. respectively. after its inventor. For refraction through a spherical interface (from medium 1 to 2 of refractive index n 1 and n 2 . totally reflecting prisms and mirage. respectively and f is the focal length of the mirror. sini /sinr = n. with a reflector of 10 metre in diameter. m= L D × fo fe where fo and fe are the focal lengths of the objective and eyepiece. 10. Approximately. f2 . is given by 7. respectively. f is positive for a converging lens. 2.. The SI unit for power of a lens is dioptre (D): 1 D = 1 m–1. The power of a lens P = 1/f. POINTS TO PONDER 1. are in contact. 344 The real image of an object placed between f and 2f from a convex lens can be seen on a screen placed at the image location. a converging corrective lens is needed. In a defective eye. f3 .Physics Lens maker’s formula 1 (n 2 − n1 ) ⎛ 1 1⎞ = − ⎜ f n1 ⎝ R1 R 2 ⎟ ⎠ R 1 and R 2 are the radii of curvature of the lens surfaces. The laws of reflection and refraction are true for all surfaces and pairs of media at the point of the incidence. because it is difficult to reconcile ourselves with an image suspended in air bl is he d . where D = 25 cm is the least distance of distinct vision and f is the focal length of the convex lens. © no N C tt E o R be T re pu 9. This focal length can be varied somewhat so that the image is always formed on the retina. is the magnification due to the eyepiece and mo is the magnification produced by the objective. respectively. a diverging corrective lens is needed. Astigmatism is corrected by using cylindrical lenses. This ability of the eye is called accommodation. For a compound microscope. if the image is focussed before the retina (myopia). f is negative for a diverging lens.5 cm. m = D/f. If the image is at infinity. 1 1 1 1 = + + +… f f1 f2 f3 The total power of a combination of several lenses is P = P1 + P 2 + P3 + … Dispersion is the splitting of light into its constituent colours. m= β fo = fe α where f0 and fe are the focal lengths of the objective and eyepiece. the magnifying power is given by m = m e × m0 where me = 1 + (D/fe ).. the effective focal length of their combination. Magnifying power m of a telescope is the ratio of the angle β subtended at the eye by the image to the angle α subtended at the eye by the object. is the image still there? This question puzzles many. if the image is focussed beyond the retina (hypermetropia). 8. and L is the distance between their focal points. If several thin lenses of focal length f1. The Ey e: The eye has a convex lens of focal length about 2. Magnifying power m of a simple microscope is given by m = 1 + (D/f ). If the screen is removed. you would need to keep the small object at 25 cm which would subtend a very small angle. 2.2 9. 3.5 cm. If the candle is moved closer to the mirror. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9. The image is at 25 cm which we can see. by what distance would the microscope have to be moved to focus on the needle again? Figures 9.63 up to the same height. he 4. 9.Ray Optics and Optical Instruments without a screen. bl 345 is 5.34(a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass-air and water-air inter face.3 9. Predict the angle of refraction in glass when the angle of incidence in water is 45º with the normal to a water -glass interface [Fig. A tank is filled with water to a height of 12. In principle.4 cm. Give the location of the image and the magnification. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. What is the refractive index of water? If water is replaced by a liquid of refractive index 1. some of which reach our eye and we see the image. how would the screen have to be moved? A 4. d . Image formation needs regular reflection/refraction. But the image does exist. For a simple microscope.34(c)]. This can be seen by the images formed in air during a laser show. Without the microscope. The screen simply diffuses these rays. Rays from a given point on the object are converging to an image point in space and diverging away. © no N C tt E o R be T re pu EXERCISES 9. Thick lenses give coloured images due to dispersion. Yet it offers magnification because we can keep the small object much closer to the eye than 25 cm and hence have it subtend a large angle. all rays from a given point should reach the same image point. This is why you do not see your image by an irregular reflecting object. Describe what happens as the needle is moved farther from the mirr or. The variety in colour of objects we see around us is due to the constituent colours of the light incident on them.4 FIGURE 9. A monochromatic light may produce an entirely different perception about the colours on an object as seen in white light. respectively. the angular size of the object equals the angular size of the image.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm.34 A small candle.1 9. say the page of a book.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. Now a lens is placed in the path of the convergent beam 1 2 cm from P. 9.5 cm can bring an object placed at 9. If the prism is placed in water (refractive index 1. and (b) a concave lens of focal length 16 cm? An object of size 3. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece? (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. What happens if the object is moved further away from the lens? What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses. What is the radius of curvature required if the focal length is to be 20 cm? A beam of light converges at a point P.10 346 bl is 9.0 mm and an eyepiece of focal length 2. and (b) at infinity? What is the magnifying power of the microscope in each case? A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.55. If an eyepiece of focal length 1. A compound microscope consists of an objective lens of focal length 2.0 cm.14 9. with both faces of the same radius of curvature. (Consider the bulb to be a point source. Double-convex lenses are to be manufactured from a glass of refractive index 1. Describe the image produced by the lens.0 cm and an eyepiece of focal length 6. (c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm.7 9. Use the mirror equation to deduce that: (a) an object placed between f and 2 f of a concave mirror produces a real image beyond 2 f.6 9.9 © no N C tt E o R be T re pu 9. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm. (b) a convex mirror always produces a virtual image independent of the location of the object. What is the separation between the two lenses? Calculate the magnifying power of the microscope.8 he d . 0 mm from the objective in sharp focus. what is the angular magnification of the telescope? (b) If this telescope is used to view the moon.) A prism is made of glass of unknown refractive index.15 9.5 A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm.Physics 9. what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106 m.33. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1. 0 cm is used. The angle of minimum deviation is measured to be 40°. A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.8 × 108 m. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. predict the new angle of minimum deviation of a parallel beam of light. and the radius of lunar orbit is 3.33). How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25 cm).11 9. A parallel beam of light is incident on a face of the prism.25 cm separated by a distance of 15 cm.13 9.12 9. Yet when we ‘see’ a virtual image.10. (a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9. The outer covering of the pipe is made of a material of refractive index 1. Is there a contradiction? (c) A diver under water. does the apparent depth increase or decrease? (e) The refractive index of diamond is much greater than that of ordinary glass.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1. Would the fisher man look taller or shorter to the diver than what he actually is? (d) Does the apparent depth of a tank of water change if viewed obliquely? If so. if they are placed 8.21 Answer the following questions: (a) You have lear nt that plane and convex mirrors pr oduce virtual images of objects. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm.17 © no N C tt E o R be T re pu FIGURE 9. we always say. looks obliquely at a fisher man standing on the bank of a lake. we are obviously bringing it on to the ‘screen’ (i. as shown in the figure.20 9. The distance between the object bl 347 is he d . Is this fact of some use to a diamond cutter? The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens.35 9.18 9. Determine the focal length of the lens. the retina) of our eye. [Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams. (b) A virtual image. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all? (b) An object 1.44. Does the answer depend on the location of the slab? (a) Figure 9. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place. Can they produce real images under some circumstances? Explain. What is the maximum possible focal length of the lens required for the purpose? A screen is placed 90 cm from an object. (b) What is the answer if there is no outer covering of the pipe? 9.0 cm apart with their principal axes coincident.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. cannot be caught on a screen.Ray Optics and Optical Instruments (d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.16 9..19 9.68.5.e.] A small pin fixed on a table top is viewed from above from a distance of 50 cm. (a) What is the magnification produced by the lens? How much is the area of each square in the virtual image ? (b) What is the angular magnification (magnifying power) of the lens ? (c) Is the magnification in (a) equal to the magnifying power in (b)? Explain. You are given prisms made of crown glass and flint glass with a wide variety of angles. From this rough data estimate the range of accommodation (i.25 mm2. (a) At what distance should the lens be held from the figure in Exercise 9.30 9. What should be the distance between the object in Exercise 9.29 9. Determine the magnification produced by the two-lens system. A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones.22 and the convex lens is 40 cm . and the size of the image. and the least converging power of the eye-lens behind the cornea is about 20 dioptres. Does short-sightedness (myopia) or long-sightedness (hyper metropia) imply necessarily that the eye has partially lost its ability of accommodation? If not.0 dioptre for distant vision. (a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass ? (b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope? A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye. the range of converging power of the eye-lens) of a normal eye. What is this defect due to? How is such a defect of vision corrected ? A man with normal near point (25 cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm..27 9. what might cause these defects of vision? A myopic person has been using spectacles of power –1. At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.25 is he d .31 348 bl 9.28 9. During old age he also needs to use separate reading glass of power + 2.Physics 9.26 © no N C tt E o R be T re pu 9.e. Suggest a combination of prisms which will (a) deviate a pencil of white light without much dispersion.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.29 in order to view the squares distinctly with the maximum possible magnifying power? (b) What is the magnification in this case ? (c) Is the magnification equal to the magnifying power in this case? Explain. Would you be able to see the squares distinctly with your eyes very close to the magnifier ? 9. (b) disperse (and displace) a pencil of white light without much deviation. For a normal eye.0 dioptres. the far point is at infinity and the near point of distinct vision is about 25 cm in front of the eye.24 9. The cornea of the eye provides a converging power of about 40 dioptres.524. Explain what may have happened.23 9. where will the final image of an object at infinity be? Light incident normally on a plane mirror attached to a galvanometer coil retraces backwards as shown in Fig.] Answer the following questions: (a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass.5 of the mirror.29 to 9. when the final image is at infinity)? (b) the final image is formed at the least distance of distinct vision (25 cm)? (a) For the telescope described in Exercise 9.34 9. Why? How much should be that short distance between the eye and eyepiece? An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.37 FIGURE 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument. what is the height of the image of the tower formed by the objective lens? (c) What is the height of the final image of the tower if it is formed at 25 cm? A Cassegrain telescope uses two mirrors as shown in Fig. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power? (d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths? (e) When viewing through a compound microscope. Such a telescope is built with the mirrors 20 mm apart.35 9. A current in the coil o produces a deflection of 3. our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing.. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm. In what sense then does a magnifying glass provide angular magnification? (b) In viewing through a magnifying glass.34 (a). 9.36.36 bl 349 is he d .32 9. How will you set up the compound microscope? A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.5 m away? 9.33.e. what is the separation between the objective lens and the eyepiece? (b) If this telescope is used to view a 100 m tall tower 3 km away.0 cm. Does angular magnification change if the eye is moved back? (c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. one usually positions one’s eyes very close to the lens.33 © no N C tt E o R be T re pu 9. What is the displacement of the reflected spot of light on a screen placed 1. 9.25 cm and an eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.Ray Optics and Optical Instruments [Note: Exercises 9.36 9. The new distance is measured to be 30. What is the refractive index of the liquid? © no N C tt E o R be T re pu FIGURE 9.50) in contact with a liquid layer on top of a plane mirror. 0 cm. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle.0 cm. The liquid is removed and the experiment is repeated.Physics 9.38 Figure 9.37 350 bl is he d .37 shows an equiconvex lens (of refractive index 1. The distance of the needle from the lens is measured to be 45. As we will see.Wave Optics Chapter Ten WAVE OPTICS 10. It explained the laws of reflection and refraction of light at an interface. the wave model could satisfactorily explain the phenomena of reflection and refraction. This is in contradiction to the prediction made by using the corpuscular model of light. Foucault carried out this experiment in 1850. It was much later confirmed by experiments where it was shown that the speed of light in water is less than the speed in air confirming the prediction of the wave model. The wave theory was not readily accepted primarily because of 351 Newton’s authority and also because light could travel through vacuum . the Dutch physicist Christiaan Huygens put forward the wave theory of light – it is this wave model of light that we will discuss in this chapter. In 1678.1 INTRODUCTION In 1637 Descartes gave the corpuscular model of light and derived Snell’s law. however. The corpuscular model predicted that if the ray of light (on refraction) bends towards the normal then the speed of light would be greater in the second medium. it predicted that on refraction if the wave bends towards the normal then the speed of light would be less in the second medium. the corpuscular model is very often attributed to Newton. This corpuscular model of light was further developed by Isaac Newton in his famous book entitled OPTICKS and because of the tremendous popularity of this book. From the wave equation. However.Physics and it was felt that a wave would always require a medium to propagate from one point to the other.5 µm. Marconi made practical applications of the Hertzian waves 352 . when Thomas Young performed his famous interference experiment in 1801. F r o m t h i s . the branch of optics in which one completely neglects the finiteness of the wavelength is called geometrical optics and a ray is defined as the path of energy propagation in the limit of wavelength tending to zero. these experiments could only be satisfactorily explained by assuming a wave model of light. Bose and G. it was much later (around 1890) that Heinrich Hertz produced radiowaves in the laboratory. In Sections 10. In this chapter we will first discuss the original formulation of the Huygens principle and derive the laws of reflection and refraction. how could light waves propagate through vacuum. Thus. many experiments were carried out involving the interference and diffraction of lightwaves. light can be assumed to approximately travel in straight lines. light waves are associated with changing electric and magnetic fields. it was firmly established that light is indeed a wave phenomenon. The wavelength of visible light was measured and found to be extremely small. Indeed. according to Maxwell. for the next 40 years or so. This was explained when Maxwell put forward his famous electromagnetic theory of light.7 we will discuss the phenomenon of polarisation which is based on the fact that the light waves are transverse electromagnetic waves. Finally in Section 10. which we had discussed in the previous chapter. Because of the smallness of the wavelength of visible light (in comparison to the dimensions of typical mirrors and lenses).5. The changing electric and magnetic fields result in the propagation of electromagnetic waves (or light waves) even in vacuum. The only major difficulty was that since it was thought that a wave required a medium for its propagation. Maxwell had developed a set of equations describing the laws of electricity and magnetism and using these equations he derived what is known as the wave equation from which he predicted the existence of electromagnetic waves* . This is the field of geometrical optics.C. changing electric field produces a time and space varying magnetic field and a changing magnetic field produces a time and space varying electric field. around the middle of the nineteenth century.4 and 10. After the interference experiment of Young in 1801. Maxwell could calculate the speed of electromagnetic waves in free space and he found that the theoretical value was very close to the measured value of speed o f l i g h t . Thus. we will discuss the phenomenon of interference which is based on the principle of superposition. * Maxwell had predicted the existence of electromagnetic waves around 1855. for example. h e p r o p o u n d e d t h a t light must be an electromagnetic wave.6 we will discuss the phenomenon of diffraction which is based on HuygensFresnel principle. the wavelength of yellow light is about 0. the wave theory seemed to be very well established. In Section 10. J. Every point on the surface starts oscillating with time. If we have a point source emitting waves uniformly in all directions. This proves. aren’t you? In school. and the candle will still be seen. light has a wavelength of about half a micrometre. Such a locus of points.Wave Optics DOES LIGHT TRAVEL IN A STRAIGHT LINE? Light travels in a straight line in Class VI. that light travels in a straight line. it bends around it and reaches points behind the obstacle. and deals with mirrors. lenses. refraction. light will not be able to bend to that extent. Then what about the primary school experiment? What happens there is that when we move any cardboard. and will not be seen on the other side. waves spread out from the point of impact. place a candle on one side and look from the other side. a point source. The speed with which the wavefront moves outwards from the source is called the speed of the FIGURE 10. Thus a micrometre size obstacle will not be able to stop a light ray. light will be able to diffract. most of it is reflected and is heard as an echo. the same phase are spheres and we have what is known as a spherical 353 wave as shown in Fig. you are shown an experiment in which you take three cardboards with pinholes in them. it can diffract and interfere. At any instant. a . The sound wave of our speech has a wavelength of about 50 cm to 1 m. At a large distance from the source. it does not do so in Class XII and beyond! Surprised. so your teacher says. etc. This is a property of a wave in general. such as a hillock. Ray optics is based on rectilinear propagation of light. One could add to the first sentence in this box : It learns how to bend as it grows up ! 10. In the present book. which is much larger than the wavelength of light. But when it comes across a larger obstacle of a few hundred metres. you cannot see the candle. Clearly. that it can bend around objects. one on ray optics and the other on wave optics. Even if one of them is displaced a little. and you are told that light travels as a wave. If it meets an obstacle of the size of a few metres. reflection. etc. The wavefronts are then the locus of points which have the same amplitude and vibrate in spherical. you can see the candle. Then you come to the chapter on wave optics. all points on such a circle are oscillating in phase because they are at the same distance from the source. and can be seen in sound waves too. If it encounters an obstacle of about this size. thus a wavefront is defined as a surface of constant phase.1(a). In optical region. If we are able to move one of the cardboards by a micrometer or less. it can bend around it and can be seen on the other side. Hence the candle cannot be seen. If the obstacle is much larger. If the flame of the candle and the three pinholes are in a straight line. 10. The energy of the wave travels in a direction perpendicular to the diverging spherical wave emanating from wavefront. a photograph of the surface would show circular rings on which the disturbance is maximum. which oscillate in phase is called a wavefront .2 HUYGENS PRINCIPLE We would first define a wavefront: when we drop a small stone on a calm pool of water. however. there are two consecutive chapters. the displacement is of the order of a few millimetres.1 (a) A wave. then Huygens principle allows us to determine the shape of the wavefront at a later time τ. and G1G2 and In a similar manner. Let us consider a diverging the source.2 F1F2 represents the spherical wavefront (with O as centre) at t = 0.2). Now. of the wave.1(b)]. we can use Huygens principle to determine the represent rays. if we know the shape of the wavefront at t = 0. FIGURE 10. Huygens could time τ. 10. The envelope of the secondary wavelets emanating from F1F2 produces the forward moving wavefront G1G2.Physics small portion of the sphere can be considered as a plane and we have what is known as a plane wave [Fig. we draw spheres of radius vτ from each point on the spherical wavefront FIGURE 10. 10. right.2 is again plane wave propagating to the spherical with point O as the centre. we obtain the new position of the construction for a wavefront at t = τ. Thus. explain the absence of the backwave. each point of the portion of the spherical wave can wavefront is the source of a secondary disturbance and the wavelets be approximated by a emanating from these points spread out in all directions with the speed plane wave. are normal to both F1F2 not satisfactory and the absence of the backwave is really justified from more rigorous wave theory. 10.2. These wavelets emanating from the wavefront are usually referred to as secondary wavelets and if we draw a common tangent to all these spheres. The backwave D1D2 does not exist. if we wish to determine the shape of the wavefront at t = τ. 10. this adhoc assumption is B1B2 … etc. Huygens argued that the amplitude of the 1 2 t = 0 and G1G2 is the secondary wavelets is maximum in the forward direction and zero in the wavefront at a later backward direction. The new wavefront shown as G1G2 in Fig. F1 F2 is the The above model has one shortcoming: we also have a backwave which plane wavefront at is shown as D D in Fig. If we now draw Huygens geometrical a common tangent to all these spheres. Huygens principle is essentially a geometrical construction. However.3 where v represents the speed of the waves in the medium. The lines A1A2. . Thus. we obtain the new position of the wavefront at a later time.1 (b) At a large distance from the shape of the wavefront at a later time. Now. a small wave and let F1F2 represent a portion of the spherical wavefront at t = 0 (Fig. shape of the wavefront for a plane wave propagating through a medium 354 (Fig. which given the shape of the wafefront at any time allows us to determine FIGURE 10. according to Huygens principle.3). 10. by making this adhoc assumption. 4. as shown in Fig. BC v1τ = AC AC (10. His book. makes fascinating reading even today. We assume a plane wavefront AB propagating in the direction A′A incident on the interface at an angle i as shown in the figure. CHRISTIAAN HUYGENS (1629 – 1695) In order to determine the shape of the refracted wavefront. respectively. we draw a sphere of radius v2τ from the point A in the second medium (the speed of the wave in the second medium is v2).2) AC AC where i and r are the angles of incidence and refraction.4 A plane wave AB is incident at an angle i on the surface PP′ separating medium 1 and medium 2. Let PP′ represent the surface separating medium 1 and medium 2. The figure corresponds to v2 < v1 so that the refracted waves bends towards the normal. Let v1 and v2 represent the speed of light in medium 1 and medium 2.1) sin r = 355 . 10. respectively. The plane wave undergoes refraction and CE represents the refracted wavefront.1 Refraction of a plane wave We will now use Huygens principle to derive the laws of refraction. He brilliantly explained the double refraction shown by the mineral calcite in this work in addition to reflection and refraction. Let CE represent a tangent plane drawn from the point C on to the sphere.3. Thus. If we now consider the triangles ABC and AEC. AE = v2 τ and CE would represent the refracted wavefront. we readily obtain sin i = and AE v 2τ = (10. mathematician and the founder of the wave theory of light.3 REFRACTION AND REFLECTION OF P LANE W AVES USING H UYGENS PRINCIPLE 10. He discovered the true geometry of Saturn’s rings. astronomer. Christiaan Huygens (1629 – 1695) Dutch physicist. Let τ be the time taken by the wavefront to travel the distance BC. He was the first to analyse circular and simple harmonic motion and designed and built improved clocks and telescopes. Treatise on light. BC = v1 τ FIGURE 10. Then.Wave Optics 10. 8) n1 Thus.5. Proceeding in an exactly similar manner we can construct a refracted wavefront as shown in Fig. n1 = and c n2 = v 2 (10.7) The above equation implies that when a wave gets refracted into a denser medium (v1 > v2) the wavelength and the speed of propagation decrease but the frequency ν (= v/λ) remains the same. Obviously.4) are known as the refractive indices of medium 1 and medium 2..6) This is the Snell’s law of refraction. v2 > v1. This prediction is opposite to the prediction from the corpuscular model of light and as later experiments showed.3) can be written as n1 sin i = n2 sin r (10. refraction. we will not have sin i c = 356 .falstad.. for i > ic.3. respectively.5) c v1 (10.e. The angle of refraction will now be greater than angle of incidence. 10. respectively and if the distance BC is equal to λ 1 then the distance AE will be equal to λ 2 ( because if the crest from B has reached C in time τ. thus.3) Demonstration of interference. then. i. http://www. then the crest from A should have also reached E in time τ ). there can not be any refracted wave. we will still have n1 sin i = n2 sin r . diffraction. Eq. The angle ic is known as the critical angle and for all angles of incidence greater than the critical angle. the speed of the light wave in the second medium (v2) will be less then the speed of the light wave in the first medium (v1). if the ray bends toward the normal).com/ripple/ λ1 v BC = = 1 AE λ2 v2 or v1 v2 λ1 = λ2 (10. however. Now.2 Refraction at a rarer medium We now consider refraction of a plane wave at a rarer medium. resonance and Doppler effect From the above equation. (10. if c represents the speed of light in vacuum. if λ1 and λ 2 denote the wavelengths of light in medium 1 and medium 2. 10.e. we get the important result that if r < i (i. Further. the prediction of the wave theory is correct. We define an angle ic by the following equation n2 (10. if i = ic then sin r = 1 and r = 90°.Physics Thus we obtain sin i v = 1 sin r v2 (10. In terms of the refractive indices. and mirrors can be understood.6.6 Reflection of a plane wave AB by the reflecting surface MN. If we now consider the triangles EAC and BAC we will find that they are congruent and therefore. The plane wave bends away from the normal.5 Refraction of a plane wave incident on a rarer medium for which v2 > v1. 10.4. These phenomena were 357 . the behaviour of prisms. The phenomenon of total internal reflection and its applications was discussed in Section 9. Let CE represent the tangent plane drawn from the point C to this sphere.6) would be equal. If v represents the speed of the wave in the medium and if τ represents the time taken by the wavefront to advance from the point B to C then the distance BC = vτ In order the construct the reflected wavefront we draw a sphere of radius vτ from the point A as shown in Fig. 10. Obviously AE = BC = vτ FIGURE 10. This is the law of reflection. the angles i and r (as shown in Fig. Once we have the laws of reflection and refraction.Wave Optics any refracted wave and the wave will undergo what is known as total internal reflection. FIGURE 10. lenses.3 Reflection of a plane wave by a plane surface We next consider a plane wave AB incident at an angle i on a reflecting surface MN. 10.3. AB and CE represent incident and reflected wavefronts. For example. this is referred to as blue shift. Clearly. For example. since the speed of light waves is less in glass. when a convex lens focusses light to form a real image. the time taken is the same as for rays travelling near the edge of the lens. The time taken between the arrival of two successive wavefronts is hence longer at the observer than it is at the source. Astronomers call the increase in wavelength due to doppler effect as red shift since a wavelength in the middle of the visible region of the spectrum moves towards the red end of the spectrum. In Fig. Thus.7(c) a plane wave is incident on a concave mirror and on reflection we have a spherical wave converging to the focal point F.7(a) we consider a plane wave passing through a thin prism. when the source moves away from the observer the frequency as measured by the source will be smaller.4 The doppler effect We should mention here that one should be careful in constructing the wavefronts if the source (or the observer) is moving.7(b) we consider a plane wave incident on a thin convex lens. In a similar manner. (b) a convex lens. although the ray going through the centre traverses a shorter path. FIGURE 10. In Fig.Physics discussed in detail in Chapter 9 on the basis of rectilinear propagation of light. we can understand refraction and reflection by concave lenses and convex mirrors.7 Refraction of a plane wave by (a) a thin prism.3. 10. (c) Reflection of a plane wave by a concave mirror. From the above discussion it follows that the total time taken from a point on the object to the corresponding point on the image is the same measured along any ray. 10. When waves are received from a source moving towards the observer. In Fig. 10. The emerging wavefront has a depression at the centre and therefore the wavefront becomes spherical and converges to the point F which is known as the focus. the lower portion of the incoming wavefront (which travels through the greatest thickness of glass) will get delayed resulting in a tilt in the emerging wavefront as shown in the figure. if there is no medium and the source moves away from the observer. then later wavefronts have to travel a greater distance to reach the observer and hence take a longer time. This is known as the Doppler effect. the central part of the incident plane wave traverses the thickest portion of the lens and is delayed the most. Here we just describe the behaviour of the wavefronts as they undergo reflection or refraction. 10. but because of the slower speed in glass. there is an apparent decrease in wavelength. 358 . the reflected and refracted light both have the same frequency as the incident frequency. intensity of light is determined by the square of the amplitude of the wave. The fractional change in frequency ∆ν/ν is given by –vradial/c. Atoms may be viewed as 359 . ∆ν ν =– ∆λ λ (for small changes in ν and λ).9)] ∆ν ∆λ vradial c ν =– λ =– EXAMPLE 10. The Doppler effect for light is very important in astronomy.1 What speed should a galaxy move with respect to us so that the sodium line at 589. A more accurate formula for the Doppler effect which is valid even when the speeds are close to that of light.Wave Optics You have already encountered Doppler effect for sound waves in Chapter 15 of Class XI textbook. Example 10.2 Solution (a) Reflection and refraction arise through interaction of incident light with the atomic constituents of matter. the Doppler shift can be expressed as: ∆ν ν =– vradial c (10. requires the use of Einstein’s special theory of relativity.6 – 589. Does the reduction in speed imply a reduction in the energy carried by the light wave? (c) In the wave picture of light.0    0. EXAMPLE 10.6 nm? Solution Since νλ = c. Explain why? (b) When light travels from a rarer to a denser medium. For velocities small compared to the speed of light.1 = + 3.0 = + 0. Thus. (10. vradial ≅ + c   589. the galaxy is moving away from us.6 nm we get [using Eq. the speed decreases. we can use the same formulae which we use for sound waves. vradial is considered positive when the source moves away from the observer.6  = 306 km/s Therefore. where vradial is the component of the source velocity along the line joining the observer to the source relative to the observer. What determines the intensity of light in the photon picture of light. For ∆λ = 589.06 × 105 m s –1 or.0 nm is observed at 589. Example 10. It is the basis for the measurements of the radial velocities of distant galaxies.2 (a) When monochromatic light is incident on a surface separating two media.9) The formula given above is valid only when the speed of the source is small compared to that of light. 8(b) shows the position of crests (solid circles) and troughs (dashed circles) at a given instant of time. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. at the point P.2 (a) 10. (b) The pattern of displacement of water molecules at an instant on the surface of water showing nodal N (no displacement) and antinodal A (maximum displacement) lines. In fact at any point on the perpendicular bisector of S1S2. which take up the frequency of the external agency (light) causing forced oscillations. waves from S1 and S2 will take the same time to travel to the point P and waves that emanate from S1 and S2 in phase will also arrive. if the displacement produced by the source S1 at the point P is given by y1 = a cos ωt then. (c) For a given frequency.Physics oscillators. (b) No. the intensity will be 4I0. the resultant intensity will be given by I = 4 I0 where I0 represents the intensity produced by each one of the individual sources. The two sources are said to (b) FIGURE 10. Thus. intensity of light in the photon picture is determined by the number of photons crossing an unit area per unit time. when this happens the two sources are said to be coherent. Thus. the frequency of scattered light equals the frequency of incident light. 10. Energy carried by a wave depends on the amplitude of the wave.8(a)]. Figure 10. in phase. Consider a point P for which S1 P = S 2 P Since the distances S1 P and S2 P are equal. not on the speed of wave propagation. the phase difference between the displacements produced by each of the waves does not change with time. and at a particular point. EXAMPLE 10. They produce two water waves.8 (a) Two needles oscillating in phase in water represent two coherent sources. I0 is proportional to a2. Indeed the entire field of interference is based on the superposition principle according to which at a particular point in the medium. You may recall that we had discussed the superposition principle in Chapter 15 of your Class XI textbook. the resultant of displacement at P would be given by y = y1 + y2 = 2 a cos ωt Since the intensity is the proportional to the square of the amplitude. 360 . the displacement produced by the source S2 (at the point P) will also be given by y2 = a cos ωt Thus. Consider two needles S1 and S2 moving periodically up and down in an identical fashion in a trough of water [Fig. the resultant displacement produced by a number of waves is the vector sum of the displacements produced by each of the waves.4 COHERENT AND INCOHERENT ADDITION OF WAVES In this section we will discuss the interference pattern produced by the superposition of two waves. Wave Optics interfere constructively and we have what is referred to as constructive interference. S2R – S1R = –2. 10. then for an arbitrary point P whenever the path difference. To summarise: If we have two coherent sources S1 and S2 vibrating in phase. ± 3λ . The two displacements are now out of phase and the two displacements will cancel out to give zero intensity. 10. S1P ~ S2P = nλ (n = 0.5λ corresponds to a phase difference of 5π. 1.10) we will have constructive interference and the resultant intensity will be 4I0.9(a)].) (10.) (10. 1.. We next consider a point Q [Fig. the amplitudes of the displacement produced by each wave are very nearly the same. In the above analysis we have assumed that the distances S1Q and S2Q are much greater than d (which represents the distance between S1 and S2) so that although S1Q and S2Q are not equal.5λ where we have used the fact that a path difference of 2.10 Locus we will have destructive interference and the resultant intensity will be of points for which zero. if the point P is such that the path difference. This is referred to as destructive interference.10) let the phase S P – S P is equal to 1 2 difference between the two displacements be φ. Now.9 (a) Constructive The waves emanating from S1 will arrive exactly two and a half cycles interference at a later than the waves from S2 [Fig... 3.10(b)]. 10..11) 2 FIGURE 10. produced by S1 is given by S1P ~ S2P = (n+ y1 = a cos ωt 361 .5 λ . We next consider a point R [Fig. the sign ~ between S1P and S2 P represents the difference between S1P and S2 P.9(a)] for which S2Q –S1Q = 2λ The waves emanating from S1 will arrive exactly two cycles earlier than the waves from S2 and will again be in phase [Fig. ±λ.. 1 ) λ (n = 0. produced by S1 is given by (b) Destructive y1 = a cos ωt interference at a point R for which the then the displacement produced by S2 will be given by path difference is y2 = a cos (ωt + 5π) = – a cos ωt 2. On the other hand. ± 2λ. 10. 3. . Thus if the displacement point Q for which the path difference is 2λ. The two displacements are once again in phase and the intensity will again be 4 I0 giving rise to constructive interference. if the displacement zero. 10. 2. 2. for any other arbitrary point G (Fig. Thus. if the displacement produced by S1 is given by y1 = a cos ωt then the displacement produced by S2 will be given by y2 = a cos (ωt – 4π) = a cos ωt where we have used the fact that a path difference of 2λ corresponds to a phase difference of 4π. Thus.9(b)] for which FIGURE 10. edu/physics/2000/applets/fourier.. ± 3π. the positions of maxima and minima will also vary rapidly with time and we will see a “time-averaged” intensity distribution. i. the positions of maxima and minima will not change with time. we will observe an average intensity that will be given by < I >= 4I 0 < cos2 ( φ/2) > (10. Indeed it is shown in Section 7.10) we will have constructive interference leading to maximum intensity.colorado. if the two needles do not maintain a constant phase difference. the displacement produced by S2 would be y2 = a cos (ωt + φ ) and the resultant displacement will be given by y = y1 + y2 = a [cos ωt + cos (ωt +φ )] = 2 a cos (φ/2) cos (ωt + φ/2) The amplitude of the resultant displacement is 2a cos (φ/2) and therefore the intensity at that point will be I = 4 I0 cos2 (φ/2) http://www. if the two needles are going up and down regularly) then the phase difference φ at any point will not change with time and we will have a stable interference pattern. If we use two sodium lamps illuminating two pinholes (Fig. ± 2 π. (10.5 INTERFERENCE EXPERIMENT 362 OF LIGHT WAVES AND YOUNG’S We will now discuss interference using light waves. Now if the two sources are coherent (i.html (10. we say that the two sources are incoherent and when this happens the intensities just add up. This is because of the fact that the light wave emitted from an ordinary source (like a sodium lamp) undergoes abrupt phase . The resultant intensity will be given by I = 2 I0 (10.2 that if φ (t ) varies randomly with time.11)] we will have destructive interference leading to zero intensity.e. then the interference pattern will also change with time and. ± 4 π.13) where angular brackets represent time averaging.Physics then. 10.… which corresponds to the condition given by Eq. the time-averaged quantity < cos2 (φ/2) > will be 1/2. 10.e. ± 5π … [which corresponds to the condition given by Eq.11) we will not observe any interference fringes. On the other hand. if the phase difference changes very rapidly with time. When the phase difference between the two vibrating sources changes rapidly with time..12) Ripple Tank experiments on wave interference If φ = 0. if φ = ± π. (10. This is indeed what happens when two separate light sources illuminate a wall. When this happens. However.14) at all points. This is also intuitively obvious because the function cos2 (φ/2) will randomly vary between 0 and 1 and the average value will be 1/2. Thus. the intensities on the screen will add up.12 Young’s arrangement to produce interference pattern. The positions of maximum and minimum intensities can be calculated by using the analysis given in Section 10. as shown in Fig. i. we must have S2P – S1P = n λ. 10. He made two pinholes S1 FIGURE 10. 10.15) 2 2 363 .. These were illuminated by another S1 and S2. 10. 2 .. 2 2  2  d   2  d  D x D x + + – – +         = 2x d (S2P ) – (S1P ) =   2  2        n = 0. waves spread out from S and fall on both S1 and S2. they will be coherent like the two vibrating needle in our water wave example [Fig.e.12(a)]. (10. as discussed in the previous section. when this happens. S1 and S2 then behave like two coherent sources because light waves coming out from S1 and S2 are derived from the same original source and any abrupt phase change in S will manifest in exactly similar phase changes in the light coming out from S1 and S2.4 where we had shown that for an arbitrary point P on the line GG′ [Fig. Thus spherical waves emanating from S1 and S2 will produce interference fringes on the screen GG′..8(a)]. The British physicist Thomas Young used an ingenious technique to “lock” the phases of the waves emanating from S1 and S2. the intensities will add pinholes that was in turn. 1. the two sources S1 and S2 will be locked in phase.12(b).11 If two sodium and S2 (very close to each other) on an opaque screen lamps illuminate two pinholes [Fig. Now. lit by a bright source. (a) (b) FIGURE 10. Thus the light waves coming out from two independent sources of light will not have any fixed phase relationship and would be incoherent.12(b)] to correspond to a maximum. Light up and no interference fringes will be observed on the screen. 10.Wave Optics changes in times of the order of 10–10 seconds. If we consider the line perpendicular to the plane of the paper and passing through O [i.13.1 cm. S2P + S1P = [(100)2 + (1. In order to determine the shape of the interference pattern on the screen we note that a particular fringe would correspond to the locus of points with a constant value of S2P – S1P. Whenever this constant is an integral multiple of λ. ranging from the structure of the eye and the mechanism of vision to the decipherment of the Rosetta stone. . In this approximation.17) Hence we will have constructive interference resulting in a bright region when x = xn = n λD . the central point O (in Fig. the fringe will be bright and whenever it is an odd integral multiple of λ/2 it will be a dark fringe. as shown in Fig. 10.13. Eq. (10.e. For example. 10.01 cm Thus if we replace S2P + S1P by 2 D. D = 100 cm. physician and Egyptologist. however. Young worked on a wide variety of scientific problems.16) becomes S2P – S1P ≈ (10.. n = 0.18) d On the other hand. 10. if the distance D is very large compared to the fringe width. Thus the fringe pattern will strictly be a hyperbola.20) d which is the expression for the fringe width.19) d Thus dark and bright bands appear on the screen. Thus 2 xd S2P – S1P = S P+S P 2 1 (10.Physics where S1S2 = d and OP = x .12) will be bright because S1O = S2O and it will correspond to n = 0. n = 0.05)2]½ + [(100)2 + (0. 10. we will have a dark region near . for d = 0. the locus of the point P lying in the x-y plane such that S2P – S1P (= ∆) is a constant. He revived the wave theory of light and recognised that interference phenomena provide proof of the wave properties of light.16) THOMAS YOUNG (1773 – 1829) If x.19) show that dark and bright fringes are equally spaced and the distance between two consecutive bright and dark fringes is given by x = xn = (n+ ) λD β = xn+1 –xn (10. ± 1. Now.. the fringes will be very nearly straight lines as shown in Fig. (10. d<<D then negligible error will be introduced if S2P + S1P (in the denominator) is replaced by 2D. Equations (10. ± 2 (10.18) and (10.. OP = 1 cm (which correspond to typical values for an interference experiment using light waves). the error involved is about 0. Obviously. we have Thomas Young (1773 – 1829) English physicist. ± 1. Such bands are called fringes. along the y-axis] then all points on this line will be equidistant from S1 and S2 and we will have a bright central fringe which is a straight line as shown in Fig.95)2]½ ≈ 200. is a hyperbola. or β = λD 364 .13.005%. ± 2. 12). which means that it is shifted by the same angle on the other side of the bisector. pierced with two small pinholes.. the mid-point Q and the point O′ of the central fringe are in a straight line. This time he saw a number of dark lines.12. If S is shifted by an angle φ to point S′. We end this section by quoting from the Nobel lecture of Dennis Gabor* The wave nature of light was demonstrated convincingly for the first time in 1801 by Thomas Young by a wonderfully simple experiment.Wave Optics FIGURE 10. He then saw two darkish lines at both sides of a bright line. which gave him sufficient encouragement to repeat the experiment.) In the double-slit experiment shown in Fig. also on the perpendicular bisector.005 mm and 0.13 Computer generated fringe pattern produced by two point source S1 and S2 on the screen GG′ (Fig. 10. This also means that the source S′. we have taken the source hole S on the perpendicular bisector of the two slits.) (Adopted from OPTICS by A. at some distance. then the central bright fringe occurs at an angle –φ. which is shown as the line SO. Ltd. This phenomenon is called * Dennis Gabor received the 1971 Nobel Prize in Physics for discovering the principles of holography. New Delhi. on the other side. then the central fringe appears at a point O′ at an angle –φ. respectively (both figures correspond to D = 5 cm and λ = 5 × 10–5 cm. He let a ray of sunlight into a dark room. Ghatak. 2000. placed a dark screen in front of it. the first clear proof that light added to light can produce darkness. a white screen. (a) and (b) correspond to d = 0.025 mm. this time with spirit flame as light source. What happens if the source S is slightly away from the perpendicular bisector. 365 . regularly spaced. then the central fringe occurs at O. If the angle S′QS is φ. Tata McGraw Hill Publishing Co. 10. Thus. and beyond this. Consider that the source is moved to some new point S′ and suppose that Q is the mid-point of S1 and S2. if the source S is on the perpendicular bisector. with a little salt in it to produce the bright yellow sodium light. (d) the source slit is moved closer to the double-slit plane.quasihome.com/interfer. Interactive animation of Young’s experiment http://vsg.14).4 What is the effect on the interference fringes in a Young’s double-slit experiment due to each of the following operations: (a) the screen is moved away from the plane of the slits. (c) the separation between the two slits is increased. each pair of points would have produced straight line fringes resulting in straight line fringes with increased intensities.html FIGURE 10.4 . Example 10. (f ) the monochromatic source is replaced by a source of white light? 366 EXAMPLE 10. Thomas Young had expected it because he believed in the wave theory of light. If we had slits instead of the point sources (Fig. What is the fringe separation when bluegreen light of wavelength 500 nm is used? Solution Fringe spacing = EXAMPLE 10. 10.3 Two slits are made one millimetre apart and the screen is placed one metre away. (e) the width of the source slit is increased.Physics interference.14 Photograph and the graph of the intensity distribution in Young’s double-slit experiment. We should mention here that the fringes are straight lines although S1 and S2 are point sources. (b) the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength.5 mm Example 10.3 D λ 1 × 5 × 10 –7 = m d 1 × 10 –3 = 5 × 10–4 m = 0. other than the one specified. the fringe will be predominantly blue. the interference pattern gets less and less sharp. and when the source is brought too close for this condition to be valid.) Solution (a) Angular separation of the fringes remains constant (= λ / d ). otherwise. there are alternate dark and bright regions just like in interference. the condition mentioned in (d) below. the fringe separation remains fixed. where λb (≈ 4000 Å) represents the wavelength for the blue colour. as S decreases (i. (d) Let s be the size of the source and S its distance from the plane of the two slits. water waves or matter waves. the central fringe is white. After a few fringes. Thus. For interference fringes to be seen.. light waves. (e) Same as in (d). be it sound waves. we do not encounter diffraction effects of light in everyday observations. The central bright fringes for different colours are at the same position. EXAMPLE 10.4 10. the condition s/S < λ/d should be satisfied. Diffraction is a general characteristic exhibited by all types of waves. however. (b) The separation of the fringes (and also angular separation) decreases. no clear fringe pattern is seen.6 DIFFRACTION If we look clearly at the shadow cast by an opaque object. interference patterns produced by different parts of the source overlap and no fringes are seen. When the source slit is so wide that the condition s/S ≤ λ/ d is not satisfied. The actual separation of the fringes increases in proportion to the distance of the screen from the plane of the two slits. Slightly farther away where S2Q–S1Q = λb = λr/2 where λr (≈ 8000 Å) is the wavelength for the red colour. As the source slit width increases. Since the wavelength of light is much smaller than the dimensions of most obstacles. Thus. the finite resolution of our eye or of optical 367 . take all parameters. to remain unchanged.Wave Optics ( In each operation. however. the fringe closest on either side of the central white fringe is red and the farthest will appear blue. the source slit is brought closer). close to the region of geometrical shadow. See. (f ) The interference patterns due to different component colours of white light overlap (incoherently). See. Therefore. the interference pattern disappears. fringe pattern gets less and less sharp. the condition mentioned in (d) below. This happens due to the phenomenon of diffraction.e. (c) The separation of the fringes (and also angular separation) decreases. the blue component will be absent and the fringe will appear red in colour. Till this happens. the fringes disappear. However. For a point P for which S2P –S1P = λb/2. there are alternate dark and bright regions. Experimental observation shown in .M.15. Indeed the colours that you see when a CD is viewed is due to diffraction effects. known as diffraction. All path differences are zero and hence all the parts of the slit contribute in phase. We want the intensity at any point P on the screen. if two points M1 and M2 in the slit plane are separated by y. From Fig. A straight line through M perpendicular to the slit plane meets the screen at C. straight lines joining P to the different points L. can only be properly understood using wave ideas. The midpoint of the slit is M. After all.21) 368 Similarly. It seems to turn around corners and enter regions where we would expect a shadow. As before. can be treated as parallel. The path difference NP – LP between the two edges of the slit can be calculated exactly as for Young’s experiment. making an angle θ with the normal MC. This calculation was made by Fresnel using integral calculus. These effects. a broad pattern with a central bright region is seen. On both sides. 10.15.Physics instruments such as telescopes or microscopes is limited due to the phenomenon of diffraction. early experimenters – including Newton – had noticed that light spreads out from narrow holes and slits. go to Fig.16). these sources are in phase. 10.1 The single slit In the discussion of Young’s experiment. and add their contributions at P with the proper phase differences.6. the path difference M2 P – M1P ≈ yθ. We are treating different parts of the wavefront at the slit as secondary sources. etc. To understand this. We now have to sum up equal. At the central point C on the screen. NP – LP = NQ = a sin θ ≈ aθ (10. each with a different phase. The main features of the diffraction pattern can be understood by simple arguments. 10. The basic idea is to divide the slit into much smaller parts. so we omit it here.N. 10. Even before Young. The diffracted light goes on to meet a screen.. coherent contributions from a large number of sources. we stated that a single narrow slit acts as a new source from which light spreads out. which shows a parallel beam of light falling normally on a single slit LN of width a. We will now discuss the phenomenon of diffraction. Because the incoming wavefront is parallel to the plane of the slit. you are hardly surprised to hear sound waves from someone talking around a corner ! When the double slit in Young’s experiment is replaced by a single narrow slit (illuminated by a monochromatic source). This gives maximum intensity at C. the intensity becoming weaker away from the centre (Fig. the angle θ is zero. etc. Then. LM and MN each of size a/2. Consider an angle θ = 3λ/2a which is midway between two of the dark fringes. one-seventh. the path difference between the two ends would be 2 2a 3λ × =λ a ×θ = 3 3 2a (10. For every point M1 in LM. . Clearly. It is easy to see why it has minima at these values of angle. 10.. In this context. divide the slit into two equal halves FIGURE 10. n = ±1. One can similarly show that the intensity is zero for θ = n λ/a . with n being any integer (except zero!). Divide the slit into three equal parts. since only one-fifth. Only the remaining one-third of the slit contributes to the intensity at a point between the two minima. The photograph and intensity pattern corresponding to it is shown in Fig. etc. Equation (10.Wave Optics Fig.. cancel each other. Contributions from the two halves of the slit LM and MN. The contributions of these two halves cancel in the same manner as described earlier.. ±3. The path difference between M1 and M2 at P = M2P – M1P = θ a/2 = λ/2 for the angle chosen.. (10.16. therefore. 3.23) The first two-thirds of the slit can therefore be divided into two halves which have a λ/2 path difference. this will be much weaker than the central maximum (where the entire slit contributes in phase). Consider first the angle θ where the path difference aθ is λ. it is FIGURE 10.15 indicates that the intensity has a central maximum at θ = 0 and other secondary maxima at θ l (n+1/2) λ/a. ±2. This means that the contributions from M1 and M2 are 180º out of phase and cancel in the direction θ = λ/a . 369 .16 Intensity distribution and photograph of fringes due to diffraction at single slit.22) Now. 10. of the slit contributes in these cases. There has been prolonged discussion about difference between intereference and diffraction among scientists since the discovery of these phenomena. It is also easy to see why there are maxima at θ (n + 1/2) λ/a and why they go on becoming weaker and weaker with increasing n . If we take the first two thirds of the slit. One can similarly show that there are maxima at (n + 1/2) λ/a with n = 2. and has minima (zero intensity) at θ l nλ/a. These become weaker with increasing n.15 The geometry of path differences for diffraction by a single slit. there is a point M2 in MN such that M1M2 = a/2.22) gives the angle at which the intensity falls to zero. Notice that the angular size of the central maximum increases when the slit width a decreases. θ ≈ λ /a . what should the width of each slit be to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern? EXAMPLE 10.θ = 10 λ a d λ a = = 0. . In the double-slit experiment. 10. and there is no specific. and the double-slit interference pattern. http://www.5 In Example 10. 10. Example 10.13(b)].17.2 mm a 5 Notice that the wavelength of light and distance of the screen do not enter in the calculation of a . The number of interference fringes occuring in the broad diffraction peak depends on the ratio d/a. we must note that the pattern on the screen is actually a superposition of single-slit diffraction from each slit or hole. and only one hole (or slit) S1 or S2. The best we can do is. λ d =2 In the double-slit interference experiment of Fig. We now have a source at S. It is just a question of usage. important physical difference between them. but if there is a large number of them.12.3. In the limit of a becoming very small. it seems that the word diffraction is more often used. is to say that when there are only a few sources. This will produce a single- 370 * Richand Feynman was one of the recipients of the 1965 Nobel Prize in Physics for his fundamental work in quantum electrodynamics. the diffraction pattern will become very flat and we will obsrve the two-slit interference pattern [see Fig. what happens if we close one slit? You will see that it now amounts to a single slit.phys.edu/~teb/optics/java/slitdiffr/ Interactive animation on single slit diffraction pattern FIGURE 10.Physics interesting to note what Richard Feynman* has said in his famous Feynman Lectures on Physics: No one has ever been able to define the difference between interference and diffraction satisfactorily. then the result is usually called interference.17 The actual double-slit interference pattern. that is the ratio of the distance between the two slits to the width of a slit. The envelope shows the single slit diffraction. 10. But you will have to take care of some shift in the pattern. It shows a broader diffraction peak in which there appear several fringes of smaller width due to double-slit interference. say two interfering sources. roughly speaking. This is shown in Fig.5 Solution We want aθ = λ .hawaii. With both filters 371 available. (ii) We calculate the interference pattern by superposing two waves originating from the two narrow slits. For example. right in front of the eye. to be able to observe good interference and diffraction patterns. of the order of 0. (i) The interference pattern has a number of equally spaced bright and dark bands. This situation also occurs when we place a converging lens after the slits and place the screen at the focus. The two or more paths from the slits to the screen were treated as parallel. With slight adjustment of the width of the bulb filament viewed slit and the parallelism of the edges. At the same angle of λ/a. . Holding two blades to Keep the slit parallel to the filament. Since the position of all the bands (except the clear diffraction central one) depends on wavelength. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit. A spectacles if you normally do. we have assumed that the screen on which the fringes are formed is at a large distance. The equipment needed can be found in most homes –– two razor blades and one clear glass electric bulb preferably with a straight filament. The diffraction pattern has a central bright maximum which is twice as wide as the other maxima.18).18 (Fig. the first null of the interference pattern occurs at an angle of λ/a. the separation d between the two slits must be of the order of a milimetre or so. Use form a single slit. Using bands. Parallel paths from the slit are combined at a single point on the screen. If f is the focal length of the lens. the size on the screen will be f λ/a. the wider fringes for red compared to blue can be seen. 10.2 mm. (iii) For a single slit of width a. they will show some colours. as the case may be. the separation of the central maximum from the first null of the diffraction pattern is λ/a .2 Seeing the single slit diffraction pattern It is surprisingly easy to see the single-slit diffraction pattern for oneself. We now compare and contrast the interference pattern with that seen for a coherently illuminated single slit (usually called the single slit diffraction pattern). The width a of each slit must be even smaller. In terms of angles. One has to hold the two blades so that the edges are parallel and have a narrow slit in between. This arrangement is often used since it gives more intensity than placing the screen far away.Wave Optics slit diffraction pattern on the screen. The intensity falls as we go to successive maxima away from the centre. In our discussion of Young’s experiment and the single-slit diffraction. One must understand that both d and a have to be quite small. Note that the lens does not introduce any extra path differences in a parallel beam.6. The centre of the central bright fringe will appear at a point which lies on the straight line SS1 or SS2. then we can easily work out the size of the central bright maximum.1 or 0. Hence. This is easily done with the thumb and forefingers FIGURE 10. on either side. 10. the pattern should be seen with its through this shows bright and dark bands. we get a maximum (not a null) for two narrow slits separated by a distance a. a filter for red or blue will make the fringes clearer. 61 λf/a .19). The bulb filament can be viewed as before to repeat Young’s experiment. If it reduces in one region. In this case the effects due to diffraction can be taken into account by considering a plane wave incident on a circular aperture followed by a convex lens (Fig. which is consistent with the principle of conservation of energy.24) 372 FIGURE 10.. it is similar to the analysis carried out to obtain the single-slit diffraction pattern. in principle.61λ f = a 2a (10. the filament plays the role of the first slit S in Fig.16. Because of diffraction effects. producing a bright fringe. 10. The primary purpose of the eyepiece is to provide magnification of the image produced by the objective.g. This is the reflection of the Sun in any shiny convex surface (e. There is no gain or loss of energy. producing a dark fringe. because of diffraction. In interference and diffraction.3 Resolving power of optical instruments In Chapter 9 we had discussed about telescopes. 10. one can cut a double slit in an aluminium foil with a blade. The angular resolution of the telescope is determined by the objective of the telescope. The stars which are not resolved in the image produced by the objective cannot be resolved by any further magnification produced by the eyepiece. A detailed analysis shows that the radius of the central bright region is approximately given by r0 ≈ 1. Taking into account the effects due to diffraction. then geometrical optics tells us that the beam will get focused to a point. If the lens is well corrected for aberrations. .19). However. 10. With some effort. The lens of the eye focuses the pattern on the screen (the retina of the eye). The analysis of the corresponding diffraction pattern is quite involved.6. Consider a parallel beam of light falling on a convex lens. it increases in another region. Do not try direct sunlight – it can damage the eye and will not give fringes anyway as the Sun subtends an angle of (1/2)º. 10.Physics In this experiment. the pattern on the focal plane would consist of a central bright region surrounded by concentric dark and bright rings (Fig. In daytime. light energy is redistributed.19 A parallel beam of light is incident on a convex lens. the beam gets focused to a spot of radius ≈ 0.22 λ f 0. the beam instead of getting focused to a point gets focused to a spot of finite area. however. a cycle bell). there is another suitable bright source subtending a small angle at the eye. 20 Real image formed by the objective lens of the microscope. Typically if λ ≈ 0.Wave Optics where f is the focal length of the lens and 2a is the diameter of the circular aperture or the diameter of the lens.61λ f a implying 0.5 µm. whichever is smaller. It can be seen from Fig.20]. FIGURE 10. it plays an important role in determining the limit of resolution of optical instruments like a telescope or a microscope. the object is placed slightly beyond f.61λ (10. Thus if.25) a Thus ∆θ will be small if the diameter of the objective is large. It is for this reason that for better resolution. 10. For the two stars to be just resolved f ∆θ ≈ r0 ≈ 0. f ≈ 20 cm and a ≈ 5 cm we have r0 ≈ 1. ∆θ ≈ Example 10. a telescope must have a large diameter objective. What is the limit of resolution of a telescope whose objective has a diameter of 100 inch? Solution A 100 inch telescope implies that 2 a = 100 inch = 254 cm. so that a real image is formed at a distance v [Fig. λ ≈ 6000Å = 6×10–5 cm then ∆θ ≈ 0.6 radians We can apply a similar argument to the objective lens of a microscope.26) D/f l 2 tan β where 2β is the angle subtended by the diameter of the objective lens at the focus of the microscope.2 µm Although the size of the spot is very small.61 × 6 × 10 –5 127 ≈ 2.6 Assume that light of wavelength 6000Å is coming from a star. In this case. 373 .20 that (10. The magnification – ratio of image size to object size – is given by m l v/f. This implies that the telescope will have better resolving power if a is large. 10.9 × 10 –7 EXAMPLE 10. Physics DETERMINE THE RESOLVING POWER OF YOUR EYE You can estimate the resolving power of your eye with a simple experiment. see figure here. the black stripes to the right of this would be more and more clearly visible. dmin. All the black stripes should be of equal width. For example. combining Eqs. Paste this pattern on a wall in a room or laboratory.22 λ  vθ = v   D   (10. (10.22 λ    m dmin = v    D  = = 1. Now watch the pattern.22 λ 2 tan β .22 λ v . Note the width d of the white stripe which separates the two regions. On the other hand. while the width of the intermediate white stripes should increase as you go from the left to the right. Let the width of the first two white stripes be 0. we get (10. etc.27) Two objects whose images are closer than this distance will not be resolved. The image of a point object will again be a diffraction pattern whose size in the image plane will be  1. find the position where you can just see some two black stripes as separate stripes.28) 374 d min = 1. D m 1.5 mm each.26) and (10. estimate the size of the speck of dust. preferably with one eye. the next two white stripes be 1 mm each. Knowing the resolution of your eye and the distance of the speck. they will be seen as one.28). and measure the distance D of the wall from your eye. Then d/D is the resolution of your eye. at the height of your eye. By moving away or closer to the wall. the diffraction effects become important. the next two 1. Make black stripes of equal width separated by white stripes. When the separation between two points in a microscopic specimen is comparable to the wavelength λ of the light.22 f λ D Now.5 mm each. Find the distance (of a speck) which you can clearly see and distinguish from a neighbouring speck. in the object plane is given by   1. let all black stripes have a width of 5 mm. All the black stripes to the left of this stripe would merge into one another and would not be distinguishable. You have watched specks of dust floating in air in a sunbeam entering through your window. The corresponding minimum separation. z a2 λ (10. The resolving power of the microscope is given by the reciprocal of the minimum separation of two points seen as distinct.30) that the resolving power can be increased by choosing a medium of higher refractive index. slit or hole) of size a illuminated by a parallel beam sends diffracted light into an angle of approximately ≈ λ/a . We may be looking at two stars or two satellites of a far-away planet. the spreading due to diffraction is smaller compared to the size of the beam.Wave Optics 1. Therefore.4 The validity of ray optics An aperture (i. In travelling a distance z. Therefore objects which are not resolved at far distance. it is good to remember that a telescope resolves whereas a microscope magnifies.31) We define a quantity zF called the Fresnel distance by the following equation zF a 2 / λ Equation (10. Thus. A telescope produces images of far objects nearer to our eye. we see that the resolving power of a microscope is basically determined by the wavelength of the light used. Such an arrangement is called an ‘oil immersion objective’.30) The product n sinβ is called the numerical aperture and is sometimes marked on the objective. A microscope.e. Eq. We thus approximately equate zλ/a with a. on the other hand. It is interesting to ask at what value of z the spreading due to diffraction becomes comparable to the size a of the aperture. (10.. 10. This gives the distance beyond which divergence of the beam of width a becomes significant. Usually an oil having a refractive index close to that of the objective glass is used.22 λ 2 n sin β (10. can be resolved by looking at them through a telescope. It becomes comparable when the distance is approximately zF . magnifies objects (which are near to us) and produces their larger image.29) If the medium between the object and the objective lens is not air but a medium of refractive index n.31) shows that for distances much smaller than zF . the spreading due to diffraction 375 . the diffracted beam therefore acquires a width zλ/a due to diffraction.29) gets modified to d min = 1. In this context. and similarly between the role of a telescope and a microscope to deal with these parameters. For distances much greater than zF .6. It can be seen from Eq. Notice that it is not possible to make sinβ larger than unity. (10.22 λ 2 sin β (10. or we may be looking at different regions of a living cell. This is the angular size of the bright central maximum. There is a likelihood of confusion between resolution and magnification. Such a wave could be described by the following equation 376 FIGURE 10.7 For what distance is ray optics a good approximation when the aperture is 3 mm wide and the wavelength is 500 nm? EXAMPLE 10.e. the size a of the aperture). we will generate a wave propagating in the +x direction (Fig. 10.Physics dominates over that due to ray optics (i. . Thus. 10. At x = ∆x. (b) The curve represents the time variation of the displacement at x = 0 when a sinusoidal wave is propagating in the +x-direction.. the time variation of the displacement will be slightly displaced to the right.31) also shows that ray optics is valid in the limit of wavelength tending to zero. diffraction spreading can be neglected for rays many metres in length.21). Example 10.7 Solution z F = a 2 λ (3 × 10 ) = 5 × 10 –7 –3 2 = 18 m This example shows that even with a small aperture. respectively when a sinusoidal wave is propagating in the +x-direction.7 POLARISATION Consider holding a long string that is held horizontally. the other end of which is assumed to be fixed. ray optics is valid in many common situations. If we move the end of the string up and down in a periodic manner.21 (a) The curves represent the displacement of a string at t = 0 and at t = ∆t. Equation (10. Rotating P1 has no effect on the transmitted beam and transmitted intensity remains constant.. the intensity transmitted 377 .e.. A polaroid consists of long chain molecules aligned in a particular direction. Thus. we have what is known as a transverse wave.Wave Optics y (x. Thus. it is often referred to as a y-polarised wave. if the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet P1. since the displacement is in the y direction.33) k represents the wavelength associated with the wave. further. In one position.34) It should be mentioned that the linearly polarised waves [described by Eqs. We had discussed propagation of such waves in Chapter 15 of Class XI textbook. if the plane of vibration of the string is changed randomly in very short intervals of time. then we have what is known as an unpolarised wave. As expected. i.t ) = a sin (kx – ωt ) (10. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. let an identical piece of polaroid P2 be placed before P1.32) where a and ω (= 2πν ) represent the amplitude and the angular frequency of the wave. Since the displacement (which is along the y direction) is at right angles to the direction of propagation of the wave. You must have seen thin plastic like sheets.34)] are all transverse waves.33) and (10.t ) = a sin (kx – ωt) (10. respectively. i. This can be easily demonstrated using a simple polaroid. it is observed that its intensity is reduced by half. In a similar manner we can consider the vibration of the string in the x-z plane generating a z-polarised wave whose displacement will be given by λ= z (x. Now. the displacement of each point of the string is always at right angles to the direction of propagation of the wave. (10. the electric field associated with a propagating light wave is always at right angles to the direction of propagation of the wave. if an unpolarised light wave is incident on such a polaroid then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules. Light waves are transverse in nature. Thus. But now rotating P1 has a dramatic effect on the light coming from P2. Also. Finally. this direction is known as the pass-axis of the polaroid. the light from the lamp is reduced in intensity on passing through P2 alone. 2π (10. Since each point on the string moves on a straight line. the wave is also referred to as a linearly polarised wave. Further. the string always remains confined to the x-y plane and therefore it is also referred to as a plane polarised wave.e. for an unpolarised wave the displacement will be randomly changing with time though it will always be perpendicular to the direction of propagation. which are called polaroids. as we rotate the polaroid P1 (or P2). P1 transmits nearly the full intensity emerging from P2 (Fig.22 (a) Passage of light through two polaroids P2 and P1. The double arrows show the oscillations of the electric vector.22). the intensity will vary as: 378 I = I 0 cos 2θ . Polaroids are also used in photographic cameras and 3D movie cameras. The transmitted polarisation is the component parallel to the polaroid axis. I = I0 cos2θ (10. Then the intensity of light after passing through second polariser P2 will be The above experiment can be easily understood by assuming that light passing through the polaroid P2 gets polarised along the pass-axis of P2. EXAMPLE 10.Physics by P2 followed by P1 is nearly zero. Notice that the light seen through a single polaroid P1 does not vary with angle. 10.8 Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids? Solution Let I0 be the intensity of polarised light after passing through the first polariser P1. Polaroids can be used to control the intensity. If the pass-axis of P2 makes an angle θ with the pass-axis of P1. The transmitted fraction falls from 1 to 0 as the angle between them varies from 0º to 90º. Thus. This is known as Malus’ law.8 Example 10. the component E cos θ (along the pass-axis of P2) will pass through P2. the intensity can be further controlled from 50% to zero of the incident intensity by adjusting the angle between the pass-axes of two polaroids. FIGURE 10. windowpanes. etc. in sunglasses. (b) Behaviour of the electric vector when light passes through two polaroids. then when the polarised beam passes through the polaroid P2. . By putting a second polaroid. When turned by 90º from this position. The above discussion shows that the intensity coming out of a single polaroid is half of the incident intensity.35) where I0 is the intensity of the polarized light after passing through P1. V. Clearly. Hence the intensity of light emerging from P3 will be π  I = I 0 cos 2θ cos 2  – θ  2  EXAMPLE 10. the incident sunlight is unpolarised. We have drawn an observer looking at 90° to the direction of the sun. Under the influence of the electric field of the incident wave the electrons in the molecules acquire components of motion in both these directions. 379 .7. 10. the transmitted intensity will be maximum when θ = π/4. The radiation scattered by the molecule is therefore represented by dots.23(a) shows. The double arrows show polarisation in the plane of the figure. which has changed its direction (having been scattered) on encountering the molecules of the earth’s atmosphere.23 (a) Polarisation of the blue scattered light from the sky.8 = I0 cos2θ sin2θ =(I0/4) sin22θ Therefore. (b) Polarisation of light reflected from a transparent medium at the Brewster angle (reflected ray perpendicular to refracted ray). As Fig. This explains the polarisation of scattered light from the sky. Raman and his collaborators in Kolkata in the 1920s. This is nothing but sunlight.Wave Optics where θ is the angle between pass axes of P1 and P2. A typical molecule is shown. The incident sunlight is unpolarised (dots and arrows). The scattering of light by molecules was intensively investigated by C. 10. FIGURE 10. The dots stand for polarisation perpendicular to the plane of the figure. Since P1 and P3 are crossed the angle between the pass axes of P2 and P3 will be (π/2 – θ ). (There is no phase relation between these two in unpolarised light). charges accelerating parallel to the double arrows do not radiate energy towards this observer since their acceleration has no transverse component. It scatters light by 90º polarised normal to the plane of the paper (dots only). Raman was awarded the Nobel Prize for Physics in 1930 for this work.1 Polarisation by scattering The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a polaroid which is rotated. It is polarised perpendicular to the plane of the figure. Let the light emitted by the laser source pass through the polariser and be incident on the surface of the prism at the Brewster’s angle of incidence iB. This can be checked by looking at the reflected light through an analyser. We have drawn a situation in which the reflected wave travels at right angles to the refracted wave. As before. Consider a related question: Is it possible that under some conditions a monochromatic beam of light incident on a surface (which is normally reflective) gets completely transmitted with no reflection? To your surprise. 10. a prism and screen as shown in the figure here.2 Polarisation by reflection Figure 10. the answer is yes. the dots and arrows indicate that both polarisations are present in the incident and refracted waves. Since we have i B+r = π/2.23(b) shows light reflected from a transparent medium. Let us try a simple experiment and check what happens. We can see that i B is related to the refractive index of the denser medium. The reflected spot will completely vanish. When unpolarised light is incident on the boundary between two transparent media. the reflected light is therefore linearly polarised perpendicular to the plane of the figure (represented by dots). water. a good polariser. the reflected wave is a totally polarised wave. i. The angle of incidence in this case is called Brewster’s angle and is denoted by i B. say. As the figure shows..e. i. the light incident on the prism is completely transmitted and no light is reflected from the surface of the prism. Arrange a laser. These move in the two directions transverse to the radiation from wave in the medium. we get from Snell’s law 380 µ= sin i B sin i B = sin r sin ( π / 2 – i B ) . Thus we have seen that when reflected wave is perpendicular to the refracted wave.7. The transmitted intensity will be zero when the axis of the analyser is in the plane of the figure.. the plane of incidence. the refracted wave.Physics A SPECIAL CASE OF TOTAL TRANSMISSION When light is incident on an interface of two media. Now rotate the polariser carefully and you will observe that for a specific alignment of the polariser. The arrows are parallel to the direction of the reflected wave.e. Motion in this direction does not contribute to the reflected wave. The oscillating electrons in the water produce the reflected wave. the reflected light is polarised with its electric vector perpendicular to the plane of incidence when the refracted and reflected rays make a right angle with each other. it is observed that some part of it gets reflected and some part gets transmitted. Example 10.9 For simplicity. and reflection at the Brewster angle.9 Unpolarised light is incident on a plane glass surface. EXAMPLE 10. This kind of light is called partially polarised. we described some associated points such as diffraction. can also be understood on the basis of Wave Optics. In order to develop a proper understanding. The rays are then perpendicular to both the wavefronts and the time of travel 381 . one of the two perpendicular components of the electric field is zero. In the next chapter. This is the Brewster’s angle for air to glass interface. one sees a maximum and a minimum of intensity but not complete darkness. When an unpolarised beam of light is incident at the Brewster’s angle on an interface of two media. if we completely remove all the light with its electric vector perpendicular to the plane of incidence and let this light be incident on the surface of the prism at Brewster’s angle. only part of light with electric field vector perpendicular to the plane of incidence will be reflected. Let us try to understand the situation. resolution. When the speed of light is independent of direction. the secondary waves are spherical. we should have tan iB = µ = 1. both components are present but one is stronger than the other. At other angles. Then we described Young’s double slit experiment which was a turning point in the study of optics. we first described how some phenomena like reflection and refraction. What should be the angle of incidence so that the reflected and refracted rays are perpendicular to each other? Solution For i + r to be equal to π/2. which add up to give the wavefront at a later time. In this special situation. we have discussed scattering of light by 90º. polarisation. and validity of ray optics.Wave Optics = sin i B = tan i B cos i B (10. Huygens’ principle tells us that each point on a wavefront is a source of secondary waves. Finally.5. SUMMARY 1. which were studied on this basis of Ray Optics in Chapter 9. you will then observe no reflection and there will be total transmission of light. When such light is viewed through a rotating analyser.D. you will see how new experiments led to new theories at the turn of the century around 1900 A. There is no stable phase relationship between the two perpendicular components since these are derived from two perpendicular components of an unpolarised beam. Huygens’ construction tells us that the new wavefront is the forward envelope of the secondary waves. We began this chapter by pointing out that there are some phenomena which can be explained only by the wave theory.36) This is known as Brewster’s law. Now by using a good polariser. This gives i B = 57°. 2. there is an interference term in addition to the sum of the individual intensities. The intensity falls to zero at angles of ± . 4. A polaroid transmits only one component (parallel to a special axis). during a measurement. before it starts to spread out due to diffraction. 4.. 5. maximum.. Diffraction phenomena define the limits of ray optics. 382 . and central bright fringe lie in a straight line. a microscope objective subtending angle 2β at the focus. Diffraction determines the limitations of the concept of light rays. as shown in Young’s experiment. The limit of the ability of microscopes and telescopes to distinguish very close objects is set by the wavelength of light. etc. POINTS TO PONDER 1. 6.g. Diffraction limits the angular resolution of a telescope to λ/D where D is the diameter. It required the insight and experiment of Huygens. A single slit of width a gives a diffraction pattern with a central λ 2λ . Polarised light can also be produced by reflection at a special angle (called the Brewster angle) and by scattering through π/2 in the earth’s atmosphere. When this kind of light is viewed through a second polaroid whose axis turns through 2π. from the sun is unpolarised. 2.Physics 3. Similarly. An extended source will destroy the fringes if it subtends angle more than λ/d at the slits. A beam of width a travels a distance a2/λ. The principle of superposition of waves applies whenever two or more sources of light illuminate the same point. rapidly and randomly. When we consider the intensity of light due to these sources at the given point. mid-point of the slits. while light was seen to travel along narrow rays. The crucial new feature of waves is interference of amplitudes from different sources which can be both constructive and destructive. 5. Waves from a point source spread out in all directions. ± a a with successively weaker secondary maxima in between. But polarisation phenomena are special to transverse waves like light waves. Natural light. called the Fresnel distance. Even a wave falling on single slit should be regarded as a large number of sources which interefere constructively in the forward direction (θ = 0). Young and Fresnel to understand how a wave theory could explain all aspects of the behaviour of light. Most interference and diffraction effects exist even for longitudinal waves like sound in air. This principle leads to the well known laws of reflection and refraction. two maxima and minima of intensity are seen. The source. is the same measured along any ray. Two stars closer than this give strongly overlapping images. which is the resolution limit of a microscope. 3. This means the electric vector takes all possible directions in the transverse plane. But this term is important only if it has a non-zero average. Young’s double slit of separation d gives equally spaced fringes of angular separation λ/d. and destructively in other directions. in a medium of refractive index n. e. which occurs only if the sources have the same frequency and a stable phase difference. The resulting light is called linearly polarised or plane polarised. will just separate two objects spaced at a distance λ /(2 n sin β ). What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray? 10. (b) Light emerging out of a convex lens when a point source is placed at its focus.Wave Optics EXERCISES 10.5. the slits are separated by 0. 10.) Light of wavelength 5000 Å falls on a plane reflecting surface. 650 nm and 520 nm.2° on a screen placed 1 m away.0 × 108 m s–1) (b) Is the speed of light in glass independent of the colour of light? If not.5 10. which of the two colours red and violet travels slower in a glass prism? 10.7 In a double-slit experiment the angular width of a fringe is found to be 0. is K units. What is the speed of light in glass? (Speed of light in vacuum is 3. 383 .9 10.4 In a Young’s double-slit experiment.1 Monochromatic light of wavelength 589 nm is incident from air on a water surface.28 mm and the screen is placed 1.8 10. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. In Young’s double-slit experiment using monochromatic light of wavelength λ.5. What is the shape of the wavefront in each of the following cases: (a) Light diverging from a point source.6 10. Determine the wavelength of light used in the experiment. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide? 10.2 10. is used to obtain interference fringes in a Young’s double-slit experiment. The wavelength of light used is 600 nm. frequency and speed of (a) reflected. What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.10 Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm. (c) The portion of the wavefront of light from a distant star intercepted by the Earth.2 cm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3. What is the intensity of light at a point where path difference is λ/3 ? A beam of light consisting of two wavelengths.3 (a) The refractive index of glass is 1.4 m away. and (b) refracted light? Refractive index of water is 1. What are the wavelength. the intensity of light at a point on the screen where path difference is λ. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.33. If both light and sound waves can bend around . depend ? 10. What is the spacing between the two slits? 10. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.14 Let us list some of the factors. (iv) wavelength. observer at rest.11 The 6563 Å Hα line emitted by hydrogen in a star is found to be redshifted by 15 Å.Physics ADDITIONAL EXERCISES 10.1º. a bright spot is seen at the centre of the shadow of the obstacle. The exact Doppler formulas for the case of light waves in vacuum are.12 Explain how Corpuscular theory predicts the speed of light in a medium. and (ii) source moving. 10. which could possibly influence the speed of wave propagation: (i) nature of the source.15 For sound waves. water. (iii) motion of the source and/or observer. however.17 Answer the following questions: (a) In a single slit diffraction experiment. does (a) the speed of light in vacuum. How does this affect the size and intensity of the central diffraction band? (b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment ? (c) When a tiny circular obstacle is placed in the path of light from a distant source. observer moving. which alternative picture of light is consistent with experiment? 10. glass or water). (ii) direction of propagation. On which of these factors. 10. the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest.13 You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction.16 In double-slit experiment using light of wavelength 600 nm. (v) intensity of the wave. Explain why? 384 (d) Two students are separated by a 7 m partition wall in a room 10 m high. strictly identical for these situations. Estimate the speed with which the star is receding from the Earth. Is the prediction confirmed by experimental determination of the speed of light in water ? If not. say. the angular width of a fringe formed on a distant screen is 0. if any. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium? 10. (b) the speed of light in a medium (say. to be greater than the speed of light in vacuum. Explain why this should be so. the width of the slit is made double the original width. What is the justification? Two towers on top of two hills are 40 km apart. Find the width of the slit. (e) Ray optics is based on the assumption that light travels in a straight line. Answer the following questions: (a) When a low flying aircraft passes overhead. (b) As you have learnt in the text. the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. Justify this by suitably dividing the slit to bring out the cancellation. which can be sent between the towers without appreciable diffraction effects? A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. we sometimes notice a slight shaking of the picture on our TV screen. It is observed that the first minimum is at a distance of 2. it was stated that the intensity is zero at angles of n λ/a. The line joining them passes 50 m above a hill halfway between the towers. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. how is it that the students are unable to see each other even though they can converse easily. Suggest a possible explanation.19 10.21 385 . 10.18 10. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. What is the justification of this principle? In deriving the single slit diffraction pattern.5 mm from the centre of the screen.20 10. What is the longest wavelength of radio waves.Wave Optics obstacles. in 1879. Towards the same period at the end of 19th century. Thomson (1856 -1940) confirmed this hypothesis. The discovery of X-rays by Roentgen in 1895. a discharge took place between the two electrodes on applying the electric field to the gas in the discharge tube. J. The colour of glow of the glass depended on the type of glass. It was found that at sufficiently low pressure of about 0. were important milestones in the understanding of atomic structure. A fluorescent glow appeared on the glass opposite to cathode. by William Crookes who later.Physics Chapter Eleven DUAL NATURE OF RADIATION AND MATTER 11. it being yellowish-green for soda glass. J. By applying mutually perpendicular electric and magnetic fields across the discharge tube. in 1870. experimental investigations on conduction of electricity (electric discharge) through gases at low pressure in a discharge tube led to many historic discoveries.1 INTRODUCTION The Maxwell’s equations of electromagnetism and Hertz experiments on the generation and detection of electromagnetic waves in 1887 strongly established the wave nature of light. Thomson in 1897. The British physicist J. suggested that these rays consisted of streams of fast moving negatively charged particles. J. J.001 mm of mercury column. The cause of this fluorescence was attributed to the radiation which appeared to be coming from the cathode. Thomson was the first to determine experimentally the speed and the specific charge [charge to mass ratio (e/m )] of the cathode ray 386 . These cathode rays were discovered. and of electron by J. The free electron is thus held inside the metal surface by the attractive forces of the ions. in 1897. A certain minimum amount of energy is required to be given to an electron to pull it out from the surface of the metal. Around the same time. This unit of energy is commonly used in atomic and nuclear physics. and suggested that they were fundamental.1 to 0. emitted negatively charged particles having small speeds. certain metals when heated to a high temperature were found to emit negatively charged particles. The minimum energy required for the electron emission from the metal surface can be supplied to the free electrons by any one of the following physical processes: 387 . the free electrons cannot normally escape out of the metal surface. Also. However. A. the value of e/m was found to be independent of the nature of the material/metal used as the cathode (emitter).14 eV ) for caesium. These values are approximate as they are very sensitive to surface impurities. named these particles as electrons. They were found to travel with speeds ranging from about 0. or the gas introduced in the discharge tube. were identical in nature. he was awarded the Nobel Prize in Physics in 1906. J. 11.2 times the speed of light (3 ×108 m/s). 1. through his theoretical and experimental investigations on conduction of electricity by gasses. Millikan (1868 -1953) performed the pioneering oil-drop experiment for the precise measurement of the charge on an electron. In 1913. If an electron attempts to come out of the metal. Millikan’s experiment established that electric charge is quantised. The presently accepted value of e/m is 1. although produced under different conditions. This minimum energy required by an electron to escape from the metal surface is called the work function of the metal. it was found that certain metals.1. Thomson. universal constituents of matter.76 × 1011 C/kg. Further. From the values of charge (e ) and specific charge (e/m ). the American physicist R. the mass (m ) of the electron could be determined.2 ELECTRON EMISSION We know that metals have free electrons (negatively charged particles) that are responsible for their conductivity. This observation suggested the universality of the cathode ray particles.1 that the work function of platinum is the highest (φ0 = 5. Consequently.Dual Nature of Radiation and Matter particles. in 1887. One electron volt is the energy gained by an electron when it has been accelerated by a potential difference of 1 volt. For his epoch-making discovery of electron. He found that the charge on an oil-droplet was always an integral multiple of an elementary charge. It is generally denoted by φ0 and measured in eV (electron volt). the metal surface acquires a positive charge and pulls the electron back to the metal. The value of e/m of these particles was found to be the same as that for cathode ray particles. so that 1 eV = 1.65 eV ) while it is the lowest (φ0 = 2. when irradiated by ultraviolet light. These observations thus established that all these particles. The values of work function of some metals are given in Table 11. The work function (φ0 ) depends on the properties of the metal and the nature of its surface.602 × 10 –19 C. J. the electron can come out of the metal surface only if it has got sufficient energy to overcome the attractive pull. Note from Table 11.602 ×10 –19 J. 30 2.28 4.65 4. When light falls on a metal surface. (iii) Photo-electric emission : When light of suitable frequency illuminates a metal surface. Light shining on the metal surface somehow facilitated the escape of free.3 PHOTOELECTRIC EFFECT 11. 11.Physics TABLE 11. 11.1 WORK FUNCTIONS OF SOME METALS Metal Cs K Na Ca Mo Pb Work function φο (eV) 2.14 2. electrons can be pulled out of the metal. current flows in the circuit (Fig. In his experimental investigation on the production of electromagnetic waves by means of a spark discharge.1). After gaining sufficient energy from the incident light.3.75 3. the electrons escape from the surface of the metal into the surrounding space.70 5. Lenard (1862-1947) observed that when ultraviolet radiations were allowed to fall on the emitter plate of an evacuated glass tube enclosing two electrodes (metal plates).65 (i) Thermionic emission: By suitably heating.17 4.2 Hallwachs’ and Lenard’s observations Wilhelm Hallwachs and Philipp Lenard investigated the phenomenon of photoelectric emission in detail during 1886-1902.1 Hertz’s observations The phenomenon of photoelectric emission was discovered in 1887 by Heinrich Hertz (1857-1894).49 4. sufficient thermal energy can be imparted to the free electrons to enable them to come out of the metal.15 5. As soon as the ultraviolet radiations were stopped. These photo(light)-generated electrons are called photoelectrons. 11. Hertz observed that high voltage sparks across the detector loop were enhanced when the emitter plate was illuminated by ultraviolet light from an arc lamp. as in a spark plug.25 Metal Al Hg Cu Ag Ni Pt Work function φο (eV) 4.3. electrons are emitted from the metal surface.20 4. some electrons near the surface absorb enough energy from the incident radiation to overcome the attraction of the positive ions in the material of the surface. the current flow also 388 . during his electromagnetic wave experiments. charged particles which we now know as electrons. (ii) Field emission : By applying a very strong electric field (of the order of 108 V m–1) to a metal. in 1888. electrons are ejected from it which are attracted towards the positive. having short wavelength. Due to negative charge. responded only to ultraviolet light.ca/site/projects/physics. These observations indicate that when ultraviolet radiations fall on the emitter plate C. The polarity of the plates C and A can be reversed by a commutator. The phenomenon is called photoelectric effect. and with frequency and intensity of incident light. Hallwachs. He observed that the zinc plate lost its charge when it was illuminated by ultraviolet light. However. Thus. From these observations he concluded that negatively charged particles were emitted from the zinc plate under the action of ultraviolet light. undertook the study further and connected a negatively charged zinc plate to an electroscope. caesium and rubidium were sensitive even to visible light. resulting in the current flow. Positive charge on a positively charged zinc plate was found to be further enhanced when it was illuminated by ultraviolet light.html Simulate experiments on photoelectric effect 11. After the discovery of the electron in 1897. these electrons were termed as photoelectrons. All these photosensitive substances emit electrons when they are illuminated by light. it became evident that the incident light causes electrons to be emitted from the emitter plate. Further. the emitted electrons are pushed towards the collector plate by the electric field.4 E XPERIMENTAL S TUDY EFFECT OF P HOTOELECTRIC Figure 11. Monochromatic light from the source S of sufficiently short wavelength passes through the window W and falls on the photosensitive plate C (emitter).kcvs. no electrons were emitted at all when the frequency of the incident light was smaller than a certain minimum value. to cause electron emission from the surface. collector plate A by the electric field. the electrons are 389 . The battery maintains the potential difference between the plates C and A. which permits ultraviolet radiation to pass through it and irradiate the photosensitive plate C. This minimum frequency depends on the nature of the material of the emitter plate. the plate A can be maintained at a desired positive or negative potential with respect to emitter C. potassium.Dual Nature of Radiation and Matter stopped. A transparent quartz window is sealed on to the glass tube. It consists of an evacuated glass/quartz tube having a photosensitive plate C and another metal plate A. Hallwachs and Lenard also observed that when ultraviolet light fell on the emitter plate. It was found that certain metals like zinc. that can be varied. Hallwachs and Lenard studied how this photo current varied with collector plate potential. sodium. cadmium. When the collector plate A is positive with respect to the emitter plate C. The electrons flow through the evacuated glass tube. etc. some alkali metals such as lithium. Thus. magnesium. The electrons are emitted by the plate C and are collected by the plate A (collector). http://www. called the threshold frequency. After the discovery of electrons. by the electric field created by the battery. the uncharged zinc plate became positively charged when it was irradiated by ultraviolet light.1 depicts a schematic view of the arrangement used for the experimental study of the photoelectric effect. light falling on the surface of the emitter causes current in the external circuit.. 11.4.1 Effect of intensity of light on photocurrent The collector A is maintained at a positive potential with respect to emitter C so that electrons ejected from C are attracted towards collector A. 11. Saturation current corresponds to the case when all the photoelectrons emitted by the emitter plate C reach the collector plate A. the photocurrent does not increase. the intensity of light is varied and the resulting photoelectric current is measured each time.1 to study the variation of photocurrent with (a) intensity of radiation. FIGURE 11. (b) frequency of incident radiation. (c) the potential difference between the plates A and C. Keeping the frequency of the incident radiation and the accelerating potential fixed. and (d) the nature of the material of plate C. The intensity of light is varied by changing the distance of the light source from the emitter. We next vary the positive potential of plate A gradually and measure the resulting photocurrent each time. 11. At some stage.2 Variation of Photoelectric current with intensity of light. It is found that the photoelectric current increases with increase in accelerating (positive) potential. This implies that the number of photoelectrons emitted per second is directly proportional to the intensity of incident radiation. Light of different frequencies can be used by putting appropriate coloured filter or coloured glass in the path of light falling on the emitter C. If we increase the accelerating potential of plate A further. The potential difference between the emitter and collector plates is measured by a voltmeter (V) whereas the resulting photo current flowing in the circuit is measured by a microammeter ( µA). FIGURE 11.2 Effect of potential on photoelectric current We first keep the plate A at some positive accelerating potential with respect to the plate C and illuminate the plate C with light of fixed frequency ν and fixed intensity I1. We now apply a negative (retarding) potential to the plate A with respect to the plate C and make it increasingly negative gradually.4. for a certain positive potential of plate A. We can use the experimental arrangement of Fig. It is found that the photocurrent increases linearly with intensity of incident light as shown graphically in Fig.Physics attracted to it.2. When the 390 . as can the potential difference V between the emitter C and the collector A.1 Experimental arrangement for study of photoelectric effect. 11. The photoelectric current can be increased or decreased by varying the potential of collector plate A with respect to the emitter plate C. The photocurrent is directly proportional to the number of photoelectrons emitted per second. This maximum value of the photoelectric current is called saturation current. The intensity and frequency of the incident light can be varied. The emission of electrons causes flow of electric current in the circuit. all the emitted electrons are collected by the plate A and the photoelectric current becomes maximum or saturates. 391 . The resulting variation is shown in Fig.4. proportional to the intensity of incident radiation. the minimum negative (retarding) potential V0 given to the plate A for which the photocurrent stops or becomes zero is called the cut-off or stopping potential. for a given frequency of the incident radiation.3 Variation of photocurrent with Photoelectric current is zero when the collector plate potential for different stopping potential is sufficient to repel even intensity of incident radiation. 11. so that Kmax = e V0 (11. FIGURE 11. the most energetic photoelectrons. but is independent of intensity of incident radiation. We note that the saturation currents are now found to be at higher values. We obtain different values of stopping potential but the same value of the saturation current for incident radiation of different frequencies. as shown graphically in Fig. the stopping potential is independent of its intensity.4. critical value of the negative potential V0 on the plate A. Thus. We suitably adjust the same intensity of light radiation at various frequencies and study the variation of photocurrent with collector plate potential.Dual Nature of Radiation and Matter polarity is reversed.1) We can now repeat this experiment with incident radiation of the same frequency but of higher intensity I2 and I3 (I3 > I2 > I1). Note from FIGURE 11. 11. The interpretation of the observation in terms of photoelectrons is straightforward. The photocurrent is found to decrease rapidly until it drops to zero at a certain sharply defined. For a particular frequency of incident radiation. In other words. 11. The stopping potential is more negative for higher frequencies of incident radiation.4 Variation of photoelectric current with collector plate potential for different frequencies of incident radiation.3 Effect of frequency of incident radiation on stopping potential We now study the relation between the frequency ν of the incident radiation and the stopping potential V0.3. The energy of the emitted electrons depends on the frequency of the incident radiations. with the maximum kinetic energy (Kmax). But the stopping potential remains the same as that for the incident radiation of intensity I1. the electrons are repelled and only the most energetic electrons are able to reach the collector A. This shows that more electrons are being emitted per second. All the photoelectrons emitted from the metal do not have the same energy. the maximum kinetic energy of photoelectrons depends on the light source and the emitter plate material. Selenium is more sensitive than zinc or copper.5 Variation of stopping potential V0 with frequency ν of incident radiation for a (i) the stopping potential V0 varies linearly with given photosensitive material. Above the threshold 392 frequency. This minimum. we need greater retarding potential to stop them completely. The graph shows that FIGURE 11. greater is the maximum kinetic energy of the photoelectrons. Consequently. (ii) there exists a certain minimum cut-off frequency ν0 for which the stopping potential is zero. but is independent of its intensity. the photoelectric emission starts instantaneously without any apparent time lag. These observations have two implications: (i) The maximum kinetic energy of the photoelectrons varies linearly with the frequency of incident radiation. The same photosensitive substance gives different response to light of different wavelengths.3). cut-off frequency ν0. the photoelectric current is directly proportional to the intensity of incident light (Fig.4 that the stopping potentials are in the order V03 > V02 > V01 if the frequencies are in the order ν3 > ν2 > ν1 . (ii) For a frequency ν of incident radiation. (i) For a given photosensitive material and frequency of incident radiation (above the threshold frequency). there exists a certain minimum cut-off frequency of the incident radiation. saturation current is found to be proportional to the intensity of incident radiation whereas the stopping potential is independent of its intensity (Fig. if frequency of the incident radiation exceeds the threshold frequency. no matter how intense the incident light is.5. If we plot a graph between the frequency of incident radiation and the corresponding stopping potential for different metals we get a straight line. (iii) For a given photosensitive material. It is different for different metals. below which no emission of photoelectrons takes place. Note that in all the above experiments. It is now known that emission starts in a time of the order of 10 – 9 s or less. Different photosensitive materials respond differently to light. lower than the cut-off frequency ν0. it is found that. (ii) For a given photosensitive material and frequency of incident radiation. is called the threshold frequency. We now summarise the experimental features and observations described in this section. the stopping potential or equivalently the maximum kinetic . This implies that greater the frequency of incident light.2). called the threshold frequency. 11. the frequency of incident radiation for a given photosensitive material. as shown in Fig. even if the incident radiation is very dim. For example.Physics Fig. 11. 11. ultraviolet light gives rise to photoelectric effect in copper while green or red light does not. 11. no photoelectric emission is possible even if the intensity is large. 11.5 PHOTOELECTRIC EFFECT OF LIGHT AND WAVE THEORY The wave nature of light was well established by the end of the nineteenth century. Further. the wave picture is unable to explain the most basic features of photoelectric emission. The phenomena of interference. but is independent of its intensity (Fig.5). light is an electromagnetic wave consisting of electric and magnetic fields with continuous distribution of energy over the region of space over which the wave is extended.4. Consequently. These expectations of the wave theory directly contradict observations (i). 11. the greater should be the energy absorbed by each electron.6 EINSTEIN’S PHOTOELECTRIC EQUATION: ENERGY QUANTUM OF RADIATION In 1905. diffraction and polarisation were explained in a natural and satisfactory way by the wave picture of light. we should note that in the wave picture. Radiation energy is built up of discrete units – the so called quanta of energy of radiation. the energy absorbed per electron per unit time turns out to be small. so that they exceed the minimum energy needed to escape from the metal surface . A threshold frequency. In this picture. In this picture. therefore. even when the incident radiation is made exceedingly dim. the absorption of energy by electron takes place continuously over the entire wavefront of the radiation. According to this picture. Albert Einstein (1879 -1955) proposed a radically new picture of electromagnetic radiation to explain photoelectric effect. the greater the intensity. the free electrons at the surface of the metal (over which the beam of radiation falls) absorb the radiant energy continuously.3. (ii) and (iii) given at the end of sub-section 11. the maximum kinetic energy of the photoelectrons on the surface is then expected to increase with increase in intensity. Each quantum of radiant energy 393 . The greater the intensity of radiation. In short. According to the wave picture of light. should not exist. no matter what the frequency of radiation is. (iv) The photoelectric emission is an instantaneous process without any apparent time lag (∼10 – 9s or less). a sufficiently intense beam of radiation (over sufficient time) should be able to impart enough energy to the electrons. Let us now see if this wave picture of light can explain the observations on photoelectric emission given in the previous section. Explicit calculations estimate that it can take hours or more for a single electron to pick up sufficient energy to overcome the work function and come out of the metal. Since a large number of electrons absorb energy. Also. This conclusion is again in striking contrast to observation (iv) that the photoelectric emission is instantaneous.Dual Nature of Radiation and Matter energy of the emitted photoelectrons increases linearly with the frequency of the incident radiation. photoelectric emission does not take place by continuous absorption of energy from radiation. the greater are the amplitude of electric and magnetic fields. 11. confirmed experimentally a few years later and provided a convincing evidence of the atomic picture of matter. and is independent of intensity of radiation. In photoelectric effect. was born in Ulm. Some of Einstein’s most significant later contributions are: the notion of stimulated emission introduced in an alternative derivation of Planck’s blackbody radiation law.3) h Equation (11.2 ) implies that photoelectric emission is possible only if h ν > φ0 or ν > ν0 . (11. Eq. This explains why. he published the general theory of relativity. • According to Eq. (11. the higher the minimum or threshold frequency ν0 needed to emit photoelectrons. This has happened because in Einstein’s picture. where (11. he developed a theory of Brownian motion. the greater is the number of electrons absorbing the energy quanta and greater.2) is known as Einstein’s photoelectric equation. Germany. is the number of electrons coming out of the metal (for ν > ν0 ). is proportional to the number of energy quanta per unit area per unit time.2). quantum statistics of a gas of massive bosons.3) shows that the greater the work function φ0. below which no photoelectric emission is possible. Albert Einstein (1879 – 1955) Einstein. for ν > ν0 . photoelectric current is proportional to intensity. he introduced the notion of light quanta (now called photons) and used it to explain the features of photoelectric effect. Note that the intensity of light of a given frequency is determined by the number of photons incident per second. The greater the number of energy quanta available. he was awarded the Nobel Prize in physics for his contribution to theoretical physics and the photoelectric effect. In 1921. ALBERT EINSTEIN (1879 – 1955) ν0 = φ0 • 394 . Increasing the intensity will increase the number of emitted electrons per second. In the second paper.3. and a critical analysis of the foundations of quantum mechanics.2) More tightly bound electrons will emerge with kinetic energies less than the maximum value. therefore. In this picture. In the first paper. Kmax depends linearly on ν. In 1916. no matter how intense the incident radiation may be or how long it falls on the surface. where h is Planck’s constant and ν the frequency of light. an electron absorbs a quantum of energy (hν ) of radiation. The third paper gave birth to the special theory of relativity. The intensity of radiation (that is proportional to the number of energy quanta per unit area per unit time) is irrelevant to this basic process. the maximum kinetic energy of the emitted photoelectrons is determined by the energy of each photon. he published three pathbreaking papers. However. Equation (11. the electron is emitted with maximum kinetic energy Kmax = hν – φ0 (11. static model of the universe which started modern cosmology. Thus. intensity of radiation as noted above. there exists a threshold frequencyν0 (= φ0/h) for the metal surface. photoelectric effect arises from the absorption of a single quantum of radiation by a single electron. We now see how this equation accounts in a simple and elegant manner all the observations on photoelectric effect given at the end of sub-section 11. • Since Kmax must be non-negative.4. in agreement with observation.Physics has energy h ν. one of the greatest physicists of all time. If this quantum of energy absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (work function φ0). In 1905. Dual Nature of Radiation and Matter In Einstein’s picture. for ν ≥ ν0 • φ h or V0 =  (11. This process is instantaneous. The particle-like behaviour of light was further confirmed. (11. whatever may be the intensity i. led to the acceptance of Einstein’s picture of photoelectric effect. the number of quanta of radiation per unit area per unit time. In 1921. each of energy h ν.e. Using Eq. Is the light quantum of energy to be associated with a particle? Einstein arrived at the important result. We can summarise the photon picture of electromagnetic radiation as follows: 395 . Eq.2). This particle was later named photon. Millikan verified photoelectric equation with great precision. Thus. A definite value of energy as well as momentum is a strong sign that the light quantum can be associated with a particle. Compton (1892-1962) on scattering of X-rays from electrons. 11. photoelectric emission is instantaneous. The successful explanation of photoelectric effect using the hypothesis of light quanta and the experimental determination of values of h and φ0. Einstein was awarded the Nobel Prize in Physics for his contribution to theoretical physics and the photoelectric effect. in agreement with values obtained from other experiments.1). Intensity only determines how many electrons are able to participate in the elementary process (absorption of a light quantum by a single electron) and.4) ν− 0  e  e This is an important result. in 1924. the photoelectric current. In 1923. In this way. independent of the nature of the material. He measured the slope of the straight line obtained for sodium. Millikan was awarded the Nobel Prize in physics for his work on the elementary charge of electricity and on the photoelectric effect.H. by the experiment of A. (11. 11. that the light quantum can also be associated with momentum (h ν/c ). he determined the value of Planck’s constant h . It predicts that the V0 versus ν curve is a straight line with slope = (h/e ). Low intensity does not mean delay in emission. since the basic elementary process is the same. for a number of alkali metals over a wide range of radiation frequencies. Using the known value of e. Millikan performed a series of experiments on photoelectric effect. the basic elementary process involved in photoelectric effect is the absorption of a light quantum by an electron. the photoelectric equation. Millikan proved the validity of Einstein’s photoelectric equation.7 PARTICLE NATURE OF LIGHT: THE PHOTON Photoelectric effect thus gave evidence to the strange fact that light in interaction with matter behaved as if it was made of quanta or packets of energy. in 1916. can be written as e V0 = h ν – φ 0. instead of disproving it. This value was close to the value of Planck’s contant (= 6. aimed at disproving Einstein’s photoelectric equation. therefore.626 × 10–34J s) determined in an entirely different context.5.. During 1906-1916. similar to that shown in Fig. on an average. with each photon having the same energy. (b) Photocurrent reduces to zero. the total energy and total momentum are conserved. and (b) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0. so that ν0 = φ0 = h 2. so that P = N E. there is only an increase in the number of photons per second crossing a given area. (iv) Photons are electrically neutral and are not deflected by electric and magnetic fields. the power P transmitted in the beam equals N times the energy per photon E. for frequencies less than this threshold frequency.14 × 1.98 × 10–19 J (b) If N is the number of photons emitted by the source per second.2 = 396 Thus. Then N= EXAMPLE 11. Thus.0 ×10–3 W. whatever the intensity of radiation may be. no photoelectrons are ejected.16 × 1014 Hz 6.1 Monochromatic light of frequency 6. Example 11. when maximum kinetic energy of the emitted photoelectrons equals the potential energy e V0 by the retarding potential V0. the energy h ν0 of the incident radiation must be equal to work function φ 0. (ii) Each photon has energy E (=hν ) and momentum p (= h ν/c). or wavelength λ.63 × 10−34 J s .Physics (i) In interaction of radiation with matter. The photon may be absorbed or a new photon may be created. photon energy is independent of intensity of radiation. Solution (a) For the cut-off or threshold frequency.0 ×1015 photons per second. radiation behaves as if it is made up of particles called photons. (v) In a photon-particle collision (such as photon-electron collision). the number of photons may not be conserved in a collision.63 × 10−34 J s EXAMPLE 11. By increasing the intensity of light of given wavelength.1 P 2. the speed of light. have the same energy E (=hν = hc/λ) and momentum p (= hν/c = h/λ).0 ×10 14 Hz is produced by a laser. and speed c. (iii) All photons of light of a particular frequency ν. Find (a) the threshold frequency for caesium. However.6 × 10−19 J = 5.98 × 10 −19 J = 5. are emitted by the source? Solution (a) Each photon has an energy E = h ν = ( 6.60 V. Einstein’s Photoelectric equation is 2. Example 11.63 ×10–34 J s) (6. The power emitted is 2. (a) What is the energy of a photon in the light beam? (b) How many photons per second.14 eV.2 The work function of caesium is 2.14 eV 6.0 ×1014 Hz) = 3.0 × 10−3 W = E 3. 30 eV) and Cs (with φ0 = 2.989 × 10 –25 J m Incident photon energy.1 and using the results of (i).6×10 –19 J/eV = 3.Dual Nature of Radiation and Matter eV0 = hν – φ 0 = hc λ or.64 eV (b) For a photoelectric device to operate.14 eV) only. E = hν = hc/λ E = (6.62×10–19 J = 1.10 × 10 –19 J = 1.63 × 10−34 J s) × (3 × 108 m/s) (0.64 eV) for any of these photosensitive materials.3 397 .75 eV).14 eV) 19. λ = hc/(eV0 + φ0 ) = = – φ0 (6. can you build a photoelectric device that operates with visible light? Solution (a) Energy of the incident photon.6 × 10 −19 J EXAMPLE 11.26 eV) for Cs (with φ0 = 2.989 × 10 –25 J m (i) For violet light. E3 = 760×10 –9 m = 2. λ EXAMPLE 11. yellow-green colour.63×10–34 J s and 1 eV = 1.26 eV (iii) For red light. λ3 = 760 nm (higher wavelength end) 1.10 × 10–19J 5. about 550 nm (average wavelength) for yellowgreen colour and about 760 nm for red colour.62×10–19 J = 2. (ii) average wavelength.89 × 10 −26 J m (2.89 × 10 −26 J m = 454 nm 2. (ii) and (iii) of (a). E1 = 390×10 –9 m = 5. It will also operate with yellow-green light (with E = 2. and (iii) red end of the visible spectrum? (Take h = 6. λ2 = 550 nm (average wavelength) 1.19 eV) photosensitive material Na (with φ0 = 2. E2 = 550×10 –9 m = 3.3 The wavelength of light in the visible region is about 390 nm for violet colour.14 eV).) (b) From which of the photosensitive materials with work functions listed in Table 11.6×10 –19J.19 eV (ii) For yellow-green light. However.2 λ= Example 11.63×10–34J s) (3×108 m/s)/λ = 1. Thus.60 eV + 2. we require incident light energy E to be equal to or greater than the work function φ0 of the material.989 × 10 –25 J m Incident photon energy. K (with φ0 = 2. it will not operate with red light (with E = 1.74 eV) 19.74 × 1. λ1 = 390 nm (lower wavelength end) 1. the photoelectric device will operate with violet light (with E = 3.989 × 10 –25 J m Incident photon energy. (a) What are the energies of photons in (eV) at the (i) violet end. 12 kg moving with a speed of 20 m s–1 is easily calculated: (11. Clearly. the de Broglie wavelength of a photon given by Eq. For example.5) is known as the de Broglie relation and the wavelength λ of the matter wave is called de Broglie wavelength. The dual aspect of matter is evident in the de Broglie relation. the de Broglie wavelength of a ball of mass 0. so should matter.) also exhibit wave-like character? In 1924.6) 398 . from Eq. in general) comes out clearly from what we have learnt in this and the preceding chapters. diffraction and polarisation. Whether a particle or wave description is best suited for understanding an experiment depends on the nature of the experiment. etc. λ is smaller for a heavier particle ( large m ) or more energetic particle (large v). The wave nature of light shows up in the phenomena of interference. On the other hand. protons. might not the particles of nature (the electrons. (11.5 ). p = hν /c Therefore. as we have seen. in the familiar phenomenon of seeing an object by our eye. A natural question arises: If radiation has a dual (wave-particle) nature. (11. radiation behaves as if it is made up of a bunch of particles – the photons.7) p ν That is. On the left hand side of Eq.8 WAVE NATURE OF MATTER The dual (wave-particle) nature of light (electromagnetic radiation. both descriptions are important. For example.5). Equation (11.5) where m is the mass of the particle and v its speed.5) equals the wavelength of electromagnetic radiation of which the photon is a quantum of energy and momentum.5) for a material particle is basically a hypothesis whose validity can be tested only by experiment. Planck’s constant h relates the two attributes. in photoelectric effect and Compton effect which involve energy and momentum transfer. λ is the attribute of a wave while on the right hand side the momentum p is a typical attribute of a particle. h c = =λ (11. For a photon. Equation (11.Physics 11. But its absorption by the rods and cones (of the retina) requires the photon picture of light. the French physicist Louis Victor de Broglie (pronounced as de Broy) (1892-1987) put forward the bold hypothesis that moving particles of matter should display wave-like properties under suitable conditions. it is interesting to see that it is satisfied also by a photon. The gathering and focussing mechanism of light by the eye-lens is well described in the wave picture. must have symmetrical character. However. If radiation shows dual aspects. (11. He reasoned that nature was symmetrical and that the two basic physical entities – matter and energy. De Broglie proposed that the wave length λ associated with a particle of momentum p is given as λ = h h = p mv (11. (invisible) ultraviolet light is continuously made to fall on a photocell installed at the doorway. It is connected to the external circuit having a high-tension battery B and microammeter (µA) as shown in the Figure.12 kg × 20 m s–1 = 2.63 × 10−34 J s h = = 2. provided they enter the hall one by one. The photocells. photo cells are used whenever it is necessary to measure the intensity of light. A person entering the door interrupts the beam falling on the photocell. In the event of breaking out of fire.40 kg m s–1 λ= 6. Light meters in photographic cameras make use of photo cells to measure the intensity of incident light. Photocurrent of the order of a few microampere can be normally obtained from a photo cell. The abrupt change in photocurrent is used to start an electric bell ringing. photoelectrons are emitted. These photoelectrons are drawn to the collector A. It is a device whose electrical properties are affected by light. In scientific work. So photocells help count the persons entering an auditorium. In burglar alarm. This completes the electric circuit through an electric bell or a siren which starts operating as a warning signal. inserted in the door light electric circuit. A person approaching a doorway may interrupt a light beam which is incident on a photocell.40 kg m s −1 p 399 . This current can be used to operate control systems and in light measuring devices. are used as automatic door opener. p = m v = 0. A photocell converts a change in intensity of illumination into a change in photocurrent. A photocell consists of a semi-cylindrical photo-sensitive metal plate C (emitter) and a wire loop A (collector) supported in an evacuated glass or quartz bulb. A part of the bulb is left clean for the light to enter it.76 × 10–34 m 2. When light of suitable wavelength falls on the emitter C. a thin layer of photosensitive material is pasted on the inside of the bulb.Dual Nature of Radiation and Matter PHOTOCELL A photocell is a technological application of the photoelectric effect. They are used for detection of traffic law defaulters: an alarm may be sounded whenever a beam of (invisible) radiation is intercepted. The abrupt change in photocurrent A photo cell may be used to start a motor which opens the door or rings an alarm. A photocell of lead sulphide sensitive to infrared radiation is used in electronic ignition circuits. light radiations fall upon the photocell. They are used in industries for detecting minor flaws or holes in metal sheets. They are used in the control of a counting device which records every interruption of the light beam caused by a person or object passing across the beam. It is also sometimes called an electric eye. In fire alarm. instead of the plate C. a number of photocells are installed at suitable places in a building. Sometimes. Photocells are used in the reproduction of sound in motion pictures and in the television camera for scanning and telecasting scenes. it is not possible to measure both the position and momentum of an electron (or any other particle) at the same time exactly. if ∆p is zero. On the other hand. the wave character of particles is significant and measurable.e. Similarly. This idea was developed by Erwin Schródinger into a fullfledged theory of quantum mechanics commonly known as wave mechanics.10) Substituting the numerical values of h.112 nm. Ordinarily. The product of ∆x and ∆p is of the order of ħ* (with ħ = h/2π). ∆x ∆p ≈ ħ (11. Eq.11) gives λ = 0. de Broglie was awarded the Nobel Prize in Physics for his discovery of the wave nature of electrons. There is always some uncertainty (∆ x ) in the specification of position and some uncertainty (∆p ) in the specification of momentum. he was awarded the Nobel Prize in Physics for his discovery of the wave nature of electrons. but then ∆p must be infinite in order that the product is non-zero. This suggests that matter waves associated with an electron could be verified by crystal diffraction experiments analogous to X-ray diffraction. A wave of definite (single) λ = 400 * A more rigorous treatment gives ∆x ∆p ≥ ħ/2.∆p = 0). it has a definite wavelength λ. In 1929. (11. i.Physics This wavelength is so small that it is beyond any measurement.9) The de Broglie wavelength λ of the electron is then λ = h h h = = p 2 mK 2 m eV (11. if an electron has a definite momentum p.12) allows the possibility that ∆x is zero. The matter–wave picture elegantly incorporated the Heisenberg’s uncertainty principle. According to the principle. charge e) accelerated from rest through a potential V. so that 2 2m p= 2 m K = 2 m eV (11. .227 nm (11.8) Now. both ∆x and ∆p are non-zero such that their product is of the order of ħ. in the sub-atomic domain. For a 120 V accelerating potential. Now. The kinetic energy K of the electron equals the work done (eV ) on it by the electric field: K=eV (11. Consider an electron (mass m.12) Equation (11. ∆x must be infinite. (i. we get 1. In 1929. This wavelength is of the same order as the spacing between the atomic planes in crystals. m.11) V where V is the magnitude of accelerating potential in volts.e. We describe the experimental verification of the de Broglie hypothesis in the next section. This is the reason why macroscopic objects in our daily life do not show wave-like properties. K = Louis Victor de Broglie (1892 – 1987) French physicist who put forth revolutionary idea of wave nature of matter. LOUIS VICTOR DE BROGLIE (1892 – 1987) 1 p2 m v2 = . by the de Broglie relation.. e. and (b) an extended wave with fixed wavelength. (b) The matter wave corresponding to a definite momentum of an electron extends all over space. the matter wave associated with the electron is not extended all over space. Figure 11. you must appreciate that a wave packet of finite extension does not have a single wavelength. By Born’s probability interpretation this means that the electron is not localised in any finite region of space.4 What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.Dual Nature of Radiation and Matter wavelength extends all over space. it is associated with an uncertainty in position (∆x ) and an uncertainty in momentum (∆p).4 401 . By de Broglie’s relation.6 shows a schematic diagram of (a) a localised wave packet. λ = h/p = 6. Then momentum p ’ = m ’ v ’ = 0. the momentum of the electron will also have a spread – an uncertainty ∆p. a spread in momentum). speed v = 5. In Chapter 12.6 (a) The wave packet description of an electron. Consequently. It is a wave packet extending over some finite region of space.0 m/s. That is. ∆ p = 0 and ∆ x → ∞.0 (m/s) p ’= 4.63 × 10 –34 J s 4. and (b) a ball of mass 150 g travelling at 30.4 × 106 (m/s) p = 4. It is built up of wavelengths spread around some central wavelength.4×106 m/s.11×10–31 kg. Example 11. speed v ’ = 30. then.92 × 10–24 kg m/s de Broglie wavelength.4×106 m/s.11×10–31 (kg) × 5. In that case ∆x is not infinite but has some finite value depending on the extension of the wave packet. The wave packet corresponds to a spread of wavelength around some central wavelength (and hence by de Broglie relation.135 nm (b) For the ball: Mass m ’ = 0.150 (kg) × 30. momentum p = m v = 9.0 m/s? Solution (a) For the electron: Mass m = 9. the de Broglie relation will be seen to justify Bohr’s postulate on quantisation of angular momentum of electron in an atom.50 kg m/s de Broglie wavelength λ’ = h/p’. Then. which is consistent with the uncertainty principle. In this case.92 × 10 –24 kg m/s λ = 0. its position uncertainty is infinite ( ∆ x → ∞ ). In general.150 kg. EXAMPLE 11. Also. FIGURE 11. This is as expected from the uncertainty principle. It can be shown that the wave packet description together with de Broglie relation and Born’s probability interpretation reproduce the Heisenberg’s uncertainty principle exactly. and a proton have the same kinetic energy.6 Example 11. de Broglie wavelength. Solution de Broglie wavelength of a moving particle. an electron. if A is the amplitude of the wave at a point. α – particle has the shortest de Broglie wavelength.4 = 6. a truly satisfactory physical understanding of the dual nature of matter and radiation has not emerged so far. and many others) struggled with this and related concepts for long. Calculate the particle’s mass and identify the particle. Which of these particles has the shortest de Broglie wavelength? Solution For a particle. EXAMPLE 11. The great founders of quantum mechanics (Niels Bohr. λ = h / 2mK For the same kinetic energy K.813 × 10–4. means.5 Then. An important milestone in this connection was when Max Born (18821970) suggested a probability interpretation to the matter wave amplitude.50 × kg m/s λ ’ = 1. there is a greater probability of the particle being found there than where the intensity is small. quite beyond experimental measurement. Probability density means probability per unit volume. say. According to this. However. the de Broglie wavelength associated with the particle is inversely proportional to the square root of their masses. ( 4 2 ) PROBABILITY INTERPRETATION TO MATTER WAVES It is worth pausing here to reflect on just what a matter wave associated with a particle. Actually. The ratio of the de Broglie wavelength of the particle to that of the electron is 1. λ = h/p Kinetic energy. |A|2 ∆V is the probability of the particle being found in a small volume ∆V around that point.63 × 10±34 J s 4. Albert Einstein. the intensity (square of the amplitude) of the matter wave at a point determines the probability density of the particle at that point. Hence. the concept of matter wave has been mathematically introduced in modern quantum mechanics with great success. Thus. an α-particle. if the intensity of matter wave is large in a certain region. A proton ( H) 1 1 is 1836 times massive than an electron and an α-particle He four times that of a proton. Despite this. Thus.Physics EXAMPLE 11.6 A particle is moving three times as fast as an electron. Example 11.5 An electron. K = p2/2m EXAMPLE 11. for the ball it is about 10–19 times the size of the proton. Still the deep physical interpretation of quantum mechanics continues to be an area of active research. having mass m and velocity v: 402 .47 ×10–34 m The de Broglie wavelength of electron is comparable with X-ray wavelengths. Davisson and L.J.7 1. 11. mass m e = h/λe ve Now. m = me   λ  v  m = (9. who observed diffraction effects with beams of electrons scattered by crystals. The experimental arrangement used by Davisson and Germer is schematically shown in Fig. Electrons FIGURE 11.7 What is the de Broglie wavelength associated with an electron.H.Dual Nature of Radiation and Matter λ= h h = p mv Mass.9 DAVISSON AND GERMER EXPERIMENT The wave nature of electrons was first experimentally verified by C. the particle.123 nm 100 The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths. or battery).675 × 10–27 kg. m = h/λv For an electron. It consists of an electron gun which comprises of a tungsten filament F. we have v/ve = 3 and λ/λe = 1. Thus.P. mass of the particle. 11. coated with barium oxide and heated by a low voltage power supply (L.227 nm V EXAMPLE 11.11×10–31 kg) × (1/3) × (1/1.7.227 nm = 0. Thomson. accelerated through a potential differnece of 100 volts? Solution Accelerating potential V = 100 V. The de Broglie wavelength λ is  λe   ve  λ = h /p = λ = 1. with this mass could be a proton or a neutron. Germer in 1927 and independently by G. in 1928.813 × 10– 4 EXAMPLE 11.T. Example 11. accelerated to a desired velocity 403 . Davisson and Thomson shared the Nobel Prize in 1937 for their experimental discovery of diffraction of electrons by crystals.813 × 10–4) m = 1.6 Then.7 Davisson-Germer electron emitted by the filament are diffraction arrangement. It has also led to the field of electron optics. or battery).11). in an experiment in 1994. The wave properties of electrons have been utilised in the design of electron microscope which is a great improvement. Also. More recently. It was noticed that a strong peak appeared in the intensity (I ) of the scattered electron for an accelarating voltage of 54V at a scattering angle θ = 50º The appearance of the peak in a particular direction is due to the constructive interference of electrons scattered from different layers of the regularly spaced atoms of the crystals. The beam is made to fall on the surface of a nickel crystal. using Eq. They are made to pass through a cylinder with fine holes along its axis. By moving the detector on the circular scale at different positions. . The detector can be moved on a circular scale and is connected to a sensitive galvanometer.edu/CMRAcfem/em. with higher resolution. From the electron diffraction measurements.nobelprize. which records the current. interference fringes were obtained with the beams of iodine molecules.167 nm 54 404 Thus. The apparatus is enclosed in an evacuated chamber. similar to that used for the wave nature of light.Physics by applying suitable potential/voltage from a high voltage power supply (H. over the optical microscope.html http://www.227 nm = 0. which are about a million times more massive than electrons. The experiment was performed by varying the accelarating voltage from 44 V to 68 V. The de Broglie wavelength λ associated with electrons. is measured by the electron detector (collector).165 nm. The intensity of the electron beam.227 V nm λ = 1. the intensity of the scattered electron beam is measured for different values of angle of scattering θ which is the angle between the incident and the scattered electron beams. scattered in a given direction. The deflection of the galvanometer is proportional to the intensity of the electron beam entering the collector.htm Development of electron microscope λ = h /p = 1. in 1989. The de Broglie hypothesis has been basic to the development of modern quantum mechanics. (11. The electrons are scattered in all directions by the atoms of the crystal. there is an excellent agreement between the theoretical value and the experimentally obtained value of de Broglie wavelength. The variation of the intensity (I ) of the scattered electrons with the angle of scattering θ is obtained for different accelerating voltages.T. producing a fine collimated beam. the wave nature of a beam of electrons was experimentally demonstrated in a double-slit experiment. the wavelength of matter waves was found to be 0.org/nobel_prizes/physics/laureates/1986/presentation-speech.unl. for V = 54 V is given by http://www. DavissonGermer experiment thus strikingly confirms the wave nature of electrons and the de Broglie relation. 6. Each photon carries an energy E (= h ν) and momentum p (= h/λ). According to this. The classical wave theory could not explain the main features of photoelectric effect. The maximum kinetic energy (1/2)m v 2 is equal to max the photon energy (hν ) minus the work function φ0 (= hν0 ) of the target metal: 2. Its picture of continuous absorption of energy from radiation could not explain the independence of Kmax on intensity. Photoelectric effect involves conversion of light energy into electrical energy. (ii) the potential difference applied between the two electrodes. and (ii) the nature of the emitter material. Reasoning that radiation and matter should be symmetrical in nature. The stopping potential is directly related to the maximum kinetic energy of electrons emitted: e V0 = (1/2) m v 2 = Kmax. 7.Dual Nature of Radiation and Matter SUMMARY 1. 8. no photoelectric emission takes place. Certain metals respond to ultraviolet light while others are sensitive even to the visible light. 3. Energy (greater than the work function (φο) required for electron emission from the metal surface can be supplied by suitably heating or applying strong electric field or irradiating it by light of suitable frequency. light is composed of discrete packets of energy called quanta or photons. Einstein’s photoelectric equation is in accordance with the energy conservation law as applied to the photon absorption by an electron in the metal. the existence of νo and the instantaneous nature of the process. Louis Victor de Broglie 405 . It follows the law of conservation of energy. it is independent of its intensity. no matter how large the intensity may be. which depend on the frequency (ν ) of incident light and not on its intensity. Radiation has dual nature: wave and particle. Photoelectric current depends on (i) the intensity of incident light. 5. characteristic of the metal. For a given frequency of incident light. The nature of experiment determines whether a wave or particle description is best suited for understanding the experimental result. The minimum energy needed by an electron to come out from a metal surface is called the work function of the metal. and (iii) the nature of the emitter material. 1 m v2 = V0 e = hν – φ0 = h (ν – ν0 ) max 2 This photoelectric equation explains all the features of the photoelectric effect. max Below a certain frequency (threshold frequency) ν 0 . introduced by Einstein. Millikan’s first precise measurements confirmed the Einstein’s photoelectric equation and obtained an accurate value of Planck’s constant h . 4. Einstein explained these features on the basis of photon picture of light. Photoelectric emission from the metal surface occurs due to absorption of a photon by an electron. Photoelectric effect is the phenomenon of emission of electrons by metals when illuminated by light of suitable frequency. The photoelectric emission is an instantaneous process and possesses certain special features. The stopping potential (Vo) depends on (i) the frequency of incident light. This led to the acceptance of particle or photon description (nature) of electromagnetic radiation. It is significantly measurable (of the order of the atomic-planes spacing in crystals) only in case of sub-atomic particles like electrons. Like molecules in a gas jar. protons. the energy required by an electron to come out of the metal is different for different electrons. Electron diffraction experiments by Davisson and Germer. Electrons with higher energy require less additional energy to come out of the metal than those with lower energies. Free electrons in a metal are free in the sense that they move inside the metal in a constant potential (This is only an approximation). The dualism of matter is inherent in the de Broglie relation which contains a wave concept (λ) and a particle concept (p). 9. The waves associated with the moving material particles are called matter waves or de Broglie waves. eV Hz m E = hν e V0= Kmax Kmax = E –φ0 φ0 ν0 λ ν0 = φ0 /h λ = h/p POINTS TO PONDER 1. it is indeed very small. as well as many later experiments. momenta). Free electrons in a metal do not all have the same energy. P. Thomson. Because of the energy distribution of free electrons in a metal. in case of macroscopic objects. (due to smallness of their masses and hence. They need additional energy to get out of the metal. The de Broglie wavelength (λ) associated with a moving particle is related to its momentum p as: λ = h/p. They are not free to move out of the metal. Work function is the least energy required by an electron to come out of the metal. This distribution is different from the usual Maxwell’s distribution that you have learnt in the study of kinetic theory of gases. The de Broglie hypothesis of matter waves supports the Bohr ’s concept of stationary orbits. 406 . The de Broglie wavelength is independent of the charge and nature of the material particle. have verified and confirmed the wave-nature of electrons. 2. the electrons have a certain energy distribution at a given temperature. 3. quite beyond measurement. but the difference has to do with the fact that electrons obey Pauli’s exclusion principle.Physics attributed a wave-like character to matter (material particles). etc. and by G. Physical Quantity Planck’s constant Stopping potential Work function Threshold frequency de Broglie wavelength Symbol Dimensions Unit Remarks h V0 [ML2 T –1] [ML2 T –3A–1] [ML2 T –2] [T –1] [L] Js V J. You will learn about it in later courses. However. commonly encountered in everyday life. 10. Observations on photoelectric effect imply that in the event of matterlight interaction.7 407 . This is not quite the same as saying that light consists of particles.14 eV. The work function of caesium metal is 2. (b) Stopping potential. How many photons (nearly) per square metre are incident on the Earth per second ? Assume that the photons in the sunlight have an average wavelength of 550 nm.5 V. What is the maximum kinetic energy of photoelectrons emitted? Monochromatic light of wavelength 632. 6. absorption of energy takes place in discrete units of hν. h p has physical significance.8 nm is produced by a helium-neon laser. arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area ). photoemission of electrons occurs. Observations on the stopping potential (its independence of intensity and dependence on frequency) are the crucial discriminator between the wave-picture and photon-picture of photoelectric effect. and (b) minimum wavelength of X-rays produced by 30 kV electrons.12 × 10–15 V s.3 11. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it.42 mW. EXERCISES 11. on the average. A 100 W sodium lamp radiates energy uniformly in all directions. What is the (a) maximum kinetic energy of the emitted electrons. the group velocity of the matter wave is physically meaningful and equals the velocity of the particle.6 11.5 The energy flux of sunlight reaching the surface of the earth is 1. The power emitted is 9. The wavelength of the sodium light is 589 nm.Dual Nature of Radiation and Matter 4. (a) What is the energy per photon associated 11. However. its phase velocity vp has no physical significance. (b) How many photons per second. The wavelength of a matter wave given by λ = 5.4 11. (a) Find the energy and momentum of each photon in the light beam. and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon? 11.2 11. and (c) maximum speed of the emitted photoelectrons? The photoelectric cut-off voltage in a certain experiment is 1.1 Find the (a) maximum frequency. In an experiment on photoelectric effect. the slope of the cut-off voltage versus frequency of incident light is found to be 4. each of energy hν. When light of frequency 6 ×1014Hz is incident on the metal surface.388 × 103 W/m2. Calculate the value of Planck’s constant. (Atomic mass of nitrogen = 14. (b) speed. 11. the stopping (cut-of f) potential of photoelectrons is 0. What is the threshold frequency for photoemission of electrons? 11. and (c) a dust particle of mass 1.00 nm. Electrons with a maximum speed of 6. When light from this spectral line is incident on the emitter. 11.0 km/s.16 An (a) (b) (c) electron and a photon each have a wavelength of 1.9 11. Find the work function of the material from which the emitter is made.2 × 1014 Hz is incident on the metal. If light of frequency 8.8 The threshold frequency for a certain metal is 3.38 V.17 (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1. (b) a ball of mass 0. Find the kinetic energy at which (a) an electron.Physics with the sodium light? (b) At what rate are the photons delivered to the sphere? 11. and the kinetic energy of electron. 11. in thermal equilibrium with matter.14 The wavelength of light from the spectral emission line of sodium is 589 nm. predict the cutoff voltage for the photoelectric emission.11 Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect.2 eV.15 What is the de Broglie wavelength of (a) a bullet of mass 0.19 What is the de Broglie wavelength of a nitrogen molecule in air at 300 K ? Assume that the molecule is moving with the root-meansquare speed of molecules at this temperature. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm? 11.40 × 10–10 m? (b) Also find the de Broglie wavelength of a neutron. would have the same de Broglie wavelength. 11. The work function for a certain metal is 4. 11. 11.060 kg moving at a speed of 1.0 m/s.13 What is the (a) momentum. 11.040 kg travelling at the speed of 1. and (b) a neutron.12 Calculate the (a) momentum.18 Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).10 Light of frequency 7. and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.0076 u) 408 . and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV. having an average kinetic energy of (3/2) k T at 300 K. Find their momenta.2 m/s ? 11. the energy of the photon.0 × 105 m/s are ejected from the surface.0 × 10–9 kg drifting with a speed of 2.21 × 1014 Hz is incident on a metal surface.3 × 1014 Hz. See answers at the end to know what ‘very high speed or energy’ means. What is the radius of the circle traced by the beam.20(b) and 11.76 × 1011C kg–1..30 × 10–4 T normal to the beam velocity. What is the wavelength associated with each γ-ray? (1BeV = 109 eV) 11.2 BeV into two γ-rays of equal energy. What is the maximum energy of a photon in the radiation ? (b) From your answer to (a). this method is known as the ‘fine beam tube’ method. The first number will tell you why radio engineers do not need to worry much about photons! The second number tells you why our eye can never ‘count photons’. Do you see what is wrong ? In what way is the formula to be modified? 11.e. given e/m for electron equals 1.21(b) take you to relativistic mechanics which is beyond the scope of this book.21 (a) A monoenergetic electron beam with electron speed of 5.23 (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.83 × 10–4 T curves the path of the electrons in a circular orbit of radius 12. a certain event is interpreted as annihilation of an electron-positron pair of total energy 10. 11. in what way is it modified ? [ Note: Exercises 11.24 In an accelerator experiment on high-energy collisions of electrons with positrons. Ignore the small initial speeds of the electrons. its e/m is given to be 1. emitting radiowaves of wavelength 500 m. (b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we 409 . even in barely detectable light. (b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not.) Determine e/m from the data.0 cm. A magnetic field of 2. i. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons.Dual Nature of Radiation and Matter ADDITIONAL EXERCISES 11.20 (a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter.] 11. (b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies.45 Å.20 × 106 m s–1 is subject to a magnetic field of 1.76 × 1011 C kg–1.25 Estimating the following two numbers should be interesting. (a) The number of photons emitted per second by a Medium wave transmitter of 10 kW power. guess what order of accelerating voltage (for electrons) is required in such a tube ? 11. and emitting light by electron capture. The specific charge of the electron.22 An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (∼10–2 mm of Hg). Which probe has greater energy? (For quantitative comparison.74 V.4 cm2. since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. Mo: 4. V03 = 0. etc. λ2= 4047 Å. If the stopping potential is –1.30 eV.95 V. respectively.26 Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal.30 Light of intensity 10–5 W m–2 falls on a sodium photo-cell of surface area 2 cm2. were performed by Millikan. λ3= 4358 Å. [Note: You will notice that to get h from the data. take the wavelength of the probe equal to 1 Å. As you have seen in Exercise 11. Take the area of the pupil to be about 0.31. estimate time required for photoelectric emission in the wave-picture of radiation. Assuming that the top 5 layers of sodium absorb the incident energy.Physics humans can perceive (∼10–10 W m–2).28 A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission. K: 2. The source is replaced by an iron source and its 427. Ni: 5. Li. K. were measured to be: V01 = 1. The work function for the metal is given to be about 2 eV.16 V. which is of the order of inter-atomic spacing in the lattice) (me=9. or electrons accelerated through appropriate voltage.2 nm line irradiates the same photo-cell. The stopping voltage is measured to be 0.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away ? 11.6 × 10–19 C).675 × 10–27 kg) 410 . The stopping voltages. (mn = 1.17 eV. who. λ4= 5461 Å. the following lines from a mercury source were used: λ1 = 3650 Å. Would a neutron beam of the same energy be equally suitable ? Explain. and the average frequency of white light to be about 6 × 1014 Hz.27 Monochromatic radiation of wavelength 640. 11. using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h. 11.32 (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. Experiments of this kind on Na.2 nm (1nm = 10–9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. What is the implication of your answer ? 11. λ5= 6907 Å. How would the photo-cell respond to a high intensity (∼105 W m–2) red light of wavelength 6328 Å produced by a He-Ne laser ? 11.] 11. V05 = 0 V Determine the value of Planck’s constant h. the threshold frequency and work function for the material.75 eV.3 V. V04 = 0.54 V. V02 = 0. you will need to know e (which you can take to be 1.28 V. Predict the new stopping voltage. 11.29 The work function for the following metals is given: Na: 2. estimate the work function of the metal. an electron beam of this energy is suitable for crystal diffraction experiments.31 Crystal diffraction experiments can be performed using X-rays.11 × 10–31 kg). In our experiment with rubidium photo-cell. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford.511 MeV.] Answer the following questions: (a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e . how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light? The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. the value of ν (and therefore.37 λ But while the value of λ is physically significant. (–1/3)e ].36 11. Determine the de Broglie wavelength associated with the electrons. Guess what might have been the order of energy of these electron beams.34 11. etc. p = h 11. USA. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses. and compare it with the mean separation between two atoms under these conditions. the electron wave-packets in a metal strongly overlap with one another.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping. Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.) are taken to be roughly the same. Why do they not show up in Millikan’s oil-drop experiment? (b) What is so special about the combination e/m? Why do we not simply talk of e and m separately? (c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures? (d) Every metal has a definite work function. Compute the typical de Broglie wavelength of an electron in a metal at 27 ºC and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10–10 m. Why ? 411 . An electron microscope uses electrons accelerated by a voltage of 50 kV. The quark structure of protons and neutrons appears at the minute length-scale of 10–15 m or less. the value of the phase speed ν λ) has no physical significance. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons? (e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations: E = h ν. This suggests that whereas molecules in an ordinary gas can be distinguished apart.) Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 ºC) and 1 atm pressure.Dual Nature of Radiation and Matter (b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 ºC).35 and 11.33 11.35 11. If other factors (such as numerical aperture. [Note: Exercises 11. electrons in a metal cannot be distintguished apart fr om one another. (Rest mass energy of electron = 0. while Huygens. Better experiments were performed in the nineteenth century to determine the speed of light. Around 1860.998×108 m/s. theoretical models about what light is made of were developed. For example. Theories and models like kinetic theory and thermodynamics were developed which quite successfully explained the various phenomena.Physics APPENDIX 11. extended Descartes′ particle model. Descartes. around 1650-60. a value of 3×108 m/s for speed of light in vacuum was arrived at. around 1660-70. In spite of this. This phenomenon could be explained only by wave theory. and explained various known properties with it. 412 . in a sense. Descartes′ proposal was merely a philosophical model. known as corpuscular theory. Maxwell proposed his equations of electromagnetism and it was realized that all electromagnetic phenomena known at that time could be explained by Maxwell’s four equations. This situation again continued for almost a century. Appropriate laws were formulated for the first four phenomena. He calculated the speed of these waves and arrived at a theoretical value of 2. With more accurate experiments. But both models could explain all the known properties of light. Newton’s particle model was firmly believed and preferred over the wave model. Bartholinus. (d) dispersion into various colours. Vast amounts of experiments were done during the period 1850-1900 on heat and related phenomena. proposed that light consists of particles. Several scientists right from the days of Galileo had tried to measure the speed of light. (c) refraction at the boundary of two media. In fact. There was nothing to choose between them. In 1887 Hertz demonstrated the generation and detection of such waves. Two models of light were also proposed in the seventeenth century. it was a natural consequence of Huygens idea of secondary wavelets emanating from every point in the path of light. Young performed his double-slit experiment and observed interference fringes. in 1669. The history of the development of these models over the next few centuries is interesting. These experiments could not be explained by assuming that light consists of particles. except one. built it up as a scientific theory. and Huygens. proposed that light consists of waves. It is therefore appropriate to summarize some observations about light that were known in the seventeenth century. The properties of light known at that time included (a) rectilinear propagation of light. We might say that while eighteenth century belonged to the particle model.1 The history of wave-particle flip-flop What is light? This question has haunted mankind for a long time. Thus wave theory of Huygens came to the forefront and Newton’s particle theory went into the background. Another phenomenon of polarisation was discovered around 1810. Soon Maxwell showed that electric and magnetic fields could propagate through empty space (vacuum) in the form of electromagnetic waves. But systematic experiments were done by scientists since the dawn of the scientific and industrial era. This established the wave theory of light on a firm footing. The close agreement of this value with the experimental value suggested that light consists of electromagnetic waves. But they had not been able to do so. Then in 1801. It was realized that diffraction was also another phenomenon which could be explained only by wave theory. This was partly because of its simplicity and partly because of Newton’s influence on contemporary physics. While building a model in any branch of science. about four centuries ago. for over one hundred years. light as waves and as particles. Newton soon after. it is essential to see that it is able to explain all the experimental observations existing at that time. These models. discovered double refraction of light in some crystals. in 1678. (b) reflection from plane and curved surfaces. an altogether different area of physics. the nineteenth century belonged to the wave model of light. Snell formulated his laws of refraction in 1621. devoid of any experiments or scientific arguments. (e) high speed. Around the same time. and this too could be naturally explained by the wave theory. was quick to explain it on the basis of his wave theory of light. in early decades of seventeenth century. They had only concluded that it was higher than the limit of their measurement. are quite opposite of each other. Somebody in those days aptly remarked that light behaves as a particle on Mondays. He derived a formula which agreed with the entire spectrum. This increased the dilemma further. This confirmed wave-particle duality as an established principle of physics. Debye explained the low temperature specific heats of solids by using the particle picture for lattice vibrations in a crystalline solid. On the one hand. Here was a principle. physicists faced with the following situation. diffraction experiments with other ‘particles’ such as neutrons and protons were performed and these too confirmed with de Broglie’s formula. Now it is regarded as a natural consequence of present theories of physics that electromagnetic radiation as well as particles of matter exhibit both wave and particle properties in different experiments. which explained all the phenomena mentioned above not only for light but also for the so-called particles. It is an ideal concept in physics. with their wavelength and frequency (wave characteristics)! In 1927 Thomson. (b) There were some phenomena such as diffraction and interference which could be explained only by the wave model but not by the particle model. like concepts of a point mass or uniform motion. velocity. Very soon. It is true that all experiments whatever agreed with de Broglie formula. All these theories mainly put wave-particle duality on a theoretical footing. It also absorbs radiation falling on it. not an outcome of a sound fundamental theory. Although the story continues. reflection. it travels as a wave. this put physicists in a dilemma. photoelectric effect. Schwinger. in separate experiments. and Heisenberg and Pauli gave it a firm footing by 1930. Wave-particle duality was at best a principle. physicists thought. while on the other hand. and Feynman. A body which absorbs all the radiation falling on it is called a black body. But there was no basic theoretical foundation for wave-particle duality. De Broglie’s proposal was merely a qualitative argument based on symmetry of nature. de Broglie proposed his theory of wave-particle duality in which he said that not only photons of light but also ‘particles’ of matter such as electrons and atoms possess a dual character. it also needs sound theoretical basis for the models proposed. it grows more and more complex and beyond the scope of this note. then we can explain the black body spectrum.Dual Nature of Radiation and Matter Every body at any temperature emits radiation of all wavelengths. Planck himself regarded these quanta. Einstein explained the photoelectric effect by assuming the particle picture of light. and sometimes even in the different parts of the same experiment. and Davisson and Germer. it needs experimental confirmation. in which the regular arrangement of atoms acted like a grating. Tomonaga. Thursdays and Saturdays. Planck hit upon a novel idea. But physics does not work that way. momentum (particle characteristics). in late 1940s. Compton’s x-ray scattering experiments from atoms also went in favour of the particle picture. and as a wave on Tuesdays. sometimes behaving like a particle and sometimes as a wave. he said. This was developed over the next two decades. A graph of the intensity of radiation emitted by a body versus wavelength is called the black body spectrum. 413 . rather than that of light. and let us be satisfied with it at the moment. He gave a formula connecting their mass. In 1907. Thus by 1923. Both these phenomena belonging to widely diverse areas of physics could be explained only by the particle model and not by the wave model. This was a confusing mixture of wave and particle pictures – radiation is emitted as a particle. But we have here the essential structure of what happened. or packets. If we assume. which could be explained by either particle model or by wave model. Wednesdays and Fridays. refraction. Their experiment was on diffraction of electrons through crystalline solids. Should we again accept the particle picture of light just to explain one phenomenon? Then what happens to the phenomena of interference and diffraction which cannot be explained by the particle model? But soon in 1905. and we don’t talk of light on Sundays! In 1924. that radiation is emitted in packets of energy instead of continuously as in a wave. Dirac developed his theory of radiation in about 1928. (c) There were some phenomena such as black body radiation. as a property of emission and absorption. produced further refinements and cleared the theory of inconsistencies which were noticed. In 1923. and is again absorbed as a particle! Moreover. (a) There were some phenomena like rectilinear propagation. showed that electrons did behave like waves with a wavelength which agreed with that given by de Broglie’s formula. and Compton scattering which could be explained only by the particle model but not by the wave model. No theory in those days could explain the complete black body spectrum! In 1900. Physics Chapter Twelve © no N C tt E o R be T re pu 12. However. Therefore. However subsequent studies on atoms. what is the structure of an atom? The first model of atom was proposed by J. But what is the arrangement of the positive charge and the electrons inside the atom? In other words. showed that the distribution of the electrons and positive charges are very different from that proposed in this model. J. Thomson in 1898. J. Thomson (1856 – 1940) revealed that atoms of different elements contain negatively charged constituents (electrons) that are identical for all atoms. as described in this chapter. the experiments on electric discharge through gases carried out by the English physicist J. According to this model. This radiation is considered to be due to oscillations of atoms bl is he d ATOMS . This model was picturesquely called plum pudding model of the atom. atoms on a whole are electrically neutral. the positive charge of the atom is uniformly distributed throughout the volume of the atom and the negatively charged electrons are embedded in it like seeds in a watermelon. an atom must also contain some positive charge to neutralise the negative charge of the electrons. In 1897. though with different intensities. enough evidence had accumulated in favour of atomic hypothesis of matter. We know that condensed matter (solids and liquids) and dense gases at all temperatures emit electromagnetic radiation in which a continuous distribution of several wavelengths is present.1 I NTRODUCTION 414 By the nineteenth century. He studied chapter. hydrogen always gives a set of lines with fixed relative position between the lines. the ‘emanation’ of thorium Ernst Rutherford (1871–1937). consisting of a single electron and a single proton. By he proposed a classic experiment of scattering of these scattering alpha-rays from α-particles by atoms to investigate the atomic structure. in 1911. J. In 1906. In 1885. as shown in bl is he 415 d ERNST RUTHERFORD (1871 – 1937) . In the early nineteenth century it was also established that each element is associated with a characteristic spectrum of radiation. we shall see that there are some serious difficulties in accepting such a model. Marsden performed some experiments. he created hydrogen. He also estimated the 12.2 ALPHA-PARTICLE SCATTERING AND RUTHERFORD’S NUCLEAR MODEL OF A TOM At the suggestion of Ernst Rutherford. nuclear model of the atom ). the radiation emitted can be considered due to individual atoms rather than because of interactions between atoms or molecules. Since hydrogen is simplest of the elements the modern theory of known. He radiation emitted by it. In one of their experiments. or excited electrically in a glow tube such as the familiar neon sign or mercury vapour light has only certain discrete wavelengths. However. the metal foils. was engaged in experiments on gas.2.Atoms and molecules. light emitted from rarefied gases heated in a flame. In contrast. Geiger and E. In such gases. emit a complex spectrum of specific wavelengths? In the classical picture of an atom. How could an atom as simple as hydrogen. According to this the entire positive charge and most of the mass of the atom is concentrated in a small volume called the nucleus with electrons revolving around the nucleus just as planets revolve around the sun. Rutherford’s nuclear model was a major step towards how we see the atom today. the average spacing between atoms is large. The explanation of the results led to the birth of approximate size of the Rutherford’s planetary model of atom (also called the nucleus. he This experiment was later performed around 1911 by Hans discovered the atomic Geiger (1882–1945) and Ernst Marsden (1889–1970. who nucleus and proposed the was 20 year-old student and had not yet earned his plenatery model of the bachelor’s degree). The details are discussed in Section atom. now known as thoron. a former research and discovered a new noble student of J. Thomson. it could not explain why atoms emit light of only discrete wavelengths. © no N C tt E o R be T re pu 12. The spectrum appears as a series of bright lines. Ernst Rutherford (1871 – This fact suggested an intimate relationship between the 1937) British physicist who did pioneering work on internal structure of an atom and the spectrum of radioactive radiation. for example. H. However. an isotope of radon. the electron revolves round the nucleus much like the way a planet revolves round the sun. Hence. governed by the interaction of each atom or molecule with its neighbours. Along with gave the wavelengths of a group of lines emitted by atomic Federick Soddy. α-particles emitted by some radioactive elements. we shall consider its spectrum in detail in this radioactivity. Johann Jakob Balmer discovered alpha-rays and (1825 – 1898) obtained a simple empirical formula which beta-rays. A typical graph of the total number of α-particles scattered at different angles. in a given interval of time. The beam was allowed to fall on a thin foil of gold of thickness 2.2 shows a schematic diagram of this experiment. The scattered alpha-particles on striking the screen produced brief light flashes or scintillations. and about 1 in 8000 deflect by more than 90º.Physics Fig.1. This force could bl is he d . 12. Many of the α-particles pass through the foil.2 Schematic arrangement of the Geiger -Marsden experiment.14% of the incident α-particles scatter by more than 1º. The dots in this figure represent the data points and the solid curve is the theoretical prediction based on the assumption that the target atom has a small. Alpha-particles emitted by a 214 83 Bi radioactive source were collimated into a narrow beam by their passage through lead bricks. These flashes may be viewed through a microscope and the distribution of the number of scattered particles may be studied as a function of angle of scattering. 12.1 Geiger -Marsden scattering experiment. The scattered alpha-particles were observed through a rotatable detector consisting of zinc sulphide screen and a microscope. FIGURE 12. it must experience a large repulsive force. dense. is shown in Fig. Only about 0. to deflect the α -particle backwards.5 MeV α -particles emitted from a 214 radioactive source at a thin metal 83 Bi foil made of gold.3. It means that they do not suffer any collisions. Rutherford argued that. © no N C tt E o R be T re pu 416 FIGURE 12. positively charged nucleus. Figure 12. they directed a beam of 5. The entire apparatus is placed in a vacuum chamber (not shown in this figure).1 × 10 –7 m. 3 can be analysed by employing Rutherford’s nuclear model of the atom.3 Experimental data points (shown by about the nucleus just as the planets dots) on scattering of α-particles by a thin foil at do around the sun. of an atom was known to be 10–10 m. it can be assumed that α -particles will suffer not more than one scattering during their passage through it.000 times larger than the size of the nucleus (see Chapter 11. Thus. for gold Z = 79. when α-particle happens to come near a nucleus. the size curve which is seen to be in good agreement with experiment. As the gold foil is very thin.Atoms be provided if the greater part of the mass of the atom and its positive charge were concentrated tightly at its centre. This is why Rutherford is credited with the discovery of the nucleus. From kinetic theory.000 to 100. 12. Rutherford’s nuclear model predicts the solid 10–14 m. In Rutherford’s nuclear model of the atom. the entire positive charge and most of the mass of the atom are concentrated in the nucleus with the electrons some distance away. 2e . the intense electric field there scatters it through a large angle. Thus. the electrons would seem to be at a distance from the nucleus of about 10. of positive charge and have the mass of the helium atom. therefore. being so light. Then the incoming α-particle could get very close to the positive charge without penetrating it. about 10. computation of the trajectory of an alpha-particle scattered by a single nucleus is enough. The charge of the gold nucleus is Ze . 12.000 times the size of the nucleus itself. This agreement supported the hypothesis of the nuclear atom. Under these assumptions. The electrons would be moving in orbits FIGURE 12. Therefore. where Z is the atomic number of the atom.1 and the nucleus to be about 10–15 m to 12. the trajectory of an alpha-particle can be computed employing Newton’s second law of motion and the Coulomb’s law for electrostatic force of repulsion between the alpha-particle and the positively 417 charged nucleus. Alphaparticles are nuclei of helium atoms and. it is reasonable to assume that it remains stationary throughout the scattering process. Since the nucleus of gold is about 50 times heavier than an α-particle. and such a close encounter would result in a large deflection.2. Rutherford’s different angles obtained by Geiger and Marsden experiments suggested the size of using the setup shown in Figs. most of an atom is empty space. Section 11. The scattering data shown in Fig. However. it is easy to see why most α -particles go right through a thin metal foil. carry two units.6 in Class XI Physics textbook).000 to 100. do not appreciably affect the α -particles. With the atom being largely empty space. © no N C tt E o R be T re pu bl is he d . The atomic electrons. 12.1 Alpha-particle trajectory d . undergoing head on collision is small. Thus. the radius of the electron’s nucleus is (10 5 orbit is 10 times larger than the radius of nucleus. This. bl is he 12. The radius of sun is taken as 7 × 108 m. so that the beam is scattered in various directions with different probabilities (Fig. in turn. b and scattering angle θ indicates that the number of α -particles are also depicted. For a large impact parameter.4).5 × 1011 m.4 Trajectory of α-particles in the The fact that only a small fraction of the coulomb field of a target nucleus. the earth would be much farther away from the sun. If the dimensions of the solar system had the same proportions as those of the atom. would the earth be closer to or farther away from the sun than actually it is ? The radius of earth’s orbit is about 1.1 © no N C tt E o R be T re pu The trajectory traced by an α -particle depends on the impact parameter. that is. the impact parameter is minimum and the α-particle rebounds back ( θ ≅ π).4). The impact parameter is the perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus (Fig. (In a beam.2.Physics The magnitude of this force is 1 (2e )( Ze ) F= (12. 12. all particles have nearly same kinetic energy. the α-particle goes nearly undeviated and has a small deflection ( θ ≅ 0).1 In the Rutherford’s nuclear model of the atom. is a powerful way to determine an upper limit to the size of the nucleus. The force is directed along the line joining the α-particle and the nucleus. the radius of the earth’s orbit would be 105 × 7 × 10 8 m = 7 × 1013 m. This is more than 100 times greater than the actual orbital radius of earth. A given beam of α -particles has a distribution of impact parameters b. If the radius of the earth’s orbit around the sun were 105 times larger than the radius of the sun. implies that the mass of the atom is concentrated in a small volume. The impact number of incident particles rebound back parameter. 418 EXAMPLE 12. The magnitude and direction of the force on an α -particle continuously changes as it approaches the nucleus and recedes away from it. FIGURE 12. Example 12. Rutherford scattering therefore. Solution The ratio of the radius of electron’s orbit to the radius of –10 –15 5 m)/(10 m) = 10 .1) 4 πε 0 r2 where r is the distance between the α -particle and the nucleus. In case of head-on collision. b of collision.) It is seen that an α -particle close to the nucleus (small impact parameter) suffers large scattering. It implies that an atom contains a much greater fraction of empty space than our solar system does. the nucleus (radius about 10–15 m) is analogous to the sun about which the electron move in orbit (radius ≈ 10–10 m) like the earth orbits around the sun. less than 3. The final energy Ef is just the electric potential energy U of the system.particle.ac.0 × 10 N m /C . Then we can write the conservation of energy E i = E f as Thus the distance of closest approach d is given by © no N C tt E o R be T re pu (2)(9. The electrostatic force of attraction. Let d be the centre-to-centre distance between the α-particle and the gold nucleus when the α-particle is at its stopping point.7 MeV or 1. and it is equal to its mechanical energy Ef when the α-particle momentarily stops.uk/camphy/nucleus/nucleus_exp. before the particle and nucleus interact.e.2 × 10 J. so that d = d = 2Ze 2 4πε 0 K The maximum kinetic energy found in α-particles of natural origin is –12 9 2 2 7. 12. we have.2. The system’s initial mechanical energy is E i.6 × 10 C. The potential energy U can be calculated from Eq.2 In a Geiger -Marsden experiment.0 × 109 Nm 2 / C2 )(1. Fe between the revolving electrons and the nucleus provides the requisite centripetal force (Fc ) to keep them in their orbits. therefore.7 MeV α-particle before it comes momentarily to rest and reverses its direction? Solution The key idea here is that throughout the scattering process.2 × 10−12 J –16 = 3. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the α-particle.6 × 10–19 C )2 Z 1. pictures the atom as an electrically neutral sphere consisting of a very small.84 × 10 Z m The atomic number of foil material gold is Z = 79. what is the distance of closest approach to the nucleus of a 7.0 × 10 m. The initial energy Ei is just the kinetic energy K of the incoming α. the total mechanical energy of the system consisting of an α-particle and a gold nucleus is conserved.) The radius of gold nucleus is. fermi) = 10–15 m.cam. the α-particle reverses its motion without ever actually touching the gold nucleus.2 he http://www-outreach. massive and positively charged nucleus at the centre surrounded by the revolving electrons in their respective dynamically stable orbits.0 × 10–14 m = 30 fm. (1 fm (i. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm.2) is EXAMPLE 12.htm Simulate Rutherford scattering experiment d 419 .Atoms Example 12. for a dynamically stable orbit in a hydrogen atom Fe = Fc mv 2 1 e2 = r 4πε0 r 2 bl –14 K= 1 (2e )( Ze ) 2Ze 2 = 4π ε0 4πε 0 d d (12.phy. Thus. Thus. d (Au) = 3. Therefore –19 with e = 1. (12. Since 1/4π ε0 = 9.2 Electron orbits The Rutherford nuclear model of the atom which involves classical concepts.1). the emitted radiation has a spectrum which contains certain specific wavelengths only.6 eV energy is required to separate a hydrogen atom into a proton and an electron.Physics Thus the relation between the orbit radius and the electron velocity is r = e2 4 πε 0mv 2 (12. Thus from Eq. A spectrum of this kind is termed as emission line spectrum and it he K = 1 e2 e2 mv 2 = and U = − 2 8πε 0r 4 π ε 0r d The kinetic energy (K ) and electrostatic potential energy (U ) of the electron in hydrogen atom are . an electron will not follow a closed orbit around the nucleus.) Thus the total energy E of the electron in a hydrogen atom is bl =− e2 8πε0r is (12.3 = 5. usually by passing an electric current through it.2 × 10 J) EXAMPLE 12. Compute the orbital radius and the velocity of the electron in a hydrogen atom.4) E = K +U = e2 e2 − 8 πε 0r 4 πε 0r © no N C tt E o R be T re pu − e2 = −2. Solution Total energy of the electron in hydrogen atom is –13.2 × 106 m/s. The velocity of the revolving electron can be computed from Eq.6 eV = –13. When an atomic gas or vapour is excited at low pressure.3 × 10 –11 m.6 × 1.2 ×10–18 J. (12. 4πε 0mr 12. Example 12.4). If E were positive. (12.1.3 ATOMIC SPECTRA 420 As mentioned in Section 12. which it emits. we have r =− (9 × 109 N m 2/C2 )(1.1 × 10–31 kg.3 It is found experimentally that 13.6 × 10 –19 J = –2.6 × 10−19 C)2 e2 =− −18 8πε0 E (2)(–2. This implies the fact that the electron is bound to the nucleus.2 × 10 −18 J 8πε 0r This gives the orbital radius The total energy of the electron is negative. each element has a characteristic spectrum of radiation.3) (The negative sign in U signifies that the electrostatic force is in the –r direction.3) with m = 9. v = e = 2. has the simplest spectrum. These dark lines correspond precisely to those wavelengths which were found in the emission line spectrum of the gas.1 nm in the violet is called Hγ. and so on.5) ⎝ λ 2 n ⎠ where λ is the wavelength.5 Emission lines in the spectrum of hydrogen.6).1 nm in the blueemission spectrum of hydrogen. there does not seem to be any resemblance of order or regularity in spectral lines. the FIGURE 12.6 Balmer series in the next line with wavelength 486. 4. Hydrogen is the simplest atom and therefore. (12. spectrum. 5. When white light passes through a gas and we analyse the transmitted light using a spectrometer we find some dark lines in the FIGURE 12. The value of R is 1. In 1885.3 nm 1 We might expect that the frequencies of the light emitted by a particular element would exhibit some regular pattern. 12.3 nm in the red is called Hα. Taking n = 3 in Eq. 656.3. Study of emission line spectra of a material can therefore serve as a type of “fingerprint” for identification of the gas. at first sight.097 × 10 7 ⎜ 2 − 2 ⎟ m –1 ⎝ λ 2 3 ⎠ 6 –1 = 1. But the spacing between lines within certain sets of the hydrogen spectrum decreases in a regular way (Fig. The spectrum emitted by atomic hydrogen is shown in Fig. the first such series was observed by a Swedish school teacher Johann Jakob Balmer (1825–1898) in the visible region of the hydrogen spectrum.097 × 107 m–1. R is a constant called the Rydberg constant. etc.5. 12. In the observed spectrum.1 Spectral series © no N C tt E o R be T re pu 1 1⎞ ⎛ 1 = 1. one obtains the wavelength of the Hα line: bl is he 421 d . the lines appear closer together and are weaker in intensity.5). Each of these sets is called a spectral series .Atoms consists of bright lines on a dark background. green is called Hβ. the third line 434.e.5).522 × 10 m i. The line with the longest wavelength. This equation is also called Balmer formula. however. This is called the absorption spectrum of the material of the gas. λ = 656. As the wavelength decreases. 12.. and n may have integral values 3. Balmer found a simple empirical formula for the observed wavelengths 1 ⎞ ⎛1 = R⎜ 2 − 2⎟ (12. 12. This series is called Balmer series (Fig. Brackett. (12. bl HYDROGEN ATOM is n = 4.6. and doubly ionised lithium) whose spectra can be represented by simple formula like Eqs.... Equations (12. after their discoverers.Physics For n = 4. one obtains the wavelength of Hβ line. (12..8. no further distinct lines appear. While the planetary system is held by gravitational force. instead only a faint continuous spectrum is seen. at λ = 364.6) Brackett series : 1 λ 1 n = 5. 12.5) may be written in terms of frequency of the light. these results are empirical and do not give any reasoning why only certain frequencies are observed in the hydrogen spectrum.9) (12. or 1 Pfund series: The Lyman series is in the ultraviolet.3.5. (12. interact by Coulomb’s Law of force. recalling that c = νλ λ c Thus. For n = ∞ .9) are useful as they give the wavelengths that hydrogen atoms radiate or absorb.5) – (12. there are some fundamental differences between the two situations.6.7.5) becomes ν = Rc ⎜ 1 ⎞ ⎛1 − ⎝ 22 n 2 ⎟ ⎠ = ν There are only a few elements (hydrogen.10) d .6 nm. one obtains the limit of the series. the nucleus-electron system being charged objects. as Lyman. consisting of a central nucleus and revolving electron is stable much like sun-planet system which the model imitates.8) (12. These are represented by the formulae: Lyman series : ⎛1 1⎞ =R ⎜ 2 − 2 ⎟ ⎝1 n ⎠ λ Paschen series : 1 1 1⎞ ⎛1 =R ⎜ 2 − 2 ⎟ ⎝3 n ⎠ λ 1 ⎞ ⎛1 =R ⎜ 2 − 2 ⎟ ⎝4 n ⎠ n = 2. (12. We know that an he (12. Paschen. Eq... However. singly ionised helium. Other series of spectra for hydrogen were subsequently discovered. etc. These are known. However. © no N C tt E o R be T re pu 1⎞ ⎛ 1 = R⎜ 2 − 2⎟ ⎝ λ 5 n ⎠ n = 6.5) – (12. This is the shortest wavelength in the Balmer series.4 BOHR MODEL OF THE 422 The model of the atom proposed by Rutherford assumes that the atom.9).7) (12.. and Pfund series. Beyond this limit..4.7. The Balmer formula Eq. and the Paschen and Brackett series are in the infrared region. Further. The energy of an accelerating electron should therefore. Thus. Clearly Rutherford model tells only a part of the story implying that the classical ideas are not sufficient to explain the atomic structure. Bohr contributed to the clarification of conceptual problems in quantum mechanics. 15 bl EXAMPLE 12. Thus.2 × 10–6 m/s. calculate the initial frequency of the light emitted by the electron revolving around a proton in hydrogen atom.4 is he 423 d . they would emit a continuous spectrum. Niels Henrik David Bohr (1885 – 1962) Danish physicist who explained the spectrum of hydrogen atom based on quantum ideas. Example 12. according to the classical electromagnetic theory. The electron would spiral inward and eventually fall into the nucleus (Fig.7 An accelerated atomic electron must spiral into the nucleus as it loses energy. the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of revolution.7). in contradiction to the line spectrum actually observed. 12. Thus.Atoms object which moves in a circle is being constantly accelerated – the acceleration being centripetal in nature. and so will the frequency of the light emitted. such an atom can not be stable. NIELS HENRIK DAVID BOHR (1885 – 1962) © no N C tt E o R be T re pu FIGURE 12.3 we know that velocity of electron moving around a proton in hydrogen atom in an orbit of radius 5. continuously decrease.2 × 106 m s−1 = 2πr 2π 5. According to the classical electromagnetic theory we know that the frequency of the electromagnetic waves emitted by the revolving electrons is equal to the frequency of its revolution around the nucleus. He gave a theory of nuclear fission based on the liquiddrop model of nucleus. As the electrons spiral inwards.6 × 10 Hz. their angular velocities and hence their frequencies would change continuously. According to classical electromagnetic theory.4 According to the classical electromagnetic theory. an accelerating charged particle emits radiation in the form of electromagnetic waves. in particular by proposing the complementary principle.3 × 10−11 m ( ) ≈ 6.3 × 10–11 m is 2.6 × 10 Hz. Solution From Example 12. the frequency of the electron moving around the proton is ν= v 2. 15 Thus the initial frequency of the light emitted is 6. But then this equation requires the radius r of the electron orbit. each atom has certain definite stable states in which it can exist. These are called the stationary states of the atom. This postulate states that the electron revolves around the nucleus only in those orbits for which the angular momentum is some integral multiple of h/2 π where h is the Planck’s constant (= 6. Niels Bohr studied in Rutherford’s laboratory for several months in 1912 and he was convinced about the validity of Rutherford nuclear model. The frequency of the emitted photon is then given by (12. That is (12. The angular momentum L is given by L = mvr Bohr’s second postulate of quantisation [Eq.12) where Ei and Ef are the energies of the initial and final states and Ei > Ef . These are : © no N C tt E o R be T re pu L = nh/2π h ν = E i – Ef L n = mvn rn = nh 2π (ii) Bohr’s second postulate defines these stable orbits. and each possible state has definite total energy.13) 424 where n is an integer. (12. rn is the radius of n th possible orbit and v n is the speed of moving electron in the n th orbit. Bohr combined classical and early quantum concepts and gave his theory in the form of three postulates. contrary to the predictions of electromagnetic theory. (12. For a hydrogen atom. Faced with the dilemma as discussed above. Eq. To calculate r. Thus the angular momentum (L ) of the orbiting electron is quantised.6 × 10–34 J s). The allowed orbits are numbered bl (i) Bohr’s first postulate was that an electron in an atom could revolve in certain stable orbits without the emission of radiant energy . Bohr.4) gives the expression to determine the energies of different energy states.11)] says that the allowed values of angular momentum are integral multiples of h /2π.11) (iii) Bohr’s third postulate incorporated into atomic theory the early quantum concepts that had been developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. Bohr’s second postulate about the angular momentum of the electron–the quantisation condition – is used. is he d . in 1913. it could not be applied to the processes at the atomic scale. When it does so. It became clear that a fairly radical departure from the established principles of classical mechanics and electromagnetism would be needed to understand the structure of atoms and the relation of atomic structure to atomic spectra. (12. a photon is emitted having energy equal to the energy difference between the initial and final states . According to this postulate.Physics It was Niels Bohr (1885 – 1962) who made certain modifications in this model by adding the ideas of the newly developing quantum hypothesis. concluded that in spite of the success of electromagnetic theory in explaining large-scale phenomena. Energy will thus be required to remove the electron from the hydrogen atom to a distance infinitely far away from its nucleus (or proton in hydrogen atom).17) yields En = − 2.13).18) can be rewritten as 13. 2. (12. (12. The total energy of the electron in the stationary states of the hydrogen atom can be obtained by substituting the value of orbital radius in Eq. m . which is called the principal quantum number of the orbit.15). From Eq.16) πme 2 Substitution of values of h . (12. En = − bl (12.3). (12. (12. the relation between vn and r n is (12.15). Since 1 eV = 1.14) depicts that the orbital speed in the nth orbit falls by a factor of n. (12. the size of the innermost orbit (n = 1) can be obtained as is 425 (12. (12. according to the values of n . From Eq. (12. Using Eq..15) he vn = 1 e2 1 n 4 πε 0 ( h 2 π) d 4πε0mrn Combining it with Eq.. we get the following expressions for v n and rn . vn = e .18) n2 Atomic energies are often expressed in electron volts (eV ) rather than joules. Eq.14) and ⎛ n 2 ⎞ h ⎞ 4 πε0 rn = ⎜ ⎟ ⎛ ⎜ ⎟ ⎝ m ⎠ ⎝ 2π ⎠ e 2 2 © no N C tt E o R be T re pu This is called the Bohr radius. a0 = ⎛ e 2 ⎞ ⎛ m ⎞ ⎛ 2π ⎞ 2 ⎛ e 2 ⎞ En = − ⎜ ⎜ ⎟ ⎜ ⎟ 4 πε ⎟ ⎝ 8 πε 0 ⎟ ⎠ ⎝n 2 ⎠ ⎝ h ⎠ ⎜ ⎝ 0⎠ r1 = h2 ε0 π me 2 h2 ε0 (12.17) Eq. Eq.29 × 10–11 m. represented by the symbol a 0.6 eV (12. ε0 and e gives a 0 = 5..18 × 10 −18 J (12. it can also be seen that the radii of the orbits increase as n 2.6 × 10–19 J.4) as or En = − me 4 2 2 8n 2 ε 0 h Substituting values.Atoms 1.19) n2 The negative sign of the total energy of an electron moving in an orbit means that the electron is bound with the nucleus. 3 . Thus. The solution of the Schrödinger wave equation. This probability can be inferred from the one-electron wave function called the orbital. we have a better understanding of the structure of atom. assigned to each orbit. This model has its place in the history of quantum mechanics and particularly in explaining the structure of an atom. contrary to the classical picture requiring an accelerating particle to radiate. it was shown by the German physicist Arnold Sommerfeld (1868 – 1951) that. An orbit of the electron in the Bohr model is the circular path of motion of an electron around the nucleus. This function depends only on the coordinates of the electron. in general elliptical. We know that energy associated with a stationary state of an electron depends on n only. Solutions of the Schrödinger wave equation assign a wave-like description to the electrons bound in an atom due to attractive forces of the protons. This orbital has no resemblance whatsoever with the orbit defined for an electron in the Bohr model. (12. (Planets move in elliptical orbits under the inverse square gravitational force of the sun. for one-electron atoms/ions. But according to quantum mechanics. We can only talk about the probability of finding an electron in a certain region of space around the nucleus. these equations continue to hold even for elliptic orbits. (12.5 A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. Bohr also introduced the idea of quantisation of angular momentum of electrons moving in definite orbits. gives information about the probability of finding an electron in various regions around the nucleus. we have m vn rn = nh/2π We are introduced to the Bohr Model of atom one time or the other in the course of physics. Assuming that Bohr’s angular momentum postulate applies to satellites just as it does to an electron in the hydrogen atom. bl is he O RBIT VS STATE (ORBITAL PICTURE) OF ELECTRON IN ATOM d . obtained for hydrogen-like atoms/ ions. we cannot associate a definite path with the motion of the electrons in an atom. It has become a milestone since Bohr introduced the revolutionary idea of definite energy orbits for the electrons.Physics The derivation of Eqs.) However.19) involves the assumption that the electronic orbits are circular. find the quantum number of the orbit of the satellite. It is therefore essential that we understand the subtle differences that exist in the two models: Bohr model is valid for only one-electron atoms/ions. an energy value. Thus it was a semi-classical picture of the structure of atom. Now with the development of quantum mechanics. Example 12. depends on the principal quantum number n in this model. For a multi-electron atom/ion. when the restriction of circular orbit is relaxed. called the wave function. this is not true.13). though orbits under inverse square force are.5 © no N C tt E o R be T re pu 426 Solution From Eq.17) – (12. EXAMPLE 12. with the electron revolving in the orbit of smallest radius.6 eV.Atoms Here m = 10 kg and rn = 8 × 106 m. That is T = 7200 s. .6) eV = 10. for n = 2.19). The atom is then said to be in an excited state. For n = 2. as the excitation of hydrogen atom increases (that is as n increases) the value of minimum energy required to free the electron from the excited atom decreases. Thus.5 d . 13.) is 427 he E XAMPLE 12. E3 = –1. Similarly. computed from Eq. E1 is –13. emitting a photon in the process. It is called the ionisation energy of the hydrogen atom. the energy E 2 is –3. The lowest state of the atom. The energy of this state (n = 1). n = (2π × 8 × 106 m)2 × 10/(7200 s × 6.64 × 10 –34 J s) 45 = 5.1 Energy levels The energy of an atom is the least (largest negative value) when its electron is revolving in an orbit closest to the nucleus i.6 eV.40 eV.2 eV.. a 0. Thus there is a continuum of energy states above E = 0 eV. From Eq. The quantum number of the orbit of satellite n = (2π rn)2 × m/(T × h). (The horizontal lines specify the presence of allowed energy states. 12.. most of the hydrogen atoms are in ground state. is an energy equal to E 2 – E1 = –3. the Bohr radius. The electron in a hydrogen atom at room temperature spends most of its time in the ground state.8 The energy level diagram for the hydrogen atom. It means that the energy required to excite an electron in hydrogen atom to its first excited state. We have the time period T of the circling satellite as 2 h. hence the energy is progressively larger in the outer orbits. Thus the velocity vn = 2 π rn/T .e. 3. or to excite the hydrogen atom from its ground state (n = 1) to second excited state (n = 3). the minimum energy required to free the electron from the ground state of the hydrogen atom is 13. Substituting the values. called the ground state.3 × 10 Note that the quantum number for the satellite motion is extremely large! In fact for such large quantum numbers the results of quantisation conditions tend to those of classical physics.40 eV – (–13. 12. To ionise a hydrogen atom an electron from the ground state. as shown in Fig. Therefore. In such situations the electron is free.09 eV energy is required.8.. the one for which n = 1. is given in © no N C tt E o R be T re pu * An electron can have any total energy above E = 0 eV. This prediction of the Bohr’s model is in excellent agreement with the experimental value of ionisation energy. From these excited states the electron can then fall back to a state of lower energy. the atom may acquire sufficient energy to raise the electron to higher energy states.09 eV. When a hydrogen atom receives energy by processes such as electron collisions. (12. and so on. bl FIGURE 12. The energy level diagram* for the stationary states of a hydrogen atom. 12. (12.19). is that of the lowest energy. At room temperature. the absolute value of the energy E is smaller.51 eV and E3 – E1 = 12.4.6 eV of energy must be supplied. The excited electron would subsequently fall back to the ground state by emission of radiation [Fig. Observe how the energies of the excited states come closer and closer together as n increases. 12.8. The electron energy was varied by subjecting the electrons to electric fields of varying strength. when an atom makes a transition from the higher energy state with quantum number n i to the lower energy state with quantum number n f (nf < n i ).9 eV. the difference of energy is carried away by a photon of frequency νif such that bl OF THE is HYDROGEN ATOM he FRANCK – HERTZ EXPERIMENT d .5 THE LINE SPECTRA 428 According to the third postulate of Bohr’s model. The wavelength of emitted radiation is: By direct measurement.625 ×10 −34 × 3 ×10 8 = = 253 nm E 4.19) and has an energy of 0 eV. In this diagram.9 × 1. (b)].9 eV or more passes through mercury. The electrons collide with the mercury atoms and can transfer energy to the mercury atoms. For this experimental verification of Bohr’s basic ideas of discrete energy levels in atoms and the process of photon emission. If an electron of having an energy of 4. This is the energy of the atom when the electron is completely removed (r = ∞) from the nucleus and is at rest.6 ×10− 19 The existence of discrete energy levels in an atom was directly verified in 1914 by James Franck and Gustav Hertz. The colliding electron’s kinetic energy would reduce by this amount. (12. an electron in mercury atom can absorb energy from the bombarding electron and get excited to the higher level [Fig (a)]. Frank and Hertz were awarded the Nobel prize in 1925. This can only happen when the energy of the electron is higher than the energy difference between an energy level of Hg occupied by an electron and a higher unoccupied level (see Figure). They studied the spectrum of mercury vapour when electrons having different kinetic energies passed through the vapour. The principal quantum number n labels the stationary states in the ascending order of energy. Franck and Hertz found that the emission spectrum of mercury has a line corresponding to this wavelength. 12. For instance. © no N C tt E o R be T re pu λ= hc 6. the difference between an occupied energy level of Hg and a higher unoccupied level is 4.Physics Fig. the highest energy state corresponds to n = ∞ in Eq. 097 × 107 m–1) obtained from the empirical Balmer formula.9 Line spectra originate in transitions between energy levels. nf = 3 and n i = 4. and Pfund series.21) is the Rydberg formula. we get R = 1. 5. the Balmer formula corresponds to nf = 2 and ni = 3. 5. absorbs a photon that has precisely © no N C tt E o R be T re pu bl is he 429 d or ν if = . if we take n f = 2 and ni = 3. In this relation. Such series were identified in the course of spectroscopic investigations and are known as the Lyman. 5. etc. and so on... Brackett. 12. for En f and Eni. But when an atom FIGURE 12. The electronic transitions corresponding to these series are shown in Fig. etc.23) If we insert the values of various constants in Eq. Since both n f and n i are integers. this immediately shows that in transitions between different atomic levels. (12.03 × 107 m–1 This is a value very close to the value (1.10) for the Balmer series.16).Atoms hν if = En i – En f Using Eq.. we get hν if = me 4 ⎛ 1 1⎞ 2 2 ⎜ 2 − 2⎟ 8 ε 0h ⎝ n f n i ⎠ me 4 ⎛ 1 1⎞ 2 3 ⎜ 2 − 2⎟ 8 ε 0h ⎝ n f n i ⎠ (12. (12. 4.22) Equation (12. The various lines in the atomic spectra are produced when electrons jump from higher energy state to a lower energy state and photons are emitted. light is radiated in various discrete frequencies. etc. Balmer.9.21) (12. The results of the Bohr’s model suggested the presence of other series spectra for hydrogen atom–those corresponding to transitions resulting from nf = 1 and ni = 2. These spectral lines are called emission lines. Paschen.23). For hydrogen spectrum.. 3. (12. for the spectrum of the hydrogen atom. 4. The Rydberg constant R is readily identified to be me 4 R = 8 ε 2h 3c 0 (12..20) (12. This agreement between the theoretical and experimental values of the Rydberg constant provided a direct and striking confirmation of the Bohr’s model. it reduces to a form similar to Eq. C. Why should the angular momentum have only those values that are integral multiples of h/2π ? The French physicist Louis de Broglie explained this puzzle in 1923. we get λ21 = 1218 Å. a series of dark spectral absorption lines appear in the continuous spectrum. The explanation of the hydrogen atom spectrum provided by Bohr’s model was a brilliant achievement.5 to nf = 1. in Chapter 11. Example 12. H. Bohr was awarded Nobel Prize in Physics.3.76 × 10 × (n i − 1) 2 2 = 913. 2. 3 …). 12. In 1922. We studied. Bohr made in his model of the atom. λ31 = 1028 Å. about the de Broglie’s hypothesis that material particles. Ln = nh/2 π.3 Å.76 × 10 −19 ⎛1 1 ⎞ − 2 ⎜ ⎝1 n i ⎟ ⎠ m EXAMPLE 12.4.Physics the same energy needed by the electron in a lower energy state to make transitions to a higher energy state.6 Using the R ydberg formula. J. hc/λif = me 4 ⎛ 1 1⎞ − 2⎟ 2 2 ⎜ 2 8ε0 h ⎝ n f ni ⎠ © no N C tt E o R be T re pu me 4 –19 J 2 = 13.6 eV = 21.6 D E BROGLIE’S EXPLANATION OF BOHR’S SECOND POSTULATE OF QUANTISATION 430 Of all the postulates.4. ten years after Bohr proposed his model. and λ51 = 951. λ41 = 974. The dark lines indicate the frequencies that have been absorbed by the atoms of the gas. n = 1.6 = 6.625 × 10−34 × 3 ×108 × n i2 m = −19 2 21.76 ×10 8ε2 0h λi 1 = hc Therefore. The wavelengths of the first four lines in the L yman series corr espond to transitions fr om ni = 2. Germer later experimentally verified the wave nature of electrons bl 0.3. also have a wave nature. We know that 21. perhaps the most puzzling is his second postulate. such as electrons. Thus if photons with a continuous range of frequencies pass through a rarefied gas and then are analysed with a spectrometer.4 ni /(ni –1) Å Substituting ni = 2. calculate the wavelengths of the first four spectral lines in the L yman series of the hydrogen spectrum. the process is called absorption.5. which greatly stimulated progress towards the modern quantum theory. Davisson and L. It states that the angular momentum of the electron orbiting around the nucleus is quantised (that is.4 Å.9134 ni2 ×10−7 m 2 (n i − 1) Solution The Rydber g formula is is he d . Bohr’s model.13)]. (12. Louis de Broglie argued that the electron in its circular orbit. In these atoms more complex electron-electron interactions are nonexistent. Thus de Broglie hypothesis provided an explanation for Bohr’s second postulate for the quantisation of angular momentum of the orbiting electron. 2πrn = 4λ. It cannot be extended even to mere two electron atoms such as helium. Waves with other wavelengths interfere with themselves upon reflection and their amplitudes quickly drop to zero. Some are: (i) The Bohr model is applicable to hydrogenic atoms. standing waves are formed when the total distance travelled by a wave down the string and back is one wavelength. the total distance is the circumference of the orbit. For an electron moving in n th circular orbit of radius rn. in particular.. However only those wavelengths survive which have nodes at the ends and form the standing wave in the string. (12. Thus. λ = h/mvn. we have This is the quantum condition proposed by Bohr for the angular momentum of the electron [Eq. we saw that this equation is the basis of explaining the discrete orbits and energy levels in hydrogen atom. 2πr n. i. 2. we know that when a string is plucked..24). where p is the magnitude of the electron’s momentum. a vast number of wavelengths are excited. The analysis of atoms with more than one electron was attempted on the lines of Bohr’s model for hydrogenic atoms but did not meet with any success. If the speed of the electron is much less than the speed of light. doubly ionised lithium. In analogy to waves travelling on a string. From Eq. Difficulty lies in the fact that each electron interacts not only with the positively charged nucleus but also with all other electrons.e. involving classical trajectory picture (planet-like electron orbiting the nucleus). It means that in a string. is he 431 d . The quantised electron orbits and energy states are due to the wave nature of the electron and only resonant standing waves can persist.5.10 illustrates a standing particle wave on a circular orbit for n = 4. From Chapter 11. where Z is the pr oton number .24) © no N C tt E o R be T re pu 2π rn = n h/mvn or m vn rn = nh/2π Figure 12. correctly predicts the gross features of the hydrogenic atoms*. must be seen as a particle wave. we have λ = h/p . where λ is the de Broglie wavelength of the electron moving in n th orbit. the frequencies of the radiation emitted or selectively absorbed. n = 1.10 A standing wave is shown on a circular orbit where four de Broglie wavelengths fit into the circumference of the orbit. singly ionised helium.Atoms in 1927. as proposed by Bohr.. the momentum is mvn. or any integral number of wavelengths. two wavelengths. In Section 12. bl FIGURE 12. * Hydrogenic atoms are the atoms consisting of a nucleus with positive charge +Ze and a single electron. 3. Thus 2π rn = nλ. (12. This model however has many limitations. Examples are hydrogen atom. particle waves too can lead to standing waves under resonant conditions. From Chapter 15 of Class XI Physics textbook. and so forth. which was reflected by the mirror on the moon and received back on the earth. suitable for delicate surgery of eye or glands in the stomach. facing the earth. LASER LIGHT Imagine a crowded market place or a railway platform with people entering a gate and going towards all directions. It does not include the electrical forces between electrons which necessarily appear in multi-electron atoms. engineering. the model is unable to explain the relative intensities of the frequencies in the spectrum. Also the average length of the packet of waves is much larger. On the other hand. There is also no phase relation between the various waves. Their footsteps are very well correlated. Since its development in 1960. The size of the reflected laser beam and the time taken for the round trip were measured. If there are N atoms in a source. medicine. surgery. we see that the light from a laser can be much stronger than that from an ordinary source. others strong. spreads very fast and the beam size increases rapidly with distance. This allowed a very accurate determination of (a) the extremely small divergence of a laser beam and (b) the distance of the moon from the earth. When astronauts of the Apollo missions visited the moon. Light coming out from an ordinary source contains a mixture of many wavelengths. See figure here. Why? Experimental observations depict that some transitions are more favoured than others. Therefore. For complex atoms we have to use a new and radical theory based on Quantum Mechanics. which provides a more complete picture of the atomic structure. each emitting light with intensity I. whereas in a laser source. In the case of laser light. This results in reducing the divergence of a laser beam substantially. There are also lasers of different power. it has entered into all areas of science and technology. then the total intensity produced by an ordinary source is proportional to NI. There are low power lasers.Physics The formulation of Bohr model involves electrical force between positively charged nucleus and electron. © no N C tt E o R be T re pu 432 bl is he d . even if it is passed through an aperture. they placed a mirror on its surface. Their footsteps are random and there is no phase correlation between them. This is similar to the difference between light emitted by an ordinary source like a candle or a bulb and that emitted by a laser. with a power of 0. such light. It has found applications in physics. The acronym LASER stands for Light Amplification by Stimulated Emission of Radiation. Bohr’s model is unable to account for the intensity variations. think of a large number of soldiers in a regulated march. biology. This means that there is better phase correlation over a longer duration of time. etc. there are lasers which can cut or weld steel. In emission spectrum of hydrogen. which serve as pointers. called pencil lasers.5 mW. Considering that N is very large. some of the visible frequencies have weak intensity. Light is emitted from a source in the form of packets of waves. (ii) While the Bohr’s model correctly predicts the frequencies of the light emitted by hydrogenic atoms. chemistry. Then scientists on the earth sent a strong laser beam. it is proportional to N 2I . Bohr’s model presents an elegant picture of an atom and cannot be generalised to complex atoms. the wavelength of each packet is almost the same. Finally. .. Atoms of each element are stable and emit characteristic spectrum. 7. The frequency (ν ) of the emitted photon is then given by hν = Ei – E f An atom absorbs radiation of the same frequency the atom emits. In Thomson’s model. (Bohr’s quantisation condition. 7. ⎝3 n ⎠ 1 ⎞ ⎛1 Brackett series: ν = Rc ⎜ 2 − 2 ⎟ .Atoms SUMMARY 1.. It provides useful information about the atomic structure. ⎝5 n ⎠ To explain the line spectra emitted by atoms. 6. 3. an electron revolves in certain stable orbits (called stationary orbits) without the emission of radiant energy. as a whole. 4. a photon is emitted having energy equal to the energy difference between the initial and final states.. n = 6. 2. Atom. When it does so. ⎝4 n ⎠ The atomic hydrogen emits a line spectrum consisting of various series. and the electrons revolve around it. 6. The frequency of any line in a series can be expressed as a difference of two terms. 1⎞ ⎛1 Pfund series: ν = Rc ⎜ 2 − 2 ⎟ . 5. in which case the electron is transferred to an orbit with a higher value of n . n = 2. an atom is a spherical cloud of positive charges with electrons embedded in it. This contradicts the stability of matter. 5. ⎝1 n ⎠ 1⎞ ⎛1 Balmer series: ν = Rc ⎜ 2 − 2 ⎟ . The spectrum consists of a set of isolated parallel lines termed as line spectrum. 5..... Niel’s Bohr proposed a model for hydrogenic (single elctron) atoms.) That is L = nh/2π. n = 4.. 4. Ei + hν = Ef bl 433 is he d .. (c) The third postulate states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy.. n = 5. 8. 6. as well as the stability of atoms. is electrically neutral and therefore contains equal amount of positive and negative charges. 4. most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus (typically one by ten thousand the size of an atom). 7. ⎝2 n ⎠ 1⎞ ⎛1 Paschen series: ν = Rc ⎜ 2 − 2 ⎟ . (b) It cannot explain the characteristic line spectra of atoms of different elements. n = 3. In Rutherford’s model. He introduced three postulates and laid the foundations of quantum mechanics: (a) In a hydrogen atom... where n is an integer called a quantum number. 3.. (b) The stationary orbits are those for which the angular momentum is some integral multiple of h/2π.. Rutherford nuclear model has two main difficulties in explaining the structure of atom: (a) It predicts that atoms are unstable because the accelerated electrons revolving around the nucleus must spiral into the nucleus. © no N C tt E o R be T re pu ⎛1 1⎞ Lyman series: ν = Rc ⎜ 2 − 2 ⎟ . In hydrogen atom the ground state energy is –13. and for circular orbits. the electron-electron electric force interaction is comparable in magnitude to the electronnucleus electrical force. This is the reason why the Bohr’s model with its planet-like electron is not applicable to many electron atoms. 5. Atoms are excited to these higher states by collisions with other atoms or electrons or by absorption of a photon of right frequency. de Broglie’s hypothesis that electrons have a wavelength λ = h/mv gave an explanation for Bohr’s quantised orbits by bringing in the waveparticle duality.6 eV. Thomson’s model is unstable electrostatically.6 eV/n 2 The n = 1 state is called ground state. For a hydrogen atom it is given by ⎛ n2 ⎞ ⎛ h ⎞ 4 πε0 rn = ⎜ ⎟ ⎜ ⎟ 2 ⎝ m ⎠ ⎝ 2 π⎠ e 2 En = − © no N C tt E o R be T re pu 1. What made Bohr quantise angular momentum (second postulate) and not some other quantity? Note. 10. POINTS TO PONDER Both the Thomson’s as well as the Rutherford’s models constitute an unstable system. the electron orbits the nucleus at only specific radii. angular momentum is a very relevant quantity. Bohr’s model is applicable only to hydrogenic (single electron) atoms. 4. The second postulate is then so natural! The orbital picture in Bohr’s model of the hydrogen atom was inconsistent with the uncertainty principle. 2.Physics 8. 3. h has dimensions of angular momentum. Higher values of n correspond to excited states (n > 1). where planet-planet gravitational forces are very small as compared to the gravitational force of the sun on each planet (because the mass of the sun is so much greater than the mass of any of the planets). It cannot be extended to even two electron atoms such as helium. Bohr’s model include only 434 bl is he me 4 2 2 2 8n ε 0 h d The total energy is also quantised: . while Rutherford’s model is unstable because of electromagnetic radiation of orbiting electrons. because the charges and distances are of the same order of magnitude. Unlike the situation in the solar system. 9. The orbits correspond to circular standing waves in which the circumference of the orbit equals a whole number of wavelengths. = –13. It was replaced by modern quantum mechanics in which Bohr’s orbits are regions where the electron may be found with large probability. Bohr laid the foundation of the quantum theory by postulating specific orbits in which electrons do not radiate. This model is also unable to explain for the relative intensities of the frequencies emitted even by hydrogenic atoms. As a result of the quantisation condition of angular momentum. . a given energy level may not correspond to just one quantum state... For example.. but has a highly non-uniform mass distribution in ..) (e) The positively charged part of the atom possesses most of the mass in ... The true picture is quantum mechanical affair which differs from Bohr model in a number of fundamental ways...... ( Thomson’s model/ Rutherford’s model... In Bohr model.2 (ii) The model incorporates many of the concepts we have learnt in classical physics....) (c) A classical atom based on .. is doomed to collapse... and s). The later is the difference between two orbital energies divided by h ...Atoms one quantum number n ... (much greater than/no different from/much less than. the atomic size in Rutherford’s model. while in ..... Choose the correct alternative from the clues given at the end of the each statement: (a) The size of the atom in Thomson’s model is ..... electrons are in stable equilibrium. electrons always experience a net force.. If the predictions of the theory or model agree with experiment. ( Thomson’s model/ Rutherford’s model. (Rutherford’s model/both the models. ( Thomson’s model/ Rutherford’s model.. But then if the Bohr model is not strictly correct. The new theory called quantum mechanics supportes Bohr’s postulate.. (Hydrogen is a solid at temperatures below 14 K.. However in quantum mechanics (more generally accepted). 6. m. Bohr’s semiclassical model based on some aspects of classical physics and some aspects of modern physics also does not provide a true picture of the simplest hydrogenic atoms... contrary to ordinary classical expectation.) Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. the frequency of revolution of an electron in its orbit is not connected to the frequency of spectral line.. 7... the two coincide as expected. n very large)..) (b) In the ground state of . © no N C tt E o R be T re pu EXERCISES 12...... (iii) The model demonstrates how a theoretical physicist occasionally must quite literally ignore certain problems of approach in hopes of being able to make some predictions. but for a pure Coulomb potential (as in hydrogen atom) the energy depends only on n... why do we bother about it? The reasons which make Bohr’s model still useful are: (i) The model is based on just three postulates but accounts for almost all the general features of the hydrogen spectrum. a theoretician then must somehow hope to explain away or rationalise the problems that were ignored along the way.. however. l .. For transitions between large quantum numbers (n to n – 1.1 12...) (d) An atom has a nearly continuous mass distribution in a ..... a state is characterised by four quantum numbers (n.....) What results do you expect? bl 435 is he d ... which help you understand the dif ference between Thomson’s model and Ruther ford’s model better. bl is he d . show that this frequency equals the classical frequency of revolution of the electron in the orbit.5 eV electron beam is used to bombard gaseous hydrogen at room temperature.4 A difference of 2. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton wer e bound by gravitational attraction. 12. which excites it to the n = 4 level.0 × 10 24 kg. scattering of α -particles at angles greater than 90°) predicted by Thomson’s model much less. it is found experimentally that for small thickness t.3 What is the shortest wavelength present in the Paschen series of spectral lines? 12. and 3 levels. What clue does this linear dependence on t provide ? (d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of α -particles by a thin foil? 12. or much greater than that predicted by Rutherford’s mode l ? (c) Keeping other factors fixed.3 eV separates two energy levels in an atom.10 In accordance with the Bohr’s model.) © no N C tt E o R be T re pu 436 ADDITIONAL EXERCISES 12. the number of α-particles scattered at moderate angles is proportional to t . For large n. or much greater than that predicted by Rutherford’s model? (b) Is the probability of backward scattering (i.6 A hydrogen atom initially in the ground level absorbs a photon. (b) Calculate the orbital period in each of these levels.e.11 Answer the following questions.13 Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n–1). (a) Is the average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model much less.9 A 12.3×10–11 m. 12.8 The radius of the innermost electron orbit of a hydrogen atom is 5. 2. What are the kinetic and potential energies of the electron in this stat e ? 12. Determine the wavelength and frequency of photon.6 eV. (Mass of earth = 6.Physics 12.5 × 10 11 m with orbital speed 3 × 10 4 m/s. What series of wavelengths will be emitted ? 12.5 The gr ound state energy of hydr ogen atom is –13.12 The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10–40.. about the same. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level? 12.7 (a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1. find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1. about the same. You will find the answer interesting. What are the radii of the n = 2 and n =3 orbits? 12. 12. it should be equally valid for the case of planetary motion also. Construct a quantity with the dimension of length from h. Then what determines the typical atomic size? Why is an atom not. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10–10m). Why then do we never speak of quantisation of orbits of planets around the sun? 12. To simulate what he might well have done before his discovery. and e will yield the right atomic size.15 The total energy of an electron in the first excited state of the hydrogen atom is about –3. Further.17 Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i. and c . it involves c. and e and confirm that its numerical value has indeed the correct order of magnitude. 12. Now. Determine its numerical value. me.16 If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature.e. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. © no N C tt E o R be T re pu bl 437 is he d . me. (b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Bohr’s great insight lay in recognising that h.14 Classically. say. thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. (a) Construct a quantity with the dimensions of length from the fundamental constants e. (a) What is the kinetic energy of the electron in this state ? (b) What is the potential energy of the electron in this state? (c) Which of the answers above would change if the choice of the zero of potential energy is changed ? 12. m e. an electron can be in any orbit around the nucleus of an atom.4 eV. an atom in which a negatively charged muon (μ –) of mass about 207me orbits around a proton].. the Planck’s constant h had already made its appearance elsewhere.Atoms 12. is 1. the mass of a carbon atom. This means the volume of a nucleus is about 10–12 times the volume of the atom. the nucleus contains most (more than 99. 12C. Does the nucleus have a structure.Physics Chapter Thirteen NUCLEI 13. the positive charge and mass are densely concentrated at the centre of the atom forming its nucleus. we shall look for answers to such questions. Kilogram is not a very convenient unit to measure such small quantities. and also associated nuclear phenomena such as radioactivity.1 INTRODUCTION In the previous chapter. Experiments on scattering of α-particles demonstrated that the radius of a nucleus was smaller than the radius of an atom by a factor of about 104. for example. Nevertheless. 13. the nucleus would be of the size of pinhead. a . We shall discuss various properties of nuclei such as their size. Therefore. In other words. an atom is almost empty. what are the constituents of the nucleus? How are these held together? In this chapter.992647 × 10–26 kg. we have learnt that in every atom. fission and fusion. mass and stability. compared to a kilogram.2 ATOMIC MASSES AND COMPOSITION OF NUCLEUS 438 The mass of an atom is very small. If an atom is enlarged to the size of a classroom. The overall dimensions of a nucleus are much smaller than those of an atom. just as the atom does? If so.9%) of the mass of an atom. Accurate measurement of atomic masses is carried out with a mass spectrometer. many striking exceptions to this rule.992647 × 10 −26 kg 12 (13.0160 u.0078 u. has two isotopes having masses 34. they occur in the same place in the periodic table of elements. Thus.Nuclei different mass unit is used for expressing atomic masses.00055 u). (In Greek. but this was ruled out later using arguments based on quantum theory. There are. The total charge of the 439 .00727 u = 1. A proton carries one unit of fundamental charge and is stable.47 u which agrees with the atomic mass of chlorine.1) = 1. The mass of a proton is = m p = 1. 2.4 and 24. defined as 1/12th of the mass of the carbon (12C) atom. According to this definition 1u = mass of one 12 12 C atom = 1.4 × 34. The other two isotopes of hydrogen are called deuterium and tritium. Even the lightest element.98 u. The relative abundances of these isotopes are 75. For example.98 + 24. Chlorine.985%. The measurement of atomic masses reveals the existence of different types of atoms of the same element. The positive charge in the nucleus is that of the protons. respectively.6 × 36. We know that the number of these electrons outside the nucleus of the atom is Z. however. Such atomic species of the same element differing in mass are called isotopes. isotope means the same place. for example. hydrogen has three isotopes having masses 1.2) This is equal to the mass of the hydrogen atom (= 1.6 per cent. which works out to be 75. which exhibit the same chemical properties.00783u). the atomic number. minus the mass of a single electron (me = 0. the average mass of a chlorine atom is obtained by the weighted average of the masses of the two isotopes. which are nearly integral multiples of the mass of a hydrogen atom. Tritium nuclei.98 100 = 35. i. This unit is the atomic mass unit (u).0141 u.46 u. do not occur naturally and are produced artificially in laboratories. and 3. is called the proton. which has a relative abundance of 99.98 u and 36. The nucleus of the lightest atom of hydrogen. All the electrons of an atom are outside the nucleus. It was earlier thought that the nucleus may contain electrons.) It was found that practically every element consists of a mixture of several isotopes. but differ in mass. the atomic mass of chlorine atom is 35. The relative abundance of different isotopes differs from element to element.e.67262 × 10 −27 kg (13. being unstable.660539 × 10−27 kg The atomic masses of various elements expressed in atomic mass unit (u) are close to being integral multiples of the mass of a hydrogen atom. the atomic number. It is. the nucleus of gold is .atomic number = number of protons N . some neutral matter. which Chadwick satisfactorily solved. respectively.4(c)] One also uses the term nucleon for a proton or a neutron. It was found that this neutral radiation could knock out protons from light nuclei such as those of helium. carbon and nitrogen. This hypothesis was verified in 1932 by James Chadwick who observed emission of neutral radiation when beryllium nuclei were bombarded with alpha-particles. (α-particles are helium nuclei. From conservation of energy and momentum. A free neutron. in addition to a proton.4(a)] [13. [13. For example. an electron and a antineutrino (another elementary particle). The clue to this puzzle. of which 79 are protons denoted by 197 79 Au and the rest118 are neutrons. Discovery of Neutron Since the nuclei of deuterium and tritium are isotopes of hydrogen.3) Chadwick was awarded the 1935 Nobel Prize in Physics for his discovery of the neutron. It decays into a proton. unlike a free proton. neutral matter in multiples of a basic unit. he was able to determine the mass of new particle ‘as very nearly the same as mass of proton’. This fact indicates that the nuclei of atoms contain. and since the atom is neutral.mass number = Z + N = total number of protons and neutrons [13. stable inside the nucleus. the charge of the nucleus is (+Ze).00866 u = 1.4(b)] 440 . The mass of a neutron is now known to a high degree of accuracy. the nuclei of deuterium and tritium must contain. But the masses of the nuclei of hydrogen. The number of protons in the nucleus of the atom is. and has a mean life of about 1000s. in addition to protons. to be discussed in a later section).neutron number = number of neutrons A . expressed in units of mass of a proton. Application of the principles of conservation of energy and momentum showed that if the neutral radiation consisted of photons. It contains 197 nucleons. they must contain only one proton each.6749×10–27 kg (13. however. The composition of a nucleus can now be described using the following terms and symbols: Z . deuterium and tritium are in the ratio of 1:2:3. exactly Z. The amount of neutral matter present in the nuclei of these isotopes. The only neutral radiation known at that time was photons (electromagnetic radiation). It is m n = 1. is approximately equal to one and two. the energy of photons would have to be much higher than is available from the bombardment of beryllium nuclei with α-particles. Thus the number of nucleons in an atom is its mass number A. therefore.Physics atomic electrons is thus (–Ze). is unstable. was to assume that the neutral radiation consists of a new type of neutral particles called neutrons. A Nuclear species or nuclides are shown by the notation Z X where X is the chemical symbol of the species. Therefore. contains one proton 3 and one neutron. If we use α-particles of higher energies than 5.0 × 10–14 m. As the atoms of isotopes have identical electronic structure they have identical chemical behaviour and are placed in the same location in the periodic table. This is understandable. Since the positive charge is confined to the nucleus. the distance of closest approach to the gold nucleus will be smaller and at some point the scattering will begin to be affected by the short range nuclear forces. Ordinary matter consisting of atoms has a large amount of empty space. For 3 example. Thus the density of nucleus is a constant. Geiger and Marsden performed their classic experiment: on the scattering of α-particles from thin gold foils. are projectiles that bombard targets made up of various elements. and differ from Rutherford’s calculations.5 MeV. instead of α-particles. 441 .5 MeV is about 4. which is 103 kg m–3. Their experiments revealed that the distance of closest approach to a gold nucleus of an α-particle of kinetic energy 5. Different nuclei are likes drop of liquid of constant density. This density is very large compared to ordinary matter. but differ from each other in their number of neutrons. We have already mentioned that chemical properties of elements depend on their electronic structure. independent of A. All nuclides with same mass number A are called isobars. The scattering of α-particle by the gold sheet could be understood by Rutherford by assuming that the coulomb repulsive force was solely responsible for scattering. which is proportional to R 3 is proportional to A. for all nuclei. Nuclides with same H and 3 2 He neutron number N but different atomic number Z. 13. Rutherford’s calculations are based on pure coulomb repulsion between the positive charges of the αparticle and the gold nucleus. It has been found that a nucleus of mass number A has a radius R = R 0 A1/3 (13. From the distance at which deviations set in. which is an isotope of hydrogen.2 × 10–15 m. At Rutherford’s suggestion. say water.3 SIZE OF THE NUCLEUS As we have seen in Chapter 12. The element gold has 32 isotopes. as we have already seen that most of the atom is empty. contains one proton and two neutrons.Nuclei The composition of isotopes of an element can now be readily explained. the sizes of nuclei of various elements have been accurately measured. The nuclei of isotopes of a given element contain the same number of protons. ranging from A =173 to A = 204. Its other isotope tritium.5) where R 0 = 1.3 × 1017 kg m–3. This means the volume of the nucleus. By performing scattering experiments in which fast electrons. for example 198 80 Hg 197 and 79 Au . the nuclides 1 are isobars. 1 H . the actual size of the nucleus has to be less than 4.0 × 10–14 m. The density of nuclear matter is approximately 2. 1 H . Rutherford was the pioneer who postulated and established the existence of the atomic nucleus. 2 Deuterium. are called isotones. nuclear sizes can be inferred. 27 × 10–26 kg −26 EXAMPLE 13.1 9.85u and A=56. there is a release of enormous amount of energy.1 Mass – Energy AND NUCLEAR BINDING ENERGY Einstein showed from his theory of special relativity that it is necessary to treat mass as another form of energy. nuclei.2 we have seen that the nucleus is made up of neutrons and protons. Example 13. 13. However. u = 9. find the nuclear density? Solution mFe = 55.2 × 10 −15 )3 56 volume = 2. Nuclear density = 13. . Einstein showed that mass is another form of energy and one can convert mass-energy into other forms of energy.27 × 10 1 mass × = (4π /3)(1.6) Here the energy equivalent of mass m is related by the above equation and c is the velocity of light in vacuum and is approximately equal to 3×108 m s–1.85.4. Einstein gave the famous mass-energy equivalence relation E = mc 2 (13. if one gram of matter is converted to energy.Physics Example 13. EXAMPLE 13. However.29 × 1017 kg m–3 The density of matter in neutron stars (an astrophysical object) is comparable to this density.4. This concept is important in understanding nuclear masses and the interaction of nuclei with one another. Therefore it may be expected that the mass of the nucleus is equal to the total mass of its individual protons and neutrons.2 Calculate the energy equivalent of 1 g of substance. Experimental verification of the Einstein’s mass-energy relation has been achieved in the study of nuclear reactions amongst nucleons. This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus. Before the advent of this theory of special relativity it was presumed that mass and energy were conserved separately in a reaction.1 Given the mass of iron nucleus as 55.4 MASS-ENERGY 13. In a reaction the conservation law of energy states that the initial energy and the final energy are equal provided the energy associated with mass is also included. They form the subject matter of the next few sections.2 Nuclear binding energy 442 In Section 13. E = 10–3 × ( 3 × 108)2 J E = 10–3 × 9 × 1016 = 9 × 1013 J Thus. electrons and other more recently discovered particles. say kinetic energy and vice-versa.2 Solution Energy. We us consider 16 8O have Mass of 8 neutrons = 8 × 1. let .3 Find the energy equivalent of one atomic mass unit.4924 × 10–10 J = 1.9979 × 108)2 kg m2/s2 = 1.6605 × 10–27 × (2.99493 u. 1u = 931. in the unbound state). we multiply it by c 2 and find that energy equivalent = 1. 8O Solution 1u = 1. The atomic mass of 16 found from mass spectroscopy experiments 8O is seen to be 15.00055 u Therefore the expected mass of 16 nucleus 8O = 8 × 2.602 × 10 −19 = 0.5 MeV/c .13691u. If one wants to break the oxygen nucleus into 8 protons and 8 neutrons. 2 16 8 EXAMPLE 13. ∆M. Using this.99053 u.3 O into its constituents is thus If a certain number of neutrons and protons are brought together to form a nucleus of a certain charge and mass.7) What is the meaning of the mass defect? It is here that Einstein’s equivalence of mass and energy plays a role. this extra energy ∆M c2. and is given by ∆M = [ Zm p + ( A − Z ) m n ] − M (13.Nuclei the nuclear mass M is found to be always less than this. express the mass defect of 16 in MeV/c 2. has to supplied.5 MeV/c 2 For 16 .00055 u) nucleus to be 15. This energy required Eb is related to the mass defect by Eb = ∆ M c 2 (13.6605 × 10–27 kg To convert it into energy units.12744 u.8) Example 13.00866 u Mass of 8 protons = 8 × 1.13691×931. For example. we get the experimental mass of 16 8O 16 Thus.13691 u = 0.5 MeV/c 2 8O = 127. the equivalent energy of the oxygen nucleus is less than that of the sum of the equivalent energies of its constituents.9315 × 109 eV = 931. an energy Eb will be released 443 . The difference in mass of a nucleus and its constituents.5 MeV or. is called the mass defect. Substracting the mass of 8 electrons (8 × 0. a nucleus which has 8 neutrons and 8 protons.00727 u Mass of 8 electrons = 8 × 0.01593 u = 16. ∆M = 0. Since the mass of the oxygen nucleus is less that the sum of the masses of its constituents (8 protons and 8 neutrons. we find that the mass of the 8 O nucleus is less than the total mass of its constituents by 0.4924 × 10 −10 eV 1. from this.5 MeV/c 2 The energy needed to separate 127. first in Joules and then in MeV. We notice the following main features of the plot: (i) the binding energy per nucleon. The property that a given nucleon .1 is a plot of the binding energy per nucleon Ebn versus the mass number A for a large number of nuclei.Physics in the process. The curve has a maximum of about 8.e. If we increase A by adding nucleons they will not change the binding energy of a nucleon inside. It will be under the influence of only some of its neighbours. we would have to supply a total energy equal to Eb. Although we cannot tear apart a nucleus in this way. We can draw some conclusions from these two observations: (i) The force is attractive and sufficiently strong to produce a binding energy of a few MeV per nucleon. Since most of the nucleons in a large nucleus reside inside it and not on the surface. i. The energy Eb is called the binding energy of the nucleus.75 MeV for A = 56 and has a value of 7. A more useful measure of the binding between the constituents of the nucleus is the binding energy per nucleon. (ii) The constancy of the binding energy in the range 30 < A < 170 is a consequence of the fact that the nuclear force is short-ranged. The binding energy per nucleon is a constant and is 444 approximately equal to pk. (ii) Ebn is lower for both light nuclei ( A <30) and heavy nuclei (A>170). A.9) We can think of binding energy per nucleon as the average energy per nucleon needed to separate a nucleus into its individual nucleons. If we separate a nucleus into its nucleons. practically independent of the atomic number for nuclei of middle mass number ( 30 < A < 170). in that nucleus: Ebn = Eb / A (13. which come within the range of the nuclear force. the change in binding energy per nucleon would be small. as a function of mass number. its binding energy would be proportional to p. Figure 13. Let the binding energy of the nucleus be pk. If any other nucleon is at a distance more than the range of the nuclear force from the particular nucleon it will have no influence on the binding energy of the nucleon under consideration. Consider a particular nucleon inside a sufficiently large nucleus. where k is a constant having the dimensions of energy. If a nucleon can have a maximum of p neighbours within the range of nuclear force. which is the ratio of the binding energy Eb of a nucleus to the number of the nucleons. is practically constant. Ebn.6 MeV FIGURE 13. to those particles.1 The binding energy per nucleon for A = 238. Ebn. the nuclear binding energy is still a convenient measure of how well a nucleus is held together. 8 fm and repulsive if they are separated by distances less 445 than 0. This means that the final system is more tightly bound than the initial system. In Section 13. Thus if a nucleus A = 240 breaks into two A = 120 nuclei. It must be strong enough to overcome the repulsion between the (positively charged) protons and to bind both protons and neutrons into the tiny nuclear volume. 13. This happens only because the nuclear force is much stronger than the coulomb force. to bind a nucleus together there must be a strong attractive force of a totally different kind. the force is A rough plot of the potential energy between two nucleons strongly repulsive.7. leads to saturation of forces in a medium or a large-sized For a separation greater nucleus.5 NUCLEAR FORCE The force that determines the motion of atomic electrons is the familiar Coulomb force.1. The binding energy per nucleon of the fused heavier nuclei is more than the binding energy per nucleon of the lighter nuclei. we have seen that for average mass nuclei the binding energy per nucleon is approximately 8 MeV. to be discussed later in Section 13. This is the energy source of sun. (iv) Consider two very light nuclei (A ≤ 10) joining to form a heavier nucleus. which is much larger than the binding energy in atoms. to be discussed later in Section 13. This implies energy would be released in the process. Many features of the nuclear binding force are summarised below. 13.4.8 fm. and for separations less than r0.2.2 Potential energy (ii) The nuclear force between two nucleons falls rapidly to of a pair of nucleons as a zero as their distance is more than a few femtometres. These are obtained from a variety of experiments carried out during 1930 to 1950. FIGURE 13. We have already seen that the constancy of binding energy per nucleon can be understood in terms of its short-range. The gravitational force is much weaker than even Coulomb force. The nuclear binding force has to dominate over the Coulomb repulsive force between protons inside the nucleus. (i) The nuclear force is much stronger than the Coulomb force acting between charges or the gravitational forces between masses.8 fm. It has very important implications for energy production through fission. has lower binding energy per nucleon compared to that of a nucleus with A = 120. Therefore. say A = 240. as a function of distance is shown in the Fig. The potential energy is a minimum at a distance r0 of about 0. the force is attractive binding energy per nucleon. which is the reason for the constancy of the than r0.Nuclei influences only nucleons close to it is also referred to as saturation property of the nuclear force. . Again energy would be released in such a process of fusion. (iii) A very heavy nucleus.3. nucleons get more tightly bound.7. This function of their separation. This means that the force is attractive for distances larger than 0. 2 (ii) β-decay in which electrons or positrons (particles with the same mass as electrons.1 Law of radioactive decay In any radioactive sample. H. ∆N have decayed. This is referred to as radioactive decay. . While studying the fluorescence and phosphorescence of compounds irradiated with visible light. Becquerel observed an interesting phenomenon. and hence is always positive. Experiments performed subsequently showed that radioactivity was a nuclear phenomenon in which an unstable nucleus undergoes a decay. Becquerel discovered radioactivity in 1896 purely by accident. but with a charge exactly opposite to that of electron) are emitted. which may have either sign. the photographic plate was developed. after several hours of exposure. The change in the number of nuclei in the sample* is dN = – ∆N in time ∆t. ∆N/∆t = λN.6. Unlike Coulomb’s law or the Newton’s law of gravitation there is no simple mathematical form of the nuclear force. he wrapped them in black paper and separated the package from a photographic plate by a piece of silver. leaving (N–∆N ) nuclei. If N is the number of nuclei in the sample and ∆N undergo decay in time ∆t then ∆N ∝N ∆t or.Physics (iii) The nuclear force between neutron-neutron. Each of these decay will be considered in subsequent sub-sections. it showed blackening due to something that must have been emitted by the compound and was able to penetrate both black paper and the silver. Here it is negative. 13. When. because out of original N nuclei.10) where λ is called the radioactive decay constant or disintegration constant. The nuclear force does not depend on the electric charge. dN is the change in N. it is found that the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample. (iii) γ-decay in which high energy (hundreds of keV or more) photons are emitted. which undergoes α. Thus the rate of change of N is (in the limit ∆t → 0) dN = – λN dt 446 * ∆N is the number of nuclei that decay.6 RADIOACTIVITY A. proton-neutron and proton-proton is approximately the same. Three types of radioactive decay occur in nature : (i) α-decay in which a helium nucleus 4 He is emitted. (13. β or γ-decay. After illuminating some pieces of uranium-potassium sulphate with visible light. 13. 14).13) which gives N( t ) = N 0 e − λ t (13. (13. named after the discoverer of radioactivity.12) gives us ln N = − λt N0 (13. population of the given species drops by a factor of 2. Henry Becquerel. we get R = λN0 e −λt or. Then dN = – ∆N.14) Note. or. The decay of radionuclides follows quite a different law. for example. Clearly. the decay count measured is ∆N.14). since you can integrate Eq. Setting t0 = 0 and rearranging Eq. (13. R = R0 e –λt (13. After a lapse of T1/2 .15) This is equivalant to the law of radioactivity decay. (13. The positive quantity R is then defined as R=– dN dt Differentiating Eq. If we test 1000 bulbs for their life (time span before they burn out or fuse). 447 . integrating both sides of the above equation. The total decay rate R of a sample is the number of nuclei disintegrating per unit time.3 Exponential decay of a radioactive species.16) FIGURE 13. we expect that they will ‘decay’ (that is. N ∫ N0 dN = − λ ∫ dt N t0 t (13. is a more direct experimentally measurable quantity and is given a specific name: activity. (13. Suppose in a time interval dt. R0 = λN0 is the decay rate at t = 0. The decay rate of a sample. The SI unit for activity is becquerel. The decay rate R at a certain time t and the number of undecayed nuclei N at the same time are related by R = λN (13. ln N − ln N0 = −λ (t – t0) (13.we get. the light bulbs follow no such exponential decay law. burn out) at more or less the same time.15) to get back Eq. rather than the number of radioactive nuclei.14).12) Here N0 is the number of radioactive nuclei in the sample at some arbitrary time t0 and N is the number of radioactive nuclei at any subsequent time t. (13. the law of radioactive decay represented by Eq.Nuclei dN = – λ dt N Now.11) or. 7 × 1010 Bq Different radionuclides differ greatly in their rate of decay. say N0 radionuclei to reduce to N0/2.g. She was the first person to be awarded two Nobel Prizes.5 × 109 y x 3.14). Each of them has lived for time t.18) Radioactive elements (e. Putting N = N0/2 and t = T1/2 in Eq.42 × 1017s 448 EXAMPLE 13.4 .5 × 109 y = 4.5 × 109 92 U ? years. There is also another unit named “curie” that is widely used and is related to the SI unit as: 1 curie = 1 Ci = 3. (13. be produced artificially in nuclear reactions. It is clear that some nuclei may live for a short time while others may live longer. and divide by the total number N0 of nuclei at t = 0.16). plutonium) which are short-lived i. we have to sum (or integrate) this expression over all times from 0 to ∞ .e.7 × 1010 decays per second = 3. Thus the total life of all these nuclei would be t λN0e–λt ∆t. λ N 0 ∫ te – λt dt τ= 0 ∞ N0 0 One can show by performing this integral that = λ ∫ te – λt dt τ = 1/λ We summarise these results with the following: T1/2 = ln 2 λ = τ ln 2 (13.17) λ λ Clearly if N0 reduces to half its value in time T1/2. Thus. however. Example 13.Physics 1 becquerel is simply equal to 1 disintegration or decay per second.14). (13. ∞ MARIE SKLODOWSKA CURIE (1867-1934) T1/2 = ln 2 = 0.. tritium. A common way to characterize this feature is through the notion of half-life.. R0 will also reduce to half its value in the same time according to Eq.16 x 107 s/y = 1.693 Marie Sklodowska Curie (1867-1934) Born in Poland. The discovery of radioactivity by Henri Becquerel in 1896 inspired Marie and her husband Pierre Curie in their researches and analyses which led to the isolation of radium and polonium elements. Another related measure is the average or mean life τ. She is recognised both as a physicist and as a chemist. have half-lives much less than the age of the universe ( ∼ 15 billion years) have obviously decayed long ago and are not found in nature. we get (13. Therefore to obtain the mean life. They can. This again can be obtained from Eq.4 The half-life of 238 undergoing α-decay is 4. The number of nuclei which decay in the time interval t to t + ∆t is R(t )∆t (= λN0e–λt∆t ).for Physics in 1903 and for Chemistry in 1911. What is the activity of 1g sample of 238 92 U Solution T1/2 = 4. (13. Half-life of a radionuclide (denoted by T1/2) is the time it takes for a sample that has initially. while the atomic number decreases by two. This difference in mass appears as kinetic energy of the products. 449 .5 Tritium has a half-life of 12. one fourth of the sample of the initial pure tritium will remain undecayed.20) X → A He Z −2Y From Einstein’s mass-energy equivalance relation [Eq.6. In general. half of the initial sample will remain undecayed after 12. EXAMPLE 13. if the initial nucleus X is at rest. In the next 12.3 × 1020 −1 N = s T1/ 2 1.21) Q is also the net kinetic energy gained in the process or.3 × 1020 atoms.693 0.4 = 0. The decay rate R is R = λN EXAMPLE 13.5 y.6)] and energy conservation. 90 Th 2 He 92 U The disintegration energy or the Q-value of a nuclear reaction is the difference between the initial mass energy and the total mass energy of the decay products.025 × 1026 atoms/kmol 238 × 10 –3 = 25.5 y undergoing beta decay. Q> 0 for exothermic processes such as α-decay. What fraction of a sample of pure tritium will remain undecayed after 25 y.2 Alpha decay A well-known example of alpha decay is the decay of uranium 4 with the emission of a helium nucleus 2 thorium 234 He 90 Th 238 92 238 92 U to U→ 234 90 Th + 4 2 He (α-decay) (13. it is clear that this spontaneous decay is possible only when the total mass of the decay products is less than the mass of the initial nucleus.19) In α-decay.42 × 1017 = 1. (13.23 × 104 Bq Example 13.693 × 25. one-half of these nuclei would have decayed. Clearly.5 13.23 × 104 s–1 = 1. For α-decay A Z Q = (mX – mY – mHe) c2 (13. the mass number of the product nucleus (daughter nucleus) is four less than that of the decaying nucleus (parent nucleus). one can check that the total mass of 234 and 4 is indeed less than that of 238 . and contains so 1g of 238 92 U 1 kmol × 6. α-decay of a parent A −4 nucleus Z X results in a daughter nucleus A Z − 2Y −4 4 + 2 (13. Solution By definition of half-life.Nuclei One k mol of any isotope contains Avogadro’s number of atoms. Hence. By referring to a table of nuclear masses. the kinetic energy of the products.5 y. (a) Calculate the energy released during the alpha decay of 238 92 U (b) Show that 238 can not spontaneously emit a proton. (b) If 238 spontaneously emits a proton. a nucleus spontaneously emits an electron (β − decay) or a positron (β+ decay).23) 450 The decays are governed by the Eqs. but one can characterize the decay by a half-life T1/2 . A common example of β − decay is 32 15 32 P → 16 S + e− + ν (13.05121 u Here the symbol Pa is for the element protactinium (Z = 91). instead.00825 u) c 2 = – (0.68 MeV to a 238 nucleus to make it emit a proton.00456 u) c 2 = (0.00825 u)(931.05079 u = 4.00260)u × c 2 = (0.00783 u 90 Th 1H Pa = 237.5 MeV/u) = 4.6 We are given the following atomic masses: 238 4 = 238.00260 u 92 U 2 He 234 1 = 234.20). 92 U 238 92 237 91 EXAMPLE 13. . therefore. the Q of the process is negative and therefore it cannot proceed spontaneously. even zero) mass compared to electrons. The energy released (a) The alpha decay of 238 92 U in this process is given by Q = (MU – MTh – MHe) c 2 Substituting the atomic masses as given in the data.25 MeV. U 92 Solution is given by Eq.00456 u) (931.04363 – 4.3 Beta decay In beta decay. We will have to supply an energy of 7.22) and that of β+ decay is 22 11 22 Na → 10 Ne + e + + ν (13. For example. we find Q = (238.00783) u × c 2 = (– 0.05079 – 234.05121 – 1. They are. so that one can never predict which nucleus will undergo decay. Neutrinos are neutral particles with very small (possiblly. a neutrino (ν) is generated.3 d and 2.6y. (13.6. since they can penetrate large quantity of matter (even earth) without any interaction.15).6 13.04363 u = 1.05079 – 237. They have only weak interaction with other particles. the decay process would be 92 U + 1 U → 237 91 Pa 1H The Q for this process to happen is = (MU – MPa – MH ) c 2 = (238.68 MeV Thus.5 MeV/u) = – 7.14) and (13. T1/2 for the decays above is respectively 14. (13. The emission of electron in β − decay is accompanied by the emission of an antineutrino ( ν ).Physics Example 13. . very difficult to detect. in β+ decay. has a long flat middle region between A = 30 and A = 170. very different.0 MeV). the less is the total mass of a bound system. 28 60 of 27 Co . the atomic number Z of the nucleus goes up by 1. a photon is emitted with energy equal to the difference in the two energy levels of the nucleus. The energy (MeV) corresponds to radiation of extremely short wavelength. it is the conversion of proton into neutron p → n + e+ + ν + (13. A > 170. In β decay. a gamma ray is emitted when a α or β decay results in a daughter nucleus in an excited state. the mass number A remains unchanged. the greater the binding energy.0 MeV. a nucleus also has discrete energy levels .4 Gamma decay Like an atom. A < 30. For the lighter nuclei region.24) (13. Consequently. In this region the binding energy per nucleon is nearly constant (8.Nuclei In both β − and β + decay. The scale of energy is. The basic nuclear process underlying β − decay is the conversion of neutron to proton − n → p + e– + ν while for β decay. 13. Typically. while in β + decay Z goes down by 1. given in Fig. A familiar example FIGURE 13. the binding energy per nucleon is less than 8.7 NUCLEAR ENERGY The curve of binding energy per nucleon Ebn.25) Note that while a free neutron decays to proton. Now.1.) Exothermic chemical reactions underlie conventional energy sources such as coal or petroleum.17 MeV and 1. This then returns to the ground state by a single photon transition or successive transitions involving more than one photon. This is what happens when a heavy nucleus decays into two or more intermediate mass fragments (fission) or when light nuclei fuse into a havier nucleus (fusion.33 MeV from the from deexcitation of the daughter deexcitation of 60 nuclei formed from β − decay 28 Ni nucleus 60 Ni . the decay of proton to neutron [Eq. however.the ground state and excited states. and for the heavier nuclei region. while the difference in nuclear energy levels is of the order of MeV. if nuclei with less total binding energy transform to nuclei with greater binding energy. as we have noted earlier. Here the energies involved are in the range of 451 . Atomic energy level spacings are of the order of eV. 13. (13. When a nucleus in an excited state spontaneously decays to its ground state (or to a lower energy state). 13.4 β-decay of 60 Ni nucleus 28 is the successive emmission of gamma rays of followed by emission of two γ rays energies 1. there will be a net energy release. shorter than the hard X-ray region. such as a nucleus.25)] is possible only inside the nucleus. since proton has smaller mass than neutron. This is the so-called gamma decay .6. 7. The energy released (the Q value ) in the fission reaction of nuclei like uranium is of the order of 200 MeV per fissioning nucleus. α-particle. There is a release of extra neutron (s) in the fission process. nuclear sources produce a million times more energy than a chemical source. Ebn for the two A = 120 fragment nuclei is about 8. On the other hand. ∴ Gain in binding energy for nucleon is about 0.Physics electron volts.1 Fission New possibilities emerge when we go beyond natural radioactive decays and study nuclear reactions by bombarding nuclei with other nuclear particles such as proton. 13. This is estimated as follows: Let us take a nucleus with A = 240 breaking into two fragments each of A = 120. generates 1014 J of energy.28). (13. compare it with burning of 1 kg of coal that gives 107 J. in a nuclear reaction. Thus for the same quantity of matter.28) Sr + 2 1 0n The fragment products are radioactive nuclei. is nuclear fission. The enormous energy released in an atom bomb comes from uncontrolled nuclear fission. Hence the total gain in binding energy is 240×0. 2½ neutrons are released per fission of uranium nucleus. they emit β particles in succession to achieve stable end products. 1 0 140 n + 235 92 U → 54 Xe + 94 38 1 0 235 92 (13. for example. neutron.9 MeV. An example of fission is when a uranium isotope 1 0 U bombarded with a neutron breaks into two intermediate mass nuclear fragments 236 144 89 1 (13. 13.6 MeV. the energy release is of the order of MeV. Then Ebn for A = 240 nucleus is about 7. The disintegration energy in fission events first appears as the kinetic energy of the fragments and neutrons.27) (13. as another example. Eventually it is transferred to the surrounding matter appearing as heat. Fission of 1 kg of uranium. We discuss some details in the next section how a nuclear reactor functions.7. A most important neutron-induced nuclear reaction is fission.2 Nuclear reactor Notice one fact of great importance in the fission reactions given in Eqs. It is a fraction since in some fission events 2 neutrons are 452 . Averagely.26) n + 235 92 U → 92 U → 56 Ba + 36 Kr + 3 0 n The same reaction can produce other pairs of intermediate mass fragments 236 133 99 1 n + 235 92 U → 92 U → 51 Sb + 41 Nb + 4 0 n Or.26) to (13. The source of energy in nuclear reactors. which produce electricity.9 or 216 MeV.5 MeV. etc. The country has mastered the complex technologies of mineral exploration and mining. our country has adopted a threestage strategy of nuclear power generation. The third stage. KAMINI is the country’s first large research reactor that uses U-233 as fuel. etc. producing still more neutrons. Bhabha (1909-1966). fuel reprocessing. with heavy water as moderator. Research reactors that have been subsequently commissioned include ZERLINA. This leads to the possibility of a chain reaction. India is now more than self-sufficient in heavy water production. besides generating power. reactor design. The main objectives of the Indian Atomic Energy programme are to provide safe and reliable electric power for the country’s social and economic progress and to be selfreliant in all aspects of nuclear technology. produced. medicine and agriculture. etc. fuel fabrication. heavy water production. The extra neutrons in turn can initiate fission processes. The first stage involves the use of natural uranium as a fuel. construction and operation. most significant in the long term. Accordingly. It used enriched uranium as fuel and water as moderator. the primary objective of a research reactor is not generation of power but to provide a facility for research on different aspects of nuclear science and technology. As the name suggests. India is currently well into the second stage of the programme and considerable work has also been done on the third — the thorium utilisation — stage. This 40 MW reactor used natural uranium as fuel and heavy water as moderator. as also adhering to stringent standards of radiological protection are the hallmark of the Indian Atomic Energy Programme. Exploration of atomic minerals in India undertaken since the early fifties has indicated that India has limited reserves of uranium. An early historic achievement was the design and construction of the first nuclear reactor in India (named Apsara) which went critical on August 4. Elaborate safety measures both in the design and operation of reactors. Pressurised Heavy Water Reactors (PHWRs) built at different sites in the country mark the accomplishment of the first stage of the programme. The Plutonium-239 obtained from reprocessing of the discharged fuel from the reactors then serves as a fuel for the second stage — the fast breeder reactors. DHRUVA and KAMINI.Nuclei INDIA’S ATOMIC ENERGY PROGRAMME The atomic energy programme in India was launched around the time of independence under the leadership of Homi J. which ushered in the technology of fuel reprocessing (separating useful fissile and fertile nuclear materials from the spent fuel of a reactor) in India. and so on. An important milestone in the first two decades of the programme was the indigenous design and construction of the plutonium plant at Trombay. They are so called because they use fast neutrons for sustaining the chain reaction (hence no moderator is needed) and. Following this was another notable landmark: the construction of CIRUS (Canada India Research U. 1956. we can get a steady energy 453 . but fairly abundant reserves of thorium.S. also breed more fissile species (plutonium) than they consume. PURNIMA (I. If the chain reaction is controlled suitably. involves using fast breeder reactors to produce fissile Uranium-233 from Thorium-232 and to build power reactors based on them. Apsara and CIRUS spurred research in a wide range of areas of basic and applied nuclear science. II and III). in some 3. as was first suggested by Enrico Fermi.) reactor in 1960. Research reactors are also an excellent source for production of a variety of radioactive isotopes that find application in diverse fields: industry. Unless the factor K is brought down very close to unity. In addition to control rods.5 shows the schematic diagram of a nuclear reactor based on thermal neutron fission. The core of the reactor is the site of nuclear . The moderators commonly used are water. 92 U These neutrons unless slowed down will escape from the reactor without interacting with the uranium nuclei. It is known experimentally that slow neutrons (thermal neutrons) are much more likely to cause fission in 235 than fast neutrons. which is what we wish it to be for steady power operation. as described here. The reaction rate is controlled through control-rods made out of neutron-absorbing material such as cadmium. it is the measure of the growth rate of the neutrons in the reactor. the operation of the reactor is said to be critical. use heavy water as moderator. the reactor will become supercritical and can even explode.29) Plutonium undergoes fission with slow neutrons. For K = 1. This is the same situation as when a marble hits head-on an identical marble at rest. When it captures a neutron. heavy water (D2O) and graphite. 454 Figure 13. the reaction rate and the reactor power increases exponentially. This is what happens in a nuclear reactor. in naturally occurring uranium is The more abundant isotope 238 92 U non-fissionable. uses water as moderator. The other Indian reactors.aspx Nuclear power plants in India 239 93 – Np → 239 94 Pu + e + ν (13. If K becomes greater than one. This ratio is called the multiplication factor.nic. which are used for power production.Physics output. when required.in/main/AllProjectOperationDisplay. light nuclei called moderators are provided along with the fissionable nuclei for slowing down fast neutrons. If the chain reaction is uncontrolled. The Apsara reactor at the Bhabha Atomic Research Centre (BARC). can be inserted into the reactor and K can be reduced rapidly to less than unity. in reactors. In fact. The explosion of the Chernobyl reactor in Ukraine in 1986 is a sad reminder that accidents in a nuclear reactor can be catastrophic. a hurdle in sustaining a chain reaction. it produces the highly radioactive plutonium through these reactions 238 92 239 – U + n → 239 92 U → 93 Np + e + ν http://www. K. as in a nuclear bomb. Chadwick’s experiments showed that in an elastic collision with hydrogen the neutron almost comes to rest and proton carries away the energy. Therefore.npcil. however. of number of fission produced by a given generation of neutrons to the number of fission of the preceeding generation may be greater than one. What one needs to do is to slow down the fast neutrons by elastic scattering with light nuclei. it is possible that the ratio. reactors are provided with safety rods which. There is. Mumbai. The average energy of a neutron produced in fission of 235 is 2 MeV. unless a very large amount of fissionable material is used for sustaining the chain reaction. Because of the use of moderator. it leads to explosive energy output. Also 92 U fast neutrons liberated in fission would escape instead of causing another fission reaction. 13. A containment vessel prevents the escape of radioactive fission products. nuclear reactors generate considerable waste products.7. The core is surrounded by a reflector to reduce leakage. fission.5 Schematic diagram of a nuclear reactor based on thermal neutron fission.29(c)] 2 1 2 1 3 + n + 3. for example. as seen from the binding energy curve in Fig.kennesaw. energy is released. The core contains a moderator 92 U to slow down the neutrons. But nuclear wastes need special care for treatment since they are radioactive and hazardous..27 MeV H+ 2 1 H → 2 He 3 1 + 4.42 MeV H+ 1 1H → 1H [13.1. The whole assembly is shielded to check harmful radiation from coming out. The steam drives turbines and generates electricity.htm A simplified online simulation of a nuclear reactor FIGURE 13. Some examples of such energy liberating nuclear fusion reactions are : 1 1 2 + e+ + ν + 0.e.29(b)] [13. 13. The fuel may be say enriched uranium (i.Nuclei http://esa21. An appropriate plan is being evolved to study the possibility of converting radioactive waste into less active and shortlived material.03 MeV H+ 2 1H → 1H + 1H 455 . These safety measures are a distinguishing feature of the Indian Atomic Energy programme. since the larger nucleus is more tightly bound. It contains the fuel elements in suitably fabricated form. The coolant transfers heat to a working fluid which in turn may produce stream. one that has greater abundance of 235 than naturally occurring uranium).29(a)] [13. The energy (heat) released in fission is continuously removed by a suitable coolant.edu/activities/nukeenergy/nuke. Elaborate safety measures. both for reactor operation as well as handling and reprocessing the spent fuel. The reactor can be shut down by means of rods (made of. cadmium) that have high absorption of neutrons.3 Nuclear fusion – energy generation in stars When two light nuclei fuse to form a larger nucleus. are required. Like any power reactor. the net effect is − 4 41 1 H + 2e → 2 He + 2ν + 6γ + 26. two deuterons combine to form a triton and a proton. fusion in the sun involves protons whose energies are much above the average energy.42 MeV energy. We can estimate the temperature at which two protons in a proton gas would (averagely) have enough energy to overcome the coulomb barrier: (3/2)k T = K 400 keV. two deuterons combine to form the light isotope of helium. which is considerably less than the estimated temperature required for fusion of particles of average energy. two protons combine to form a deuteron and a positron with a release of 0. The proton-proton (p. It can be shown. Clearly. Thus.7 MeV of energy. When fusion is achieved by raising the temperature of the system so that particles have enough kinetic energy to overcome the coulomb repulsive behaviour. four hydrogen atoms combine to form an release of 26. The star begins to collapse under its own gravity . As the hydrogen in the core gets depleted and becomes helium.02 MeV 2 1 1 3 H+1 H→2 He + γ + 5. Thermonuclear fusion is the source of energy output in the interior of stars. the fuel in the sun is the hydrogen in its core. in which case two light helium nuclei unite to form ordinary helium nucleus.30) e + + e – → γ + γ + 1. which gives T ~ 3 × 109 K. that the barrier height for two protons is ~ 400 keV. For fusion to take place.7 MeV (13. since they are both positively charged particles. therefore.29c). it is called thermonuclear fusion. They. and is higher for nuclei with higher charges. In reaction (13. If we consider the combination 2(i) + 2(ii) + 2(iii) +(iv). the two nuclei must come close enough so that attractive short-range nuclear force is able to affect them.Physics In the first reaction. The height of the barrier depends on the charges and radii of the two interacting nuclei.31) 4 2 Thus. He atom with a 456 Helium is not the only element that can be synthesized in the interior of a star. The fusion reaction in the sun is a multi-step process in which the hydrogen is burned into helium.5×107 K. the first three reactions must occur twice.42 MeV H+ 1 1H → 1H (i) (ii) (iii) (13.86 MeV (iv) He + 3 2 He → 2 He + 1 H + 1 H For the fourth reaction to occur. the core starts to cool. for example. The interior of the sun has a temperature of 1.29(b)]. they experience coulomb repulsion. must have enough energy to overcome this coulomb barrier. In reaction [13. p) cycle by which this occurs is represented by the following sets of reactions: 1 1 2 + e+ + ν + 0.49 MeV 3 2 4 1 1 + 12. However.7 MeV − 4 − or (4 1 1 H + 4e ) → ( 2 He + 2e ) + 2ν + 6γ + 26. 13.43. the aim is to generate steady power by heating the nuclear fuel to a temperature in the range of 108 K. It is estimated that a nuclear arsenal sufficient to destroy every form of life on this planet several times over is in position to be triggered by the press of a button. If this temperature increases to about 108 K. At these temperatures. which will raise the core temperature. 457 .4 Controlled thermonuclear fusion The natural thermonuclear fusion process in a star is replicated in a thermonuclear fusion device. In controlled fusion reactors. Such a nuclear holocaust will not only destroy the life that exists now but its radioactive fallout will make this planet unfit for life for all times.1 cannot be so produced.000 tons of TNT. the fuel is a mixture of positive ions and electrons (plasma). The age of the sun is about 5×109 y and it is estimated that there is enough hydrogen in the sun to keep it going for another 5 billion years. this time of helium nuclei into carbon.7. 1945 an atomic device was used in warfare for the first time. The US dropped an atom bomb on Hiroshima. The explosion was equivalent to 20.Nuclei which increases the temperature of the core. This kind of process can generate through fusion higher and higher mass number elements. If successful. 13.000 were killed and 69. NUCLEAR HOLOCAUST In a single uranium fission about 0. Scenarios based on theoretical calculations predict a long nuclear winter. After that.000 tons of TNT. fusion takes place again. But elements more massive than those near the peak of the binding energy curve in Fig.9×235 MeV (≈200 MeV) of energy is liberated. On August 6.000 inhabitants. Super-explosions equivalent to 10 megatons of explosive power of TNT were tested in 1954. This energy is equivalent to about 20. Japan. If each nucleus of about 50 kg of 235U undergoes fission the amount of energy involved is about 4 × 1015J. Of this number 66. The outer envelope of the sun will expand. as the radioactive waste will hang like a cloud in the earth’s atmosphere and will absorb the sun’s radiation. since no container can stand such a high temperature. enough for a superexplosion. High temperature conditions for fusion reactions can be created by exploding a fission bomb. fusion reactors will hopefully supply almost unlimited power to humanity. Instantly the radioactive products devastated 10 sq km of the city which had 3. Such bombs which involve fusion of isotopes of hydrogen. deuterium and tritium are called hydrogen bombs. more than 67% of the city’s structures were destroyed. turning it into the so called red giant. the hydrogen burning will stop and the sun will begin to cool and will start to collapse under gravity.000 were injured. Several countries around the world including India are developing techniques in this connection. The challenge is to confine this plasma. Uncontrolled release of large nuclear energy is called an atomic explosion. It is in these sense that a nuclear reaction is an example of massenergy interconversion. This is strictly speaking.. we can equally well say that the difference in the total mass of atoms or molecules.] In nuclear reactions (e. Since.g. we say that the difference in the total mass of nuclei on the two sides get converted into energy or vice-versa. since proton number and neutron number are conserved in a nuclear reaction.. incorrect. chemical binding energy also gives a negative contribution (mass defect) to the total mass of an atom or molecule. even this is not strictly true in the realm of very high energies – what is strictly conserved is the total charge and total ‘baryon number’. Eq. Explain. 458 EXAMPLE 13. However.7 Answer the following questions: (a) Are the equations of nuclear reactions (such as those given in Section 13. in what way is mass converted into energy (or vice-versa) in a nuclear reaction? (c) A general impression exists that mass-energy interconversion takes place only in nuclear reaction and never in chemical reaction. the number of protons and the number of neutrons are the same on the two sides of the equation. In a nuclear reaction. (c) From the point of view of mass-energy interconversion. on the two sides of the chemical reaction gets converted into energy or vice-versa.g. (b) We know that the binding energy of a nucleus gives a negative contribution to the mass of the nucleus (mass defect). The energy released or absorbed in a chemical reaction can be traced to the difference in chemical (not nuclear) binding energies of atoms and molecules on the two sides of a reaction. the number of protons and the number of neutrons are both separately conserved in a nuclear reaction. strictly speaking. the mass defects involved in a chemical reaction are almost a million times smaller than those in a nuclear reaction. [Actually. Thus. 13. Solution (a) A chemical equation is balanced in the sense that the number of atoms of each element is the same on both sides of the equation. in what sense are they balanced on both sides? (b) If both the number of protons and the number of neutrons are conserved in each nuclear reaction. a chemical reaction is similar to a nuclear reaction in principle. Now. However.7) ‘balanced’ in the sense a chemical equation (e. Since binding energy contributes to mass. 2H2 + O2→ 2 H2O) is? If not.26). A chemical reaction merely alters the original combinations of atoms.Physics Example 13. the number of atoms of each element is not necessarily conserved in a nuclear reaction. But the total binding energy of nuclei on the left side need not be the same as that on the right hand side.7 . the total rest mass of neutrons and protons is the same on either side of a reaction. The difference in these binding energies appears as energy released or absorbed in a nuclear reaction. We need not pursue this matter here.This is the reason for the general impression. elements may be transmuted. (which is incorrect ) that mass-energy interconversion does not take place in a chemical reaction. Nuclei SUMMARY 1. An atom has a nucleus. The nucleus is positively charged. The radius of the nucleus is smaller than the radius of an atom by a factor of 104. More than 99.9% mass of the atom is concentrated in the nucleus. On the atomic scale, mass is measured in atomic mass units (u). By definition, 1 atomic mass unit (1u) is 1/12th mass of one atom of 12C; 1u = 1.660563 × 10–27 kg. A nucleus contains a neutral particle called neutron. Its mass is almost the same as that of proton The atomic number Z is the number of protons in the atomic nucleus of an element. The mass number A is the total number of protons and neutrons in the atomic nucleus; A = Z+N; Here N denotes the number of neutrons in the nucleus. A 2. 3. 4. A nuclear species or a nuclide is represented as Z X , where X is the chemical symbol of the species. Nuclides with the same atomic number Z, but different neutron number N are called isotopes. Nuclides with the same A are isobars and those with the same N are isotones. Most elements are mixtures of two or more isotopes. The atomic mass of an element is a weighted average of the masses of its isotopes. The masses are the relative abundances of the isotopes. 5. A nucleus can be considered to be spherical in shape and assigned a radius. Electron scattering experiments allow determination of the nuclear radius; it is found that radii of nuclei fit the formula R = R0 A1/3, where R0 = a constant = 1.2 fm. This implies that the nuclear density is independent of A. It is of the order of 1017 kg/m3. 6. Neutrons and protons are bound in a nucleus by the short-range strong nuclear force. The nuclear force does not distinguish between neutron and proton. 7. The nuclear mass M is always less than the total mass, Σm, of its constituents. The difference in mass of a nucleus and its constituents is called the mass defect, ∆ M = ( Z m p + ( A – Z ) mn ) – M Using Einstein’s mass energy relation, we express this mass difference in terms of energy as ∆Eb = ∆M c2 The energy ∆Eb represents the binding energy of the nucleus. In the mass number range A = 30 to 170, the binding energy per nucleon is nearly constant, about 8 MeV/nucleon. 8. Energies associated with nuclear processes are about a million times larger than chemical process. 9. The Q-value of a nuclear process is Q = final kinetic energy – initial kinetic energy. Due to conservation of mass-energy, this is also, Q = (sum of initial masses – sum of final masses)c2 10. Radioactivity is the phenomenon in which nuclei of a given species transform by giving out α or β or γ rays; α-rays are helium nuclei; 459 Physics β-rays are electrons. γ-rays are electromagnetic radiation of wavelengths shorter than X-rays; 11. Law of radioactive decay : N (t) = N(0) e–λt where λ is the decay constant or disintegration constant. The half-life T1/2 of a radionuclide is the time in which N has been reduced to one-half of its initial value. The mean life τ is the time at which N has been reduced to e–1 of its initial value T1/2 = ln 2 λ = τ ln 2 235 92 12. Energy is released when less tightly bound nuclei are transmuted into more tightly bound nuclei. In fission, a heavy nucleus like into two smaller fragments, e.g., 235 92 U breaks U+ n → 1 0 133 51 Sb + 99 41 Nb + 4 1 0n 13. The fact that more neutrons are produced in fission than are consumed gives the possibility of a chain reaction with each neutron that is produced triggering another fission. The chain reaction is uncontrolled and rapid in a nuclear bomb explosion. It is controlled and steady in a nuclear reactor. In a reactor, the value of the neutron multiplication factor k is maintained at 1. 14. In fusion, lighter nuclei combine to form a larger nucleus. Fusion of hydrogen nuclei into helium nuclei is the source of energy of all stars including our sun. Physical Quantity Atomic mass unit Symbol Dimensions [M] Units u Remarks Unit of mass for expressing atomic or nuclear masses. One atomic mass unit equals 1/12th of the mass of 12C atom. Disintegration or decay constant Half-life λ T1/2 [T –1] [T ] s–1 s Time taken for the decay of one-half of the initial number of nuclei present in a radioactive sample. Time at which number of nuclei has been reduced to e–1 of its initial value Measure of the activity of a radioactive source. Mean life τ [T ] s Activity of a radioactive sample R [ T–1 ] Bq POINTS TO PONDER 1. 2. The density of nuclear matter is independent of the size of the nucleus. The mass density of the atom does not follow this rule. The radius of a nucleus determined by electron scattering is found to be slightly different from that determined by alpha-particle scattering. 460 Nuclei This is because electron scattering senses the charge distribution of the nucleus, whereas alpha and similar particles sense the nuclear matter. 3. After Einstein showed the equivalence of mass and energy, E = mc 2, we cannot any longer speak of separate laws of conservation of mass and conservation of energy, but we have to speak of a unified law of conservation of mass and energy. The most convincing evidence that this principle operates in nature comes from nuclear physics. It is central to our understanding of nuclear energy and harnessing it as a source of power. Using the principle, Q of a nuclear process (decay or reaction) can be expressed also in terms of initial and final masses. The nature of the binding energy (per nucleon) curve shows that exothermic nuclear reactions are possible, when two light nuclei fuse or when a heavy nucleus undergoes fission into nuclei with intermediate mass. For fusion, the light nuclei must have sufficient initial energy to overcome the coulomb potential barrier. That is why fusion requires very high temperatures. Although the binding energy (per nucleon) curve is smooth and slowly varying, it shows peaks at nuclides like 4He, 16O etc. This is considered as evidence of atom-like shell structure in nuclei. Electrons and positron are a particle-antiparticle pair. They are identical in mass; their charges are equal in magnitude and opposite. ( It is found that when an electron and a positron come together, they annihilate each other giving energy in the form of gamma-ray photons.) In â--decay (electron emission), the particle emitted along with electron is anti-neutrino ( ν ). On the other hand, the particle emitted in β+decay (positron emission) is neutrino (ν). Neutrino and anti-neutrino are a particle-antiparticle pair. There are anti particles associated with every particle. What should be antiproton which is the anti particle of the proton? A free neutron is unstable ( n → p + e – + ν ). But a similar free proton decay is not possible, since a proton is (slightly) lighter than a neutron. 4. 5. 6. 7. 8. 9. 10. Gamma emission usually follows alpha or beta emission. A nucleus in an excited (higher) state goes to a lower state by emitting a gamma photon. A nucleus may be left in an excited state after alpha or beta emission. Successive emission of gamma rays from the same nucleus (as in case of 60Ni, Fig. 13.4) is a clear proof that nuclei also have discrete energy levels as do the atoms. 11. Radioactivity is an indication of the instability of nuclei. Stability requires the ratio of neutron to proton to be around 1:1 for light nuclei. This ratio increases to about 3:2 for heavy nuclei. (More neutrons are required to overcome the effect of repulsion among the protons.) Nuclei which are away from the stability ratio, i.e., nuclei which have an excess of neutrons or protons are unstable. In fact, only about 10% of knon isotopes (of all elements), are stable. Others have been either artificially produced in the laboratory by bombarding α, p, d, n or other particles on targets of stable nuclear species or identified in astronomical observations of matter in the universe. 461 Physics EXERCISES You may find the following data useful in solving the exercises: e = 1.6×10–19C N = 6.023×1023 per mole 9 2 2 1/(4πε0) = 9 × 10 N m /C k = 1.381×10–23J 0K–1 1 MeV = 1.6×10–13J 1 year = 3.154×107 s mH = 1.007825 u m( He ) = 4.002603 u 13.1 4 2 1 u = 931.5 MeV/c2 mn = 1.008665 u me = 0.000548 u and 7 have respective (a) Two stable isotopes of lithium 6 3 Li 3 Li abundances of 7.5% and 92.5%. These isotopes have masses 6.01512 u and 7.01600 u, respectively. Find the atomic mass of lithium. (b) Boron has two stable isotopes, 10 and 11 . Their respective 5B 5B masses are 10.01294 u and 11.00931 u, and the atomic mass of and 11 . boron is 10.811 u. Find the abundances of 10 5 B 5B 20 21 22 The three stable isotopes of neon: 10 Ne, 10 Ne and 10 Ne have respective abundances of 90.51%, 0.27% and 9.22%. The atomic masses of the three isotopes are 19.99 u, 20.99 u and 21.99 u, respectively. Obtain the average atomic mass of neon. 13.2 13.3 13.4 Obtain the binding energy (in MeV) of a nitrogen nucleus given m 14 7 N =14.00307 u ( ) ( 14 7 N , ) Obtain the binding energy of the nuclei MeV from the following data: ) = 55.934939 u m ( 56 26 Fe 56 26 Fe and 209 83 Bi in units of m ( 209 ) = 208.980388 u 83 Bi 13.5 A given coin has a mass of 3.0 g. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity assume that the coin is entirely made of 63 atoms (of mass 62.92960 u). 29 Cu Write nuclear reaction equations for (i) α-decay of – + 226 88 13.6 Ra P (ii) α-decay of (iv) β -decay of (vi) β -decay of 120 54 242 94 Pu Bi (iii) β -decay of (v) β -decay of 13.7 13.8 32 15 – 210 83 11 6 C + 97 43 Tc 462 Xe A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to a) 3.125%, b) 1% of its original value? The normal activity of living carbon-containing matter is found to be about 15 decays per minute for every gram of carbon. This activity arises from the small proportion of radioactive 14 present with the 6C 12 stable carbon isotope 6 C . When the organism is dead, its interaction with the atmosphere (which maintains the above equilibrium activity) ceases and its activity begins to drop. From the known half-life (5730 years) of 14 , and the measured activity, the age of the specimen 6C can be approximately estimated. This is the principle of 14 dating 6C (vii) Electron capture of Nuclei used in archaeology. Suppose a specimen from Mohenjodaro gives an activity of 9 decays per minute per gram of carbon. Estimate the approximate age of the Indus-Valley civilisation. 13.9 necessary to provide a radioactive source Obtain the amount of 60 27 Co of 8.0 mCi strength. The half-life of 60 is 5.3 years. 27 Co is 28 years. What is the disintegration rate of 13.10 The half-life of 90 38 Sr 15 mg of this isotope? 13.11 Obtain approximately the ratio of the nuclear radii of the gold isotope 197 and the silver isotope 107 . 79 Au 47 Ag 13.12 Find the Q-value and the kinetic energy of the emitted α-particle in and (b) 220 . the α-decay of (a) 226 88 Ra 86 Rn ) = 226.02540 u, Given m ( 226 88 Ra m ( 222 ) = 220.01137 u, 86 Rn 13.13 The radionuclide 11 6 m ( 222 ) = 222.01750 u, 86 Rn m ( 216 ) = 216.00189 u. 84 Po 11 C decays according to C→ 11 5 B + e +ν : T1/2 =20.3 min + The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m ( 11 ) = 11.011434 u and m ( 11 ) = 11.009305 u, 6C 6B calculate Q and compare it with the maximum energy of the positron emitted. – 23 13.14 The nucleus 10 Ne decays by β emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that: 23 m ( 10 Ne ) = 22.994466 u 23 m ( 11 Na ) = 22.089770 u. 13.15 The Q value of a nuclear reaction A + b → C + d is defined by Q = [ mA + mb – mC – md]c2 where the masses refer to the respective nuclei. Determine from the given data the Q-value of the following reactions and state whether the reactions are exothermic or endothermic. 3 2 2 (i) 1 1 H+1 H →1 H+1 H (ii) m m m m 12 6 20 4 C+12 6 C →10 Ne+2 He Atomic masses are given to be 2 (1 H ) = 2.014102 u 3 ( 1 H ) = 3.016049 u ( 12 ) = 12.000000 u 6C 20 ( 10 Ne ) = 19.992439 u nucleus into two equal 13.16 Suppose, we think of fission of a 56 26 Fe 28 fragments, 13 . Is the fission energetically possible? Argue by Al 56 working out Q of the process. Given m ( 26 Fe ) = 55.93494 u and 28 m ( 13 Al ) = 27.98191 u. 13.17 The fission properties of 239 are very similar to those of 235 . The 94 Pu 92 U average energy released per fission is 180 MeV. How much energy, 463 Physics in MeV, is released if all the atoms in 1 kg of pure 239 undergo 94 Pu fission? 13.18 A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much 235 did it contain initially? Assume that the reactor operates 92 U 80% of the time, that all the energy generated arises from the fission and that this nuclide is consumed only by the fission process. of 235 92 U 13.19 How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as 2 1 3 H+ 2 1 H → 2 He +n +3.27 MeV 13.20 Calculate the height of the potential barrier for a head on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.) 13.21 From the relation R = R0 A1/3, where R 0 is a constant and A is the mass number of a nucleus, show that the nuclear matter density is nearly constant (i.e. independent of A). 13.22 For the β+ (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K–shell, is captured by the nucleus and a neutrino is emitted). A A e+ + Z X → Z− Y +ν 1 Show that if β+ emission is energetically allowed, electron capture is necessarily allowed but not vice–versa. ADDITIONAL EXERCISES 13.23 In a periodic table the average atomic mass of magnesium is given as 24.312 u. The average value is based on their relative natural abundance on earth. The three isotopes and their masses are 24 12 Mg 26 (23.98504u), 25 (24.98584u) and (25.98259u). The natural Mg Mg 12 12 abundance of 24 is 78.99% by mass. Calculate the abundances 12 Mg of other two isotopes. 13.24 The neutron separation energy is defined as the energy required to remove a neutron from the nucleus. Obtain the neutron separation 41 from the following data: energies of the nuclei 20 Ca and 27 13 Al m ( 40 ) = 39.962591 u 20 Ca 41 m ( 20 Ca ) = 40.962278 u ) = 25.986895 u m ( 26 13 Al m ( 27 ) = 26.981541 u 13 Al 13.25 A source contains two phosphorous radio nuclides 32 (T1/2 = 14.3d) 15 P ( T = 25.3d). Initially, 10% of the decays come from 33 . and 33 P 15 15 P 1/2 How long one must wait until 90% do so? 13.26 Under certain circumstances, a nucleus can decay by emitting a particle more massive than an α-particle. Consider the following decay processes: 464 223 88 14 Ra → 209 82 Pb + 6 C Nuclei 223 88 4 Ra → 219 86 Rn + 2 He Calculate the Q-values for these decays and determine that both are energetically allowed. 13.27 Consider the fission of 238 by fast neutrons. In one fission event, 92 U no neutrons are emitted and the final end products, after the beta 99 and 44 decay of the primary fragments, are 140 Ru . Calculate Q 58 Ce for this fission process. The relevant atomic and particle masses are m ( 238 ) =238.05079 u 92 U m ( 140 ) =139.90543 u 58 Ce 99 m ( 44 Ru ) = 98.90594 u 13.28 Consider the D–T reaction (deuterium–tritium fusion) 2 1 4 H+ 3 1 H → 2 He + n (a) Calculate the energy released in MeV in this reaction from the data: m(2 )=2.014102 u 1H ) =3.016049 u m(3 1H (b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.) 13.29 Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. 13.6. You are given that m (198Au) = 197.968233 u m (198Hg) =197.966760 u FIGURE13.6 465 Physics 13.30 Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of 235U in a fission reactor. 13.31 Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e. conversion to electric energy) of thermal energy produced in a reactor was 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of 235U to be about 200MeV. 466 Chapter Fourteen SEMICONDUCTOR ELECTRONICS: MATERIALS, DEVICES AND SIMPLE CIRCUITS 14.1 INTRODUCTION Devices in which a controlled flow of electrons can be obtained are the basic building blocks of all the electronic circuits. Before the discovery of transistor in 1948, such devices were mostly vacuum tubes (also called valves) like the vacuum diode which has two electrodes, viz., anode (often called plate) and cathode; triode which has three electrodes – cathode, plate and grid; tetrode and pentode (respectively with 4 and 5 electrodes). In a vacuum tube, the electrons are supplied by a heated cathode and the controlled flow of these electrons in vacuum is obtained by varying the voltage between its different electrodes. Vacuum is required in the inter-electrode space; otherwise the moving electrons may lose their energy on collision with the air molecules in their path. In these devices the electrons can flow only from the cathode to the anode (i.e., only in one direction). Therefore, such devices are generally referred to as valves. These vacuum tube devices are bulky, consume high power, operate generally at high voltages (~100 V ) and have limited life and low reliability. The seed of the development of modern solid-state semiconductor electronics goes back to 1930’s when it was realised that some solidstate semiconductors and their junctions offer the possibility of controlling the number and the direction of flow of charge carriers through them. Simple excitations like light, heat or small applied voltage can change the number of mobile charges in a semiconductor. Note that the supply Physics and flow of charge carriers in the semiconductor devices are within the solid itself, while in the earlier vacuum tubes/valves, the mobile electrons were obtained from a heated cathode and they were made to flow in an evacuated space or vacuum. No external heating or large evacuated space is required by the semiconductor devices. They are small in size, consume low power, operate at low voltages and have long life and high reliability. Even the Cathode Ray Tubes (CRT) used in television and computer monitors which work on the principle of vacuum tubes are being replaced by Liquid Crystal Display (LCD) monitors with supporting solid state electronics. Much before the full implications of the semiconductor devices was formally understood, a naturally occurring crystal of galena (Lead sulphide, PbS) with a metal point contact attached to it was used as detector of radio waves. In the following sections, we will introduce the basic concepts of semiconductor physics and discuss some semiconductor devices like junction diodes (a 2-electrode device) and bipolar junction transistor (a 3-electrode device). A few circuits illustrating their applications will also be described. 14.2 CLASSIFICATION OF METALS, CONDUCTORS SEMICONDUCTORS AND 468 On the basis of conductivity On the basis of the relative values of electrical conductivity (σ ) or resistivity ( ρ = 1/σ ), the solids are broadly classified as: (i) Metals: They possess very low resistivity (or high conductivity). ρ ~ 10–2 – 10–8 Ω m σ ~ 102 – 108 S m–1 (ii) Semiconductors: They have resistivity or conductivity intermediate to metals and insulators. ρ ~ 10–5 – 106 Ω m σ ~ 105 – 10–6 S m–1 (iii) Insulators: They have high resistivity (or low conductivity). ρ ~ 1011 – 1019 Ω m σ ~ 10–11 – 10–19 S m–1 The values of ρ and σ given above are indicative of magnitude and could well go outside the ranges as well. Relative values of the resistivity are not the only criteria for distinguishing metals, insulators and semiconductors from each other. There are some other differences, which will become clear as we go along in this chapter. Our interest in this chapter is in the study of semiconductors which could be: (i) Elemental semiconductors: Si and Ge (ii) Compound semiconductors: Examples are: • Inorganic: CdS, GaAs, CdSe, InP, etc. • Organic: anthracene, doped pthalocyanines, etc. • Organic polymers: polypyrrole, polyaniline, polythiophene, etc. Most of the currently available semiconductor devices are based on elemental semiconductors Si or Ge and compound inorganic Semiconductor Electronics: Materials, Devices and Simple Circuits semiconductors. However, after 1990, a few semiconductor devices using organic semiconductors and semiconducting polymers have been developed signalling the birth of a futuristic technology of polymerelectronics and molecular-electronics. In this chapter, we will restrict ourselves to the study of inorganic semiconductors, particularly elemental semiconductors Si and Ge. The general concepts introduced here for discussing the elemental semiconductors, by-and-large, apply to most of the compound semiconductors as well. On the basis of energy bands According to the Bohr atomic model, in an isolated atom the energy of any of its electrons is decided by the orbit in which it revolves. But when the atoms come together to form a solid they are close to each other. So the outer orbits of electrons from neighbouring atoms would come very close or could even overlap. This would make the nature of electron motion in a solid very different from that in an isolated atom. Inside the crystal each electron has a unique position and no two electrons see exactly the same pattern of surrounding charges. Because of this, each electron will have a different energy level. These different energy levels with continuous energy variation form what are called energy bands. The energy band which includes the energy levels of the valence electrons is called the valence band. The energy band above the valence band is called the conduction band. With no external energy, all the valence electrons will reside in the valence band. If the lowest level in the conduction band happens to be lower than the highest level of the valence band, the electrons from the valence band can easily move into the conduction band. Normally the conduction band is empty. But when it overlaps on the valence band electrons can move freely into it. This is the case with metallic conductors. If there is some gap between the conduction band and the valence band, electrons in the valence band all remain bound and no free electrons are available in the conduction band. This makes the material an insulator. But some of the electrons from the valence band may gain external energy to cross the gap between the conduction band and the valence band. Then these electrons will move into the conduction band. At the same time they will create vacant energy levels in the valence band where other valence electrons can move. Thus the process creates the possibility of conduction due to electrons in conduction band as well as due to vacancies in the valence band. Let us consider what happens in the case of Si or Ge crystal containing N atoms. For Si, the outermost orbit is the third orbit (n = 3), while for Ge it is the fourth orbit (n = 4). The number of electrons in the outermost orbit is 4 (2s and 2p electrons). Hence, the total number of outer electrons in the crystal is 4N. The maximum possible number of electrons in the outer orbit is 8 (2s + 6p electrons). So, for the 4N valence electrons there are 8N available energy states. These 8N discrete energy levels can either form a continuous band or they may be grouped in different bands depending upon the distance between the atoms in the crystal (see box on Band Theory of Solids). At the distance between the atoms in the crystal lattices of Si and Ge, the energy band of these 8N states is split apart into two which are separated by an energy gap Eg (Fig. 14.1). The lower band which is 469 Physics completely occupied by the 4N valence electrons at temperature of absolute zero is the valence band. The other band consisting of 4N energy states, called the conduction band, is completely empty at absolute zero. BAND THEORY OF SOLIDS Consider that the Si or Ge crystal contains N atoms. Electrons of each atom will have discrete energies in different orbits. The electron energy will be same if all the atoms are isolated, i.e., separated from each other by a large distance. However, in a crystal, the atoms are close to each other (2 to 3 Å) and therefore the electrons interact with each other and also with the neighbouring atomic cores. The overlap (or interaction) will be more felt by the electrons in the outermost orbit while the inner orbit or core electron energies may remain unaffected. Therefore, for understanding electron energies in Si or Ge crystal, we need to consider the changes in the energies of the electrons in the outermost orbit only. For Si, the outermost orbit is the third orbit (n = 3), while for Ge it is the fourth orbit (n = 4). The number of electrons in the outermost orbit is 4 (2s and 2p electrons). Hence, the total number of outer electrons in the crystal is 4N. The maximum possible number of outer electrons in the orbit is 8 (2s + 6p electrons). So, out of the 4N electrons, 2N electrons are in the 2N s-states (orbital quantum number l = 0) and 2N electrons are in the available 6N p-states. Obviously, some p-electron states are empty as shown in the extreme right of Figure. This is the case of well separated or isolated atoms [region A of Figure]. Suppose these atoms start coming nearer to each other to form a solid. The energies of these electrons in the outermost orbit may change (both increase and decrease) due to the interaction between the electrons of different atoms. The 6N states for l = 1, which originally had identical energies in the isolated atoms, spread out and form an energy band [region B in Figure]. Similarly, the 2N states for l = 0, having identical energies in the isolated atoms, split into a second band (carefully see the region B of Figure) separated from the first one by an energy gap. At still smaller spacing, however, there comes a region in which the bands merge with each other. The lowest energy state that is a split from the upper atomic level appears to drop below the upper state that has come from the lower atomic level. In this region (region C in Figure), no energy gap exists where the upper and lower energy states get mixed. Finally, if the distance between the atoms further decreases, the energy bands again split apart and are separated by an energy gap Eg (region D in Figure). The total number of available energy states 8N has been re-apportioned between the two bands (4N states each in the lower and upper energy bands). Here the significant point is that there are exactly as many states in the lower band (4N ) as there are available valence electrons from the atoms (4N ). Therefore, this band ( called the valence band ) is completely filled while the upper band is completely empty. The upper band is called the conduction band. 470 Devices and Simple Circuits The lowest energy level in the conduction band is shown as EC and highest energy level in the valence band is shown as EV .2(a). called the is partially empty or when the conduction and valance bands overlap. It may be large. are depicted in Fig. When there conduction band. the resistance of such materials is low or the conductivity is high. FIGURE 14. These different situations. 14. small.Semiconductor Electronics: Materials. Therefore. (b) insulators and (c) semiconductors. depending upon the material. is overlap electrons from valence band can called the valence band. 14.2 and discussed below: Case I: This refers to a situation. 14. as shown in Fig. This situation makes a large number of electrons available for electrical conduction. When the valence band is partially empty.1. completely filled energy states. consists of closely spaced easily move into the conduction band. One can have a metal either when the conduction band FIGURE 14. electrons from its lower level can move to higher level making conduction possible. as shown in Fig. The upper band. consists of infinitely large number of closely spaced energy states. or zero. Above EC and below EV there are a large number of closely spaced energy levels. 471 . The lower band.1 The energy band positions in a is partially filled and the balanced band semiconductor at 0 K.2 Difference between energy bands of (a) metals. The gap between the top of the valence band and bottom of the conduction band is called the energy band gap (Energy gap E g ). This vacancy with the effective positive electronic charge is called a hole.1) Silicon or Germanium with e h where n is called intrinsic carrier concentration. Because of the small band gap. Each atom is surrounded by four nearest neighbours. ne is equal to the number of holes. a large band gap Eg exists (Eg > 3 eV).56. the holes also move. This is the case of insulators. from which the free electron (with charge –q ) has come out leaves a vacancy with an effective charge (+q ). There are no electrons in the conduction band. more thermal energy becomes available to these electrons and some of these electrons may break–away (becoming free electrons contributing to conduction). 472 which. The hole behaves as an apparent free particle with effective positive charge. the resistance of semiconductors is not as high as that of the insulators.66 Å. As the temperature increases.Physics Case II: In this case. 14.2(c). Case III: This situation is shown in Fig.3 Three-dimensional diaelectrons. 5. and therefore no electrical conduction is possible.3 which overemphasises the covalent bond. 14. The two shared electrons can be assumed to shuttle back-andforth between the associated atoms holding them together strongly.3.4 schematically shows the 2-dimensional representation of Si or Ge structure shown in Fig.2(b). n = n = ni (14. The neighbourhood.3 INTRINSIC SEMICONDUCTOR We shall take the most common case of Ge and Si whose lattice structure is shown in Fig. Note that the energy gap is so large that electrons cannot be excited from the valence band to the conduction band by thermal excitation. That is mond-like crystal structure for Carbon. every Si or Ge atom tends to share one of its four valence electrons with each of its four nearest neighbour atoms. and also to take share of one electron from each such neighbour. respective lattice spacing a equal i Semiconductors posses the unique property in to 3. These structures are called the diamond-like structures. Such a situation arises at low temperatures. conductors and semiconductors. These electrons (though small in numbers) can move in the conduction band. Here a finite but small band gap (Eg < 3 eV) exists. . These shared electron pairs are referred to as forming a covalent bond or simply a valence bond. 14. nh. In the section which follows you will learn the conduction process in semiconductors.5(a). In its crystalline structure. In this section we have made a broad classification of metals. apart from electrons. 14. In intrinsic semiconductors.43 and 5. Figure 14. It shows an idealised picture in which no bonds are broken (all bonds are intact). the number of free FIGURE 14. 14. We know that Si and Ge have four valence electrons. Hence. as shown in Fig. at room temperature some electrons from valence band can acquire enough energy to cross the energy gap and enter the conduction band. The thermal energy effectively ionises only a few atoms in the crystalline lattice and creates a vacancy in the bond as shown in Fig. 14. The total current. +4 symbol the crystal. 14. 14. An electron from the covalent bond at site 2 may jump to the vacant site 1 (hole).5(a)] is not involved in this process of hole motion. The recombination occurs due to an electron colliding with a hole. leaving a hole at its site indicating an 473 apparent movement of the hole from site 1 to site 2. Remember that the motion of hole is only a convenient way of FIGURE 14. Ie under an applied electric field. The movement of holes can be visualised as shown in Fig.Semiconductor Electronics: Materials. Devices and Simple Circuits Suppose there is a hole at site 1 as shown in Fig.5(a). the hole is at site 2 and the site 1 has now an electron. a simultaneous process of recombination occurs in which the electrons recombine with the holes. At equilibrium. Therefore.5(b).4 Schematic two-dimensional describing the actual motion of bound electrons. apparently. after such a jump. (a) (b) FIGURE 14. the rate of generation is equal to the rate of recombination of charge carriers. The electron from the lower left hand covalent bond (site 2) goes to the earlier hole site1. these holes move towards negative potential giving the hole current. (b) Simplified representation of possible thermal motion of a hole.2) It may be noted that apart from the process of generation of conduction electrons and holes. 14.5 (a) Schematic model of generation of hole at site 1 and conduction electron due to thermal energy at moderate temperatures. The free electron moves completely independently as conduction electron and gives rise to an electron current. . indicates inner cores of Si or Ge. Under the action of an electric field. Note that the electron originally set free [Fig. I is thus the sum of the electron current Ie and the hole current Ih: I = Ie + Ih (14. Ih. the hole has moved from site 1 to site 2. representation of Si or Ge structure showing whenever there is an empty bond anywhere in covalent bonds at low temperature (all bonds intact). Thus. . (b) At T > 0 K. Hence. number of free electrons for conduction in Ge and Si are significant but negligibly small for C.6(b). in the second. say. Example 14.6 (a) An intrinsic semiconductor at T = 0 K behaves like insulator. the conductivity of the semiconductor is increased manifold. Phosphorous (P). a few parts per million (ppm). There are two types of dopants used in doping the tetravalent Si or Ge: (i) Pentavalent (valency 5). etc. It is the thermal energy at higher temperatures (T > 0K). Si or Ge lie. some electrons are shown in the conduction band. Why is C insulator while Si and Ge intrinsic semiconductors? EXAMPLE 14. As such.e. energy required to take out an electron from these atoms (i. of a suitable impurity is added to the pure semiconductor. The deliberate addition of a desirable impurity is called doping and the impurity atoms are called dopants. Therefore. Hence. When a small amount. Hence there is a necessity of improving their conductivity. A necessary condition to attain this is that the sizes of the dopant and the semiconductor atoms should be nearly the same.1 C. Such materials are known as extrinsic semiconductors or impurity semiconductors. This can be done by making use of impurities.Physics An intrinsic semiconductor will behave like an insulator at T = 0 K as shown in Fig. like Arsenic (As). These thermally excited electrons at T > 0 K. Such a material is also called a doped semiconductor. respectively. which excites some electrons from the valence band to the conduction band. The filled circles ( ) represent electrons and empty fields ( ) represent holes. 14. third and fourth orbit.1 Solution The 4 bonding electrons of C. FIGURE 14. 14. followed by Si and highest for C. Here. 474 . no important electronic devices can be developed using these semiconductors.6(a). The dopant has to be such that it does not distort the original pure semiconductor lattice. 14. Si and Ge have same lattice structure. four thermally generated electron-hole pairs. Antimony (Sb). It occupies only a very few of the original semiconductor atom sites in the crystal. ionisation energy Eg ) will be least for Ge. the energy-band diagram of an intrinsic semiconductor will be as shown in Fig.4 EXTRINSIC SEMICONDUCTOR The conductivity of an intrinsic semiconductor depends on its temperature. but at room temperature its conductivity is very low. partially occupy the conduction band. These have come from the valence band leaving equal number of holes there. Si or Ge belongs to the fourth group in the Periodic table and. the number of holes would get reduced further. Sb. For example. Hence in an extrinsic . Thus.7. On the other hand. the pentavalent and trivalent dopants in Si or Ge give two entirely different types of semiconductors as discussed below. But the rate of recombination of holes would increase due to the increase in the number of electrons.72 eV for germanium and about 1. four of its electrons bond with the four silicon neighbours while the fifth remains very weakly bound to its parent atom. As a result. and (b) Commonly used the ionisation energy required to set this schematic representation of n-type material electron free is very small and even at room which shows only the fixed cores of the temperature it will be free to move in the substituent donors with one additional effective lattice of the semiconductor. This is in contrast to the energy required to jump the forbidden band (about 0. This is because the four electrons participating in bonding are seen as part of the effective core FIGURE 14. like Indium (In).) doped for tetravalent Si or Ge giving nof the atom by the fifth electron. we choose the dopant element from nearby fifth or third group. In a doped semiconductor the total number of conduction electrons ne is due to the electrons contributed by donors and those generated intrinsically. while the total number of holes nh is only due to the holes from the intrinsic source.1 eV for silicon) at room temperature in the intrinsic semiconductor. The number of electrons made available for conduction by dopant atoms depends strongly upon the doping level and is independent of any increase in ambient temperature. expecting and taking care that the size of the dopant atom is nearly the same as that of Si or Ge. Aluminium (Al). the number of free electrons (with an equal number of holes) generated by Si atoms. with proper level of doping the number of conduction electrons 475 can be made much larger than the number of holes.05 eV for silicon. Boron (B). energy required is ~ 0. etc. Interestingly. When an atom of +5 valency element occupies the position of an atom in the crystal lattice of Si. and 0.01 eV for germanium. etc. therefore. As a result type semiconductor. We shall now discuss how the doping changes the number of charge carriers (and hence the conductivity) of semiconductors. to separate this electron from its atom. P. Devices and Simple Circuits (ii) Trivalent (valency 3). (i) n-type semiconductor Suppose we dope Si or Ge with a pentavalent element as shown in Fig. increases weakly with temperature.Semiconductor Electronics: Materials. the pentavalent dopant is donating one extra electron for conduction and hence is known as donor impurity. the positive charge and its associated extra electron. Thus.7 (a) Pentavalent donor atom (As. 14. In. Similarly. For doped in tetravalent Si or Ge p-type semiconductors. ne >> nh (14. for such a material. Since the neighbouring Si atom in the lattice wants an electron in place of a hole.4) p-type material which shows only the fixed core of the Note that the crystal maintains an overall charge neutrality substituent acceptor with as the charge of additional charge carriers is just equal and one effective additional opposite to that of the ionised cores in the lattice. therefore. So the conduction band will have most electrons 476 coming from the donor impurities. the recombination process will reduce lattice giving p-type semiconthe number (ni )of intrinsically generated electrons to ne. 14. Thus. leaving a vacancy or hole at its own site. etc. Therefore. Note that the trivalent foreign atom becomes effectively negatively charged when it shares fourth electron with neighbouring Si atom. because of the abundance of associated hole. The dopant has one valence electron less than Si or Ge and. We ductor. So the bond between the fourth neighbour and the trivalent atom has a vacancy or hole as shown in Fig.8 (a) Trivalent are minority carriers. we have. negative charge and its In extrinsic semiconductors. known as n -type semiconductors. the dopant atom of p-type material can be treated as core of one negative charge along with its associated hole as shown in Fig.9(a). this atom can form covalent bonds with neighbouring three Si atoms but does not have any electron to offer to the fourth Si atom. indirectly helps to reduce the intrinsic concentration of minority carriers. majority current carriers. In the energy band diagram of n-type Si semiconductor. B. the holes are the majority carriers and electrons FIGURE 14. as shown in Fig. 14. additional energy states due to donor impurities (ED ) and acceptor impurities (EA ) also exist. which become the majority carriers. Thus the hole is available for conduction. B etc. Hence. The semiconductor’s energy band structure is affected by doping. for p-type semiconductors schematic representation of nh >> ne (14. These holes are in addition to the intrinsically generated holes while the source of conduction electrons is only intrinsic generation.) with trivalent impurity are called p-type semiconductors.3) (ii) p-type semiconductor This is obtained when Si or Ge is doped with a trivalent impurity like Al. electrons become the majority carriers and holes the minority carriers. (b) Commonly used have. 14. For n-type semiconductors. by adding a large number of current carriers of one type. These semiconductors are. At room temperature. Al. It is obvious that one acceptor atom gives one hole. . therefore. the minority carriers produced thermally have more chance of meeting majority carriers and thus getting destroyed.8. Therefore. In the case of extrinsic semiconductors. extrinsic semiconductors doped acceptor atom (In. most of the donor atoms get ionised but very few (~10–12) atoms of Si get ionised. the donor energy level ED is slightly below the bottom EC of the conduction band and electrons from this level move into the conduction band with very small supply of energy. the dopant.Physics semiconductor doped with pentavalent impurity. an electron in the outer orbit of an atom in the neighbourhood may jump to fill this vacancy.8(b). Solution Note that thermally generated electrons (ni ~1016 m–3 ) are negligibly small as compared to those produced by doping. Therefore.9(b). (Alternately. FIGURE 14. With very small supply of energy an electron from the valence band can jump to the level EA and ionise the acceptor negatively. 14. Given that ni =1.5) Though the above description is grossly approximate and hypothetical.7 eV. 1. Calculate the number of electrons and holes. (b) p-type semiconductor at T > 0K. the acceptor energy level EA is slightly above the top EV of the valence band as shown in Fig.2 477 .4 eV.) At room temperature. Sn also is a group IV element but it is a metal because the energy gap in its case is 0 eV. insulators and semiconductors (extrinsic and intrinsic) in a simple manner.5 × 1016 m–3. Since nenh = ni2. Si and Ge. For C (diamond).2 Suppose a pure Si crystal has 5 × 1028 atoms m–3. The electron and hole concentration in a semiconductor in thermal equilibrium is given by nenh = ni2 (14.1 eV and 0. The difference in the resistivity of C. it helps in understanding the difference between metals. most of the acceptor atoms get ionised leaving holes in the valence band. Example 14. Si and Ge depends upon the energy gap between their conduction and valence bands.5 × 109 m–3 EXAMPLE 14. ne ≈ ND. Electrons rise up and holes fall down when they gain external energy.9 Energy bands of (a) n-type semiconductor at T > 0K. Thus at room temperature the density of holes in the valence band is predominantly due to impurity in the extrinsic semiconductor. we can also say that with very small supply of energy the hole from level EA sinks down into the valence band. The number of holes nh = (2. It is doped by 1 ppm concentration of pentavalent As.Semiconductor Electronics: Materials. Devices and Simple Circuits for p-type semiconductor. the energy gaps are 5.25 × 1032 )/(5 ×1022 ) ~ 4. respectively. Similarly. 478 current (Fig. it leaves behind an ionised acceptor (negative charge) which is immobile. in a p-type semiconductor.1 p-n junction formation Consider a thin p-type silicon (p-Si) semiconductor wafer. This motion of charge carries gives rise to diffusion current across the junction. an electron on p-side of the junction moves to n-side and a hole on nside of the junction moves to p-side. 14.10) starts.phy-astr. Due to the positive space-charge region on n-side of the junction and negative space charge region on p-side of the junction. The thickness of depletion region is of the order of one-tenth of a micrometre. a layer of positive charge (or positive space-charge region) on n-side of the junction is developed. and n. This ionised donor (positive charge) is immobile as it is bonded to the surrounding atoms. Due to this field. the concentration of holes is more than the concentration of electrons.10 p-n junction drift current. Thus a FIGURE 14. and due to the concentration gradient across p-.5. transistor. it leaves behind an ionised donor on n-side. part of the p-Si wafer can be converted into n-Si. etc. A clear understanding of the junction behaviour is important to analyse the working of other semiconductor devices.edu/hbase/solids/pnjun. holes diffuse from p-side to n-side (p → n) and electrons diffuse from n-side to p-side (n → p). We will now try to understand how a junction is formed and how the junction behaves under the influence of external applied voltage (also called bias). Two important processes occur during the formation of a p-n junction: diffusion and drift. Formation and working of p-n junction diode .sides. By adding precisely a small quantity of pentavelent impurity. 14.gsu. the concentration of electrons (number of electrons per unit volume) is more compared to the concentration of holes.region.5 p-n JUNCTION A p-n junction is the basic building block of many semiconductor devices like diodes.Physics 14. and n. Similarly. which is opposite in direction to the diffusion formation process. During the formation of p-n junction. This space-charge region on either side of the junction together is known as depletion region as the electrons and holes taking part in the initial movement across the junction depleted the region of its free charges (Fig. We know that in an n-type semiconductor. When an electron diffuses from n → p. a layer of negative charge (or negative space-charge region) on the p-side of the junction is developed. As the holes continue to diffuse. when a hole diffuses from p → n due to the concentration gradient. The wafer now contains p-region and n-region and a metallurgical junction between p-.10). http://hyperphysics. As the electrons continue to diffuse from n → p. The motion of charge carriers due to the electric field is called drift. There are several processes by which a semiconductor can be formed. an electric field directed from positive charge towards negative charge develops.html 14. A p-n junction diode is symbolically represented as shown in Fig. The junction will behave as a discontinuity for the flowing charge carriers. As the diffusion process continues. EXAMPLE 14. the space-charge regions on either side of the junction extend.1 p-n junction diode under forward bias When an external voltage V is applied across a semiconductor diode such that p-side is connected to the positive terminal of the battery and n-side to the negative terminal [Fig. The loss of electrons from the n-region and the gain of electron by the p-region causes a difference of potential across the junction of the two regions.3 Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction? Solution No! Any slab. In a p-n junction under equilibrium there is no net current. FIGURE 14. This process continues until the diffusion current equals the drift current. (This is because the resistance of the depletion region – a region where there are no charges – is very high compared to the resistance of n-side and p-side. howsoever flat. 14.12 (a) Semiconductor diode.11(a) and (b). Figure 14.13(a)]. 14. 14.6 SEMICONDUCTOR DIODE A semiconductor diode [Fig.12(b). Since this potential tends to prevent the movement of electron from the n region into the p region. will have roughness much larger than the inter-atomic crystal spacing (~2 to 3 Å) and hence continuous contact at the atomic level will not be possible.12(a)] is basically a p-n junction with metallic contacts provided at the ends for the application of an external voltage. (b) Symbol for p-n junction diode. FIGURE 14. it is said to be forward biased.11 shows the p-n junction at equilibrium and the potential across the junction. The direction of arrow indicates the conventional direction of current (when the diode is under forward bias). The n material is thus positive relative to the p material.11 (a) Diode under equilibrium (V = 0). Example 14. and p material has acquired electrons. The situation of p-n junction diode under equilibrium (without bias) is shown in Fig. The n-material has lost electrons.) The direction of the applied voltage (V ) is opposite to the 479 .Semiconductor Electronics: Materials. Devices and Simple Circuits Initially. Thus a p-n junction is formed. The polarity of this potential is such as to oppose further flow of carriers so that a condition of equilibrium exists. It is a two terminal device.3 14. (b) Barrier potential under no bias.6. The equilibrium barrier potential can be altered by applying an external voltage V across the diode. thus increasing the electric field strength and hence drift current. diffusion current is large and drift current is small. 14. The applied voltage mostly drops across the depletion region and the voltage drop across the p-side and n-side of the junction is negligible. 14. it is often called a barrier potential. The current . The diode reverse current is not very much dependent on the applied voltage. they will be swept to its majority zone. the barrier height increases and the depletion region widens due to the change in the electric field. This process under forward bias is known as minority carrier injection. Likewise. The total diode forward current is sum of hole diffusion current and conventional current due to electron diffusion. and (3) High voltage battery. The effective barrier height under forward bias is (V0 – V ). electrons from n-side cross the depletion region and reach p-side (where they are minority carries).13 (a) p-n junction diode under forward bias. The direction of applied voltage is same as the direction of barrier potential. If we increase the applied voltage significantly.6. Due to the applied voltage.14). [Fig. (2) Low battery voltage. So the current will be small. As a result. The applied voltage mostly drops across the depletion region. 14. decreases enormously compared to the diode under forward bias. If the applied voltage is small. The electric field direction of the junction is such that if electrons on p-side or holes on n-side in their random motion come close to the junction. This motion of charged carriers on either side gives rise to current.15(a)]. the injected holes on n-side diffuse from the junction edge of n-side to the other end of n-side (Fig. This drift of carriers gives rise to current. 14. As a result. the barrier height will be reduced and more number of carriers will have the required energy. Similarly. the barrier potential will be reduced only slightly below the equilibrium value. holes from p-side cross the junction and reach the n-side (where they are minority carries). it is said to be reverse biased [Fig. Thus the current increases. This suppresses the flow of electrons from n → p and holes from p → n.Physics built-in potential V0.15(b)]. the minority carrier concentration increases significantly compared to the locations far from the junction. 480 When an external voltage (V ) is applied across the diode such that n-side is positive and p-side is negative.14. the depletion layer width decreases and the barrier height is reduced [Fig. (b) Barrier potential (1) without battery. FIGURE 14.13(b)]. The magnitude of this current is usually in mA. the injected electrons on p-side diffuse from the junction edge of p-side to the other end of p-side. 14. Due to this concentration gradient.2 p-n junction diode under reverse bias FIGURE 14. Even a small voltage is sufficient to sweep the minority carriers from one side of the junction to the other side of the junction. Thus. diffusion current.14 Forward bias minority carrier injection. on each side. This is quite low because it is due to the motion of carriers from their minority side to their majority side across the junction. 14. At the junction boundary. The drift current is of the order of a few µA. The effective barrier height under reverse bias is (V0 + V ). The drift current is also there under forward bias but it is negligible (µA) when compared with current due to injected carriers which is usually in mA. and only a small number of carriers in the material—those that happen to be in the uppermost energy levels—will possess enough energy to cross the junction. The current under reverse bias is essentially voltage independent upto a critical reverse bias voltage.16 Experimental circuit arrangement for studying V-I characteristics of a p-n junction diode (a) in forward bias . Once it exceeds the rated value. the variation of current as a function of applied FIGURE 14. the value of the current is noted. You can see in Fig. 14.16(c) that in forward FIGURE 14. we use a milliammeter since the expected current is large (as explained in the earlier section) while a micrometer is used in reverse bias to measure the current.. (b) in reverse bias. 14. of voltages.e. if the forward current exceeds the rated value. . (i. A graph between V and I is obtained as in Fig. (c) Typical V-I 481 characteristics of a silicon diode. This can happen even for the diode under forward bias. If the reverse current is not limited by an external circuit below the rated value (specified by the manufacturer) the p-n junction will get destroyed. The battery is connected under reverse bias. Devices and Simple Circuits is not limited by the magnitude of the applied voltage but is limited due to the concentration of the minority carrier on either side of the junction. The circuit arrangement for studying the V-I characteristics of a diode.16(a) and (b). Note that in forward bias measurement. the diode gets destroyed due to overheating. For different values reverse bias.Semiconductor Electronics: Materials. When V = Vbr. to the diode through a potentiometer (or reheostat) so that the (b) Barrier potential under applied voltage to the diode can be changed. the diode reverse current increases sharply.15 (a) Diode voltage) are shown in Fig. known as breakdown voltage (Vbr ). Even a slight increase in the bias voltage causes large change in the current. 14.16(c). For the diode in reverse bias.6) Example 14. This special action of the diode is discussed later in Section 14.8 V. After the characteristic voltage. the current suddenly increases. This voltage is called the threshold voltage or cut-in voltage (~0. This property is used for rectification of ac voltages as discussed in the next section. rrb = 10 V/1µA= 1.4 The V-I characteristic of a silicon diode is shown in the Fig.7 V rf b = ∆V/∆I = 0.0 × 107 Ω EXAMPLE 14. at I = 20 mA. the current is very small (~µA) and almost remains constant with change in bias. the current first increases very slowly. It is called reverse saturation current.Physics bias. The general purpose diode are not used beyond the reverse saturation current region. (a) From the curve. 14.17 482 Solution Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin.17. I = 10 mA. Calculate the resistance of the diode at (a) ID = 15 mA and (b) VD = –10 V. the diode current increases significantly (exponentially). I = –1 µA. we can calculate the resistance using Ohm’s law.8. However. for special cases. almost negligibly. FIGURE 14. even for a very small increase in the diode bias voltage.4 . The above discussion shows that the p-n junction diode primerly allows the flow of current only in one direction (forward bias). V = 0. at very high reverse bias (break down voltage). The forward bias resistance is low as compared to the reverse bias resistance.7 V for silicon diode).1V/10 mA = 10 Ω (b) From the curve at V = –10 V. we define a quantity called dynamic resistance as the ratio of small change in voltage ∆V to a small change in current ∆I: rd = ∆V ∆I (14. Therefore. till the voltage across the diode crosses a certain value. V = 0.2V for germanium diode and ~0. For diodes. 14. The n-side of the diodes are connected together and the output is taken between this common point of diodes and the midpoint of the secondary of the transformer.19(a). So for a full-wave rectifier the secondary of the transformer is provided with a centre tapping and so it is called centre-tap transformer. Thus.18. though still varying.18(b). This property is used to rectify alternating voltages and the circuit used for this purpose is called a rectifier. Each diode rectifies only for half the cycle. Since the rectified output of this circuit is only for half of the input ac wave it is called as half-wave rectifier. again we get the output voltage. (Note that there is another circuit of full wave rectifier which does not need a centre483 tap transformer but needs four diodes. is called a half-wave rectifier. (b) Input ac voltage and output get an output voltage. the diode is reverse-biased and it does not conduct.) Therefore. The secondary of a transformer supplies the desired ac voltage across terminals A and B. When A is negative. 14. is restricted to only one direction and is said to be rectified. As can be seen from Fig. (The reverse breakdown voltage of the diode must be sufficiently higher than the peak ac voltage at the secondary of the transformer to protect the diode from reverse breakdown. in the positive half-cycle of ac there FIGURE 14.18 (a) Half-wave rectifier is a current through the load resistor RL and we circuit.7 APPLICATION OF JUNCTION DIODE AS A RECTIFIER From the V-I characteristic of a junction diode we see that it allows current to pass only when it is forward biased. as shown in Fig. shown in Fig. Such rectifier circuit. voltage waveforms from the rectifier circuit. as shown in Fig. the output between their common terminals and the centretap of the transformer becomes a full-wave rectifier output.14. 14. but the two do so for alternate cycles. So if an alternating voltage is applied across a diode the current flows only in that part of the cycle when the diode is forward biased. Thus. The reverse saturation current of a diode is negligible and can be considered equal to zero for practical purposes. Here the p-side of the two diodes are connected to the ends of the secondary of the transformer. the output voltage. Devices and Simple Circuits 14. a pulsating voltage will appear across the load only during the half cycles of the ac input during which the diode is forward biased. When the voltage at A is positive.Semiconductor Electronics: Materials. the diode is forward biased and it conducts. whereas there is no current in the negative halfcycle. Hence. it is known as full-wave rectifier.) Suppose the input voltage to A . If an alternating voltage is applied across a diode in series with a load. gives output rectified voltage corresponding to both the positive as well as negative half of the ac cycle.19(c) the voltage rectified by each diode is only half the total secondary voltage. The circuit using two diodes. In the next positive half-cycle. In the next half-cycle of rectified output it again gets charged to the peak value (Fig. In the course of the ac cycle when the voltage at A becomes negative with respect to centre tap. The output voltage obtained by using capacitor input filter is nearer to the peak voltage of the rectified voltage. this is a more efficient circuit for getting rectified voltage or current than the halfwave rectifier The rectified voltage is in the form of pulses of the shape of half sinusoids. This type of filter is most widely 484 used in power supplies. If there is no external load. at that instant.Physics with respect to the centre tap at any instant is positive. the voltage at B would be positive. during this positive half cycle we get an output current (and a output voltage across the load resistor RL) as shown in Fig.19(b). When the voltage across the capacitor is rising. . it remains charged to the peak voltage of the rectified output. So. 14.20). To make the time constant large value of C should be large.14. it gets discharged through the load and the voltage across it begins to fall. so they are (c) Output waveform across the called filters. (b) Input wave forms given to the circuits appear to filter out the ac ripple diode D1 at A and to the diode D2 at B. capacitor in filtering.14. Since these additional circuit. So capacitor input filters use large capacitors. Thus. In this part of the cycle diode D1 would not conduct but diode D2 would. Though it is unidirectional it does not have a steady value. diode D1 gets forward biased and conducts (while D2 being reverse biased is not conducting). voltage at B being out of phase will be negative as shown in Fig. One can also use an inductor in series with RL for FIGURE 14. The rate of fall of the voltage across the capacitor depends upon the inverse product of capacitor C and the effective resistance RL used in the circuit and is called the time constant. and give a pure dc voltage.19(c). we get output voltage during both the positive as well as the negative half of the cycle. Hence. Obviously.19 (a) A Full-wave rectifier the same purpose. When there is a load. load RL connected in the full-wave Now we shall discuss the role of rectifier circuit. it gets charged. giving an output current and output voltage (across RL ) during the negative half cycle of the input ac. To get steady dc output from the pulsating voltage normally a capacitor is connected across the output terminals (parallel to the load RL). It is clear that. there is a large change in the current. Zener voltage remains constant. The symbol for Zener diode is shown in Fig. Note that after the breakdown voltage Vz. . Zener.8 SPECIAL PURPOSE p-n JUNCTION DIODES In the section. even though current through the Zener diode varies over a wide range. Devices and Simple Circuits FIGURE 14.sides of the junction. These electrons account for high current observed at the breakdown.21 Zener diode.8.1 Zener diode It is a special purpose semiconductor diode. It is seen that when the applied reverse bias voltage(V ) reaches the breakdown voltage (Vz ) of the Zener diode. 14.20 (a) A full-wave rectifier with capacitor filter.21(b). (b) I-V electric field is known as internal field emission or field characteristics. (b) Input and output voltage of rectifier in (a). The FIGURE 14. 14. In other words. It is designed to operate under reverse bias in the breakdown region and used as a voltage regulator. we shall discuss some devices which are basically junction diodes but are developed for different applications. We know that reverse current is due to the flow of electrons (minority carriers) from p → n and holes from n → p. The I-V characteristics of a Zener diode is shown in Fig. Let us understand how reverse current suddenly increases at the breakdown voltage. a large change in the current can be produced by almost insignificant change in the reverse bias voltage. then the electric field strength is high enough to pull valence electrons from the host atoms on the p-side which are accelerated to n-side. ionisation. named after its inventor C. depletion region formed is very thin (<10–6 m) and the electric field of the junction is extremely high (~5×106 V/m) even for a small reverse bias voltage of about 5V. As the reverse bias voltage is increased. 14.21(a). and n. emission of electrons from the host atoms due to the high (a) symbol. When the reverse bias voltage V = Vz. 14. This property of the Zener diode is used for regulating supply voltages so that they are constant. the electric field at the junction becomes significant. Due to this.Semiconductor Electronics: Materials. The electric field required for field ionisation is of 485 the order of 106 V/m. Zener diode is fabricated by heavily doping both p-. its rectified output also fluctuates. increase/ decrease of the voltage drop across Rs without any change in voltage across the Zener diode.0 – 6. Example 14.22 Zener diode as DC voltage across the Zener diode. if the input voltage decreases. What should be the value of series resistor RS? Solution The value of RS should be such that the current through the Zener diode is much larger than the load current. (ii) Light emitting diodes (LED) which convert electrical energy into light. Thus any increase/ voltage regulator decrease in the input voltage results in.. Note that slight variation in the value of the resistor does not matter.0 mA and the unregulated input is 10. If the input voltage increases. the current through Rs and Zener diode also increases.5 In a Zener regulated power supply a Zener diode with VZ = 6.22.e. In this section. therefore. The voltage drop across Rs decreases without any change in the FIGURE 14. All these devices are called optoelectronic devices. 14. Choose Zener current as five times the load current. we use a Zener diode.Physics Zener diode as a voltage regulator We know that when the ac input voltage of a rectifier fluctuates.5 14.0 V is used for regulation. So. IZ = 20 mA. To get a constant dc voltage from the dc unregulated output of a rectifier. we learn about semiconductor diodes in which carriers are generated by photons (photo-excitation). This is to have good load regulation. The unregulated dc voltage (filtered output of a rectifier) is connected to the Zener diode through a series resistance Rs such that the Zener diode is reverse biased.0V/(24 × 10–3) A = 167 Ω. (iii) Photovoltaic devices which convert optical radiation into electricity (solar cells). a series resistor of 150 Ω is appropriate.8. The total current through RS is. The load current is to be 4. Similarly. We have to select the Zener diode according to the required output voltage and accordingly the series resistance Rs. The voltage drop across R S is 10. This gives RS = 4.0 V. what is important is that the current IZ should be sufficiently larger than IL.0 V. Thus the Zener diode acts as a voltage regulator. the current through Rs and Zener diode also decreases. EXAMPLE 14. The circuit diagram of a voltage regulator using a Zener diode is shown in Fig.0 = 4. how a semiconductor diode behaves under applied electrical inputs.2 Optoelectronic junction devices We have seen so far. i. We shall study the functioning of the following optoelectronic devices: (i) Photodiodes used for detecting optical signal (photodetectors). 24 mA. Zener voltage remains constant even though the current through the Zener diode changes. 486 . This increases the voltage drop across Rs without any change in the voltage across the Zener diode. This is because in the breakdown region. The nearest value of carbon resistor is 150 Ω. It is operated under reverse bias.23 (a) An illuminated photodiode under reverse bias . For the fabrication of LEDs. On illumination.6 The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~µA). thermal excitation. Devices and Simple Circuits (i) Photodiode A Photodiode is again a special purpose p-n junction diode fabricated with a transparent window to allow light to fall on the diode.23(a) and a typical I-V characteristics in Fig. When an external load is connected. semiconductors like GaAs. When the photodiode is illuminated with light (photons) with energy (hν) greater than the energy gap ( E g ) of the semiconductor. 14. Thus photodiode can be used as a photodetector to detect optical signals. electrons and holes are separated before they recombine. Hence. It depends on semiconductor and the method of fabrication of the p-n junction. n >> p).6 n′ = n + ∆n p′ = p + ∆ p Here n′ and p ′ are the electron and hole concentrations * at any particular illumination and n and p are carriers concentration when there is no illumination.Semiconductor Electronics: Materials. Electrons are collected on n-side and holes are collected on p-side giving rise to an emf. 487 . GaAsGaP are used in which radiative recombination dominates. to create an e-h pair. Due to electric field of the junction. the majority carrier density ( n ) is considerably larger than the minority hole density p (i. the * Note that. FIGURE 14. (b) I-V characteristics of a photodiode for different illumination intensity I4 > I3 > I2 > I1. Remember ∆n = ∆p and n >> p.). Obviously.. The diode is fabricated such that the generation of e-h pairs takes place in or near the depletion region of the diode. if a reverse bias is applied. let the excess electrons and holes generated be ∆n and ∆p. The circuit diagram used for the measurement of I-V characteristics of a photodiode is shown in Fig. Example 14. respectively: EXAMPLE 14. The direction of the electric field is such that electrons reach n-side and holes reach p-side. we spend some energy (photoexcitation. Therefore when an electron and hole recombine the energy is released in the form of light (radiative recombination) or heat (non-radiative recombination). The magnitude of the photocurrent depends on the intensity of incident light (photocurrent is proportional to incident light intensity). What is the reason then to operate the photodiodes in reverse bias? Solution Consider the case of an n-type semiconductor. then electron-hole pairs are generated due to the absorption of photons. current flows. It is easier to observe the change in the current with change in the light intensity. etc.23(b). 14.e. (ii) Fast action and no warm-up time required. When the forward current of the diode is small. But the threshold voltages are much higher and slightly different for each colour. At the junction boundary the concentration of minority carriers increases compared to the equilibrium concentration (i. typically around 5V.4 µm to 0. In general. Hence. yellow.. 488 . we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current.8 eV).7 µm.e. Photons with energy equal to or slightly less than the band gap are emitted. when there is no bias). The reverse breakdown voltages of LEDs are very low. The diode is encapsulated with a transparent cover so that emitted light can come out.4 (Eg ~ 1. (iv) Long life and ruggedness.. Further increase in the forward current results in decrease of light intensity. electrons are sent from n → p (where they are minority carriers) and holes are sent from p → n (where they are minority carriers). The compound semiconductor Gallium Arsenide – Phosphide (GaAs1–x Px ) is used for making LEDs of different colours. (v) Fast on-off switching capability.6 (ii) Light emitting diode It is a heavily doped p-n junction which under forward bias emits spontaneous radiation. So care should be taken that high reverse voltages do not appear across them. ∆n/n ) would be much less than that in the minority carriers (i. LEDs that can emit red. EXAMPLE 14. i. The V-I characteristics of a LED is similar to that of a Si junction diode.. green and blue light are commercially available. These LEDs find extensive use in remote controls.Physics fractional change in the majority carriers (i. When the diode is forward biased. photodiodes are preferably used in the reverse bias condition for measuring light intensity.9 eV) is used for red LED. Thus at the junction boundary on either side of the junction. LEDs are biased such that the light emitting efficiency is maximum. On recombination. etc.e.6 P0.e. The semiconductor used for fabrication of visible LEDs must at least have a band gap of 1. ∆p/p). burglar alarm systems. Extensive research is being done for developing white LEDs which can replace incandescent lamps. GaAs (Eg ~ 1.e. optical communication.4 eV) is used for making infrared LED. from about 3 eV to 1. LEDs have the following advantages over conventional incandescent low power lamps: (i) Low operational voltage and less power. intensity of light increases and reaches a maximum. orange. excess minority carriers are there which recombine with majority carriers near the junction. (iii) The bandwidth of emitted light is 100 Å to 500 Å or in other words it is nearly (but not exactly) monochromatic.. the intensity of light emitted is small. the energy is released in the form of photons. GaAs0.8 eV (spectral range of visible light is from about 0. As the forward current increases. 25(a) a photocurrent IL flows through the load.0 to 1.04 eV).1 eV). (ii) separation of electrons and holes due to electric field of the depletion region. Devices and Simple Circuits (iii) Solar cell A solar cell is basically a p-n junction which generates emf when solar radiation falls on the p-n junction. (iv) availability of the raw FIGURE 14. (ii) high optical absorption (~104 cm–1). Any light with photon energies (b) I-V characteristics of a solar cell.43 eV). Note that sunlight is not always illuminated p-n junction solar cell. The other side of p-Si is coated with a metal (back contact). CdTe (Eg = 1.8 eV). 14. (iii) the electrons reaching the n-side are collected by the front contact and holes reaching p-side are collected by the back contact. when light falls on.45 eV). metallic grid occupies only a very small fraction of the cell area (<15%) so that light can be incident on the cell from the top. The important criteria for the selection of a material for solar cell fabrication are (i) band gap (~1. A p-Si wafer of about 300 µm is taken over which a thin layer (~0.25 (a) A typical material.3 µm) of n-Si is grown on one-side by diffusion process. A typical I-V characteristics of a solar cell is shown in the Fig.24.25(b).Semiconductor Electronics: Materials. separation and collection— (i) generation of e-h pairs due to light (with hν > Eg ) close to the junction. A simple p-n junction solar cell is shown in Fig. On the top of n-Si layer. it is due to the following three basic processes: generation. 14. Solar cells are made with semiconductors like Si (Eg = 1. This is because a solar cell does not draw current but supplies the same to the load. When an external load is connected as shown in the Fig. etc. Electrons are swept to n-side and holes to p-side. Solar cells are used to power electronic devices in satellites and space vehicles and also as power supply to some calculators. CuInSe2 (Eg = 1. required for a solar cell. The generation of emf by a solar cell. and (v) cost. 14.5 eV are ideal materials for solar cell fabrication. Thus p-side becomes positive and n-side becomes negative giving rise to photovoltage. (iii) electrical conductivity. . This acts as a front contact. The solar cell. Semiconductors with band gap close to 1. (b) Cross-sectional view. GaAs (Eg = 1. Note that the I – V characteristics of solar cell is drawn in the fourth quadrant of the coordinate axes. greater than the bandgap will do.24 (a) Typical p-n junction grid) is deposited. It works on the same principle (photovoltaic effect) as the photodiode. Production of low-cost photovoltaic cells for large-scale solar energy is a topic 489 for research. except that no external bias is applied and the junction area is kept much larger for solar radiation to be incident because we are interested in more power. metal finger electrode (or metallic FIGURE 14. Hence. That transistor was a point-contact transistor. But over the years new types of transistors were invented and to differentiate it from the new ones it is now called the Bipolar Junction Transistor (BJT). we can use only the high energy component of the solar energy for photo-conversion and a significant part of energy will be of no use. For photo-excitation. most of the solar radiation will be absorbed on the top-layer of solar cell and will not reach in or near the depletion region. For effective electron-hole separation. GaAs is better (in spite of its higher band gap) than Si because of its relatively higher absorption coefficient.S. often the word transistor 490 . In fact. U. Even now. hν > Eg.26 The maxima is near 1. semiconductor with band gap ~1. The first junction transistor consisting of two back-to-back p-n junctions was invented by William Schockley in 1951.4 eV). it was known simply as transistor. If we choose materials like CdS or CdSe (Eg ~ 2. 14.5 eV or lower is likely to give better solar conversion efficiency. The question arises: why we do not use material like PbS (Eg ~ 0. we want the photo-generation to occur in the junction region only. due to the junction field.5 eV.A.Physics Example 14.4 eV) which satisfy the condition hν > Eg for ν maxima corresponding to the solar radiation spectra? If we do so.53 eV. As long as only the junction transistor was known.H.1 eV while for GaAs it is ~1. EXAMPLE 14.7 Why are Si and GaAs are preferred materials for solar cells? Solution The solar radiation spectrum received by us is shown in Fig. Brattain of Bell Telephone Laboratories. FIGURE 14.7 14.26. Bardeen and W. Silicon has Eg ~ 1.9 JUNCTION TRANSISTOR The credit of inventing the transistor in the year 1947 goes to J. 491 . The schematic representations of an n-p-n and a p-n-p configuration are shown in Fig. there are two types of transistors. Devices and Simple Circuits is used to mean BJT when there is no confusion. we shall use the word transistor for BJT without any ambiguity. First we shall see what gives the transistor its amplifying capabilities. that there is a formation of depletion region acorss the junction. 14. When the transistor is biased in this way it is said to be in active state. (ii) p-n-p transistor : Here two segments of p-type semiconductor (termed as emitter and collector) are separated by a segment of n-type semiconductor (termed as base).We represent the voltage between emitter and base as VEB and that between the collector and the base as VCB. 14. 14. In FIGURE 14.1 Transistor: structure and action A transistor has three doped regions forming two p-n junctions between them. This situation is shown in Fig. For understanding the action of a transistor. But later its use as a switch acquired equal importance.27(a).28.27(a). It is of moderate size and heavily doped. 14. (i) n-p-n transistor : Here two segments of n-type semiconductor (emitter and collector) are separated by a segment of p-type semiconductor (base). we have to consider the nature of depletion regions formed at these junctions. 14. Basically. A brief description of the three segments of a transistor is given below: • Emitter : This is the segment on one side of the transistor shown in Fig. We shall study both these functions and the ways the transistor is biased to achieve these mutually exclusive functions. The transistor can be used in two distinct ways. The biasing of the transistor is done differently for different uses. as shown in Fig. The charge carriers move across different regions of the transistor when proper voltages are applied across its terminals. In the schematic symbols used for representing p-n-p and n-p-n transistors [Fig. it was invented to function as an amplifier. In case of a transistor depletion regions are formed at the emitter base-junction and the basecollector junction. where VCC and VEE are used for creating the respective biasing.Semiconductor Electronics: Materials. Obviously. The transistor works as an amplifier. The collector side is moderately doped and larger in size as compared to the emitter. It is very thin and lightly doped. Since our study is limited to only BJT.27(b)] the arrowhead shows the direction of conventional current in the transistor. • Base: This is the central segment. All the three segments of a transistor have different thickness and their doping levels are also different.9. 14.27. and (b) Symbols for n-p-n and p-n-p transistors.27 (a) Schematic representations of a n-p-n transistor and p-n-p transistor. with its emitter-base junction forward biased and the base-collector junction reverse biased. a device which produces a enlarged copy of a signal. • Collector: This segment collects a major portion of the majority carriers supplied by the emitter. It supplies a large number of majority carriers for the current flow through the transistor. We have seen earlier in the case of a p-n junction. The base is made thin so that most of the holes find themselves near the reverse-biased base-collector junction and so cross the junction instead of moving to the base FIGURE 14. and VCC. but most of it is diverted to adjacent reverse-biased base-collector junction and the current coming out of the base becomes a very small fraction of the current that entered the junction. The current entering into the emitter from outside is equal to the emitter current IE. Let us see now the paths of current carriers in the transistor with emitter-base junction forward biased and base-collector junction reverse biased. p-n-p transistor and (b) n-p-n transistor.28(a) that the emitter current is the sum of collector current 492 and base current: .28 Bias Voltage applied on: (a) terminal. In circuits. the power supply between the base and the emitter is represented as VBB and that between collector and emitter as VCC. We see that the emitter current IE = Ih + Ie but the base current IB << Ih + Ie. 14. these holes. These majority carriers enter the base region in large numbers. respectively. So the two power supplies are represented as VEE. The large number of holes entering the base from the emitter swamps the small number of electrons there.Physics Fig. It is interesting to note that due to forward bias a large current enters the emitter -base junction. The holes in the base could move either towards the base terminal to combine with the electrons entering from outside or cross the junction to enter into the collector and reach the collector terminal. So the majority carriers there would be few. 14. The base current is thus a small fraction of the emitter current. because a major part of IE goes to collector instead of coming out of the base terminal. where emitter is the common terminal. It is obvious from the above description and also from a straight forward application of Kirchhoff’s law to Fig. As the base collector-junction is reversebiased. base is a common terminal for the two power supplies whose other terminals are connected to emitter and collector. which will be holes in a p-n-p transistor and electrons in an n-p-n transistor.28. The base is thin and lightly doped. In a p-n-p transistor the majority carriers in the base are electrons since base is of n-type semiconductor. Similarly the current emerging from the base terminal is IB and that from collector terminal is IC. respectively. which appear as minority carriers at the junction. If we represent the hole current and the electron current crossing the forward biased junction by Ih and Ie respectively then the total current in a forward biased diode is the sum Ih + Ie. The heavily doped emitter has a high concentration of majority carriers. can easily cross the junction and enter the collector. B or C) is common to both the input and the output.2 Basic transistor circuit configurations and transistor characteristics In a transistor. The description about the paths followed by the majority and minority carriers in a n-p-n is exactly the same as that for the p-n-p transistor. The arrowhead in the emitter shows the direction of the conventional current.29. The variation of the base current IB with the base-emitter voltage VBE is called the input characteristic. as shown in Fig. in a circuit the input/output connections have to be such that one of these (E. 14. You will see that the output characteristics are controlled by the input characteristics. 14. Common Collector (CC) The transistor is most widely used in the CE configuration and we shall restrict our discussion to only this configuration. Base (B) and Collector (C). Thus in a p-n-p transistor the current enters from emitter into base whereas in a n-p-n transistor it enters from the base into the emitter.28(b) the electrons are the majority carriers supplied by the n-type emitter region. 493 . we shall confine our discussion to such transistors only. Devices and Simple Circuits IE = IC + IB (14. the transistor can be connected in either of the following three configurations: Common Emitter (CE). Common Base (CB). 14.29 Circuit arrangement for studying the input and output characteristics of n-p-n transistor in CE configuration. They cross the thin p-base region and are able to reach the collector to give the collector current. In Fig. To study the input characteristics of the transistor in CE configuration. This implies that the collector current changes with the base current.Semiconductor Electronics: Materials. a curve is plotted between the base current IB against the base-emitter voltage VBE.7) We also see that IC ≈ IE. the variation of the collector current IC with the collector-emitter voltage VCE is called the output characteristic.9. viz.28. Similarly. 14. With p-n-p transistors the polarities of the external power supplies are to be inverted. Emitter (E). From the above description we can conclude that in the active state of the transistor the emitter-base junction acts as a low resistance while the base collector acts as a high resistance. Accordingly.. only three terminals are available. The FIGURE 14. But the current paths are exactly opposite. Therefore. But the direction of motion of electrons is just opposite to that of the current. IC . Since more commonly used transistors are n-p-n Si transistors. The input and the output characteristics of an n-p-n transistors can be studied by using the circuit shown in Fig. Our description of the direction of motion of the holes is identical with the direction of the conventional current. Common emitter transistor characteristics When a transistor is used in CE configuration. the input is between the base and the emitter and the output is between the collector and the emitter. . We are interested to obtain the input characteristic when the transistor is in active state.7 V. and (b) Typical output increases IC also increases. input characteristics for various values of VCE will give almost identical curves. the reverse bias across the basecollector junction is high most of the time.6 to 0.30(a).30 (a) Typical input proportionately.30(b). Since VCE = VCB + VBE and for Si transistor VBE is 0. its value varies with the operating current in the transistor:  ∆VBE  ri =   ∆I B  V (14. (i) Input resistance (ri ): This is defined as the ratio of change in baseemitter voltage (∆VBE) to the resulting change in base current (∆IB ) at constant collector-emitter voltage (VCE ). So the collector -emitter voltage V CE is kept large enough to make the base collector junction reverse biased. This shows that when I B characteristics. Therefore. The output characteristic is obtained by observing the variation of IC as VCE is varied keeping IB constant. VCE must be sufficiently larger than 0. The plot of IC versus characteristics. 14. As a consequence. both hole current from the emitter region and the electron current from the base region will increase. As a consequence both I B and I C will increase FIGURE 14. VCE for different fixed values of IB gives one output characteristic. the input characteristics may be obtained for VCE somewhere in the range of 3 V to 20 V. Since the transistor is operated as an amplifier over large range of VCE.7 V. 14.8) CE 494 The value of ri can be anything from a few hundreds to a few thousand ohms. This is dynamic (ac resistance) and as can be seen from the input characteristic. So there will be different output characteristics corresponding to different values of IB as shown in Fig. Hence. Since the increase in VCE appears as increase in VCB. The linear segments of both the input and output characteristics can be used to calculate some important ac parameters of transistors as shown below. The input characteristics of a transistor is as shown in Fig.Physics collector-emitter voltage VCE is kept fixed while studying the dependence of IB on VBE. it is enough to determine only one input characteristics. It is obvious that if VBE is increased by a small amount. its effect on IB is negligible. Semiconductor Electronics: Materials. the values of both βdc and βac are nearly equal. for most calculations βdc can be used. when the transistor is in saturation state.8 From the output characteristics shown in Fig. the transistor is in the saturation state and the current is controlled by the supply voltage VCC (=VCE) in this part of the characteristic. When VCE is more than that required to reverse bias the base-collector junction.0 mA. (Choose characteristics for IB= 30 and 20 µA. Here IC = 4. The reciprocal of the slope of the linear part of the output characteristic gives the values of ro.9) B The output characteristics show that initially for very small values of VCE . βdc = Example 14. Devices and Simple Circuits (ii) Output resistance (ro): This is defined as the ratio of change in collector-emitter voltage (∆VCE) to the change in collector current (∆IC) at a constant base current IB. Consider any two characteristics for two values of IB which lie above and below the given value of IC .8 495 .11) IB Since IC increases with IB almost linearly and IC = 0 when IB = 0. calculate the values of β ac and β dc of the transistor when VCE is 10 V and IC = 4. If we simply find the ratio of IC and IB we get what is called dc β of the transistor. IC (14.30(b). IC increases very little with VCE. Hence. This also explains why the resistance at the initial part of the characteristic.  ∆VCE  ro =   ∆I C  I (14. (iii) Current amplification factor (β ): This is defined as the ratio of the change in collector current to the change in base current at a constant collector-emitter voltage (VCE) when the transistor is in active state.0 mA. is very low.) At VCE = 10 V we read the two values of IC from the graph. IC increases almost linearly.10) CE This is also known as small signal current gain and its value is very large. 14. The output resistance of the transistor is mainly controlled by the bias of the basecollector junction. This happens because the base-collector junction is not reverse biased and the transistor is not in active state. Then I EXAMPLE 14. βac =  C   ∆I B  V  ∆I  (14. In fact. Both βac and βdc vary with VCE and IB (or IC) slightly. βdc = C  ∆I B  V IB CE For determining βac and βdc at the stated values of VCE and IC one can proceed as follows. Solution  ∆I  βac =  C  . So. The high magnitude of the output resistance (of the order of 100 kΩ) is due to the reverse-biased state of this diode. 5 mA Therefore. In the case of Si transistor. (b) Transfer characteristic. active region and saturation. we get VBB = IBRB + VBE (14.9. (i) Transistor as a switch We shall try to understand the operation of the transistor as a switch by analysing the behaviour of the base-biased transistor in CE configuration as shown in Fig.0) mA = 1. Hence Vo = VCC When Vi becomes greater than 0.Physics ∆IB = (30 – 20) µA = 10 µA. βac = 1.6 V. When the transistor is used in the cutoff or saturation state it acts as a switch. as long as input Vi is less than 0.31(a).8 14. the biasing of the E-B and B-C junction and the operation region namely cutoff.5 mA/ 30 µA = 150 and for IC = 3. So. for IC = 4.6 V the transistor is in active state with 496 some current IC in the output path and the output Vo decrease as the . either estimate the value of IB corresponding to IC = 4. 14. ∆IC = (4.31 (a) Base-biased transistor in CE state and current IC will be zero. βdc =(150 + 150) /2 = 150 EXAMPLE 14. it has to operate in the active region. βdc = 4.0 mA and IB = 20 µA βdc =3.0 mA at VCE = 10 V or calculate the two values of βdc for the two characteristics chosen and find their mean.5 – 3.5 mA/ 10 µA = 150 For determining βdc. Therefore.13) We shall treat VBB as the dc input voltage Vi and VCE as the dc output voltage VO. the transistor will be in cut off FIGURE 14. we have Vi = IBRB + VBE and Vo = VCC – ICRC. configuration. CC and CE). (14.0 mA / 20 µA = 150 Hence.3 Transistor as a device The transistor can be used as a device application depending on the configuration used (namely CB. Applying Kirchhoff’s voltage rule to the input and output sides of this circuit.12) and VCE = VCC – ICRC. On the other hand for using the transistor as an amplifier.5 mA and IB = 30 µA. Let us see how V o changes as V i increases from zero onwards. As mentioned earlier we have confined only to the CE configuration and will be concentrating on the biasing and the operation region to understand the working of a device. The switching circuits are designed in such a way that the transistor does not remain in active state. Thus the linear portion of the active region of the transistor can be exploited for the use in amplifiers. When the transistor is not conducting it is said to be switched off and when it is driven into saturation it is said to be switched on.10). We have. If the VBB voltage has a fixed value corresponding to the mid point of the active region.0 V.31(b)]. IC increases almost linearly and so V o decreases linearly till its value becomes less than about 1. [also called the transfer characteristics of the base-biased transistor (Fig. (14. 14. With further increase in Vi the output voltage is found to decrease further towards zero though it may never become zero.32) is given by AV = – RC ∆ IC / RB ∆IB = –βac(RC /RB ) (14. As long as Vi is low and unable to forward-bias the transistor. This shows that if we define low and high states as below and above certain voltage levels corresponding to cutoff and saturation of the transistor. If we consider ∆Vo and ∆Vi as small changes in the output and input voltages then ∆Vo/∆Vi is called the small signal voltage gain AV of the amplifier.14) where βac is equal to ∆ IC/ ∆IB from Eq. then we can say that a low input switches the transistor off and a high input switches it on. we can say that a low input to the transistor gives a high output and a high input gives a low output. So. Vo is high (at VCC ). We can express the voltage gain AV in terms of the resistors in the circuit and the current gain of the transistor as follows. the change becomes non linear and transistor goes into saturation state. the circuit will behave as a CE amplifier with voltage gain ∆Vo/ ∆Vi . 14. Transistor as an amplifier (CE configuration) is discussed in detail in the next section.Semiconductor Electronics: Materials. (ii) Transistor as an amplifier For using the transistor as an amplifier we will use the active region of the Vo versus Vi curve. from Vi = IBRB + VBE ∆Vi = RB ∆IB + ∆VBE But ∆VBE is negligibly small in comparison to ∆IBRB in this circuit. If we plot the Vo vs Vi curve. 497 . Let us see now how the transistor is operated as a switch. the voltage gain of this CE amplifier (Fig. Devices and Simple Circuits term ICRC increases. It is negative because the output is VCC – ICRC and not ICRC. If Vi is high enough to drive the transistor into saturation. Alternatively. That is why as input voltage of the CE amplifier increases its output voltage decreases and the output is said to be out of phase with the input. Beyond this. Vo = VCC – ICRC Therefore. we see that between cut off state and active state and also between active state and saturation state there are regions of non-linearity showing that the transition from cutoff state to active state and from active state to saturation state are not sharply defined. ∆Vo = 0 – RC ∆ IC Similarly. very near to zero. With increase of Vi . The slope of the linear part of the curve represents the rate of change of the output with the input. then Vo is low. the input loop gives VBB = VBE + IB RB When vi is not zero.11). We can measure the ac variations across the input and output terminals by blocking the dc voltages by large capacitors. we get VBE + vi = VBE + IB RB + ∆IB (RB + ri ) The change in VBE can be related to the input resistance ri [see Eq. The operating values of VCE and IB determine the operating point. which is similar to the βdc defined in Eq.15) (14. (14. amplifiers are used to amplify alternating signals. Now let us superimpose an ac input signal vi (to be amplified) on the FIGURE 14. producing in turn corresponding change in the value of VO.ICRC would also remain constant.17) ∆I B i b which is also known as the ac current gain Ai . In general.Physics 14. The dc voltage VCE = VCC .4 Transistor as an Amplifier (CE-Configuration) To operate the transistor as an amplifier it is necessary to fix its operating point somewhere in the middle of its active region. The CE-transistor amplifier.9. if we first assume that vi = 0. of the amplifier.32. As a consequence the collector current also will have sinusoidal variations superimposed on the value of IC. In the discription of the amplifier given above we have not considered any ac signal. as ∆I i βac = c = c (14. Then applying Kirchhoff’s law to the output loop. 14.32 A simple circuit of a bias VBB (dc) as shown in Fig.8)] and the change in IB. then the base current will have sinusoidal variations superimposed on the value of IB. If a small sinusoidal voltage with amplitude vs is superposed on the dc base bias by connecting the source of that signal in series with the VBB supply. The change in Ic due to a change in IB causes a change in VCE and the voltage drop across the resistor RL because VCC is fixed. Usually βac is close to βdc in the linear region of the output characteristics. (14. The working of an amplifier can be easily understood. we get Vcc = VCE + IcRL Likewise.16) 498 . If we fix the value of VBB corresponding to a point in the middle of the linear part of the transfer curve then the dc base current IB would be constant and corresponding collector current IC will also be constant. output is taken between the collector and the ground. Hence vi = ∆IB (RB + ri ) = r ∆ IB The change in IB causes a change in Ic. We define a parameter βac. (14. The Si transistor has βdc = 250 and RB = 100 kΩ. However it should be realised that transistor is not a power generating device. the VBB supply can be varied from 0V to 5. In this range. Assume that when the transistor is saturated.8V VCE = VCC – ICRC IC = VCC/RC = 5. In the saturation range.10). we get ac power gain. (b) determine V1 when the transistor is ‘switched on’. (14. Note that the transistor is in active state when IB varies from 0. RC = 1 KΩ. Hence. we get v0 = ∆VCE = –βac RL ∆IB The voltage gain of the amplifier is Av = v 0 ∆VCE = vi r ∆I B (14. (14. The energy for the higher ac power at the output is supplied by the battery. (c) find the ranges of V1 for which the transistor is ‘switched off’ and ‘switched on’. From Eq.Semiconductor Electronics: Materials. VCC = 5.9 In Fig. Solution Given at saturation VCE = 0V. Mathematically Ap = βac × Av (14.8V.8V The value of input voltage below which the transistor remains cutoff is given by VIL = 0.0V. Between 2.18) r The negative sign represents that output voltage is opposite with phase with the input voltage.0V and 0. Example 14.0kΩ = 5.31(a).6V. Calculate (a) the minimum base current.8V Between 0. it will be in ‘switched on’ state.0V/1.0mA to 20mA. VIH = 2. 14. VCE = 0V and VBE = 0.0 V.0V.0 mA/250 = 20µA The input voltage at which the transistor will go into saturation is given by VIH = VBB = IBRB +VBE = 20µA × 100 kΩ + 0. I C ≤ βI B . Devices and Simple Circuits These changes can be given by Eq.6V. the transistor will be in the ‘switched off’ state. Therefore the power gain Ap can be expressed as the product of the current gain and voltage gain.19) Since βac and Av are greater than 1. VBE = 0.9 499 .0 mA Therefore IB = IC/β = 5. IC = βIB is valid. for which the transistor will reach saturation. =– βac R L EXAMPLE 14. Here we have also seen the voltage gain Av.15) as ∆VCC = ∆VCE + RL ∆IC = 0 or ∆VCE = –RL ∆IC The change in VCE is the output voltage v0. From the discussion of the transistor characteristics you have seen that there is a current gain βac in the CE configuration.8V = 2.8V and 5. 0 kΩ is 2.VBE ) /IB. From Eq. Suppose the current amplification factor of the transistor is 100. (Refer to Fig. and fall (or built up) of current Ic and Ie due to the Let us try to understand how oscillations inductive coupling. RB = (VBB . The dc collector current IC = 100×0. apart from the resonant circuit for obtaining oscillation at a particular frequency.10 14. we have seen that a sinusoidal input is given which appears as an amplified signal in the output. For understanding the oscillator action. Solution The output ac voltage is 2. we get ac output without any external input signal. Note that the coils T2 and T1 are wound on the same core and hence are inductively coupled through their mutual inductance. The signal current through the base is.0 – 0. In an oscillator.0/2000 = 1.0 V.0 mA. This means that an external input is necessary to sustain ac signal in the output for an amplifier. A portion of the output power is returned back (feedback) to the input in phase with the starting power (this process is termed positive feedback) as shown in Fig. Assuming VBE = 0.6 V.33). In other words. the audio signal voltage across the collector resistance of 2.6 )/0. an amplifier is taken.0 V. So. As in an amplifier.0 mA/100 = 0. Different types of oscillators essentially use different methods of coupling the output to the input (feedback network). EXAMPLE 14.10 mA. we consider the circuit shown in Fig. 14. (c) Rise used have been omitted for simplicity.010 = 0. 14.Physics Example 14. 500 are built.5 Feedback amplifier and transistor oscillator In an amplifier. Also calculate the dc drop across the collector resistance. therefore given by iB = iC /β = 1.16. the ac collector current iC = 2.10 = 14 kΩ. The feedback can be achieved by inductive coupling (through mutual inductance) or LC or RC networks.0 V if the dc base current has to be 10 times the signal current.10 = 10 mA.10 For a CE transistor amplifier.010 mA. The dc base current has to be 10× 0. To attain this. the output in an oscillator is self-sustained. the base-emitter junction is forward biased FIGURE 14. Detailed biasing circuits actually oscillator and (b) Tuned collector oscillator. 14.33(a).33(b) in which the feedback is accomplished by inductive coupling from one coil winding (T 1) to another coil winding (T2).9. What should be the value of RB in series with VBB supply of 2. RB = (2.14. Suppose switch S1 is put on to .33 (a) Principle of a transistor while the base-collector junction is reverse amplifier with positive feedback working as an biased. the tank or tuned circuit is connected in the collector side.20)  2π LC  In the circuit of Fig.34(b). 14. a decrease of collector current causes the magnetic field to decay around the coil T2. 14. RC-oscillator. Without continued feedback. This current flows through the coil T2 where terminals are numbered 3 and 4 [Fig. the magnetic field around T2 ceases to grow. ν=   14. this current (in T1. Since there is no further change in collector current. 14. time-varying voltage or current. the transistor is driven to saturation. A typical analogue signal is shown in Figure. As soon as the field becomes static.33(c)(ii)]. 0V) and ‘1’ (say. it is known as tuned collector oscillator. It is convenient to use binary numbers to represent such signals. As a result of this positive feedback. 501 . emitter current) also increases from X´ to Y´ [Fig. Fig. Such signals are called Digital Signals.33(b). Such signals are called continuous or analogue signals. That is. This current does not reach full amplitude instantaneously but increases from X to Y. This means that maximum collector current is flowing and can increase no further. The time for change from saturation to cut-off and back is determined by the constants of the tank circuit or tuned circuit (inductance L of coil T2 and C connected in parallel to it). it will be known as tuned base oscillator. Devices and Simple Circuits apply proper bias for the first time.33(c)(i)]. the emitter current begins to fall. the transistor has reverted back to its original state (when the power was first switched on). If the tuned circuit is on the base side. a surge of collector current flows in the transistor. collector current decreases from Y towards Z [Fig.34(b) shows a pulse waveform in which only discrete values of voltages are possible. This means that both IE and IC cease to flow. A binary number has only two digits ‘0’ (say. In digital circuits only two values (represented by 0 or 1) of the input and output voltage are permissible.34(a). the signal (current or voltage) has been in the form of continuous. There are many other types of tank circuits (say RC) or feedback circuits giving different types of oscillators like Colpitt’s oscillator. The resonance frequency (ν ) of this tuned circuit determines the frequency at which the oscillator will oscillate. Obviously. 14. Therefore. 1  (14.10 DIGITAL ELECTRONICS AND LOGIC GATES In electronics circuits like amplifiers. Thus. 14. as shown in Fig. and then back to saturation. The whole process now repeats itself. 14. Consequently.33(b)]. there will be no further feedback from T2 to T1. 5V). then to cut-off. T1 is now seeing a decaying field in T2 (opposite from what it saw when the field was growing at the initial start operation). 14. The current in T2 (collector current) connected in the collector circuit acquires the value Y when the transistor becomes saturated.33(c)(i)]. In digital electronics we use only these two levels of voltage as shown in Fig. This causes a further decrease in the emitter current till it reaches Z′ when the transistor is cut-off. The inductive coupling between coil T2 and coil T1 now causes a current to flow in the emitter circuit (note that this actually is the ‘feedback’ from input to output). oscillators. [14.Semiconductor Electronics: Materials. introduced to you in earlier sections. Hence. Hartley oscillator. However. Each logic gate is indicated by a symbol and its function is defined by a truth table that shows all the possible input logic level combinations with their respective output logic levels. FIGURE 14.Physics This section is intended to provide the first step in our understanding of digital electronics. digital watches.36. robots. The commonly used symbol together with the truth table for this gate is given in Fig. 502 .35 (a) Logic symbol. It produces a ‘1’ output if the input is ‘0’ and vice-versa. 14. (b) Truth table of NOT gate.1 Logic gates A gate is a digital circuit that follows curtain logical relationship between the input and output voltages. This is why it is also known as an inverter. (b) Digital signal. The switch is placed either in the ON or OFF position to activate the light. We shall restrict our study to some basic building blocks of digital electronics (called Logic Gates) which process the digital signals in a specific manner. When the light is OFF the output value is ‘0’. (i) NOT gate This is the most basic gate. Input A 0 1 Output Y 1 0 (b) FIGURE 14. the output is high. OR. The five common logic gates used are NOT. with one input and one output. it produces an inverted version of the input at its output. if any of the input is high. Truth tables help understand the behaviour of logic gates. These logic gates can be realised using semiconductor devices.10. they are generally known as logic gates — gates because they control the flow of information.34 (a) Analogue signal. NAND. The logic symbol and truth table are shown in Fig. 14. Logic gates are used in calculators. AND. the output value is ‘1’. When the light is ON. that is. The inputs are the position of the light switch. and in telecommunications.35. The light is either ON or OFF depending on the switch position. 14. The output Y is 1 when either input A or input B or both are 1s. Therefore. industrial control systems. NOR. (ii) OR Gate An OR gate has two or more inputs with one output. computers. A light switch in your house can be used as an example of a digital circuit. That is. Semiconductor Electronics: Materials. B = 0. Devices and Simple Circuits Input A 0 0 1 1 B 0 1 0 1 (b) Output Y 0 1 1 1 FIGURE 14. EXAMPLE 14. Example 14. B = 0.37. B = 1. B = 1. Hence Y = 1 • For t2 to t3. (b) Truth table of AND gate. 14. B = 0. The logic symbol and truth table for this gate are given in Fig.11 FIGURE 14.11 Justify the output waveform (Y) of the OR gate for the following inputs A and B given in Fig. The output Y of AND gate is 1 only when input A and input B are both 1. A = 0.37 (iii) AND Gate An AND gate has two or more inputs and one output. Apart from carrying out the above mathematical logic operation.38 Input A 0 0 1 1 B 0 1 0 1 (b) Output Y 0 0 0 1 FIGURE 14. 14. A = 1. A = 1. Hence Y = 0 • For t1 to t2. 503 . B = 0. B = 1.37. Hence Y = 1 • For t4 to t5. this gate can be used for modifying the pulse waveform as explained in the following example. A = 1. A = 0. Hence Y = 1 • For t3 to t4. Hence Y = 1 Therefore the waveform Y will be as shown in the Fig. 14. Hence Y = 0 • For t5 to t6. A = 0. Hence Y = 1 • For t > t6.38 (a) Logic symbol. Solution Note the following: • At t < t1.36 (a) Logic symbol (b) Truth table of OR gate. A = 0. Based on the above. Hence Y = 0 Hence Y = 0 Hence Y = 1 Hence Y = 0 Hence Y = 0 Hence Y = 0 Hence Y = 0 can be drawn EXAMPLE 14. 0. B B B B = = = = 1. (b) Truth table of NAND gate. • For t5 to t6. Sketch the output waveform obtained from AND gate. • For t3 to t4. Figure 14.12 Take A and B input waveforms similar to that in Example 14. B = 0. Solution • For t ≤ t1.13 .11.13 Sketch the output Y from a NAND gate having inputs A and B given below: Solution • For t < t1. B = 0. Hence Hence Hence Hence Y Y Y Y = = = = 0 1 1 1 504 EXAMPLE 14. 0. B = 0. • For t1 to t2.39 (iv) NAND Gate This is an AND gate followed by a NOT gate. the output waveform for AND gate as given below. • For t4 to t5. If inputs A and B are both ‘1’. A = 0. B = 1.40 shows the symbol and truth table of NAND gate.Physics Example 14. Example 14.16 and 14. The gate gets its name from this NOT AND behaviour. B = 0. Input Output A 0 0 1 1 B 0 1 0 1 (b) Y 1 1 1 0 FIGURE 14. • For t2 to t3.12 FIGURE 14. NAND gates are also called Universal Gates since by using these gates you can realise other basic gates like OR. 1. • For t3 to t4. AND and NOT (Exercises 14. 0.17). 1. A = 0.40 (a) Logic symbol. the output Y is not ‘1’. A A A A = = = = 1. A = 0. • For t1 to t2. A = 1. B = 1. • For t > t6. A = 1. 0. A = 0. B = 1. A = 1. • For t2 to t3. A NOT. The word 505 .18 and 14. Apart from this.Semiconductor Electronics: Materials.operation applied after OR gate gives a NOT-OR gate (or simply NOR gate).e. OR.. Inspite of the miniaturisation introduced by the discovery of transistors. B = 1. C etc. Devices and Simple Circuits • For t4 to t5. The symbol and truth table for NOR gate is given in Fig. NOR gates are considered as universal gates because you can obtain all the gates like AND. Such a circuit is known as Integrated Circuit (IC). A = 1. neither one input nor the other is ‘1’.19). 14.41 (v) NOR Gate It has two or more inputs and one output. The most widely used technology is the Monolithic Integrated Circuit. and connect them by soldering wires in the desired manner. The concept of fabricating an entire circuit (consisting of many passive components like R and C and active devices like diode and transistor) on a small single block (or chip) of a semiconductor has revolutionised the electronics technology. R. i.11 INTEGRATED CIRCUITS The conventional method of making circuits is to choose components like diodes. B = 1. • For t > t6. L. (b) Truth table of NOR gate. B = 0. Input Output A 0 0 1 1 B 0 1 0 1 (b) Y 1 0 0 0 FIGURE 14.42. Its output Y is ‘1’ only when both inputs A and B are ‘0’. NOT by using only NOR gates (Exercises 14.13 FIGURE 14. such circuits were less reliable and less shock proof. A = 0. transistor.42 (a) Logic symbol. 14. • For t5 to t6. A = 0. such circuits were still bulky.. Hence Y = 0 Hence Y = 1 Hence Y = 1 EXAMPLE 14. You must have heard of the microprocessor. the number of circuit components or logic gates). in effect. it varies linearly with the input. monos means single and lithos means stone. Medium Scale Integration.e. like keeping track of what keys are pressed. the ICs are termed as Small Scale Integration. and it would be cheaper to throw it away than to park it”. In the 1970s. The miniaturisation that made the modern personal computer possible could never have happened without the IC. FIGURE 14. connecting wires – all in one package. pointed out that the memory capacity of a chip (IC) approximately doubled every one and a half years. televisions. more efficient and less expensive than present day computers. LSI (logic gates < 1000). resistors. The explosive growth in the semiconductor industry and computer technology is best expressed by a famous quote from Gordon Moore: “If the auto industry advanced as rapidly as the semiconductor industry.43 The casing and Depending on nature of input signals. The microprocessor is an IC that processes all information in a computer. The IC was first invented by Jack Kilky at Texas Instruments in 1958 and he was awarded Nobel Prize for this in 2000. They contain circuits such as logic gates. i.. Over the years. This is popularly known as Moore’s law.Physics monolithic is a combination of two greek words. One of the most useful linear IC’s is the operational amplifier. This. the complexity of ICs has increased while the size of its features continued to shrink. running programmes. means that the entire circuit is formed on a single silicon crystal (or chip). MSI (logic gates < 100). The linear IC’s process analogue signals which change smoothly and continuously over a range of values between a maximum and a minimum.. Depending upon the level of integration (i. a Rolls Royce would get half a million miles per gallon. partly removed to reveal the connections coming out from the ‘chip’ to the pins that enable it to make external connections. ICs are electronic devices that contain many transistors.43 shows a chip in its protective plastic case. the entire Information Technology (IT) industry hinges on semiconductors. In the past five decades. The number of transistors per chip has risen exponentially and each year computers are becoming more powerful.e. The output is more or less directly proportional to the input. In fact ICs are found in almost all electrical devices like cars. co-founder of INTEL. VLSI (logic gates > 1000). Figure 14. SSI (logic gates < 10).000 MHz) and would be much smaller. The chip dimensions are as small as 1mm × 1mm or it could even be smaller. a dramatic miniaturisation in computer technology has made modern day computers faster and smaller. yet cheaper than the year before. Thus. capacitors. and Very Large Scale Integration. FASTER AND SMALLER: THE FUTURE OF COMPUTER TECHNOLOGY The Integrated Chip (IC) is at the heart of all computer systems. It is intimated from current trends that the computers available in 2020 will operate at 40 GHz (40. The technology of fabrication is very involved but large scale industrial production has made them very inexpensive. ICs are produced on a piece of semiconductor crystal (or chip) by a process called photolithography. IC’s can be connection of a ‘chip’. CD players. cell phones etc. The digital IC’s process signals that have only two values. 506 . Gordon Moore. games etc. Large Scale Integration. grouped in two categories: (a) linear or analogue IC’s and (b) digital IC’s. 8. The number of electrons (ne ) is equal to the number of holes (nh ) in intrinsic conductors. metal or semiconductor. Semiconductors are elemental (Si. a ‘depletion layer’ is formed consisting of immobile ion-cores devoid of their electrons or holes. Ge) as well as compound (GaAs. while for metals Eg ≈ 0. 3. Holes are essentially electron vacancies with an effective positive charge. Such semiconductors are known as extrinsic semiconductors. while p-type Si or Ge can be obtained by doping with trivalent atom (acceptors) like B.Semiconductor Electronics: Materials. There are some special purpose diodes. 14. 13. Hence. Metals have low resistivity (10–2 to 10–8 Ω m). for semiconductors Eg is 0. Further. In forward bias (n-side is connected to negative terminal of the battery and p-side is connected to the positive). ne >> nh while in p-type semiconductors nh >> ne. The extent of conductivity depends upon the energy gap (Eg ) between the top of valence band (EV ) and the bottom of the conduction band EC. P. insulators have very high resistivity (>108 Ω m–1). 6. etc.). produce a change in the current flowing in a semiconductor. For insulators Eg > 3 eV. All energy levels in the valence band are filled while energy levels in the conduction band may be fully empty or partially filled. 5. These are of two types (n-type and p-type). Devices and Simple Circuits SUMMARY 1. while semiconductors have intermediate values of resistivity. nenh = ni2 in all cases. 15. With the help of a capacitor or a suitable filter. When such a junction is made.2 eV to 3 eV. By changing the external applied voltage. The electrons from valence band can be excited by heat.. the barrier is decreased while the barrier increases in reverse bias. 12. 2. 7. 507 . The electrons in the conduction band are free to move in a solid and are responsible for the conductivity. junction barriers can be changed. 4. In n-type semiconductors. n-type semiconducting Si or Ge is obtained by doping with pentavalent atoms (donors) like As. In etc. forward bias current is more (mA) while it is very small (µA) in a p-n junction diode. etc. 9. Lattice structure and the atomic structure of constituent elements decide whether a particular material will be insulator. a dc voltage can be obtained. Semiconductors are the basic materials used in the present solid state electronic devices like diode. Pure semiconductors are called ‘intrinsic semiconductors’. The number of charge carriers can be changed by ‘doping’ of a suitable impurity in pure semiconductors. etc. This is responsible for a junction potential barrier. Diodes can be used for rectifying an ac voltage (restricting the ac voltage to one direction). transistor. There are two distinct band of energies (called valence band and conduction band) in which the electrons in a material lie. The presence of charge carriers (electrons and holes) is an ‘intrinsic’ property of the material and these are obtained as a result of thermal excitation. the material possesses an overall charge neutrality. ICs. Valence band energies are low as compared to conduction band energies. CdS. Al. 10. 11. light or electrical energy to the conduction band and thus. p-n junction is the ‘key’ to all semiconductor devices. Sb. There are some special circuits which handle the digital data consisting of 0 and 1 levels. POINTS TO PONDER 1. (b) Solar cells which convert photon energy into electricity. where RC and RB are respectively RB  vi  the resistances in collector and base sides of the circuit. When you see a picture in which EC or EV are drawn as straight lines. 19. This property has been used to obtain voltage regulation. (c) Light Emitting Diode and Diode Laser in which electron excitation by a bias voltage results in the generation of light. NAND. after a certain voltage. v  R configuration is: Av =  o  = β C . The central block (thin and lightly doped) is called ‘Base’ while the other electrodes are ‘Emitter’ and ‘Collectors’. In modern day circuit. p-n junctions have also been used to obtain many photonic or optoelectronic devices where one of the participating entity is ‘photon’: (a) Photodiodes in which photon excitation results in a change of reverse saturation current which helps us to measure light intensity. then they should be respectively taken simply as the bottom of conduction band energy levels and top of valence band energy levels. The important digital circuits performing special logic operations are called logic gates. The plot between IC and VCE for fixed IB is called output characteristics while the plot between IB and VBE with fixed VCE is called input characteristics. The voltage gain of a transistor amplifier in common emitter  ∆I  21. The energies are the overall averages. ro =   ∆I   C CE IB current amplification factor. Transistor can be used as an amplifier and oscillator. In reverse bias. 24. Transistor is an n-p-n or p-n-p junction device. β =  C   ∆I B  V CE 20. The important transistor parameters for CE-configuration are: BE input resistance. an oscillator can also be considered as a self-sustained amplifier in which a part of output is fed-back to the input in the same phase (positive feed back). When the transistor is used in the cutoff or saturation state. the current suddenly increases (breakdown voltage) in a Zener diode. 23. and NOR gates. it acts as a switch. The transistors can be connected in such a manner that either C or E or B is common to both the input and output. Common Collector (CC) and Common Base (CB). many logical gates or circuits are integrated in one single ‘Chip’. This forms the subject of Digital Electronics. 508 . 18. In fact. NOT. 22. AND. The emitter-base junction is forward biased while collector-base junction is reverse biased.Physics 16. These are known as Intgrated circuits (IC). This gives the three configurations in which a transistor is used: Common Emitter (CE). ri =   ∆I B  V  ∆V   ∆VCE  output resistance. Zener diode is one such special purpose diode. 17. These are: OR. The energy bands (EC or EV ) in the semiconductors are space delocalised which means that these are not located in any specific place inside the solid. This is known as Barkhausen's Criteria. For example. In general. (b) they move across the junction by the potential difference. Devices and Simple Circuits 2. which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. the base region is both narrow and lightly doped.Semiconductor Electronics: Materials. β′ and after amplification its value A(vo. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C. (d) All the above. Which of the statements given in Exercise 14. For stable oscillations. Carbon. If a fraction β′ is feedback. (Eg)Si and (Eg)Ge. If the feedback voltage is in opposite phase (negative feedback). (b) Electrons are minority carriers and pentavalent atoms are the dopants. In compound semiconductors. However.3 14. silicon and germanium have four valence electrons each. the gain is less than 1 and it can never work as oscillator.β′ ) should be equal to Vo. the voltage feedback (Vfb) from the output voltage (Vo) should be such that after amplification (A) it should again become Vo.4 509 . 3. in ideal GaAs the ratio of Ga:As is 1:1 but in Ga-rich or As-rich GaAs it could respectively be Ga1.2 14. In transistors. 5. the negative feedback also reduces noise and distortion in an amplifier which is an advantageous feature.1 In an n-type silicon. then Vfb = Vo. This means that the criteria for stable oscillations to be sustained is A β′ = 1.9 or Ga0. It will be an amplifier with reduced gain.1 As0. We have described an oscillator as a positive feedback amplifier. holes diffuse from the p-region to n-region because (a) free electrons in the n-region attract them. Which of the following statements is true? (a) (Eg)Si < (Eg)Ge < (Eg)C (b) (Eg)C < (Eg)Ge > (Eg)Si (c) (Eg)C > (Eg)Si > (Eg)Ge (d) (Eg)C = (Eg)Si = (Eg)Ge In an unbiased p-n junction. emitter in CE-configuration) will not be able to reach the collector. the change in relative stoichiometric ratio can also change the type of semiconductor. otherwise the electrons or holes coming from the input side (say.1 is true for p-type semiconductos. the n-type or p-type semiconductors are obtained by introducing ‘dopants’ as defects.9 As1. the feedback is in the same phase (positive feedback). (c) hole concentration in p-region is more as compared to n-region. the presence of defects control the properties of semiconductors in many ways. (c) Holes are minority carriers and pentavalent atoms are the dopants.1. 4. In an oscillator. 14. EXERCISES 14. (d) Holes are majority carriers and trivalent atoms are the dopants. In elemental semiconductors (Si or Ge). emitter and collector regions should have similar size and doping concentrations. (d) Both the emitter junction as well as the collector junction are forward biased. the voltage gain (a) remains constant for all frequencies. (c) The emitter junction is forward biased and collector junction is reverse biased. 14. 14. (d) None of the above. Its hole mobility is much smaller than electron mobility and independent of temperature. (d) None of the above. the audio signal voltage across the collected resistance of 2 kΩ is 2 V. Given that ni = 1. 14. . (b) The base region must be very thin and lightly doped.8 eV. (b) is high at high and low frequencies and constant in the middle frequency range.6 For transistor action. The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. (c) lowers the potential barrier.10 Two amplifiers are connected one after the other in series (cascaded). If the input signal is 0. if the base resistance is 1 kΩ. (b) reduces the majority carrier current to zero.9 For a CE-transistor amplifier. 14.7 For a transistor amplifier. Can it detect a wavelength of 6000 nm? 14. 14. 14. which of the following statements are correct: (a) Base. Is the material n-type or p-type? 14. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Suppose the current amplification factor of the transistor is 100.01 volt.Physics When a forward bias is applied to a p-n junction. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration ni is given by Eg   n i = n 0 exp  –  2k B T   510 where n0 is a constant.5 ADDITIONAL EXERCISES 14.5 × 1016 m–3. it (a) raises the potential barrier.12 The number of silicon atoms per m3 is 5 × 10 28.13 In an intrinsic semiconductor the energy gap Eg is 1. what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency. calculate the output ac signal.8 In half-wave rectification. (c) is low at high and low frequencies and constant at mid frequencies. find the input signal voltage and base current.11 A p-n photodiode is fabricated from a semiconductor with band gap of 2. Calculate the number of electrons and holes.2eV. 18 Write the truth table for circuit given in Fig. 14.16 Write the truth table for a NAND gate connected as given in Fig. 14.45. which consist of NAND gates.47 below consisting of NOR gates and identify the logic operation (OR.46 14.44. V is the voltage across the diode and is positive for forward bias and negative for reverse bias. Devices and Simple Circuits 14. the current I can be expressed as  eV  – 1 I = I 0 exp   2k B T  where I0 is called the reverse saturation current. If for a given diode I0 = 5 × 10–12 A and T = 300 K. Identify the logic operation carried out by the two circuits. NOT) which this circuit is performing.45 Hence identify the exact logic operation carried out by this circuit.44 14. FIGURE 14.17 You are given two circuits as shown in Fig.14 In a p-n junction diode. then (a) What will be the forward current at a forward voltage of 0.6 V? (b) What will be the increase in the current if the voltage across the diode is increased to 0. k B is the Boltzmann constant (8. and I is the current through the diode. FIGURE 14.7 V? (c) What is the dynamic resistance? (d) What will be the current if reverse bias voltage changes from 1 V to 2 V? 14.Semiconductor Electronics: Materials.46. AND. 511 . Show that circuit (a) acts as OR gate while the circuit (b) acts as AND gate. 14.6×10 –5 eV/K) and T is the absolute temperature. 14.15 You are given the two circuits as shown in Fig. 14. FIGURE 14. 48 512 . AND.48 consisting of NOR gates only. 14. AND. Identify the logic operations (OR. B = 1 then A and B inputs of second NOR gate will be 0 and hence Y=1.47 (Hint: A = 0.) 14. NOT) performed by the two circuits. FIGURE 14. NOT gates and find the correct one.19 Write the truth table for the circuits given in Fig. Similarly work out the values of Y for other combinations of A and B.Physics FIGURE 14. Compare with the truth table of OR. to meet the growing demands in terms of speed and complexity of information. Modern communication has its roots in the 19th and 20th century in the work of scientists like J. production and deduction of amplitude modulation. For communication to be successful. every communication system has three essential elements- . F. The aim of this chapter is to introduce the concepts of communication. it is essential that the sender and the receiver understand a common language. namely the mode of communication. Man has constantly made endeavors to improve the quality of communication with other human beings. Morse. Languages and methods used in communication have kept evolving from prehistoric to modern times.C.B. G.1 INTRODUCTION Communication is the act of transmission of information. We can hope to see many more accomplishments in the coming decades. the need for modulation. Marconi and Alexander Graham Bell. Every living creature in the world experiences the need to impart or receive information almost continuously with others in the surrounding world. 15. The pace of development seems to have increased dramatically after the first half of the 20th century.2 ELEMENTS OF A COMMUNICATION SYSTEM Communication pervades all stages of life of all living creatures. It would be worthwhile to look at the major milestones in events that promoted developments in communications.1.Chapter Fifteen COMMUNICATION SYSTEMS 15. as presented in Table 15. Irrespective of its nature. Bose. 1835 Invention of telegraph by Samuel F.R.C. Event The reporting of the delivery of a child by queen using drum beats from a distant place to King Akbar. ARPANET was a project undertaken by the U.(Alexander Bain) ARPANET .Physics TABLE 15. 1876 1895 1936 1955 1968 1975 1989-91 514 . It resulted in tremendous growth of messages through post offices and reduced physical travel of messengers considerably. defence department.S. (wireless) First television broadcast by BBC The idea of FAX transmission was patented by Alexander Bain in 1843. It allowed file transfer from one computer to another connected to the network.B. It meant a giant leap – from an era of communication using wires to communicating without using wires. Fiber optical systems are superior and more economical compared to traditional communication systems. Remarks It is believed that minister Birbal experimented with the arrangement to decide the number of drummers posted between the place where the queen stayed and the place where the king stayed.the first internet came into existence(J. Morse and Sir Charles Wheatstone Telephone invented by Alexander Graham Bell and Antonio Meucci Jagadis Chandra Bose and Guglielmo Marconi demonstrated wireless telegraphy.1 SOME MAJOR MILESTONES IN THE HISTORY OF COMMUNICATION Year Around 1565 A. WWW may be regarded as the mammoth encyclopedia of knowledge accessible to everyone round the clock throughout the year. Perhaps the most widely used means of communication in the history of mankind.D. Licklider) Fiber optics developed at Bell Laboratories Tim Berners-Lee invented the World Wide Web. Television broadcast(John Logi Baird) First radio FAX transmitted across continent. a channel may be in the form of wires or cables connecting the transmitter and the receiver or it may be wireless. There are two basic modes of communication: point-to-point and broadcast.Communication System transmitter. It would be easy to understand the principles underlying any communication. Depending upon the type of communication system. The block diagram shown in Fig. Moreover. medium/channel and receiver.1 depicts the general form of a communication system. the transmitter is located at one place. If the output of the information source is a non-electrical signal like a voice signal. The receiver has the task of operating on the received signal. FIGURE 15.3 B ASIC TERMINOLOGY USED COMMUNICATION SYSTEMS IN ELECTRONIC By now. noise. Radio and television are examples of broadcast mode of communication. 15. When a transmitted signal propagates along the channel it may get distorted due to channel imperfection. there are a large number of receivers corresponding to a single transmitter. It reconstructs a recognisable form of the original message signal for delivering it to the user of information. in the broadcast mode.1 Block diagram of a generalised communication system. The purpose of the transmitter is to convert the message signal produced by the source of information into a form suitable for transmission through the channel. the receiver is located at some other place (far or near) separate from the transmitter and the channel is the physical medium that connects them. 515 . a transducer converts it to electrical form before giving it as an input to the transmitter. we have become familiar with some terms like information source. In point-to-point communication mode. Telephony is an example of such a mode of communication. transmitter. if we get ourselves acquainted with the following basic terminology. communication takes place over a link between a single transmitter and a receiver. receiver. In contrast. noise adds to the transmitted signal and the receiver receives a corrupted version of the transmitted signal. channel. 15. etc. In a communication system. There are several coding schemes useful for digital communication. 2. JAGADIS CHANDRA BOSE (1858 – 1937) 516 * In BCD. ** It is a character encoding in terms of numbers based on English alphabet since the computer can only understand numbers. They are essentially single-valued functions of time. Bose also invented highly sensitive instruments for the detection of minute responses by living organisms to external stimulii and established parallelism between animal and plant tissues. Bose published three papers in the British magazine. 1895. The source generating the noise may be located inside or outside the system. 4 in the decimal system are written as 0000. 0010. Sound and picture signals in TV are analog in nature. (vi) Attenuation: The loss of strength of a signal while propagating through a medium is known as attenuation. (iv) Transmitter: A transmitter processes the incoming message signal so as to make it suitable for transmission through a channel and subsequent reception. (iii) Noise: Noise refers to the unwanted signals that tend to disturb the transmission and processing of message signals in a communication system.Physics (i) Transducer: Any device that converts one form of energy into another can be termed as a transducer. Jagadis Chandra Bose (1858 – 1937) He developed an apparatus for generating ultrashort electro-magnetic waves and studied their quasioptical properties. (v) Receiver: A receiver extracts the desired message signals from the received signals at the channel output. They employ suitable combinations of number systems such as the binary coded decimal (BCD)*. temperature. (ii) Signal: Information converted in electrical form and suitable for transmission is called a signal. force. Signals can be either analog or digital. American Standard Code for Information Interchange (ASCII)** is a universally popular digital code to represent numbers. His invention was published in the ‘Proceedings of The Royal Society’ on 27 April 1899 over two years before Marconi’s first wireless communication on 13 December 1901. letters and certain characters. 1. All other analog signals can be fully understood in terms of their sine wave components. For example the numbers 0. Digital signals are those which can take only discrete stepwise values. ‘0’ corresponds to a low level and ‘1’ corresponds to a high level of voltage/ current. Sine wave is a fundamental analog signal. . 0001. displacement. Binary system that is extensively used in digital electronics employs just two levels of a signal. Analog signals are continuous variations of voltage or current. He was said to be the first to employ a semiconductor like galena as a selfrecovering detector of electromagnetic waves. we usually come across devices that have either their inputs or outputs in the electrical form. ‘The Electrician’ of 27 Dec. An electrical transducer may be defined as a device that converts some physical variable (pressure. etc) into corresponding variations in the electrical signal at its output. 0011 and 0100. 1000 would represent eight. In electronic communication systems. 3. a digit is usually represented by four binary (0 or 1) bits. frequency range 300 Hz to 3100 Hz is considered adequate. Modulation: The original low frequency message/information signal cannot be transmitted to long distances because of reasons given in Section 15. which acts as a carrier of the information. To transmit music. As will be explained later. Range: It is the largest distance between a source and a destination up to which the signal is received with sufficient strength. 15. Amplification is done at a place between the source and the destination wherever signal strength becomes weaker than the required strength.7. The energy needed for additional signal strength is obtained from a DC power source. 517 . 15. This is the reverse process of modulation. This process is known as modulation. A repeater.2. Demodulation: The process of retrieval of information from the carrier wave at the receiver is termed demodulation. picture or computer data. (viii) (ix) (x) (xi) (xii) FIGURE 15. For speech signals. Amplification is necessary to compensate for the attenuation of the signal in communication systems. amplifies and retransmits it to the receiver sometimes with a change in carrier frequency. Therefore speech signal requires a bandwidth of 2800 Hz (3100 Hz – 300 Hz) for commercial telephonic communication.Communication System (vii) Amplification: It is the process of increasing the amplitude (and consequently the strength) of a signal using an electronic circuit called the amplifier (reference Chapter 14). Bandwidth: Bandwidth refers to the frequency range over which an equipment operates or the portion of the spectrum occupied by the signal.4 BANDWIDTH OF SIGNALS In a communication system. A communication satellite is essentially a repeater station in space. information contained in the low frequency message signal is superimposed on a high frequency wave. FM and PM. abbreviated as AM. picks up the signal from the transmitter. at the transmitter. Therefore. The type of communication system needed for a given signal depends on the band of frequencies which is considered essential for the communication process. Repeaters are used to extend the range of a communication system as shown in Fig. there are several types of modulation. music. Each of these signals has different ranges of frequencies. the message signal can be voice.2 Use of repeater station to increase the range of communication. Repeater: A repeater is a combination of a receiver and a transmitter. Physics an approximate bandwidth of 20 kHz is required because of the high frequencies produced by the musical instruments. The International Telecommunication Union (ITU) administers the present system of frequency allocations. for practical purposes. Optical communication using fibers is performed in the frequency range of 1 THz to 1000 THz (microwaves to ultraviolet). free space and fiber optic cable. Video signals for transmission of pictures require about 4. 3 ν 0 .3 Approximation of a rectangular wave in terms of a the bandwidth.5 BANDWIDTH OF TRANSMISSION MEDIUM 518 Similar to message signals. different types of transmission media offer different bandwidths. and fundamental (ν0 ) + second harmonic (2ν0 ) + third harmonic (3ν0 ). we have considered only analog signals. the contribution from higher harmonics can be neglected.3.2 MHz of bandwidth. are shown in the same figure to illustrate this fact.. fundamental (ν0 ) + second harmonic (2ν0 ). nν0 where n is an integer extending to infinity and ν0 = 1/T0. This is so because the higher the harmonic. Communication through free space using radio waves takes place over a very wide range of frequencies: from a few hundreds of kHz to a few GHz. One can show that this rectangular wave can be decomposed into a superposition of sinusoidal waves of frequencies ν0.. 2 ν 0 . The audible range of frequencies extends from 20 Hz to 20 kHz. However. 15.. A TV signal contains both voice and picture and is usually allocated 6 MHz of bandwidth for transmission. received waves are a distorted version of the transmitted one. fundamental sine wave and its harmonics. the information is not lost and the rectangular signal is more or less recovered. less is its contribution to the wave form.. 4ν0 . Spectrum allocations are arrived at by an international agreement. 3ν0. In the preceeding paragraph. If the bandwidth is large enough to accommodate a few harmonics. which implies an infinite bandwidth. thus limiting FIGURE 15. Such cables are normally operated below 18 GHz. 15. This range of frequencies is further subdivided and allocated for various services as indicated in Table 15. The fundamental (ν0 ). 2ν0. 4 ν0. . which offers a bandwidth of approximately 750 MHz. An optical fiber can offer a transmission bandwidth in excess of 100 GHz.. The commonly used transmission media are wire. It is clear that to reproduce the rectangular wave shape exactly we need to superimpose all the harmonics ν 0 .2. As a result. Coaxial cable is a widely used wire medium. Digital signals are in the form of rectangular waves as shown in Fig. 2 GHz VHF (very high frequencies) TV UHF (ultra high frequencies) TV Mobile to base station Base station to mobile Uplink Downlink Comments 15. For such antennas. The maximum range of coverage depends on the transmitted power and frequency (less than a few MHz). The mode of propagation is called surface wave propagation and the wave glides over the surface of the earth.Communication System TABLE 15.925-6.425 GHz 3. Several factors influence the propagation of em waves and the path they follow.2 SOME IMPORTANT WIRELESS COMMUNICATION FREQUENCY BANDS Service Standard AM broadcast FM broadcast Television Frequency bands 540-1600 kHz 88-108 MHz 54-72 MHz 76-88 MHz 174-216 MHz 420-890 MHz Cellular Mobile Radio Satellite Communication 896-901 MHz 840-935 MHz 5. ground based vertical towers are generally used as transmitting antennas. at lower frequencies). the strength of the wave keeps on decreasing.6.6 PROPAGATION OF ELECTROMAGNETIC WAVES In communication using radio waves. In standard AM broadcast. which travel through the space and reach the receiving antenna at the other end. At this point. 519 .. ground has a strong influence on the propagation of the signal.7-4.1 Ground wave To radiate signals with high efficiency. A wave induces current in the ground over which it passes and it is attenuated as a result of absorption of energy by the earth. the antennas have large physical size and they are located on or very near to the ground.e. The attenuation of surface waves increases very rapidly with increase in frequency.3. As the em wave travels away from the transmitter. it is also important to understand the composition of the earth’s atmosphere as it plays a vital role in the propagation of em waves. 15. the antennas should have a size comparable to the wavelength λ of the signal (at least ~ λ/4). A brief discussion on some useful layers of the atmosphere is given in Table 15. an antenna at the transmitter radiates the Electromagnetic waves (em waves). At longer wavelengths (i. the details of which are given in Table 15.4. Ionisation occurs due to the absorption of the ultraviolet and other high-energy radiation coming from the sun by air molecules. merges with F2 at night Day and night Frequencies most affected VHF (up to several GHz) Reflects LF. The ionosphere is further subdivided into several layers. . Close to the earth. The phenomenon of bending of em waves so that they are diverted towards the earth is similar to total internal reflection in optics*. However.Physics TABLE 15. even though the molecular concentration is very high.3.3 DIFFERENT LAYERS OF ATMOSPHERE AND THEIR INTERACTION WITH THE PROPAGATING ELECTROMAGNETIC WAVES Name of the stratum (layer) Troposphere D (part of stratosphere) E (part of Stratosphere) F1 (Part of Mesosphere) F2 (Thermosphere) Approximate height over earth’s surface 10 km P A R T S O F I O N O S P H E R E Exists during Day and night Day only Day only Daytime. 250-400 km during daytime 15. absorbs MF and HF to some degree Helps surface waves.6. there occurs a peak of ionisation density. at some intermediate heights. particularly at night 65-75 km 100 km 170-190 km 300 km at night. The ionospheric layer acts as a reflector for a certain range of frequencies (3 to 30 MHz). 520 * Compare this with the phenomenon of mirage. long distance communication can be achieved by ionospheric reflection of radio waves back towards the earth. the radiation intensity is low so that the ionisation is again low. This mode of propagation is called sky wave propagation and is used by short wave broadcast services. Electromagnetic waves of frequencies higher than 30 MHz penetrate the ionosphere and escape. The degree of ionisation varies with the height. 15. These phenomena are shown in the Fig. At great heights the solar radiation is intense but there are few molecules to be ionised.2 Sky waves In the frequency range from a few MHz up to 30 to 40 MHz. It extends from a height of ~ 65 Km to about 400 Km above the earth’s surface. The density of atmosphere decreases with height. reflects HF Partially absorbs HF waves yet allowing them to reach F2 Efficiently reflects HF waves. The ionosphere is so called because of the presence of a large number of ions or charged particles. 4 Sky wave propagation.5 Line of sight communication by space waves.5. FIGURE 15. At these frequencies.6.Communication System FIGURE 15.3.3 Space wave Another mode of radio wave propagation is by space waves. If the transmitting antenna is at a height hT. Because of line-of-sight nature of propagation. Space waves are used for line-of-sight (LOS) communication as well as satellite communication. 15. 521 . 15. At frequencies above 40 MHz. then you can show that the distance to the horizon dT is given as . 15. direct waves get blocked at some point by the curvature of the earth as illustrated in Fig. communication is essentially limited to line-of-sight paths. the antennas are relatively smaller and can be placed at heights of many wavelengths above the ground. where R is the radius of the earth (approximately 6400 km). With reference to Fig. A space wave travels in a straight line from transmitting antenna to the receiving d T = 2Rh T antenna. If the signal is to be received beyond the horizon then the receiving antenna must be high enough to intercept the line-of-sight waves.5 the maximum line-of-sight distance dM between the two antennas having heights hT and hR above the earth is given by dM = 2Rh T + 2Rh R (15. dT is also called the radio horizon of the transmitting antenna.1) where hR is the height of receiving antenna. The layer nomenclature is given in Table 15. . No signal. is a single frequency sinusoid.1 A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna is 50 m. Message signals are also called baseband signals.Physics Television broadcast. Solution EXAMPLE 15. but it spreads over a range of frequencies called the signal bandwidth. microwave links and satellite communication are some examples of communication systems that use space wave mode of propagation. as delivered by the source of information. which essentially designate the band of frequencies representing the original signal. What is the maximum distance between them for satisfactory communication in LOS mode? Given radius of earth 6.6 summarises the various modes of wave propagation discussed so far.4 × 106 m. Figure 15. Let us find what factors prevent us from doing so and how we overcome these factors. Example 15.6 Various propagation modes for em waves. FIGURE 15. Suppose we wish to transmit an electronic signal in the audio frequency (AF) range (baseband signal frequency less than 20 kHz) over a long distance directly.1 dm = 2 × 64 × 105 × 32 + 2 × 64 × 105 × 50 m = 64 × 10 × 10 + 8 × 10 × 10 m = 144 × 102 × 10 m = 45. in general.5 km 2 3 15.7 MODULATION AND ITS NECESSITY 522 As already mentioned. the purpose of a communication system is to transmit information or message signals. the effective power radiated by a long wavelength baseband signal would be small. any of the three parameters. 15. This implies that for the same antenna length. A sinusoidal carrier wave can be represented as c(t ) = Ac sin (ωct + φ) (15. viz Ac. i. such a long antenna is not possible to construct and operate. Ac is the amplitude. Obviously. we need high powers and hence this also points out to the need of using high frequency transmission. All these signals will get mixed up and there is no simple way to distinguish between them. Suppose many people are talking at the same time or many transmitters are transmitting baseband information signals simultaneously. increasing frequency. This antenna should have a size comparable to the wavelength of the signal (at least λ/4 in dimension) so that the antenna properly senses the time variation of the signal. The above arguments suggest that there is a need for translating the original low frequency baseband message or information signal into high frequency wave before transmission such that the translated signal continues to possess the information contained in the original signal. ωc ( = 2πνc ) is the angular frequency and φ is the initial phase of the carrier wave.3 Mixing up of signals from different transmitters Another important argument against transmitting baseband signals directly is more practical in nature.7. In doing so. 15.7 (a) Sinusoidal. of the carrier wave can be controlled by the message or 523 . Therefore. For an electromagnetic wave of frequency 20 kHz. The carrier wave may be continuous (sinusoidal) or in the form of pulses as shown in Fig. This points out towards a possible solution by using communication at high frequencies and allotting a band of frequencies to each message signal for its transmission. For a good transmission. During the process of modulation. and a process known as modulation which attaches information to it. the wavelength λ is 15 km.2 Effective power radiated by an antenna A theoretical study of radiation from a linear antenna (length l ) shows that the power radiated is proportional to (l/λ)2 .7. Hence direct transmission of such baseband signals is not practical.e. ωc and φ. We can obtain transmission with reasonable antenna lengths if transmission frequency is high (for example. then λ is 300 m).2) where c(t) is the signal strength (voltage or current). FIGURE 15. the power radiated increases with decreasing λ. and (b) pulse shaped signals. Hence. if ν is 1 MHz..7.7. there is a need of translating the information contained in our original low frequency baseband signal into high or radio frequencies before transmission. we take the help of a high frequency signal. 15.1 Size of the antenna or aerial For transmitting a signal. known as the carrier wave. we need an antenna or an aerial.Communication System 15. 15. 15. (15.3) Ac   Note that the modulated signal now contains the message signal. (b) a modulating signal. (d) frequency modulation.8 AMPLITUDE MODULATION In amplitude modulation the amplitude of the carrier is varied in accordance with the information signal. and pulse position (denoting the time of rise or fall of the pulse amplitude) as shown in Fig. Here we explain amplitude modulation process using a sinusoidal signal as the modulating signal.8(c). In this chapter. This results in three types of modulation: (i) Amplitude modulation (AM). (b) pulse duration modulation (PDM) or pulse width modulation (PWM). and (c) pulse position modulation (PPM). the significant characteristics of a pulse are: pulse amplitude.3).in/?sub=59&brch=163&sim=259&cnt=358 Modulation and Demodulation FIGURE 15. pulse duration or pulse Width.8 Modulation of a carrier wave: (a) a sinusoidal carrier wave. different types of pulse modulation are: (a) pulse amplitude modulation (PAM).vlab. (c) amplitude modulation. (ii) Frequency modulation (FM) and (iii) Phase modulation (PM). Let c(t) = Ac sin ωct represent carrier wave and m(t) = Am sin ωmt represent the message or the modulating signal where ωm = 2πfm is the angular frequency of the message signal. 15. Similarly.7(b). 524 cm (t ) = Ac sin ωct + µ Ac sin ωm t sin ωct (15. Hence. This can also be seen from Fig. we can write. as shown in Fig. we shall confine to amplitude modulation on ly.Physics information signal. http://iitg. From Eq. and (e) phase modulation.co.8. 15.4) . The modulated signal cm (t ) can be written as cm (t) = (Ac + Am sin ωmt) sin ωct   A = Ac 1 + m sin ωm t  sin ωc t (15. Communication System Here µ = Am/Ac is the modulation index.2 A message signal of frequency 10 kHz and peak voltage of 10 volts is used to modulate a carrier of frequency 1 MHz and peak voltage of 20 volts. we can write cm (t) of Eq. (b) the side bands produced. cm (t ) = Ac sin ω c t + µ Ac cos(ω c − ωm ) t − µ Ac FIGURE 15. 15.9. Solution (a) Modulation index =10/20 = 0.10.9 PRODUCTION OF AMPLITUDE MODULATED WAVE Amplitude modulation can be produced by a variety of methods. (15. The frequency spectrum of the amplitude modulated signal is shown in Fig. Using the trignomatric relation sinA sinB = ½ (cos(A – B) – cos (A + B). different stations can operate without interfering with each other.9 A plot of amplitude versus ω for an amplitude modulated signal. A conceptually simple method is shown in the block diagram of Fig.10 Block diagram of a simple modulator for obtaining an AM signal. 525 . Determine (a) modulation index.µ is kept ≤ 1 to avoid distortion. Example 15.4) as cos(ω c + ωm )t (15.2 15.5) 2 2 Here ωc – ωm and ωc + ωm are respectively called the lower side and upper side frequencies. known as side bands. in practice. The modulated signal now consists of the carrier wave of frequency ωc plus two sinusoidal waves each with a frequency slightly different from. 15. EXAMPLE 15. As long as the broadcast frequencies (carrier waves) are sufficiently spaced out so that sidebands do not overlap. FIGURE 15.5 (b) The side bands are at (1000+10 kHz)=1010 kHz and (1000 –10 kHz) = 990 kHz. This signal x (t) = Am sinωmt + Ac sinωct is passed through a square law device which is a non-linear device which produces an output y (t ) = B x (t ) + Cx 2 (t ) (15.8) where the trigonometric relations sin A = (1 – cos2A)/2 and the relation for sin A sin B mentioned earlier are used.7) = BAm sin ωmt + BAc sin ωct 2 2 2 C Am C Am C Ac 2 – cos 2ωm t – cos 2ω c t + Ac 2 2 2 + CAm Ac cos (ωc – ωm ) t – CAm Ac cos (ωc+ ωm ) t + (15. 15. .11 Block diagram of a transmitter. FIGURE 15.Physics Here the modulating signal Am sin ωmt is added to the carrier signal Ac sin ωct to produce the signal x (t). 2ωc.8). It is to be mentioned that the modulated signal cannot be transmitted as such. (15.6) where B and C are constants. there is a dc term C/2 Am 2 ( ) and sinusoids of frequencies ωm. to facilitate further processing. The receiving antenna is therefore to be followed by an amplifier and a detector. the carrier frequency is usually changed to a lower frequency by what is called an intermediate frequency (IF) stage preceding the detection. 15.10 this signal is passed through a band pass filter* which rejects dc and the sinusoids of frequencies ωm . ωc – ωm and ωc + ωm. 2ωm. 15. In addition. The modulator is to be followed by a power amplifier which provides the necessary power and then the modulated signal is fed to an antenna of appropriate size for radiation as shown in Fig. ωc – ωm and ωc + ωm.11. As shown in Fig.10 DETECTION OF AMPLITUDE MODULATED WAVE The transmitted message gets attenuated in propagating through the channel. y (t )= BAm sin ωmt + BAc sin ωct 2 2 2 2 +C   Am sin ωm t + Ac sin ωct + 2 Am Ac sin ωm t sin ωct   (15. Thus. (15. The detected signal may not be strong enough to be made use of and hence is required 526 * A band pass filter rejects low and high frequencies and allows a band of frequencies to pass through. ωc. 2 2 + Ac In Eq. 2ωm and 2 ωc and retains the frequencies ωc. The output of the band pass filter therefore is of the same form as Eq.5) and is therefore an AM wave. 15.13 is passed through a rectifier to produce the output shown in (b). 15. it is worth taking a glance at some of the communication systems (see the box) that in recent times have brought major changes in the way we exchange information even in our day-to-day life: 527 . In the present chapter we have discussed some basic concepts of communication and communication systems. The quantity on y-axis can be current or voltage. the signal is passed through an envelope detector (which may consist of a simple RC circuit). a simple method is shown in the form of a block diagram in Fig. FIGURE 15. The modulated signal of the form given in (a) of fig. These and other exciting developments are taking place everyday. We just saw that the modulated carrier wave contains the frequencies ωc and ωc ± ωm. 15. Detection is the process of recovering the modulating signal from the modulated carrier wave. So far we have restricted our discussion to some basic communication systems. A block diagram of a typical receiver is shown in Fig. In order to retrieve m (t ).Communication System to be amplified. In order to obtain the original message signal m(t ) of angular frequency ωm . Before we conclude this chapter. This envelope of signal (b) is the message signal.12 Block diagram of a receiver.12 FIGURE 15.13 Block diagram of a detector for AM signal.13. Other forms of modulation and digital communication systems play an important role in modern communication. We have also discussed one specific type of analog modulation namely Amplitude Modulation (AM). This procedure is called handover or handoff. Everyone belonging to the chat group gets the message instantaneously and can respond rapidly. it is necessary for the mobile user to be transferred to another base station. (iii) World Wide Web (WWW) – Computers that store specific information for sharing with others provide websites either directly or through web service providers. to the extent that the consumer does not even notice it. Customers view images and receive all the information about various products or services of companies through their websites. This process is carried out very rapidly. Hypertext is a powerful feature of the web that automatically links relevant information from one page on the web to another using HTML (hypertext markup language). Yahoo! etc. Facsimile (FAX) It scans the contents of a document (as an image. NGO’s (Non-Government Organisations) and individuals can post information about their activities for restricted or free use on their websites. Several search engines like Google. Mobile telephones operate typically in the UHF range of frequencies (about 800-950 MHz). Mobile telephony The concept of mobile telephony was developed first in 1970’s and it was fully implemented in the following decade. When a mobile receiver crosses the coverage area of one base station. Each cell contains a low-power transmitter called a base station and caters to a large number of mobile receivers (popularly called cell phones). Each cell could have a service area of a few square kilometers or even less depending upon the number of customers. (v) Chat – Real time conversation among people with common interests through typed messages is called chat. (ii) File transfer – A FTP (File Transfer Programmes) allows transfer of files/software from one computer to another connected to the Internet. It permits communication and sharing of all types of information between any two or more computers connected through a large and complex network. (iv) E-commerce – Use of the Internet to promote business using electronic means such as using credit cards is called E-commerce. This information becomes accessible to the users. help us in finding information by listing the related websites. It was started in 1960’s and opened for public use in 1990’s. companies. These signals are then sent to the destination (another FAX machine) in an orderly manner using telephone lines.Physics ADDITIONAL INFORMATION The Internet It is a system with billions of users worldwide. With the passage of time it has witnessed tremendous growth and it is still expanding its reach. We can write a letter and send it to the recipient through ISP’s (Internet Service Providers) who work like the dispatching and receiving post offices. Its applications include (i) E mail – It permits exchange of text/graphic material using email software. the signals are reconverted into a replica of the original document. 528 . The central concept of this system is to divide the service area into a suitable number of cells centred on an office called MTSO (Mobile Telephone Switching Office). Government departments. Note that FAX provides image of a static document unlike the image provided by television of objects that might be dynamic. Goods are dispatched or services are provided by the company through mail/courier. They can do on-line shopping from home/office. not text) to create electronic signals. At the destination. ωc and (ω c + ωm ).Communication System SUMMARY 1. 7. The radiated signals propagate as electromagnetic waves and the mode of propagation is influenced by the presence of the earth and its atmosphere. Pulse Duration Modulation (PDM) or Pulse Width Modulation (PWM) and Pulse Position Modulation (PPM). 4. which is referred to as the bandwidth of the system. Space waves are used for line-of-sight communication and satellite communication. 8. Electronic communication refers to the faithful transfer of information or message (available in the form of electrical voltage and current) from one point to another point. the range dT is given by 2Rh T where R is the radius of the earth. Sky wave propagation takes place up to frequency of about 30 MHz. some characteristic of the carrier signal like amplitude. Two important forms of communication system are: Analog and Digital. If an antenna radiates electromagnetic waves from a height hT. Amplitude modulated signal contains frequencies (ω c – ω m ). The information to be transmitted is generally in continuous waveform for the former while for the latter it has only discrete or quantised levels. which are necessary for satisfactory transmission of the information contained in the signal. they are called Amplitude Modulated (AM). Every message signal occupies a range of frequencies. Near the surface of the earth. is carried out using a rectifier and an envelope detector. 3. transmission channel and receiver are three basic units of a communication system. Long distance communication between two points on the earth is achieved through reflection of electromagnetic waves by ionosphere. they are superimposed on a high frequency carrier signal by a process known as modulation. signals are radiated into space using devices called antennas. electromagnetic waves propagate as surface waves. Such waves are called sky waves. 10. For transmission over long distances. AM detection. In modulation. Transmitter. 2. 9. followed by a band pass filter. which is the process of recovering the modulating signal from an AM waveform. Low frequencies cannot be transmitted to long distances. 13. then 11. 529 . The bandwidth of a message signal refers to the band of frequencies. Amplitude modulated waves can be produced by application of the message signal and the carrier wave to a non-linear device. any practical communication system permits transmission of a range of frequencies only. Above this frequency. Therefore. frequency or phase varies in accordance with the modulating or message signal. Frequency Modulated (FM) or Phase Modulated (PM) waves. Similarly. Correspondingly. Pulse modulation could be classified as: Pulse Amplitude Modulation (PAM). 6. 5. 12. electromagnetic waves essentially propagate as space waves. Surface wave propagation is useful up to a few MHz frequencies. 4 15. (ii). 3. (ii) and (iii) but not (iv) (d) All of (i). (b) Sky waves.6 . 15. noise gets added to the signal anywhere between the information source and the receiving end.3 15.1 Which of the following frequencies will be suitable for beyond-thehorizon communication using sky waves? (a) 10 kHz (b) 10 MHz (c) 1 GHz (d) 1000 GHz 15. How much service area can it cover if the receiving antenna is at the ground level? A carrier wave of peak voltage 12V is used to transmit a message signal. new frequencies called sidebands are generated on either side (higher and lower than the carrier frequency) of the carrier by an amount equal to the highest modulating frequency.Physics POINTS TO PONDER 1. In the process of transmission of message/ information signal. Which of the above statements are true? (a) (i) and (ii) only (b) (ii) and (iii) only (c) (i). and (iv) can utilize decimal as well as binary systems. Is it possible to retrieve the message by transmitting (a) only the side bands. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%? A modulating signal is a square wave. (d) Space waves. Can you think of some sources of noise? In the process of modulation. (c) Surface waves. Digital signals (i) do not provide a continuous set of values. What will happen if µ > 1? 2.14. (iii) and (iv).2 Frequencies in the UHF range normally propagate by means of: (a) Ground waves. EXERCISES 15. (b) only one side band? In amplitude modulation. modulation index µ ≤ 1 is used.5 530 15. (iii) can utilize binary system. as shown in Fig. 15. (ii) represent values as discrete steps. Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for line-of-sight communication? A TV transmitting antenna is 81m tall. µ.14 15. only the upper sideband of an AM wave is transmitted. What would be the value of µ if the minimum amplitude is zero volt? Due to economic reasons.Communication System FIGURE 15. Show that if a device is available which can multiply two signals. the maximum amplitude is found to be 10V while the minimum amplitude is found to be 2V.7 15. Determine the modulation index. (i) Sketch the amplitude modulated waveform (ii) What is the modulation index? For an amplitude modulated wave. but at the receiving station. then it is possible to recover the modulating signal at the receiver station. 531 .8 The carrier wave is given by c (t ) = 2 sin(8πt ) volts. there is a facility for generating the carrier. Physics APPENDICES APPENDIX A 1 THE GREEK ALPHABET © no N C tt E o R be T re pu APPENDIX A 2 COMMON SI PREFIXES AND SYMBOLS FOR MULTIPLES AND SUB-MULTIPLES 532 bl is he d . Appendices Answers APPENDIX A 3 SOME IMPORTANT CONSTANTS © no N C tt E o R be T re pu OTHER USEFUL CONSTANTS 533 bl is he d . v → f (but never beyond f while m → 0).25 cm give ue = –5 cm.47 cm.6 9.8 cm.5. v = 6. Area = 2. image is virtual.3 9.5 9. ngw = 1.7 cm nga = 1. image is erect and virtual.5 cm (not at infinity as one might wrongly think). vO = (15 – 5) cm = 10 cm.5 cm. Note that when the object is placed at the focus of the concave lens (21 cm). As u → ∞. virtual) (a) f = +20 cm. the image is located at 10.59 cm. v → ∞. v = 8. | ue |= v = –54 cm.1 9.5 cm from the lens on its right side. 1.Physics ANSWERS 9. u = +12 cm (object on right. As u → ∞.6 m2 n ≅ 1.10 9. Magnification = 5/9.51.12 = 25 +1 = 11.144.2 cm O 11 2.8 9. A diverging lens of focal length 60 cm (a) v e = –25 cm and fe = 6. for u < f. v → f (but never beyond) while m → 0. The size of the image is 5. As u → f..6181 i. n wa = 1. The image is real. Magnifying power = 13. Image is real and at 7.e.33 ≅ 0.32.4 cm. i. where r is the radius (in m) of the largest circle from which light comes out and ic is the critical angle for water -air inter face. Angular magnification of the eye-piece for image at 25 cm 25 cm = 2. r = 0. the size of the image is 2. (b) f = –16 cm.5 Separation = 9. Magnifying power = 20 (b) u O = – 2.9 9. inverted and magnified..2 9.11 9.75 . which gives sin r = 0.4 © no N C tt E o R be T re pu 9. Image is real and at 48 cm from the lens on its right side.7 cm.7 9.53 and Dm for prism in water ≅ 10º R = 22 cm Here the object is virtual and the image is real.5 cm.e. Magnifying power = 88 534 bl is CHAPTER 9 he d . v = 7. It is diminished to a size 1. r ~ 38 º.33.0 cm. 1. f O = u O = – 2. 27cm .8 × tan i c and sin ic = 1/1. 68 which gives i c takes place when i > 5 9° or when r < rmax = 31°. (c) Taller (d) The apparent depth for oblique viewing decreases from its value for near -normal viewing. There is no contradiction. 21.Answers 9. much less than that of glass.42. It can be seen with an explicit ray diagram that the answer is independent of the location of the slab (for small angles of incidence).5). Therefore. which gives i max © no N C tt E o R be T re pu 9. For fixed distance s between object and screen.5° and i ′ = 53. there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe. Convince yourself by drawing an appropriate ray diagram. 24° to 90°. all incident rays of angles in the range 0 < i < 60° will suffer total internal reflections in the pipe. The convex lens of the eye does just that.16 9.7 cm. The critical angle of diamond is about 24°.13 9.4 cm max max 9.17 (sin i /sin r ) = 1. ~ 60°.19 9. u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror). (a) Rays converging to a point ‘behind’ a plane or convex mirror are reflected to a point in front of the mirror on a screen. Thus. Total internal reflection (a) sin ic′ = 1. (b) When the reflected or refracted rays are divergent. The pin appears raised by 5.5° < i < 90°) will suffer total internal reflections. Note.0 cm. a plane or convex mirror can produce a real image if the object is virtual. u < 0 (object on left) (b) f > 0.20 bl 535 is he d .5°. f < u < 0 to deduce the desired result. i = 90° will have r = 36. i c ′ = sin (1/1. fmax = 0.68 . Thus . to ensure that light entering the diamond is totally reflected from many faces before getting out–thus producing a sparkling effect. the lens equation does not give a real solution for u or v if f is greater than s/4. Now . The divergent rays can be converged on to a screen by means of an appropriate converging lens. Now. the screen here is not located at the position of the virtual image.14 9. Apply mirror equation and the condition: (a) f < 0 (concave mirror). which it is in practice. the image is virtual.44/1.5° which is greater than ic ′.68) = 36. Convince yourself of this fact by drawing ray diagrams for differ ent positions of the observer. The virtual image here serves as an object for the lens to produce a real image. A skilled diamondcutter exploits the larger range of angles of incidence (in the diamond). (e) Refractive index of a diamond is about 2. In other words. ′ = 59°. all incident rays (in the range 53.) –1 (b) If there is no outer coating.15 24. 150 cm (a) Angular magnification = 1500 (b) Diameter of the image = 13. much larger than that of ordinary glass (about 1.75 m. (If the length of the pipe is finite.18 9. u1 = – ∞. This image becomes a real object for the second lens: f 2 = + 30 cm. f1 = 30 cm. Because of the adjustments involved for so many colours. The notion of effective focal length. (b) To disperse without deviation. f1 = 30 cm and u1 = – ∞ . gives v1= 120 cm. and the separation between the lenses). f2 = – 20 cm which gives v 2 = − © no N C tt E o R be T re pu 9. f 2 = –20 cm. v2 = – 220 cm. Magnitude of magnification due to the first (convex) lens is 3.Physics 9. the answer depends on which side of the lens system the parallel beam is incident. The parallel incident beam appears to diverge from a point 216 cm from the centre of the two-lens system . u 2 = + (30 – 8) cm = + 22 cm which gives. Net magnitude of magnification = 0. (b) u1 = – 40 cm. give v1 = – 20 cm.652 Size of the image = 0. u 2 = + (120 – 8) cm = +112 cm (object virtual). choose. therefore.e.21 (a) (i) Let a parallel beam be the incident from the left on the convex lens first. these are not meant to be precise arrangements for the purpose required. the angle of refraction r at the first face is (60°–i c). ic = sin–1 (1/1.4962. i ~ 30° Two identical prisms made of the same glass placed with their bases on opposite sides (of the incident white light) and faces touching (or parallel) will neither deviate nor disperse. say. try flint glass prisms of greater and greater angle) so that deviations due to the two prisms are equal and opposite.524) ~ 41° Therefore.98 cm If the refracted ray in the prism is incident on the second face at the critical angle ic . but will mearly produce a parallel displacement of the beam. give v1 = + 30 cm. (ii) Let the parallel beam be incident from the left on the concave lens first: f1 = – 20 cm. does not seem to be meaningful for this system.. r = 19° sin i = 0. Further we do not have a simple lens equation true for all u (and v) in terms of a definite constant of the system (the constant being determined by f1 and f 2. Now. This image becomes a virtual object for the second lens.22 9. Clearly. (The flint glass prism angle will still be smaller than that of crown glass because flint glass has higher refractive index than that of crown glass). v2 = – 420 cm. the first prism to be of crown glass. 536 bl 112 × 20 cm 92 is he d . and take for the second prism a flint glass prism of suitably chosen refracting angle (smaller than that of crown glass prism because the flint glass prism disperses more) so that dispersion due to the first is nullified by the second. (a) To deviate without dispersion. The parallel incident beam appears to diverge from a point 416 cm on the left of the centre of the two-lens system. increase the angle of flint glass prism (i. u2 = – (20 + 8) cm = – 28 cm which gives.23 Magnitude of magnification due to the second (concave) lens = 20/92. Objects at infinity produce virtual image at 100 cm (using spectacles). v = – 90 cm. but those in the horizontal plane can get the required extra convergence due to refraction by the curved surface of the cylindrical lens if the curvature of the cylindrical surface is chosen appropriately. The defect (called astigmatism) arises because the curvature of the cornea plus eye-lens refracting system is not the same in different planes.28 (a) Closest distance = 4 1 cm ≈ 4.Answers 9. Minimum angular magnification = (25/5) = 5 bl 537 is he d ⎡ 1 3⎤ + ⎢ ⎣ 25 5 ⎥ ⎦ −1 = 25 cm 16 . Myopia arises when the eye-ball from front to back gets too elongated..29 (a) 1 1 1 + = v 9 10 i.27 9. The far point of the person is 100 cm.2 cm 6 9. so sharp images of vertical lines can be formed on the retina. The defect can be corrected by using a cylindrical lens with its axis along the vertical. Clearly .e. No. while his near point may have been normal (about 25 cm). the person uses the ability of accommodation of his eye-lens. the eye lens may also lose some of its ability of accommodation. those which are (or whose images using the spectacles are) between 100 cm and 25 cm. parallel rays in the vertical plane will suffer no extra refraction. [The eye-lens is usually spherical i. In practice.e. the focal length should be 9. To focus an object at the near point (u = –25 cm). the eye uses its least converging power = (40+20) dioptres = 60 dioptres. so horizontal lines appear blurred. hypermetropia arises when it gets too shortened. the person needs converging lens of power + 2 dioptres.24 To see objects at infinity .e. the ‘defect’ is called presbyopia and is corrected in the same manner as hypermetropia. This gives a rough idea of the distance between the retina and cornea-eye lens: (5/3) cm. When the eyeball has the normal length but the eye lens loses partially its ability of accommodation (as happens with increasing age for any normal eye). The range of accommodation of the eye-lens is roughly 20 to 24 dioptres. Magnitude of magnification = 90/9 = 10. in addition. But the curvature is insufficient in the horizontal plane.. on the retina (v = 5/3 cm). To view closer objects i. The power of the eye lens then is (64 – 40) dioptres = 24 dioptres. The near point of the person recedes to 50 cm. This ability usually gets partially lost in old age (presbyopia).. To view objects at 25 cm clearly. the curvature in the vertical plane is enough. a person may have normal ability of accommodation of the eyelens and yet may be myopic or hypermetropic. 9.25 © no N C tt E o R be T re pu 9.] In the present case. has the same curvature on different planes but the cornea is not spherical in case of an astigmatic eye. Farthest distance = 5 cm (b) Maximum angular magnification = [25/(25/6)] = 6.26 corresponding to a converging power of 64 dioptres. 5 Yes.5u − 1 u = 1 10 9. (b) Magnitude of magnification = (25/ |u|) = 3. magnification of the objective is given by vO |u O | 538 = (|u O |/ f O ) − 1 bl 1 is he d .. magnification magnitude is |(v/ u )| and magnifying power is (25/ |u|).] (c) First. the angles subtended at the eye by the first object and its image are not equal. However. (d) Angular magnification of eye-piece is [( 25/fe ) + 1] ( fe in cm ) which increases if f e is smaller. one can increase this limit by a factor of 10 or so. The closer object has larger angular size than the same object at 25 cm.e. grinding lens of very small focal length is not easy.14 cm. The latter is the ratio of the angular size of the object (which is equal to the angular size of the image even if the image is magnified) to the angular size of the object if placed at the near point (25 cm). The effect is negligible if the image is at a very large distance away.25 / 1) = 2. Thus. (a) Even though the absolute image size is bigger than the object size.5 9. More important. The magnifier helps in the following way: without it object would be placed no closer than 25 cm.5 u + 1 (6. if you decrease focal length.8 (c) No.32 i. (a) Maximum magnifying power is obtained when the image is at the near point (25 cm) u = – 7. you cannot get a magnifying power of more than 3 or so with a simple convex lens. (b) Yes. So. Further. with it the object can be placed much closer. in practice. magnification of an image by a lens and angular magnification (or magnifying power) of an optical instrument are two separate things. using an aberration corrected lens system.5. are the two quantities equal. the angular size of the image is equal to the angular size of the object.Physics Each square in the virtual image has an area 10 × 10 × 1 mm2 2 2 = 100 mm = 1 cm (b) Magnifying power = 25/9 = 2. [Note: When the eye is separated from the lens . Magnification = v = +2. the magnifying power (when the image is produced at 25 cm) is equal to the magnitude of magnification. it decreases a little because the angle subtended at the eye is then slightly less than the angle subtended at the lens. It is in this sense that angular magnification is achieved. (c) Magnifying power = 3.30 9. aberrations ( both spherical and chromatic ) become more pronounced. Only when the image is located at the near point |v| = 25 cm. u = – 6 cm | v| = 15 cm The virtual image is closer than the normal near point (25 cm) and cannot be seen by the eye distinctly.31 © no N C tt E o R be T re pu 2. Assume microscope in normal use i. If we position our eyes on the eye-ring and the area of the pupil of our eye is greater or equal to the area of the eye-ring. Angular magnification of the eye-piece + 1= 6 5 Magnification of the objective = = 30 6 =5 25 9. Using the mirror formula. The separation between the objective and the eye-piece should be (7. Angle subtended by the image produced by the objective = h h = f O 140 Equating the two. it is an ideal position for our eyes for viewing.6 fe ⎢ 25 ⎣ ⎦ (a) fO + fe = 145 cm (b) Angle subtended by the tower = (100/3000) = (1/30) rad.67 cm. The focal length of smaller mirror is 70 mm.36 The image formed by the larger (concave) mirror acts as virtual object for the smaller (convex) mirror.25 which gives uO= –1. So |u O | is small. our eyes will collect all the light refracted by the objective. bl 539 is he d . image is formed at 315 mm from the smaller mirror.35 fO ⎡ f ⎤ 1 + O ⎥ = 33. When you view through a microscope by placing your eyes on one end. The microscope is used for viewing very close object. image at 25 cm. (e) The image of the objective in the eye-piece is known as ‘eyering’.34 (b) m = 9. 9. Further the object should be placed 1.the ideal distance between the eyes and eye-piece is usually built-in the design of the instrument.17) cm = 11. | u e|= (25/6) cm = 4. The precise location of the eye-ring naturally depends on the separation between the objective and the eye-piece.7 cm.e.5 cm. (a) m = ( fO/ fe ) = 28 9. Height of the final image (magnitude) = 28 cm.5 + 4.5 cm from the objective to obtain the desired magnification. we shall not collect much of the light and also reduce our field of view.5 cm.33 © no N C tt E o R be T re pu 1 1 1 − = 5u O u O 1. Therefore.. Parallel rays coming from the object at infinity will focus at a distance of 110 mm from the lar ger mirror.17 cm. All the rays from the object refracted by objective go through the eye-ring.Answers |u O | is slightly greater than fO. h = 4. and which is large when so is fO. The distance of virtual object for the smaller mirror = (110 – 20) = 90 mm. If we place our eyes too close to the eye-piece. (c) Magnification (magnitude) of the eye-piece = 6. v0= 7. 86 × 105 m s–1 6563 10.12 In corpuscular (particle) picture of refraction. v = c ( λ' − λ) = v c λ λ 3 × 108 × 15 = 6. speed same as incident light) λ = 589 nm.17 mm 0.3 © no N C tt E o R be T re pu 10. the violet component of white light travels slower than the red component.6 10. [When no particular wavelength or colour of light is specified. v > c. i. nv > n r.56 mm tan –1(1.10 40 m 10. we may take the given refractive index to refer to yellow colour.33 CHAPTER 10 10.28 × 10– 3 4 × 1. 540 bl (b) Refracted light: (frequency same as the incident frequency) ν = 5. Hence d = 18.e.e.2 10. 6 × 1014 Hz.3o 10. n = 1.Physics 9.4 (a) 2. Therefore. ν = 5. The refractive index.26 × 10 m s . and hence the speed of light in a medium. This results in an increase in the normal component of the velocity but the component along the surface is unchanged.09 × 1014Hz 8 –1 v = (c/n) = 2. frequency. Therefore.15° (b) 1. c = 3.5) ~ 56. d/1.. λ = (v/ν ) = 444 nm (a) Spherical (b) Plane (c) Plane (a small area on the surface of a large sphere is nearly planar).5 10.11 Use the formula λ ′ – λ = i. particles of light incident from a rarer to a denser medium experience a force of attraction normal to the surface.0 × 108 m s–1 (b) No. Since n > 1.5 = tan 7°. is he d .8 K/4 (a) 1.4 cm.1 (a) Reflected light: (wavelength.2 ×10 – 2 × 0.] Now we know violet colour deviates mor e than red in a glass prism. 45° 10.09 × 1014 Hz. depends on wavelength.7 10. This means c sin i = v sin r or v c = sin i sin r = n .00 × 108 m s–1 10.37 9. λ= 1.38 The reflected rays get deflected by twice the angle of rotation of the mirror.4 m = 600 nm 10.9 5000 Å. etc. Next draw the locations of this same wavefront after a time t in the presence of the mirror.4 × 10 m.Answers The prediction is opposite to the experimental results ( v < c ). (c) Waves dif fracted from the edge of the cir cular obstacle interfere constructively at the centre of the shadow producing a bright spot. we cannot expect Doppler formulas for sound to be identical for (i) and (ii). (d) For diffraction or bending of waves by obstacles/apertures by a large angle. This fact is a basic axiom of Einstein’s special theory of relativity.. diffraction is by a small angle. and in the absence of the mirr or . however. there is clearly nothing to distinguish between (i) and (ii).16 3. You will get two ar cs symmetrically located on either side of the mirror. For light waves in vacuum. e. 10. Here only the relative motion between the source and the observer counts and the relativistic Doppler formula is the same for (i) and (ii). the centre of the reflected wavefront (the image of the object) is seen to be at the same distance from the mirror as the object. (iii) independent of the motion of the source relative to the medium but depends on the motion of the observer relative to the medium. they are not identical physically since the motion of the observer relative to the medium is different in the two situations. water waves.14 (a) The speed of light in vacuum is a universal constant independent of all the factors listed and anything else. –4 10. once again like for sound waves. (ii) independent of the direction of propagation for isotropic media. The wave picture of light is consistent with the experiment.] 10.15 Sound waves require a medium for propagation. This is also true for other waves. (v) independent of intensity. (iv) depends on wavelength. note the surprising fact that it is independent of the relative motion between the source and the observer.13 With the point object at the centre. 10. 10. (b) The intensity of interference fringes in a double-slit arrangement is modulated by the diffraction pattern of each slit. draw a circle touching the mirr or. Thus even though the situations (i) and (ii) may correspond to the same relative motion (between the source and the observer). In particular.17 (a) The size reduces by half according to the relation: size ~ λ/d. sound waves. Using simple geometry. Here © no N C tt E o R be T re pu bl 541 is he d .g. For light propagation in a medium. the two situations are not identical and we should expect the Doppler formulas for this case to be different for the two situations (i) and (ii). [For high intensity beams. This is a plane section of the spherical wavefront from the object that has just reached the mirror. the size of the latter should be comparable to wavelength. Therefore. Intensity increases four fold.). the situation is mor e complicated and need not concern us here. If the size of the obstacle/aperture is much too large compared to wavelength. (b) Dependence of the speed of light in a medium: (i) does not depend on the nature of the source (wave speed is determined by the properties of the medium of propagation. Physics the size is of the order of a few metres. This is clearly wrong. The angle θ = n λ/a = λ/a ′.14 (a) 6.13 (a) 5.34 eV = 0. (b) If we use the same formula with V = 107 V. sound waves can bend around the partition while light waves cannot.112 nm J = 0.88 × 109 m s–1.686 × 10 J = 4.5 4 × 1021 photons/m2 s 11.51 eV (a) 6.8 2. The wavelength of light is about 5 × 10–7 m.0 × 10–23 m he (c) 0. Each of the smaller slits sends zero intensity in the direction θ.5 cm. so is any linear combination of y 1 and y 2.3 11.1 11.0 × 1020 photons/s 11. Typical sizes of apertures involved in ordinary optical instruments are much larger than the wavelength of light.18 11.20 542 (b) 1.17 11.04 × 10–24 kg m s–1 (b) 6.145 nm (a) 6.4 × 10–19 J (b) 3 × 1016 photons/s (a) 3. the above formula for kinetic energy (m v 2/2) is valid only when –25 is 10.33 × 107 m s–1 .5 eV = 2.95 × 10 –25 –35 (a) 7. high intensity laser beams) and non-linear effects are important. we get v = 1.164 nm 11.34 V (c) 344 km/s (a) 0. 10.0 V 11.7 × 10 m (b) 1.24 × 1018 Hz (b) 0.11 eV (b) 3. v = 1. Thus.7 (a) 3.19 0. say.4 11.50 × 106 m s–1 (b) 3.21 Divide the single slit into n smaller slits of width a ′ = a/n .38 × 10–19 J = 2.15 (a) 1.12 (a) 4.16 11. v = [(2eV/ m )]1/2 . The combination gives zero intensity as well.1 × 10 11. Justification based on what is explained in (d). If y 1 and y 2 are solutions of the wave equation.028 nm (a) Use eV = (m v 2 /2) i. while sound waves of.16 eV = 3.05 × 10–27 kg m/s (c) 0.041 nm (b) 0.6 6. since nothing can move with a speed greater than the speed of light (c = 3 × 108 m s–1). (c) 3.2 11.63 × 10 kg m/s (for both) –21 –2 (b) 0. © no N C tt E o R be T re pu CHAPTER 11 11.34 μ eV 11.73 × 1014 Hz 11. . the situation is far more complicated and need not concern us here.11 2. Actually.63 m/s 11.20 (a) Interference of the direct signal received by the antenna with the (weak) signal reflected by the passing aircraft.92 × 10–24 kg m s–1 11.46 × 10–19J (b) 0.24 keV (c) 1. because ν < νo 11. 1.e. (b) Superposition principle follows from the linear character of the (differential) equation governing wave motion.2 nm.236 neV d 10.9 No. When the amplitudes are large (e.14 × 10–19J.174 × 10 eV λ = h/p = h/(hν/c) = c/ν 0.18 12..19 11.10 4. (e) 10.g.3 m. 1 kHz frequency have wavelength of about 0.59 × 10–34 J s 11..78 × 10 m –28 –32 bl J = 4.54 × 10–19J 1. λ = 6328Å h corresponds to ν = 4.43 × 10–16 m .602 × 10–10J to get λ = 2.40 eV. implies relativistic domain. a 20 MeV electron moves at relativistic speed. negligible error involved in ignoring the existence of a minimum quantum of energy (photon) and treating the total energy of a radio wave as continuous. As explained above. The relativistic formula is or R = m0 v / e B 1 – v 2 / c 2 ( 11. At very high speeds when ( v/c ) is comparable to (though always less than) 1.7 × 1014 Hz < ν0 .98 × 10–28J.26 φ0 = h ν – e V0 = 6.23 (a) 27.51 MeV. E ∼ 4 × 10 –19J. using the given data (e/m) = 1. v (for 10 MeV kinetic energy) = 0. it is large enough for us never to ‘sense’ or ‘count’ individual photons by our eye. Though this number is not as large as in (a) above.999 c.1 × 1.21 (a) 22. ⎛ v2 ⎞ where the relativistic mass m is given by m = m 0 ⎜1 − 2 ⎟ ⎝ c ⎠ © no N C tt E o R be T re pu R = p / eB = mv / e B 2 m0 is called the rest mass of the particle.2 eV. E = (h c/ λ ) = 3. 11.7 × 10 –19 J = 4.27 Use e V0 = h ν – φ0 for both sources. The photo-cell will not respond howsoever high be the intensity of laser light.7 cm (b) No. we come to the relativistic domain where the following formulae are valid: Relativistic momentum p = m v Total energy E = m c 2 Kinetic energy K = m c 2 – m 0 c 2 . ν0 = φ0 = 1. Number of photons emitted per second = 104J s–1/3. therefore. 11. and the number of photons emitted per second in a radio beam is enormously large. Use this value to obtain for the second source V0 = 1. 11. Photon flux corresponding to minimum intensity = 10–10 W m–2/4×10 –19J = 2.6 keV (b) of the order of 30 kV 11. From the data on the first source. K or energy ≥ m 0c 2 (rest mass energy).5 × 10 8 m–2 s–1 Number of photons entering the pupil per second = 2. Thus a kinetic energy of 10 MeV. 11. These relations also imply: 2 4 1/2 E = (p 2c 2 + m 0 c) Note that in the relativisitc domain when v/ c is comparable to 1.5 × 108 × 0.4 × 10–4 s–1 = 104 s–1.22 We have e V = (m v /2) and R = (m v / e B) which gives (e/m) = (2V/ R 2 B 2 ). There is. φ0 = 1. (b) For ν = 6 × 1014 Hz. Consequently. being much greater than electron’s rest mass energy. 11.Answers (v/c) << 1.0 × 1015 Hz. Using relativistic formulas. The rest mass energy of electron is about 0.98 × 10–28J ∼ 3 × 1031 s–1 We see that the energy of a radiophoton is exceedingly small. the non-relativistic formula R = (m 0v/ e B ) is not valid.25 (a) For λ = 500 m.73 × 10 11 C kg –1 .50 V. bl ) 543 is he d − 1/ 2 .24 Use λ = (hc/E) with E = 5. Therefore. energy of the radiation is not continuously shared by all the electrons in the top layers. Using e = 1. energy absorbed per second per electron = 2 × 10–9/1017 = 2 × 10–26 W Time requir ed for photoelectric emission = 2 × 1.31. Implication: Experimentally. and ν0 (Ni).0 × 1014 Hz (threshold frequency). but less than ν0 (Mo). for same K. electron’s energy = 150 eV. 11. he ) × d . The interatomic spacing is about a hundred times gr eater. A neutron beam of 150 eV energy is therefore not suitable for diffraction experiments. Consequently. photoelectric current from Na and K will increase in proportion to intensity. φ0 = h ν0 = 2. Mo and Ni will not give photoelectric emission.6. 11.31 For λ = 1 Å. Slope of the plot is found to be 4.6 × 10–19J/2 × 10–26W = 1. incident power is uniformly absorbed by all the electrons continuously. A photon is either not absorbed.45 × 10–10 m [Use λ = (h / 3 m k T ) ] which is comparable to interatomic spacing in a crystal. but this does not affect the result regarding Mo and Ni.5 year. h h = Thus. and ν0 (K). intensity of radiation increases. therefore for the same energy. The slope of the plot is (h/ e) and its intercept on the ν-axis is ν0.Physics 11. wavelength of neutron = 1/ 1838. h = 6. The fifth point corresponds to ν < ν0 . there is no photoelectric emission and therefore no stopping voltage is required to stop the current.15 × 10–15 V s.64 × 10–34 J s (standard value h = 6. The first four points lie nearly on a straight line which intercepts the ν-axis at ν0 = 5.6 × 107 s which is about 0. 11. Effective atomic area ~10–20m2 Number of electrons in 5 layers = 5 × 2 ×10−4 m2 = 1017 10−20 m2 © no N C tt E o R be T re pu 11. However. or absorbed by an electron nearly instantly.4 keV. Now (mn /me) = 1838.32 (a) Incident power = 10–5 W m–2 × 2 × 10–4 m2 = 2 × 10–9 W In the wave picture.33×10 m. photoelectric emission is observed nearly instantaneously ( ∼10 – 9 s): Thus. 544 (b) λ = 1. a photon has much greater energy than an electron. If the laser is brought closer.29 It is found that the given incident frequency ν is greater than ν 0 (Na). for the same wavelength. ener gy comes in discontinuous ‘quanta’. and absorption of energy does not take place gradually.6 –10 –12 10 m = 2. Thus. 11.30 Assume one conduction electron per atom.6 × 10–19 C.626 × 10–34 J s).11 V. the wave picture is in gross disagreement with experiment. In the photon-picture. Rather. λ= bl is ( (150 eV) as in Ex. λ decreases with m as p 2mK (1/ m ). photon’s energy = 12.28 Obtain V0 versus ν plot. for the same incident radiation.37 (a) Quarks are thought to be confined within a proton or neutron by forces which grow stronger if one tries to pull them apart.5 × 10–12 m λ ( yellow light) = 5.6)2 × 10–26 ∼ 9 × (6. for electric and magnetic fields. so a high energy neutron beam should be first thermalised before using it for diffraction.989 × 10–10 J = 1.9 × 10–7 m Resolving Power (RP) is inversely proportional to wavelength. absolute value of ν of a matter wave of an electron has no direct physical meaning.4 × 10–9 m. therefore.511 × 1. Consequently.36 Using the same formula as in Exercise 11. They occupy a continuous band of levels. RP of an electron microscope is about 105 times that of an optical microscope. We find r >> λ . Thus. ions have no chance to do so because of collisions with gas molecules and recombination. electrons knocked off from different levels come out with different energies. m = He bl 545 is he d . The phase speed νλ is likewise not physically significant.63) 2 × 10–22.01 × 105 Pa.Answers Clearly. In practice.63 × 10–19 kg m s–1 λ 10–15 m Use the relativistic formula for energy: E 2 = c 2 p 2 + m 02 c 4 = 9 × (6. respectively. Therefore. Not all electrons in the metal belong to this level. the second term (rest mass energy) being negligible. E = 1. observable charges are still integral multiples of e. (e) The absolute value of energy E (but not momentum p ) of any particle is arbitrary to within an additive constant. 11. ions have a chance to reach their respective electrodes and constitute a current.34 p = 6.24 BeV. differences in other (geometrical) factors can change this comparison somewhat. Thus.63)2 × 10–22 + (0. 11.35 Use λ = 4 ×10 3 kg 6 ×1023 3m kT This gives λ = 0. thermal neutrons are a suitable probe for diffraction experiments.35. seems that though fractional charges may exist in nature. The group speed given by h . 11. from (a) and (b) above.63 × 10–34 Js = 6. At ordinary pressures. (b) Both the basic relations e V = (1/2 ) m v 2 or e E = m a and e B v =m v 2/r.73 × 10 –10 m. r = 3.2 × 10–9 m which is much greater than the given inter -electron separation. 11. and m separately but by the combination e/m . It. (d) Work function merely indicates the minimum energy requir ed for the electron in the highest level of the conduction band to get out of the metal. electron energies from the accelerator must have been of the or der of a few BeV. show that the dynamics of electrons is determined not by e. Hence. λ = 6. while λ is physically significant. 11. h = © no N C tt E o R be T re pu dν dE d ⎛ p2 ⎞ p = = ⎜ ⎟= d (1/ λ ) dp dp ⎝ 2m ⎠ m is physically meaningful.33 λ = 5. Mean separation r = (V/N)1/3 = (kT/p) 1/3 For T = 300 K. (c) At low pressures. p = 1. the α-particle won’t bounce back in even in a head-on collision. (d) In Thomson’s model.1 © no N C tt E o R be T re pu 12.6 eV. 4.11 × 10 –15 s.2 The nucleus of a hydrogen atom is a proton. It is similar to a football colliding with a tenis ball at rest.13 ν= 3 This is much greater than the estimated size of the whole universe! 1⎤ me 4 me 4 (2n − 1) ⎡ 1 − 2⎥ = 2 3 ⎢ 2 3 2 (4π) ε 0 (h / 2π) ⎣ (n − 1) n ⎦ (4π) ε 0 (h / 2π)3 n 2 (n − 1)2 me 4 32 π ε (h / 2π)3 n 3 3 2 0 546 For large n.1 × 1015 Hz.11 (a) About the same. 12. 12.2 × 1029 m. Because the scattering particle is more massive than the target nuclei (proton). the a0 = G = radius of the first Bohr orbit is given by a 0 12.6 × 1074 12. If we consider the atom bound by the gravitational m ee 2 2 2 force (Gmp me/r ). we should replace (e /4 π ε0 ) by Gm p m e. Rutherford’s model (e) Both the models 12. 12. 1. 12.64 × 10–27 kg. (b) Much less. (c) It suggests that the scattering is predominantly due to a single collision. –27.09 × 10 6 m/s.52 × 10–16 s. In Rutherford’s model. and hence linearly with thickness.6 × 1014 Hz 12.77 × 10–10 m 12. because the chance of a single collision increases linearly with the number of target atoms.2 eV 12. Thus. whereas the mass of an incident α-particle is 6. 7. 12.18 × 106 m/s. 2 Gm pm e he d .10 2.7 (a) 2. The mass of it is 1. ν ≅ bl is (h / 2π)2 ≅ 1.Physics CHAPTER 12 (a) No different from (b) Thomson’s model.4 5.6 9. The observed average scattering angle can be explained only by considering multiple scattering. 4.22 × 10–15 s. 3.12 The first orbit Bohr’s model has a radius a0 given by 4 π ε 0 (h / 2π)2 . Balmer series: 656 nm. Rutherford’s model (c) Rutherford’s model (d) Thomson’s model. most of the scattering comes through a single collision and multiple scattering effects can be ignored as a first approximation. 12.27 × 105 m/s (b) 1.3 820 nm. there would be no large-angle scattering.9 Lyman series: 103 nm and 122 nm. 1. That is.8 2.7 × 10 – 8 m. So it is wrong to ignore multiple scattering in Thomson’s model.12×10–10 m.67 × 10 –27 kg.5 13. a single collision causes very little deflection. kinetic energy does not change. In terms of the Bohr’s quantisation postulate. For such large values of n. for all practical purposes. (b) The quantity 12. We note that keeping other factors fixed. r = Ze 2 m n 2 8 π ε 0r These relations have nothing to do with the choice of the zero of potential energy. Its value is + 3.4 eV. (b) Using V = – 2T . would alter if a dif ferent zer o of the potential energy is chosen. Its me 2 value is 0. Now. of course. this corresponds to a very large value of n (of the order of 10 70).6 × 207) eV ≅ – 2. In Bohr model v = n (h / 2π) . mvr = nh and which give T= mv 2 Ze 2 = 4 π ε 0r 2 r © no N C tt E o R be T re pu rm = re m e 0. and mr me 4 n (h / 2π) = 4 π ε 0 (h / 2π)2 2 r = n . Its value 0 ⎝ ⎠ is 2.53 × 10 –10m – of the order of atomic sizes. 12.14 (a) The quantity ⎜ 4 π ε mc 2 ⎟ has the dimensions of length. 12. Therefore. = 2.(Note that the dimensional arguments cannot. and the total energy of the state. choosing the zero of potential energy at infinity we have V = – (Z e2/4 π ε 0 r) which gives V = –2T and E = T + V = – T (a) The quoted value of E = – 3.56 × 10–13 m Eμ= = – (13.) he d . For example. however.4 eV is based on the customary choice of zero of potential energy at infinity. This gives νc = 2 32 π3 ε 0 (h / 2π)3 n 3 2 π mr 2 me 2 which is same as ν for large n. the kinetic energy of the electron in this state is + 3. consider the levels continuous.82 ×10 –15 m – much smaller than the typical atomic size. The potential energy.17 All that is needed is to replace m e by mμ in the formulas of the Bohr model. the differences in the successive energies and angular momenta of the quantised levels of the Bohr model are so small compared to the energies and angular momenta respectively for the levels that one can.16 Angular momenta associated with planetary motion are incomparably large relative to h .8 keV bl 547 is 4πε 0 (h / 2π)2 has the dimensions of length. potential energy of the electron is = – 6.15 In Bohr’s model.8 eV (c) If the zero of potential energy is chosen differently.4 eV independent of the choice of the zero of potential energy. Using E = – T .53 ×10 = mm 207 Ee m m me 13 4 π ε 0 2 2 1 Ze 2 mv 2 = . angular momentum of the earth in its orbital motion is of the order of 1070h . ⎛ ⎞ e2 12.Answers Orbital frequency νc = (v/2 π r ). tell us that we should use 4π and h/2π in place of h to arrive at the right size. r ∝ (1/m) and E ∝ m. 7 MeV 8.65 T years 4224 years 7. the positron carries maximum energy.90 MeV. E α = 6.584 × 1025 MeV or 2. E α = 4. Q = 4.11 13. Q = Ed+ Ee + Eν + The daughter nucleus is too heavy compared to e and ν.79 MeV.41 MeV.6 (a) 6.93 MeV.8 13.13. maximum kinetic energy of the electron (max E e ) = Q = 4.18 u 104. Note me has been cancelled.16 Q = m ( 56 26 Fe – 2m 26 ) ( 28 13 Al = 26.6 × 10−13 J g −1 235 he 222 86 Rn + 4 He 2 ii) 242 94 4 Pu → 238 U+ 2 He 92 d . so it carries negligible energy (E d ≈0). If we 11 use atomic masses. Q = 0.12 (a) 5 T years (b) 6.13 11 6 13.2 13. 23 10 13. endothermic (ii) Q = 4.85 MeV C → 11 B +e+ +ν + Q 6 11 6 N Q=⎡ ⎣m N ( C) – m ( B) – m 11 6 e ⎤c2.37 MeV. 80. hence maximum Ee ≈ Q).877 ×1010 Bq or 2. Q = ⎡ ⎣m N 1 0 Ne – m N 1 1 Na – m e ⎦ c . where the masses used are masses of nuclei and not of atoms as in Exercise 23 23 ⎤ 2 13. and this is practically all energy Q.4 13. As in Exercise 13.Physics CHAPTER 13 13.536 × 10 MeV 548 13. zero). not possible.3 13.13 Ci 1.37 MeV.13. If the kinetic energy (E ν) carried by the neutrino is minimum (i.10 13.941 u (b) 19. 7. ) 13. we have to add 6me in case of C and 5me in 11 case of B.29 MeV (a) Q = 4.126 ×10– 6 g 7.535×1012J i) 226 88 Ra → iii) v) 32 15 11 6 P → 32 S+ e – + ν 16 11 + iv) vi) 210 83 97 43 B → 210 Po+ e – +ν 84 97 42 C → 5 B +e + ν Tc → Mo+ e +ν + © no N C tt E o R be T re pu 13.17 4.961 MeV.. exothermic 13.7 13. 13.5 13. Using atomic masses Q = ⎡ ⎣m 10 Ne – m 11 Na ⎦ c .1% 20.03 MeV.84 MeV 1.15 (i) Q = – 4. Using given masses. ⎦ where the masses used are those of nuclei and not of atoms.1 13.62 MeV.14 23 23 23 – ⎤ 2 Ne → 1 1 Na + e +ν + Q . Hence Q = ⎡ ⎣m ( 11 6 C –m ) ( 11 6 2 B – 2m e ⎤ ⎦c ) Using given masses.9%.9 13.e.18 Energy generated per gram of bl ( ) ( ) (Note m e has been doubled) ( 235 92 U = is ) ( ) vii) 120 54 Xe+ e+ → 120 I +ν 53 6 × 1023 × 200 × 1.23 (b) Q = 6. 06MeV 13. 26 12 Mg :11. required to overcome Coulomb repulsion = 480.26 For 14 6 C emission 223 88 Q = [m N ( = [m ( 4 2 Ra) − m N ( 2 09 82 209 82 Pb) − m N ( 6 C)]c 14 14 2 2 22 3 88 Ra) − m ( Pb ) − m ( 6 C)]c = 31.2 × 1.98MeV 1 40 99 2 13.36MeV.8 × 3. ν (γ 3 ) = 1.627 × 10 Hz .19 About 4.284MeV.0 KeV = 7.59 MeV 2 3 4 2 (b) K.995 × 1020Hz .960MeV ( ) ( ) ν (γ 1) = 2.24 Neutron separation energy Sn of a nucleus A −1 A 2 Sn = ⎡ ⎣m N ( Z X) + m n − m N ( Z X)⎤ ⎦c A Z X is 41 27 From given data .1 MeV 92 U) + m n − m ( 58 Ce) − m ( 44 Ru)]c 13.3%.85 MeV For 223 219 4 2 He emission.7% 13.632 × 1020 Hz 20 bl 549 ( A Z X – Z me – m ) ( A Z –1 2 Y – ( Z – 1 )m e – m e ⎤ ⎦c ) is ( X) – m ( ) he 13.154 × 1016 × 235 g 13 1.68 ×10–14 –23 3 × 1.9 × 10 y 13.381 × 10 (as k = 1.Answers The amount of = 235 92 U consumed in 5y with 80% on-time = 1544 kg 5 × 0.E.68×10–14 J = 3kT ∴T = 7.381 × 10–23 J K –1 ) = 1.28 (a) Q = [m ( 1 H) + m ( 1 H) − m ( 2 He) − m n ]c = 17. Q = [m ( 88 Ra) − m ( 86 Rn) − m ( 2 He)]c = 5. Hence the result.85 ×109 K (required temperature) – 13.22 Consider the competing processes: d .23 25 12 A A A A ⎤ 2 ⎡ ⎤ 2 Q2 = ⎡ ⎣m N ( Z X ) + m e – m N ( Z –1 Y )⎦ c = ⎣ m ( Z X ) – m ( Z –1 Y )⎦ c This means Q1 > 0 implies Q2 > 0 but Q 2 > 0 does not necessarily mean Q1 > 0. Sn ( 13 Al) = 13.29 Kmax β1– = 0.20 360 KeV 13.27 Q = [m ( 2 38 = 231. Kmax β2 = 0. A Z X→ A Z –1 + Y + e + νe + Q1 (positron capture) e– + A Z A X→Z –1 Y + ν e + Q2 (electron capture) Q1 = ⎡ ⎣m N =⎡ ⎣m N =⎡ ⎣m A Z N A Z –1 2 Y – me ⎤ ⎦c ( A Z X –m ) ( A Z –1 2 Y – 2m e ⎤ ⎦c ) © no N C tt E o R be T re pu 13.6 × 10 235 92 The initial amount of 4 U = 3088 kg.0 keV 480.25 209 d 13. Mg : 9. Sn ( 20 Ca) = 8. ν (γ 2 ) = 0. 95 × 10 . showing almost infinite dynamic resistance in the reverse bias.2 (c) Decimal system implies continuous set of values 550 bl 1⎡ 2 2 (N − N A ) + (N D − N A ) + 4n i ⎤ ⎥ ⎣ D ⎦ 2⎢ is he d The energy released in fusion of 1 kg of hydrogen is about 8 times that of the energy released in the fusion of 1 kg of uranium.336 Ω (d) For both the voltages. (b) AND CHAPTER 15 15.4 14.3 14. (b) 2.2 14. A Y 0 1 1 0 14.9 14. IB= 10 μA 2V No (hν has to be greater than Eg ). 100 Hz for full-wave vi = 0.17 (a) AND (b) OR 14.7 14.97 A.16 NOT .Physics 13.076 × 10 kg .10 14.0629 A. (d) Consult Table 15.14 (a) 0. 2 He Energy released in fusion of 1kg of hydrogen = 39 ×10 (b) Energy released in fission of 1kg of 235 92 26 26 1 MeV U = 5. the current I will be almost equal to I0.13 About 1 × 10 14. (c) (c) 50 Hz for half-wave.01 V . ne. 14.2 15.6 14. nh = 4.nh = n i2 14. ne = 5 (c) (d) (c) (c) (c) (b).75 × 10 .30 (a) Note that in the interior of Sun.1 14. (c) 0.11 14. 22 9 ne ≈ 4.12 © no N C tt E o R be T re pu Solving these equations.8 14.18 OR gate 14. n-type since ne >> nh For charge neutrality ND – NA = ne – nh .1×10 MeV CHAPTER 14 14.19 (a) NOT.31 3. 1 GHz and 1000 GHz will penetrate.3 (b) 10 kHz cannot be radiated (antenna size).1 15.5 14. 4 13. four 1 H nuclei combine to form one 4 nucleus releasing about 26 MeV of energy per event. 15.6 (a) © no N C tt E o R be T re pu (b) μ = 0. 15.Answers 15. for simplicity. By multiplying the two signals. clearly. Ac M 1 + M 2 12 3 With M2 = 0. No. the received signal be A1 cos (ωc + ωm ) t The carrier A c cos ωc t is available at the receiving station. Service area will be A = π d 2 = T 7 15.4 × 106 = 3258 km2. we get A1 Ac cos (ωc + ωa ) t cos ωc t = = A1 Ac ⎡cos ( 2ωc + ωm ) t + cos ωmt ⎤ ⎦ 2 ⎣ If this signal is passed through a low-pass filter.8 Let. m =1. irrespective of M1 .75 = Am Ac Am = 0. the maximum amplitude is M1 = A c + A m while the minimum amplitude is M2 = A c – Am.75 × 12 = 9V. 2 bl 551 is he d . Hence the modulation index is m = Am M – M2 8 2 = 1 = = . we can record the modulating signal A1 Ac cos ωmt .5 15.5 μ = 0.4 22 × 162 × 6.7 Since the AM wave is given by (A c + A m sinωm t ) cos ωc t. Serway and Jerry S.W.A. 1 – Mechanics: (Kittel. Harris Benson. John Wiley (1996). Rogers. Addison – Wesley (1967). H.M. R. Norton (1989). e. Tata McGraw Hill (1977). Addision – Wesley (1965). Crawford) Vol. DC Heath and Co. 19 University Physics. you may like to consult one or more of the following books. John Murray (2000). 16 Conceptual Physics. Lectur es on Physics (3 volumes).V. Weber. Morton M. A. Weygand. b. 5 – Statistical Physics (F. University Physics (Sears and Zemansky’s). Addison—Wesley (2004). J. Hans C.P. MIR Publishers. Brown (1994). Alonso and E. Freedman. Manning. Problems in Elementary Physics. Kane. M. Arnold-Heinemann (1984). Tom Duncan.W. Young and R. Physics for the Inquiring Mind. 20 General Physics. Finn. G. B. 10 College Physics. (1999).D. (1971). is he d . 18 University Physics. Nelkon and P. 7th Edition John Wily (2004). 3 – Waves and Oscillations (Frank S. Parker. 11th Edition. 552 21 Physics. Berkeley Physics Course (5 volumes) McGraw Hill (1965). Some of these books however are more advanced and contain many more topics than this book. bl For additional reading on the topics covered in this book. G. c. Crummet and Arthur B. Robert Resnick and Jearl Walker. M. John Wiley (1988). Wm. Cleveland.C. Tata McGraw Hill (1978). Shalnov. Faughn. R. Arnold-Heinemann (1987). a. 4 – Quantum Physics (Wichmann) 9 12 13 14 15 1 2 Or dinary Level Physics. Princeton University Press (1960).A. Paul G. Advanced Level Physics . d. NCER T (1967).Physics BIBLIOGRAPHY TEXTBOOKS © no N C tt E o R be T re pu 3 4 5 6 7 8 Advanced Physics . M. W illiam P. K. E. (1965) Indian Edition. Vol. Myakishev and V. Vol. 6th Edition. Physics Advanced Level. Feynman. W. PSSC Physics Course. White and G.M. Harcourt Brace and Co. Fundamentals of Physics. Physics. Raymond A. Addision—Wesley (1998). Hewitt. 2 – Electricity and Magnetism (E. Purcell) Vol. Krivchenkov. 17 College Physics. Bukhovtsa.L. David Halliday. 11 Physics: Foundations and Frontiers. V. Jim Breithampt. Patrick Fullick. Stanley Thornes Publishers (2000).F. Sternheim and Joseph W.M. Knight and Ruderman) Vol. Heinemann (2000). Abbott. Gamow and J. Ohanian. Reif ) Fundamental University Physics. Western. W.. Feynman. Narlikar. Bloomfield. Senior Physics.D.H. Buffa. B. Niels Bohr. Physics Matters: An Introduction to Conceptual Physics. Redish. How Things Work: The Physics of Everyday Life. F. Senior Physics. Wilson and Anthony J. Barnett. The Universe and Dr. Three… Infinity. New York (1966). Hoyle and J.P. 28 Essentials of Physics. Gamow. Atomic Theory and the Description of Nature. Einstein. G. L. Perelman. (1961). Thirty years that Shook Physics. that many of these books are written at a level far beyond the level of the present book. Part – II. W. MIR Publishers (1987). Power of Ten. Kikoin. Landau and A. Gamow. G. W. Kikoin and A. Gamow. Physics can be Fun. Johnson. The Flying Circus of Physics with Answer. John Wiley (2005). you may like to read some of the following books. Hazen. Y. James Trefil and Robert M. Tompkins in paperback. G. Physics in your Kitchen Lab. Feynman. Book 1: Physical Bodies Book 2: Molecules Book 3: Electrons Book 4: Photons and Nuclei. Philip Morrison and Eames. Co. University of Chicago Press (1930). The Physics—Astronomy Frontier. Patrick J. John Wiley (2005). T ime Inc. Bantam books (1986). Walker. 1 2 3 4 5 6 7 8 Mr . (Indian Edition) Oxford and IBH Pub. Heisenberg. Prentice Hall (1997). Cambridge University Press (1967).I. John Wiley and Sons (1977). One. F. GENERAL BOOKS For instructive and entertaining general reading on science. Cambridge (1934). Understanding Basic Mechanics. Cummings. New York (1962). Freeman (1985). The Meaning of Relativity. Part – I. © no N C tt E o R be T re pu 9 10 11 12 13 14 15 16 bl is he 553 d . College Physics. MIR Publishers (1986). Surely You’re Joking. MIR Publishers (1985). Double Day. Kikoin. (1965). Cutnell and Kenneth W. A. K. The Physical Principles of Quantum Theory. MIR Publishers (1988). Einstein. John Wiley (2005). Remember however. R. Priscilla W. Mr. Louis A. John Wiley (2004). Understanding Physics. Fr eeman (1980). Physics for Everyone (series).K. Cooney. MIR Publisher (1978). C. Cambridge University Press (1996).H. John Wiley (1995). Viking Inc.Bibliography Answers 22 23 24 25 26 27 Advanced Physics. Bekhovtsev. Two.K. I. Laws and Edward F.V. Keith Gibbs. I. J.K. Jerry D. John D. Kitaigorodski. Reif. Physics INDEX Absorption spectra AC current AC Generator AC voltage applied to a capacitor applied to a resistor applied to an inductor applied to a series LCR circuit Accelerators in India Accommodation of eye Activity of radioactive substances Additivity of charges Alpha decay Alpha particle scattering Ammeter Ampere Amperes circuital law Amplification Amplitude modulation Analog signal AND gate Andre. Ampere Angle of deviation of incidence of reflection of refraction Angular magnification Apparent depth Area element vector Astigmatism Atomic mass unit number spectra Attenuation Aurora Boriolis Band gap Bandwidth of signal © no N C tt E o R be T re pu 330 355 355 355 341 318 26 337 439 440 420 516 139 470 517 421 233 224 233 241 234 237 244 142 336 447 8 449 415 165 155 147 517 524 502 503 148 Bandwidth of transmission medium Bar magnet as solenoid Barrier potential Base Becquerel Beta decay Binding energy per nucleon Biot-Savart law Bohr magneton Bohr radius Bohr’s model of atom Bohr’s postulates Brewster’s angle Brewster’s law C. Volta Capacitance Capacitive reactance Capacitive circuit Capacitor parallel plate in parallel in series Cartesian sign convention Cassegrain telescope Cells in parallel in series Chain reaction Channel Charging by induction Charles August de Coulomb Chromatic aberration Ciliary muscles Coercivity Coherent source Collector Colour code of resistors Combination of lenses is bl 554 he 518 174 176 479 491 448 450 444 143 163 425 422 424 380 381 53 73 241 252 74 79 78 311 342 110 114 113 452 515 6 11 333 337 195 360 491 103 328 d .A. sources waves. nature damping spectrum Electron emission Electrostatic analog potential shielding Electrostatics of conductors Electromotive force (emf ) Emission spectra Emitter Energy bands generation in stars levels stored in a capacitor Equipotential surfaces Excited state Experiments of Faraday & Henry Extrinsic semiconductor Eye Farad Faraday’s law of Induction Fast breeder reactor Ferromagnetism Field due to infinite plane sheet due to uniformly charged thin spherical shell Field emission Flemings left hand rule Flux leakage Focal length Force between two parallel currents Forward bias Franck-Hertz experiment 218 394 1 93 27 77 18 20 19 23 25 72 105 bl is © no N C tt E o R be T re pu 501 502 328 28 28 31 29 332 270 358 98 185 6 he 274 275 218 280 387 180 53 69 1 67 110 421 491 469 455 427 80 60 427 205 474 336 75 207 453 193 38 39 388 261 311 154 479 428 d 555 .Index Combination of resistors series parallel Composition of nucleus Concave mirror Conduction band Conductivity Conductors Conservation of charge Conservative force Continuous charge distribution Control rods Convex mirror Coulomb Coulomb’s law Critical angle Curie temperature Curie Current amplification factor density loop as a magnetic dipole sensitivity of galvanometer Cut-off voltage/Stopping potential Cyclotron frequency Davisson & Germer Experiment de Broglie relation wavelength explanation Decay constant Detection of amplitude modulated wave Diamagnetism Dielectrics Dielectric constant strength Diffraction single slit Digital electronics signal Dioptre Dipole moment moment vector in uniform electric field physical significance Dispersion by a prism Displacement current Doppler effect Drift velocity Earth’s magnetism Earthing 107 108 438 312 469 97. 468 5 8 51 32 454 312 11 10 320 194 448 94 495 97 160 165 391 140 141 403 398 398 430 446 526 192 71 76 74 367 368 Eddy currents Einstein’s photoelectric equation Electric charge current dipole displacement field field. physical significance field due to a system of charges field lines flux susceptibility Electrical energy Electromagnetic waves. 206 force on a current carrying conductor 135 force 133 hysteresis 195 inclination 187 intensity 190 meridian 186 moment of a current loop 158 moment 178 permeability 190 potential energy 178 susceptibility 190 torque 178 Magnetisation 189 Majority carriers 476 Mass defect 443 number 440 energy relation 442 Maxwell’s equations 273 Mean life 448 Meter bridge 120 Michael Faraday 208 Microscope 339 compound 340 Microwaves 281 Minority carriers 476 Mirage 321 Mirror equation 314 Mobility 100 Moderator 454 © no N C tt E o R be T re pu 252 238 494 5 505 361 361 363 110 472 427 337 441 441 439 270 490 182 115 556 bl is he d . Gauss Kirchhoff’s rules 364 483 96 283 451 33 37 181 35 186 4 427 519 134 448 483 388 220 274 472 187 353 337 418 246 219 220 222 6 6 Lateral shift 317 Law of radioactive decay 447 of reflection 357 of refraction 356 LC oscillations 255 Least distance of distinct vision 336 Lenz’s law 210 Lens maker’s formula 326 Light emitting diode 488 Limitations of Ohm’s law 101 Linear charge density 32 magnification/Magnifying power 339 Logic gates 502 Lorentz force 134 Magnetic declination 186 dipole 177 dipole moment of a revolving electron 162 field 132 field lines 175 field on the axis of a circular current loop 145 flux 182. Maxwell Junction transistor K. Ohm Gamma rays decay Gauss’s law its applications in magnetism Gaussian surface Geographic meridian Gold leaf electroscope Ground state wave H. Lorentz Half life Half-wave rectifier Hallwachs’ and Lenard’s observations Henry Hertz Experiment Holes Horizontal component of earth’s magnetic field Huygen’s Principle Hypermetropia Impact parameter Impedence diagram Inductance mutual self Induction of charge Inductive circuit reactance Input resistance of a transistor Insulators Integrated circuits (IC) Interference constructive destructive fringes Internal resistance Intrinsic semiconductor Ionisation energy Iris Isobars Isotones Isotopes J.Physics Fringe width Full-wave rectifier G.C.F.A.S. 522 525 137 212 163 454 336 504 336 440 516 72 505 174 502 491 475 442 442 451 452 445 455 457 452 441 457 375 95 95 322 502 163 495 192 195 143 Polarisation 71. 356 Relation between field and potential 61 Relaxation time 98 Rententivity 195 Repeater 517 Resistance 95 Resistivity 96. 76 76 237 237 487 388 399 388 389 395 2 377 478 491 10 72 bl is he 557 d . 468 of some materials 102 Resolving power 373 of eye 374 Resonance 248 Sharpness 249 Resonant frequency 248 Reverse bias 480 © no N C tt E o R be T re pu 11. validity of 375 Rayleigh scattering 335 Rectifier 483 Red shift 358 Reflection of light 310 Refraction 318 of a plane wave 355 Refractive index 317. 376 by reflection 380 by scattering 379 Polarity of charge 2 Polaroid 378 Potential 53 due to an electric dipole 55 due to a point charge 54 due to a system of charges 57 energy difference 53 energy for a system of charges 61 energy of a dipole 66 energy of a single charge 64 energy of a system of two charges 65 energy 52 Potentiometer 122 Power (electrical) 106 factor 252 in ac circuit 252 of lens 327 Pressurised heavy water reactors 453 Primary coil Principal focus 311 Principle of superposition 15 Principle quantum number 425 Prism formula 331 Production of amplitude modulated wave 525 Properties of electric charge 8 p-type semi conductor 476 Q factor/quality factor 250 Quanta of energy 393 Quantisation of charge 8 Radio waves 281 Radioactivity 446 Rainbow 333 Ray optics.Index Modulation index Motion in a magnetic field Motional emf Moving coil galvanometer Multiplication factor (fission) Myopia NAND gate Near point Neutrons Noise Non-polar molecules NOR gate North pole NOT gate n-p-n transistor n-type semi conductor Nuclear binding energy density energy fission force fusion holocaust reactor size winter Numerical aperture Ohm Ohm’s law Optical fibers OR gate Orbital magnetic moment Output resistance of a transistor Paramagnetism Permanent magnets Permeability of free space Permittivity of free space of medium Phasors diagram Photodiode Photoelectric effect Photocell Photoelectric emission Photoelectrons Photon Pith ball Plane polarised wave p-n Junction p-n-p transistor Point charge Polar molecules 517. 356 489 151 174 521 421 422 422 422 422 332 310. 311 163 32 341 103 135 388 456 326 392 Tokamak Toroid Torque on a current loop on a dipole Total internal reflection Transducer Transformer Step-down Step-up Transistor as a switch as an amplifier oscillator common emitter configuration Truth table Uncertainty Principle Unpolarised wave Ultraviolet rays Valence band Van de Graaf f Generator Velocity selector Visible rays Voltage Regulator Voltage sensitivity of a galvanometer Voltmeter Volume charge density Wattless current Wavefront plane spherical Wheatstone bridge Work function X rays Young’s experiment Zener diode breakdown 153 152 157 31 319 516 259 261 261 © no N C tt E o R be T re pu 558 bl is he 496 497 500 493 502 400 377 282 469 83 140 282 486 165 165 32 252 353 354 354 118 394 283 362 485 485 d .Physics Right hand rule Root mean square (rms) or effective current voltage Roget’s spiral Rutherford’s model of atom Saturation current Scattering of light Secondary wavelet Semiconductors diode elemental compound Shunt resistance Signal Sky wave Snell’s law Solar cell Solenoid South pole Space wave Spectral series Brackett Fund Lyman Paschen Spectrum of light Spherical mirror Spin magnetic moment Surface charge density Telescope Temperature dependence of resistivity Tesla Thermionic emission Thermonuclear fusion Thin lens formula Threshold frequency 149 235 235 156 415 390 335 354 469 479 468 468 164 516 520 317.