st em. 1s Se .200 PHYSIC 08.P CS UPL LBPHYSI ICS @ IM MSP Teo odorick B Barry Rica ardo Ma anguerra a [LEC [ CTURE NOT ] N TES] The e extent of ele ectricity and magnetism in human lives has le d n ead to numerous disco overies and innovation This seme d ns. ester, let us, together, re ediscover electricity an magnetism… nd PHYSICS 13 LECTURE NOTES |2 UN ONE: ELECTR NIT : ROSTATI ICS PART ONE ELECTRIC CHARGE P E: C E OUTLINE E 1. Electric Charge Cha arge Quantiza ation Cha arge Conserv vation 2. Conducto and Insula ors ators The Electroscope e Cha arging by Conduction Cha arging by Ind duction 3. Coulombs Law and the Electric For s e rce 4. The Electri Field ic Elec ctric Dipoles 5. Electric Fie Lines eld 6. Motion of Point Charge in Electric es Fields 4. . 5. . 2. . OBJECTIVES At the end of this c e chapter, you must be u able to: t 1. . Define and explain the concept e ctric charge and its of the elec properties s; Differentia a conduc ate ctor from an insulato or; Enumerate and explaiin the charging processes; different c Define Co oulomb’s Law w; Define the Electric Field and e Electric fie lines; and eld Illustrate th motion of charges he f in an elec ctric field. 3. . 6. . THE ELEC CTRIC PHENOM MENON • Humans are extremely de ependent on electricity. n • The study of electricity da ates back wa before the first electric lamp glowe ay e c ed. The history of electricity re eaches back to the Ancient Greeks, w k when attract tion and repu ulsion are • observed wh hen “rubbed amber” are placed near common m materials. • Indeed, the word “electr ric” comes fro the Gree word for am om ek mber “elektro on” • We begin ou examinatio of electric with ELEC ur on city CTROSTATICS – the study of charges at rest. 1. ELECTR CHARGE RIC • Basic element of Electricity • An intrinsic p property of th fundamen particles that make up matter he ntal • Three types: ▫ Posit tive Charge ▫ Neg gative Charge ▫ Neu utral Charge ATOMIC MODEL C • onsists of atom ms All matter co • Atoms are m made of a nucleus (neutro and proto and on on) electron/(s) revolving aro ound the nuc cleus. (Orbital Model) • Protons are p positively cha arged and Neutrons are neutrally n See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N PHYSICS 13 LECTURE NOTES |3 charged. e Electrons are negatively charged. Pa article ELECT TRON, ePRO OTON, p NEU UTRON Mass 9.11 x 10-31 kg 9 g 1.67 x 10-27 kg g 1.67 x 10-27 kg g Charg ge - 1.6 x 10-19 C 0 + 1.6 x 10-19 C None • • The electron is 2000 less m n massive as a proton, yet they possess t t the same am mount of cha arge e (though opp posite in sign) THE FUND DAMENTAL LA OF ELECTROSTATICS AW • Like charges repel, unlike charges attract s e CHARGE QUANTIZATI E ION • All observab charges in nature occur in discrete packets or iin ble n e integral amo ounts of the fu undamental unit of charg e. ge • Any charge Q occurring in nature ca be written an Q = + Ne, where N is an integer. • For large syst tems, howev ver, N is usually very large and charge appears to b continuou just as be us, air appears t be continu to uous even th hough air con nsists of many discrete mo y olecules. • To give an everyday example of N, charging a pla astic rod typiically require a transfer of 1010 or es o more electro to the rod ons d. CHARGE CONSERVAT E TION • When 2 obje ects are rubb bed: ▫ The object left w an excess of electron becomes ne with s egatively cha arged. ▫ The other object is left lacking electrons, thus become positively c t g t es charged. The net char rge of the two objects rem o mains constant; that is, ch harge is conserved. • • This is known as the law o conservatio of charge this is one o the fundam of on e, of mental laws of o nature. • Even in certa interactio ain ons, where ch harged partic cles are crea ated and ann nihilated, the amount of charges th are prod hat duced and de estroyed is eq qual, so there is conserva e ation. T E THE UNIT OF CHARGE • Coulomb (abbr. C) is the SI unit of charge. e ctric current, the ampere (A). • It is defined iin terms of the unit of elec • The Coulomb (C) is the a amount of ch harge flowing through a w in one se g wire econd when the e mpere. current in the wire is 1 Am • The fundame ental unit of charge, e, is related to th coulomb b he by: -19 C e = 1.602177 x 10 7 • For Physics 13, we will use e = 1.60 x 10-19 C e EXAMPLE 1.1: E A charge of magnitu ude 50 nC (1 nC = 10-9C) c can be prod duced in the laboratory by simply rubb y bing two objects t together. How ma electrons must be tran any nsferred to produce this charge? c Hint: (Use Q=Ne) e See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N PHYSICS 13 LECTURE NOTES |4 2. CONDUCTORS AND INSULATORS • Conductors ▫ are materials, where electrons are free to move about the entire material (ex. Cu and other metals) • Insulators ▫ are materials, where electrons are bound to a nearby atom, rendering no motion (ex. Wood and glass) • Ion ▫ An atom where electron/(s) is/are added or removed. Normally, a conductor is electrically neutral due to a balance between positive and negative charges. So in order to create a net charge, free electrons are added or removed from the lattice. A macroscopic object can be… Net Charge Electrically Neutral Positively Charged Negatively Charged • • Property p = ep > ep < eProcess None Remove electron Add electron Only electrons can be transferred due to the atomic structure, and the minimal amount of energy required. Protons are bound by very “strong forces” so their removal is very hard to accomplish. THE ELECTROSCOPE • The Electroscope ▫ Is a device for detecting electric charges • The Diverging Leaves: ▫ Two gold “leaves” diverge when a charge is placed near or in contact with the bob. ▫ The leaves return to normal, when charges are no longer present in the bob CHARGING BY INDUCTION AND CONDUCTION • • By Conduction- charging by contact ▫ Implements an effective transfer of electrons By Induction – charging without contact, only by placing objects close to each other ▫ Implements only motion of charges within a material How to produce a NET charge? RUB!!! • • ARBITRARY, BARI-ABLE • Question1: When a glass rod is rubbed by silk, which of the two materials acquire a net positive charge? • Answer 1: Any of the two, as long as the other gets the opposite. We can not know for certain which charge is which. We can only arbitrarily assign a charge • Question 2: If you walk across a rug and scuff electrons from your feet, are you negatively or positively charged? • Answer 2: You are positively charge, since electrons were scuffed off/from your feet! See more @ teodorickbarry.multiply.com PHYSICS DIVISION PHYSICS 13 LECTURE NOTES |5 CHARGING BY INDUC CTION CHARGING BY INDUC CTION VIA GR ROUNDING Ground – A v very large co onductor tha can supply an unlimited amount of charge (such as the at y d h • earth, extrem mely negatively charged) ) 3. COULO OMB’S LAW A AND THE ELEC CTROSTATIC F FORCE • The Fundamental Law of electrostatic is not enou f cs ugh to chara acterize fully t the “Electric Phenomenon” • Being able to quantify an measure t o nd the phenome enon is a mu ust. • Cavendish w the one w was who first expe erimentally verify the Univ versal Law of Gravitation. • A similar exp periment of C Coulomb gav the mathe ve ematical desc cription/quantification of the fundamenta law of elec al ctrostatics • • Coulomb's la develope in the 178 by French physicist Ch aw, ed 80s h harles August de Coulom tin mb. The magnitude of the ele ectrostatic for rce between two point ch harges is dire ectly proporti ional to the magnitud of each charge and inversely pro des oportional to the square o the distance of between the charges. e This has the s same form as Newton’s Th s hird Law of Motion: M The electric force exerted by the two objects on one another have the sam magnitud but d o o me de opposite in d direction • • • Formula: ▫ ▫ ▫ F is t the magnitud of the elec de ctrostatic for rce exerted, iin N (F is -, if a attractive; F is +, if repu ulsive) |q1|, |q2| are the magnitud of the cha de arges, in C r is th separation of the charges, in m he ROSTATIC CO ONSTANT ELECTR PERMITTIVIT OF FREE SP TY PACE See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N PHYSICS 13 LECTURE NOTES |6 FORCE E EXERTED BY A SYSTEM OF C CHARGES If you need t find the ne force exert • to et ted on a cha arge by a gro oup or system of m forces, we need to imple ement vector addition! r Because forc are vecto they supe • ces ors, erimpose! 4. ELECTR FIELD RIC • pt duced to circ cumvent the conceptual Field-concep was introd dilemmas of action-at-af -distance forc ces. • An electric c charge will se et-up an ELEC CTRIC FIELD in space surro n ounding it. • This Electric F Field will exer an electric force on any electrical o rt y objects in it. • This concept was first introduced by M t Michael Faraday. • Strength/Magnitude: • The ratio of the e electric force on a charge at a point t the magniitude of the charge e e to c plac ced at that p point. Direction: the electric F Force on a ch harge at a point to the direction of th he • The direction of t charge placed a that point. at . • ELECTRIC FIELD PROP C PERTIES • • • Symbol: Type: Derived, Vector SI Units (duall): N/C or V/m m Formula: Special D Details: 1. Electric Field is a ve ector field* 2. Electric Field propa agates through space at the speed of light o 3. The fo orce that an e electric field exerts, acts o test charg on ges, not on th charge that caused it. he . 4. We do not need a actual “tes charge” to calculate th electric fie o an st o he eld! • • Since the ele ectric field is a vector, it a also follows th superposition principle. he This only mea that if the are a gro ans ere oup of charges and a test charge was made to ap t s pproach these charge the test charge will ex es, xperience a force exerted by the net e f d effect of the individual ele ectric fields s set-up by the group of ch e harges! See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N PHYSICS 13 LECTURE NOTES |7 THE ELECTRIC FIELD AND COULOMB’S LAW • Since Force and the Field are both vectors they follow superposition principle! • • Coulomb’s Law: Obtaining the Electric Field Note: We do not need to know the magnitude of the test charge to calculate the electric field, all we need to know is the magnitude of the charge that produces it, and where we will measure the electric field! ELECTRIC DIPOLES • Electric dipoles are systems composed of two equal and opposite charges q, separated by a small distance L. • Electric dipole moment describes the strength and orientation of electric dipoles. 5. ELECTRIC FIELD LINES • We can picture the electric field by drawing lines to indicated its direction. • At any given point, the field vector E is tangent to the lines, because they show the direction of the force exerted on a positive test charge! • At any point near the positive charge, the electric field points radially away from the charge. • Similarly the electric field lines converge toward a point occupied by a negative charge! See more @ teodorickbarry.multiply.com PHYSICS DIVISION PHYSICS 13 LECTURE NOTES |8 DRAWING RULES /End: 1. Begin/ Field lines be egin at positiv charge (o infinity) and end at neg ve or d gative charge (or infinity) es metry: 2. Symm Lines are dra awn symmetr rically enterin or leaving an isolated c ng charge ber: 3. Numb The number of lines entering/leaving the charge is proportiona to the mag al gnitude of that charge 4. Densit of the Lines ty s: The number of lines per u area perp unit pendicular to the lines, at any point is proportiona to the o t al magnitu ude of the field at that po oint 5. When Far: When very fa from the so ar ource of the field, the line are spaced equally an radially, as if they es nd s om source of cha arge come fro a single s 6. Star-crossed: o ey ndicate two d directions of the field at that Field lines do not cross, because if the do, they in hich is imposs sible! point wh 6. MOTIO OF CHARG IN ELECT ON GES TRIC FIELDS • When a part ticle with cha arge q is plac ced in an Electric Field E, it experience a force qE. Via the es 2nd Law of Newton, the p particle will ex xperience an acceleratio n on: Example e: • An electron is projected into a uniform horizontal electric field (E = 1000 N/C) with a horizontal m velocity (v0= 2 x 106 m/s) in the directiion of the field. How far d does the elec ctron travel before it is mentarily to r rest? brought mom See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N PHYSICS 13 LECTURE NOTES |9 PART TWO: THE ELEC P CTRIC FIELD D OUTLIN NE 1. Electric Field of Discr rete Charge Dis stributions 2. Electric Field of Cont tinuous Charge Dis stributions 3. Electric Flux and Gauss’s Law 4. Disconti inuity of En 5. Charge and Field at Conductor Sur rfaces 6. Derivatio of Coulom on mb’s Law from m Ga auss’s Law an vice-versa nd a OBJECTIVES At th end of this chapter, yo should he s ou be able to: a 1.Ca alculate the e electric field o different of char rge distributio using techniques ons prese ented; 2. St tate, use, app Gauss’s Law in ply theo oretical and p practical problems; and 3. Re elate Gauss’s Law and Co s oulomb’s Law. . 1. ELECTR FIELD OF DISCRETE CH RIC HARGE DISTRIBUTIONS 1. System m: Use superposition principle 2. Electr Dipole ric axis pole at a The electric field on the a of the dip x away is in the direc ction of the dipole moment and has point a great distance the magnitude e EXAMPLE E: A molec cule of water vapor cause an electric field in the surrounding s es c s space as if it were an elec ctric -30 C-m. dipole. It dipole mom ts ment has a m magnitude p = 6.2 x 10 What is t the magnitud of the electric field at a distance z = 1.1 nm from the molec de m cule on its dip pole axis? ANS: 8.4 x 107 N/C 2. ELECTR FIELD OF CONTINUOUS CHARGE DISTRIBUTIONS RIC S Properties co onvention for continuous charge distri r ibutions Name Charge Linear ch harge density y Surface charge dens sity Volume charge dens sity Symbol q λ σ ρ SI Unit C C/m C/m2 C/m3 See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 10 DISTRIBUTION LINE CHA ARGE CON NFIGURATION N FIELD FO ORMULA (1)At the e perpend dicular bisector of a finite line e charge (2) At the e perpend dicular bisector of an infinite lin ne charge CHARGE ED RING z is the distance of P from the center of the disk R is the radius of the charged diisk e e DISTRIBUTION CHARGE ED DISK CON NFIGURATION N FIELD FO ORMULA z is the distance of P from the center of the disk R is the radius of the charged diisk e e INFINTIE PLANE O OF CHARGE E If we let the charge disk e expand to infinit we find tha ty, at electric field normal to th he the plan is, ne See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 11 SPHERICAL SHELL OF F CHARGE E DISTRIBUTION SOLID SP PHERE OF CHARGE CO ONFIGURATIO ON FIE FORMULA ELD A SOLID SP PHERE OF CHARGE See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 12 DISTRIBUTION CHARGE ED CYLINDE ER CO ONFIGURATIO ON FIE FORMULA ELD A 3. GAUSS LAW AND ELECTRIC FLU S’S D UX The qualitative descriptio of the elec on ctric field usin electric fie lines, disc ng eld cussed in Cha apter one, is relate to a mathematical equation known as Gauss’s Law. ed Gauss’s Law: ◦ Nam med after Kar Friedrich Gauss (1777-18 rl 855) ◦ Presents an Alter rnative to the difficult Cou e ulomb’s Law ◦ Uses geometry/symmetry to s s simplify calcu ulations ◦ A re elation betwe een the elect field at al points in the surface and the total ch tric ll e d harge enclosed within t the surface See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 13 GAUSSIA SURFACE AN etical closed surface Any hypothe Can be any shape, but the most usef ones are those that miimics the sha ful t ape and symm metry of the problem at hand. ELECTRIC FLUX C Mathematic quantity th correspo cal hat onds to the nu umber of field lines cro ossing a surfa ace Which for a s surface perp pendicular to E is defined as the product of th magnitud of the field E and the area A he de d a Φ is Elec ctric Flux, in Nm2/C E is Elect Field, in N tric N/C A is perp pendicular to the area of the Gaussian surface, in m2 o n θ is the s smallest angle between E and A EXAMPLES: 1. If the e electric field in a region h a magnitude of 2.0x103 N/C has 0 directed towards the right as show in the figu d e wn ure, what is the value of the elec ctric flux passiing through a rectangula Gaussian su ar urface of cross sec ctional area 0.0314 m2? 2. What is the value o the electric flux passing through the of c g e n 0 ct rectangular Gaussian surface if it is angled 50o with respec to the electric field? MORE NO OTES ON ELEC CTRIC FLUX: Electric Flux iis zero, if E an A are perp nd pendicular to each o other (IMAG GINE THIS!) When an E-fiield vector e enters the surf face, it has a negative va alue. When an E-fiield vector le eaves the sur rface, it has a positive value! See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 14 FINDING THE NET FLUX (Φnet) G X QUALITA ATIVE DESCRIP PTION OF GA AUSS’S LAW The net f flux through a surface e any equals the ch harge enclos sed over permittivity of fre ee-space (ε0) EXAMPLE ve of C, 5.9 Given fiv charges o values q1 = q4 = +3.1 nC q2 = q5 = -5 nC, and q3 = -3.1 nC, find the net electric flux through the Gaussian surface S. e Refer to the figure. ATIONS OF GA AUSS’S LAW APPLICA Gauss’s Law is valid for any distributio of charges and for any closed surfa w on s y ace Gauss’s Law can be used in two way w d ys: 1. If we know the c e charge distrib bution and we want to find E. e 2. If we know E and we want to find the cha e d o arge distribution causing E E. 4. DISCO ONTINUITY OF En F When E1 passes through the sheet, it experiences discontinuity y as it exits as E2 See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 15 5. CHARGE AND FIELD AT CONDUC D CTOR SURFAC CES Electrost tatic Equilibrium harges that a able to move around the conduct are m tor. Conductors have free ch If there is an electric field inside the conductor, there will be a net force on this charges causing d n s a momentar electric cu ry urrent (CHAPT FIVE). TER However, un nless there is a source of e energy to ma aintain this cu urrent, the ch harges will me erely redistribute it tself to nullify the field cre y eated inside the conducto t or This is known as Electrosta Equilibriu atic um! CHARGE AND FIELD E S harge in a co onductor will only reside o the surface on e. VIA GAUSS’S LAW, the ch The consequ uence of this finding is tha no charge shall exist INSIDE the con at nductor, henc no ce field shall exiist INSIDE the conductor. (A finding we have seen earlier, see S e SHELLS) Thus conduc ctors provide effective shiielding! You will see m more of this in the problem set! Research ab bout Faraday Pail! y’s 6. DERIVA ATION OF GA AUSS’S LAW F FROM COULO OMB’S LAW AND VICE VERSA A Apply G Gauss’ Law Use E = F 0 F/q k, ulomb’s Law Manipulate, collect k derive Cou TRY TO DE ERIVE GAUSS’S LAW FROM CO OULOMB’S LAW BY REVERSING THIS PROCES W G SS See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 16 PART THREE: THE ELECTRIC POTENTIAL OUTLINE 1. Potential Difference Continuity of V Units Potential and Electric Field Lines 2. Potential due to a System of Point Charges 3. Finding the Electric Field from the Potential General Relation between E and V 4. V of Continuous Charge Distributions 5. Equipotential Surfaces The Van de Graff Generator Dielectric Breakdown OBJECTIVES After this chapter you should be able to: 1. Define and differentiate electric potential difference, electric potential, and electrostatic potential energy. 2. Calculate the potential difference between two points, given the electric field in the region. 3. Define of the electron-volt (eV) energy and the conversion factor between eV and the joule. 4. Calculate the electric potential of discrete and continuous charge distributions __________________________________________________________________ THE CONSERVATIVE ELECTRIC FORCE Electric Force between two charges is directed along the line of charges and depends on the inverse square of their separation (this is the same as the gravitational force between two mass, Recall Physics 3) Like Gravitational Force, electric force is conservative! Now, when we say that a force is conservative, there is always a potential energy function U associated with it! ELECTRIC POTENTIAL If we place a test charge q0 in an electric field, its potential energy is proportional to q0. The potential energy per unit charge is a function of the position in space of the charge and its called ELECTRIC POTENTIAL! See more @ teodorickbarry.multiply.com PHYSICS DIVISION P H Y S I C S 1 3 L E C T U R E N O T E S | 17 1. POTEN NTIAL DIFFERENCE DEFINITIO POTENTIA DIFFERENC ON: AL CE For a finiite displacem ment from po a to point b, the chan oint nge in potent is tial 1. 2. Notes: The potentia difference Vb-Va is the n al negative of the work done per unit ch harge by the electric field on a po ositive test ch harge when it moves from point a to p t m point b. ∆V is also the positive wo done per charge that you must do against the electric field to e ork o d move the ch harge from a to b CONCEP RECALL: PT You might be confused r e right now, so let’s clear things up!!! Potential Ene ergy the c capacity for doing work w which arises from position or configura n ation in a forc field. ce An object m moves against the field t U inc creases. Wor is done aga rk ainst the field d An object m moves with the same direc e ction of the field U de ecreases. The work is done by the field e d. ANOTHER CONVENTIO TO MAKE IT EASIER ON E If a test char rge (+) move against the direction of es e o electric field, its U increa ases. If a test char rge(+) move with the sa es ame direction as the n electric field, its U decrea ases he The function V is called th electric potential or just the potential. See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 18 Like the elec ctric field, the potential V is a function of position. e The electric f field is a vec ctor function, whereas the electric pot e tential is a scalar function n! As with pote ential energy U, only chan nges in the po otential V are important. e We are agaiin, free to cho oose the pot tential to be zero at any c z convenient p point. If the electric potential and potential energy of a test charge are chosen t be zero at the c to t same point t they are related by: UITY OF V CONTINU In chapter tw we saw t wo, that the electric field is dis scontinuous b σ/ε0 at a p by point where there is a surface dens σ. sity The potentia function, on the other h al n hand, is continuous everyw where in space. UNITS OF V F ential (V) is po otential energ per unit charge, hence the SI Unit iis gy Electric Pote 1 J/C = 1 V (Volt, after A Volta) A. nd hysics, the un used for energy is calle the electro nit ed on-volt (eV). eV In atomic an nuclear ph converts to J Joule by 1 eV = 1.6 x 10-19 CV = 1.6 x 10-19 J 6 For example an electron moving from the negati terminal t the positive terminal of a 12-V e, n m ive to e f car battery g gains potentiial energy of 12eV. POTENTIA AND ELEC AL CTRIC FIELD LIN NES If we place a positive tes charge q0 iin an electric field E, and release it, it a st c accelerates in the direction of E E. As kinetic en nergy of the c charge increases, its pote ential energy decreases. The charge t therefore mo oves toward a region of lo ower potentia just as a m al, mass falls tow ward a region of low gravitatio wer onal potentia energy. al “ELECTRIC FIELD LINES PO OINT IN THE DIRECTION OF DECREASING ELECTRIC PO G OTENTIAL” 2. POTEN NTIAL DUE TO A SYSTEM OF POINT CHAR F RGES Definition: Th Potential is zero at an infinite distan he s nce from the point charge e. The electric potential at a distance r f from a point charge q at the origin is calculated as: t a This potentia is known as Coulomb po al s otential. It is positive or ne p egative depe ending on the sign of e the charge q q. The potentia energy U of a test charg q0 placed a distance r from the po charge q is al ge d oint This is the ele ectrostatic po otential energ of the two gy o-charge syst tem relative to U= 0 at inf finite separation EXAMPLE E 1. What is the electric potential at a distance r = 0.529x10-10 m from a p c proton? (This is the averag ge e distance between a proton an electron in a H atom) What is t the potential energy of th electron a he and the proto at this sep on paration? See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 19 2. In nuc clear fission, a uranium-235 nucleus ca aptures a neu utron and splits apart into two lighter nuclei. n Sometim the two fiission produc are a bariium nucleus (charge +56e and a kryp mes cts e) pton nucleus (charge s +36e). g e parated by r = 14.6 x10-15 m. Calcu 0 ulate the Assuming that these two nuclei are point charges sep potentia energy of this two-charg system in electron-volts. al ge al oint everal point charges is the sum of the potentials du to c e ue The potentia at some po due to se each charge separately. (Superpositiion Principle following the force and the field) e e MORE Examples are give in the exam en mple listings. 3. FINDIN THE ELECT NG TRIC FIELD FRO THE POTE OM ENTIAL: THE GENERAL RELA G ATION BETWEE E AND V EN If we know th potential, we can use it to calculate the electr field using this relation he , ric Conversely w can know the potential from the electric field a was shown earlier in this we w e as n chapter. CONTINUOUS CHARGE DIS S STRIBUTION 4. V OF C The potentia of continuo charge d al ous distribution ca be obtained using techniques furni an ished by calculus. ae s stribution shall be given, lo osing empha on asis Only formula and speciial properties for each dis the techniqu used to de ue erive them. DISTRIBUTIO ON INFINITE LINE CHARGE CONFIG GURATION and POTENTIA FORMULA AL See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 20 CHARGED RING CHARGED DISK INFINITE PLA ANE The potential V at a point P with a dis stance x (eith her to left or to the right) fro om the infinite plane Where V0 is the potentia s al at x=0 ecessarily zero o! V0 is not ne SPHERICAL SHELL The potential inside the hell e spherical sh of charge is constant See more @ teod o dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 21 The potential outside th he spherical sh of charge hell e depends o r distance on from the su urface WHEN TH FIELD IS ZER HE RO… It is a commo mistake th when the field is zero, the potentia is also zero on hat e , al o. When the fie is zero, it o eld only means th the potential at that llocation is hat constant/unchanging… leading to a “change of potential” equal to zero!!! 5. EQUIPOTENTIAL LIN AND SURF NES FACES Equipotentia Line al are lines drawn in an electric field such th that all the points on the line are at the c hat me sam potential. Equipotentia Surface al is a s surface, all p points of whic are at the same potential. ch Equipotentia Lines and S al Surfaces are a always perpe endicular to the electric f field lines! Movement a along an equ uipotential lin requires no work because such mo ne o ovement is alw ways perpendicular to the elec ctric field. METALS A AND THE EQU UIPOTENTIAL How much w work would it take to drag over the surface of a g conducting metal, a pos sitive charge q from Point A to Point B? t ? ONE! Answer is NO Because me etals are equiipotential volumes and th have hey equipotentia surfaces. Se Gauss’s La for explanation! al ee aw See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 22 VAN DE GRAFF GENERATOR How does a Van de Graff Generator works? Topics that can explain: 1. Potential 2. Electric Field Lines 3. Conductor Property DIELECTRIC BREAKDOWN Happens when non-conducting material become ionized when exposed to very high electric fields and become conductors Dielectric Strength The magnitude of the electric field for which dielectric breakdown occurs in a material Emax,air = 3 x 106 V/m = 3MN/C Arc Discharge The discharge through the conducting air resulting from dielectric breakdown. An example is the electric shock you receive when you touch the metal door knob after walking across a rug on a dry day. See more @ teodorickbarry.multiply.com PHYSICS DIVISION P H Y S I C S 1 3 L E C T U R E N O T E S | 23 UNIT TW ELE WO: ECTRIC E ENERGY AND C Y CAPACI ITANCE OUTLINE 1. Electrosta Potential Energy atic 2. Capacita ance and Ca apacitors 3. The Stora age of Electrical Energy and Electro ostatic Field Energy 4. Combina ations of Cap pacitors: Parallel and Series Capacitors d 5. Dielectric cs: Energy Sto ored in the presence of a Dielectric ar Dielectric: 6. Molecula View of a D Magnitude of the bound charge d The Piezoel lectric Effect OBJ JECTIVES At th end of th chapter, you should he his be able to: a 1. De efine Electros static Energy; 2. Define Capac citance and apply the d cept to ex xplain how capacitors conc work k; 3. Ex xplain how electrostatic energy is c store ed; 4. Define a Dielectric and how diele ectrics store electrostat tic energy; and 5. Solve problems that ca alculate for acitance, electrostatic c energy, capa mag gnitude of th bound an the free he nd char rge; INTRODU UCTION When we briing a point c charge q from a far away to a region where other charges are present, m y r e WE MUST DO WORK qV, w O where V is the potential at the final po a osition due to other charg in the o ges vicinity. as tic The work done is stored a electrostat potential energy. The electrost tatic potentia energy of a system of charges is th total work needed to assemble al he a a system. When a cha arge is placed on an isolated conduct d tor, the poten ntial of the conductor inc creases. The ratio of t the charge to the potentiial is called th capacitance of the conductor. o he When a cha arge is placed on an isola ated conduc ctor, the pote ential of the conductor in ncreases. (Why???) The ratio of t the charge to the potentiial is called th capacitance of the conductor. o he A useful dev vice for storing charge and energy is th capacitor. g he A capacitor consists of 2 conductors, closely spac ced but insula ated from ea ach other. When a cap pacitor is atta ached to a s source of potential differe ence (such a a battery), the two as conductors c carry equal a and opposite charges. e The ratio of the magnitude of the charge on either condu e uctor to the potential difference between the conductors is called the capacitanc of the cap e s e ce pacitor. CAPACITORS have m many uses! 1. The flash attachment is your cam t mera uses a capacitor to store the o energy need ded to provid the sudde flash of ligh de en ht. 2. Tuning c circuits of c communicatiion devices such as t the radios, televisions, a and cellular phones allo owing them to operate at certain frequencies. 3. The defibrillator used by paramed dics in revivin near-dea patients ng ath uses capacitors to store charges in o order to relea the right amount of ase See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 24 shock acitor was th Leyden Ja a glass con he ar, ntainer lined inside and o with gold foil. out The first capa ROSTATIC PO OTENTIAL ENER RGY 1. ELECTR The Electrost tatic Potentia Energy of a system of point charge is the work needed to bring the al p es k charges from an infinite s m separation to their final po o ositions! The electrost tatic potentia energy U o a system of n point cha al of o arges is: Where Vi is th potential at the location of the ith charge due to all the oth charges. he her This is also tru for continu ue uous charge distributions and system o conducto of ors. 2. CAPACITANCE al o ntial at infinit of a single isolated c ty) conductor ca arrying a The potentia (relative to zero poten charge Q is proportional to the ch s harge Q, an depends on the size and shape of the nd s e conductor. In ge eneral, the la arger the con nductor, the greater the amount of charge it can carry for a giv ven potentia al. For example the potential of a spher e, rical conduct of radius R carrying a charge Q is tor The ratio of t the charge Q to the pote ential V of an isolated co n onductor is ca alled its capa acitance C: ce capacity to st tore charge f a given p for potential diffe erence. Capacitanc is the measure of the c Since the po otential is alw ways proportiional to the charge, this ratio does N c NOT depend on either Q or V, but o only on the siz and shape of the cond ze e ductor. The capacitance of a spherical c conductor is The SI Unit o capacitan of nce is the co oulomb per volt, which is called a farad (F) after Michael v s r Faraday: 1 F = 1 C/V Since the farad is a rathe large unit, submultiples such as the microfarad (1 μF = 10-6 F) or the er , e d are ed. picofarad (1pF = 10-12F) a often use Since capac citance is in f farads and R is in meters, we can see from the las slide, that the SI unit st t for permittivity of free spa ace εo can also be written as farad pe meter. n er See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 25 HOW BIG IS THAT??? G 1. Find th radius of a spherical conductor tha has a capacitance of 1 farad. he at Answer, 8.99 x 109 m, which is abo 1400 time the radius of the earth!!! Whoa!!! , out es 2. A sphe of capac ere citance C1 carries a char rge of 20μC. If the charge is increased to 60μC, what is the e d new cap pacitance C2? C1 = C2. The capacit tance does n depend o the charg nor the po not on ge otential differ rence. If the charge is c tripled, t the potential also triples, hence, ratio is preserved Capacitan d. nce depends only on the size and s e shape of the conduc ctor and of th capacitor he r! TORS CAPACIT Is a syste of two co em onductors carrying equal but opposite charges. e A capacitor is usually charged by tra ansferring a charge Q from c m one conduc ctor to the oth leaving o her one of the co onductors wit th a charge of +Q and the other with –Q Q. The capacita ance of this d device is def fined to be Q/V* Q In general, to calculate t o the capacita ance, 1. W place equ and oppo We ual osite charges on the cond s ductors, then n 2. Fiind the pote ential differe ence V by first finding the electric field E betw ween them. PARALL PLATE CAPACITORS LEL ates are the m most common capacitor types. Parallel pla In practice the plate may be thin meta e, es e allic foils tha are at separa ated and insu ulated from o one another by a thin plas film. b stic The capac citance of the parallel pla capacitor is: e ate rs When a cap pacitor is co onnected to a battery (a shown as above), charge is transferred fr rom one con nductor to the other t tween the co onductors eq quals the until the potential difference bet potentia difference across the battery termin al nals. The amount of charge transferred is Q = CV PARALLE PLATOS EL 1. A para allel-plate ca apacitor has square plate of side 10c separated by 1mm. es cm d (a) Calc culate the ca apacitance o this device. of (b) If this capacitor is charged to 12 V, how m s s much charge is transferred d? Ans (a. 8 88.5 pF b. 1.06 nC) 2. How la arge would t the plates ha ave to be for the capacit r tance to be 1F if they are to be sepa e arated by 1mm? Ans (1.13 x 108 m2, wh 3 hich correspo onds to a squ uare with a si ide of 10.6km m) See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 26 CYLINDR RICAL CAPAC CITORS A cylindricall capacitor c consists of a s small conduc cting cylinde or wire of r er radius r1 and a larger, concentric c cylindrical she of radius r2. ell A coaxial cable, such a that used for cable television can be though of as a cylindrical as d t ht c capacitor. 3. THE ST TORAGE OF ELECTRICAL EN NERGY When a ca apacitor is being charged, positiv charge is ve transferred from the ne egatively ch harged cond ductor to th he positively charged condu uctor. Work must th herefore be d done to char rge a capacitor. Some of this work is stored as electros static potenti energy. ial When a sm mall amount of charge ∆q is mov t e ved from th he negative co onductor to the positive c conductor, its potential energy is incre s eased by ∆U = V(∆q), U where V is th potential d he difference be etween the conductors! c The potentia energy inc al crease, is stor red in the ca apacitor via the field. Usiing C = Q/V, we can express this e energy in a v variety of way ys: Example es: 1. A 15-μ capacitor is charged to 60V. How m μF o much energy is stored in t y the capacito (Ans 0.027 or? 7J) 2. How m much energy is stored in the capac n citor if it 24.5μC of charg was transf ge ferred to the positive e conduct using a po tor otential of 10 00V? (Ans 0.0 00245J) PP EXAM MPLE A paralle el-plate capa acitor with sq quare plates 14cm on a side and separated by 2.0mm is conn s nected to a batter and charged to 12V. ry The batt tery is then diisconnected from the capacitor and the plate sep paration is increased to 3.5mm. 3 Question ns: 1. What is the charge on the cap e pacitor? 2. How m much energy was originally stored in th capacitor? y he 3. By how much is the energy increased when the plate se w e n eparation is c changed? THE ELEC CTRIC FIELD EN NERGY In the process of charging a capacito an electric field is prod or, duced betwe een the plate es. quired to cha arge the cap pacitor can be thought o as the wo required to create of ork t The work req the electric f field. That is, we c can think of t the energy st tored in a ca apacitor as e energy stored in the elec d ctric field, called ELECT TROSTATIC FIE ENERGY ELD The quantity that charac y cterizes electr rostatic field energy is called ENERGY DENSITY (ue), which is the energy stored per un volume of space cover nit red by the fie given by eld, y See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 27 PP EXAM MPLE PART TW WO 4. Calcu ulate the ene ergy density ue when the p plate separation is 2.0mm m 4. COMB BINATIONS OF CAPACITORS Two or more capacitors a often use in combin are ed nation. Types of Com mbinations 1. PARAL LLEL he tes s cted by a conducting When th upper plat of the two capacitors are connec wire and are therefo at a com d ore mmon poten ntial Va, and the lower plates are p connect ted together and are at a common p r potential Vb. When in parallel, t n the potentia difference Va-Vb is t al e the same across all a capacitors! 2. SERIES S When t two capacit tors are connected so that the c charge on the two capacitors are equa al. n potential diff ference is th sum of the individual potential he e When in series, the p differenc in each o the capac ces of citors! PA ARALLEL CAPACITORS SERIES CAPACITORS . Notes for C CAPACITORS IN PARALLEL 1. The e voltag ge (pote ential difference) across each ca apacitor is the same and is eq qual to the pote ential difference of th source he The charge stored in each e ca apacitor totals the ch harge sto ored in all of the capacit tors in the circuit e As you increas the numb of se ber ca apacitors in the circuit, you inc crease Ceq. Ce is always greater than the eq ind dividual capacitances o the of ca apacitors in th circuit! he Note for CAPAC es CITORS IN SERIES 1. 1 The char rges stored across all the capacito in the circ ors cuit are the same. The pote ential differe ence across each c capacitor totals the potential difference o source. of As you add capac citors in the Ceq decrease es. circuit, C Ceq is always less than the s al nces of the individua capacitan capacito ors. 2. 2 2. 3. 3 3. 4. 4 4. See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 28 CTRICS 5. DIELEC Dielectric Any non n-conducting material (i.e an insulato Examples are air, glas paper, g e. or). s ss, or wood d. When a Dielectric is iinserted in th space bet he tween the ca apacitor… The cap pacitance is increased by a factor κ that is c s characteristic of the dielectric, a fact exp perimentally discovered b M.Faraday d by y. Reasons for t the Increase… son for this increase is that the electric field be t etween the plates is The reas WEAKEN NED by the diielectric. Thus, for a given cha arge on the plates, the po p otential differ rence is redu uced and the Cap pacitance (C = Q/V) is inc C creased! κ (Kappa) Called the dielectric constant CASE 1: A. Charg on the pla ge ates did not change whe the dielect is inserted c en tric d B. This is only the cas if the cap se, pacitor is charged and the removed from the en charging source (in t case: the battery) bef g this e fore the inser rtion of the dielectric. d C. Equation to use? “All the equa ations found in the previo slide”, an noting ous nd that Q = Q0 CASE 2: A. Charg will chang accordin to Q = κQ0 (Q, is the total charge after the ge ge ng Q dielectric has been inserted) if th dielectric is inserted w he while the batt tery is still ted! connect B. This ha appens beca ause, the ba attery will sup pply more ch harge to the plates to maintain the originall potential dif n fference! C. Equation to use? “All the equ uations found in the previious slide, ex d xcept V = cause V = V0) κV0 (bec e, citance chan nges according to C = κC0 C In either case the capac See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 29 Dielectric Exercises: 88.5 pF capa acitor of “PAR RALLEL PLATO #1” is filled with a diele OS d ectric consta κ=2. ant 1. The 8 (a) Find the new cap pacitance. (b) Find the charge o the capa on acitor with the dielectric in place if the capacitor is attached to a 12-V e e t battery. (THIS IS CASE 2: Ans: (a) 177 pF; (b) 2. .12nC) 2. The capacitor in the abo e ove exercise is charged to 12V wit d thout the diielectric and is then d disconne ected from the battery. T The dielectric κ=2 is then in c nserted. Find the new vallues for (a) The c charge Q, (b) the v voltage V, an nd (c) the c capacitance C. e (THIS IS C CASE 1: Ans: ( Q =1.06nC (b) V = 6V (c) 177 pF) (a) C; V; ENERGY STORED IN TH PRESENCE OF A DIELEC HE CTRIC Recall: the e energy densit ty Now, in the p presence of a dielectric, the energy density becomes d CULAR VIEW OF THE DIELECTRIC 6. MOLEC As we all kn now by now the dielec w, ctric weakens the electric field betw ween the pla ates of a capacitor. This was seen in t the equation in the slide entitled “The Increase an The Decre ns e nd ease”. But the true r reason behin this pheno nd omena and the resulting e t equations is t that: The dielectric pro oduces a fielld opposite to the field pr roduced by t the capacito plates! or Now, Wh does the d hy dielectric pro oduce an ele ectric field ins side? 1. Molecules in the dielectric are neutral, but it doesn’t mean that they are not affe n d n ected by electric fields! s contain positiive and nega ative charge that respon to the field es nd ds. 2. Because this molecules c THE ATOMIC MODEL… … Atoms can be thoug in this man ght nner: The nucleus is ap pproximated as a sphere and is at the “geometric” center of th atom e he The electrons are distributed as an electron cloud, and some of these clo d , ouds are distr ributed as sph heres. POLARITY AND N NONPOLARITY Y If the nu ucleus and the electron cloud are concentric, the n atom m/molecule is NONPOLAR because th dipole mo he oment is zero ucleus and th electron cloud are not concentric, the he c t If the nu atom m/molecule is POLAR and there is a net dipole mo d oment See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 30 When there is no field, the dipole mom ments are randomly orien nted. When there is an externa field: al AR s: For POLA Molecules The dipole moment aligns ac ccording to th direction of the field. he For NON NPOLAR Mole ecules The field induces a dipole m moment within the molec cule, and the align acco ey ording to field the f The Effec of the Pola ct arization and the Bound C Charge The effect of the polariza ation of a ho omogeneous dielectric in a parallel plate is the creation of a surface ch harge on the dielectric faces near the conductor. e The surface charge on t the dielectric is called a BOUND CHA c ARGE becau it is boun to the use nd molecules o the dielect of tric and can nnot move about like the free charge on the conducting e e capacitor pllates! These bound charges produce an Electric Fie n eld opposite to the field produced by the e d conducting capacitor plates! Thus if we ad all the ele dd ectric fields e existing in the capacitor, the field be e etween the plates are p reduced! Illustratio of Field Re on eduction The Magnitu ude of the Bo ound Charge is given by: See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 31 If κ=1 (Meaning there is no dielectric), the bound charge density, σb, is zero If κ=∞ (Meaning a conducting slab is inserted between the plate), the bound charge density, σb, is equal to the free charge density, σf! PIEZOELECTRIC EFFECT In certain crystals that contain polar molecules such as quartz, tourmaline, and topaz, a mechanical stress applied to the crystal produces polarization of the molecules! As we all know, again, these polarization mean an electric field is produced, thus a potential difference across the crystals! USES OF PIEZOELECTRIC EFFECT Transducers in microphone, phonograph pickups, and vibration-sensing devices See more @ teodorickbarry.multiply.com PHYSICS DIVISION P H Y S I C S 1 3 L E C T U R E N O T E S | 32 UNIT THREE: ELECTR CIRC RIC CUITS OUTLINE 1. Current a and Motion o Charges of OBJ JECTIVES he s At th end of this chapter, it is expected that you would b able to: be 1. De efine steady-state current and relate t this to the motion of charges; n 2. Define a ma D aterial’s resis stance and relate the three circuital p e parameters: stance, Voltage, and C Current via Resis Ohm Law; m’s 3. De efine electric energy, ele c ectromotive force and enume e erate their ap pplications; 2. Resistanc and Ohm’s Law ce 3. Energy in Electric Circ n cuits: EMF and Ba atteries 4. Combina ations of Resi istors: Series Resis stors and Para allel Resistors s 4. Le earn the different comb binations of resist tors in a DC C Circuit and s solve for the equivalent resista ance of a giv ven circuit; 5. Use Kirchhoff’s Rules in solving the tances of a complex equivalent resist uit; circu and 6. So olve problem involving r ms resistors and capa acitors in com mbination. 5. Kirchhoff Rules: Junc f’s ction, Loops and Measu uring Devices s Ammeters, Voltmeters, and Ohmme eters 6. RC Ci ircuits: Capacitor Disc charging an nd a onservation Energy Co Capacitor in Chargin ng a UCTION INTRODU When we tu on a ligh we conne the wire filament in the light bulb across a potential urn ht, ect difference th causes el hat lectric charg to flow thro ge ough the wire e! Very much like the way a pressure d difference in a garden ho causes w ose water to flow through w the hose electric charg constitutes an ELECTRIC CURRENT. ge C The flow of e being in con nducting wire but the electron beam in a video monitor es, m o Usually, we think of currents as b and an electron mo oving around the nucleus of the H atom can also be considered as an ELECTRIC d CURRENT! ENT 1. CURRE AND THE MOTION OF CHARGES ELECTRIC CU URRENT The rate of flow of electric charge through a crossw tional area. sect See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 33 The SI Unit of current is the ampere (A f e A): 1 A = 1 C/s Convent tions: 1. The direction of current is considered to be the dir n s rection of flow of positive charge. e ablished befo it as know that free electrons are the particle that actua move ore wn e es ally 2. This was esta in a conducting wire! ons he n ent. 3. Thus, electro move in th direction OPPOSITE to the direction of the curre TANCE AND O OHM’s LAW 2. RESIST Current in the conductor e Is dr riven by an e electric field E inside the conductor t that exerts a force qE on the free n charges c ? What about Electrostatic Equilibrium? Since E is in t the direction of the force on a positive charge, it is in the direct e s tion of the cu urrent! Figure shows a wire segm s ment Electric Field points in the direction of lower poten d e f ntial. (Va>Vb) Assume that ∆L is small enough so t t that we may consider th y he electric field E to be cons stant in that s segment al points A and B is given by y The potentia difference V between p Again, we u use V rather than ∆V for the potent r tial differenc ce (which in this case is a potential decr rease/drop) to simplify th he notation. The ratio of the potential drop to the c current is call led the resistance of the s segment. Definition of Resistance: f the volt per a ampere, is ca alled an ohm (Ω): 1 Ω= 1V m V/A The SI Unit of resistance, t OHM’s L LAW For many materials (Term med OHMIC), the resistan nce does not depend on the voltage drop or t n e the current. OHM’s Law: For O Ohmic materials, the pote ential drop across a segm a ment is propo ortional to the current: e *For non-ohm materials the resistan mic s, nce depends on the curre I, so V is n proportional to I. s ent not See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 34 1. 2. 3. Ohm’s Law is not a fundamenta law of nature like N al Newton’s La aws or the Laws of Thermodyna amics. Ohm’s Law is simply an e empirical description of a property sha ared by many materials. y Examples: nt hat tential drop a across the wi ire? (Ans: A wire of resistance 3 carries a curren of 1.5A. Wh is the pot 4.5V) creases, what happens to the resistanc of a mate t o ce erial? If current inc If potential d difference inc creases, wha happens to the resistance of a mate at o erial? NCE AND RES SISTIVITY RESISTAN The resistanc of a cond ce ducting wire is found to be proportiona to the leng wire and inversely e al gth proportional to its cross-se ectional area a: esistivity of the conducting material. It unit is called the ohm-m ts meter (Ω-m) ρ is called re Example es: 1. A Nichrome wire ( =10-6 Ω-m) has a r radius of 0.65 5mm. What le ength of wire is needed to obtain e t a resistance of 2.0 Ω)? 2. Calculate th resistance per unit leng of a 14-ga he gth auge coppe wire. er RESISTAN NCE AND RES SISTORS Carbon, whiich has a relatively high electrical re esistivity, is used in resist tors found in electric equipment. Resistors and their com mbinations, a are used in circuit to cessary resista ance to man nipulate the electrical provide nec property of certain circuits, netw f works, and complex electrical de evices. OWER ENERGY IN ELECTRIC CIRCUITS: PO See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 35 EXAMPLE POWER E A 12-Ω re esistor carries a current of 3 A. Find the power dissi s e ipated in this resistor. s Solution: : P = I2R = (3A)2 (12 Ω) = 108 W Other Po ower Example es: A wire of resistance 5 Ω carries a current of 3A for 6s. A (a) How much powe is put into th wire? (45W er he W) (b) How much therm energy is p mal produced? ( (270 J) EMF AND BATTERIES D maintain a ste eady current in a conductor, we nee a constan supply of electrical ed nt e In order to m energy Devices that can supply electrical en t nergy are called source o emf (electro of omotive forc ce) Examples of sources of e emf are a bat ttery (conver chemical energy to electrical ener rts rgy), and generator (c converts mec chanical ene ergy to electrical energy) OES ORK? HOW DO “SOURCE OF EMF” WO A source of emf does wo on the ch ork harge passing through it by raising the potential energy of e the charge. The work per unit charge is called the emf ξ,of the source. e e e volt, the sam as the unit of the poten me ntial differenc ce. The unit of emf is still the v E A SIMPLE RESISTIVE CIRCUIT 1. 2. 3. 4. 5. Symbols used d What does the source of EMF maintain? What is Vab, Vac, Vdb, a , and Vcd?’ What is the d direction of th current I? he What is the p power delive ered by the so ource of EMF and the pow dissipate at the resis F wer ed stor? IDEAL AN REAL BATT ND TERIES Inside the ba attery, charge flows from a region of lo potentiall to a region of high poten ow ntial, so it gains potent energy tial IDEAL BATTER RY Maintain a constant potential differenc ns ce between its two terminals, n t indepen ndent of the f flow rates of the charge b between the em. The pote ential difference betwee the termin en nals of an ide battery is equal in eal magnitu ude to the em of the battery. mf REAL BATTERY Y tery termina als, called TERMINAL T The potential difference across the batt VOLTAG is not simp equal to the emf of the battery. GE, ply e See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 36 Consider this simple circu s uit: The circuit co onsists of the following: Re Battery and a Resistor eal r If the curren is varied b varying th resistance and the te nt by he e erminal volta age is measu ured, the terminal volt tage is found to decreas slightly as the current increases, ju as if there were a d se ust e small resistan nce inside the battery e REAL BAT TTERIES GETTING REAL ss emf, since th internal re he esistance The terminall voltage of the battery is always les than the e reduces the emf! Real batteries ha ave very sma internal re all esistances, meaning there is a small difference e betw ween the terminal voltage and the em unless the is a very la mf, ere arge current. Malf functioning b batteries hav very large internal resis ve e stances! Whiich results to very low term minal voltage es Batteries are often rated in Ampere-h e hours (A•h), which is the total charge they can deliver w 1 A•h= 1 C/s (3600s) = 36 C s 600 The total ene ergy stored in the battery (W) is the to charge times the emf n y otal f! W = Qξ I’M REAL BATTERY L An 11-Ω resistor is connected acr ross a battery of emf 6V and internal r y a resistance 1 Ω Ω. Find the following a) The current b) The te erminal voltage of the ba attery c) The power deliver red by the em source mf d) The power deliver red to the ext ternal resistor r e) The power dissipated by the battery’s internal resistance f) If the b battery is rate at 150 A•h, how much energy doe it store? ed h es See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 37 4. COMB BINATIONS OF RESISTORS Two or more resistors can be used in c n combinations. b g y The analysis of a circuit can often be simplified by replacing two or more resistors by a single equivalent resistor that c carries the sa ame current with the sam potential drop as the original me e resistors. mbinations: Types of Com SERIES Resisto ors PARALLEL Re esistors 1. 2. STORS SERIES RESIS PARAL LLEL RESISTORS No otes for Resistors in Series 1. All resistors h have the s same current flowing through them The total pote e ential drop is the sum of the ind m dividual pote ential dro in each o the resistor ops of rs 2. 2 1. 1 Notes for Resistors in P r Parallel The curre splits as it leaves the ent t point of separ ration of connection (terme ed as a junction) and rejo oins as it reaches the other jun nction The pote ential drop in any of the n resistors a equal in v are value. The cur rrent in an ny of the junction is the sum of all the currents through all the paths out leading o and in the junction. 2. 3. 3 See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 38 5. KIRCH HHOFF’s RULES S There are m many circuits such as the one given above e that can no be analy ot yzed by me erely replacing the resistors by e equivalent res sistors. Thus we dev vise another technique that can do so with breeze! e apply them in two nto We will utilize Kirchhoff’s Rules and a different con nfigurations. Two Kirchhoff’s Rules: 1. Junction Rule e 2. Loop Rule Two Con nfigurations: 1. Single Loops 2. Multiple Loops KIRCHHOFF’s JUNCTION R s RULE At a junction p any point in the c circuit, the current can divide, the sum of the currents into m the j junction mus equal the s st sum of the cu urrents out of the junction f n. KIRCHHOFF’s LOOP RULE s Whe any close en ed-circuit loop is traversed all potentiial gains mus be equal to all the d, st pote ential drops Or, a changes in potentials m all must add up to zero! p HOW TO APPLY KIRCHOFF’s RULES O S 1. Draw a sketch of the circuit 2. Choo ose a directio o current in each br on t ranch of the circuit, and label the c e d currents in th circuit he diagram Add plus or minus sig m. gns to indica ate the high and low- potential sid hdes of each resistor, h capacito or source of emf. or 3. Replace any combination of re esistors in series or parallel with the equ uivalent resistance. 4. Apply the junction rule to each junction wh y h here the curre divides. ent 5. Apply the loop rule until you ob y e btain as man equations as the unkno ny owns. 6. Solve the equation to obtain t ns the values of the unknowns. 7. Check your result b assigning a potential o zero to one point in the circuit and uses the valu of the k by of e e ues currents found to determine the p potentials at other points in the circuit t. S MEASURING DEVICES The devices that measur electric cu re urrent, poten ntial differenc and resis ce, stance are called the ammeter, vo oltmeter, and the ohmme d eter respectiv vely. They are combined into a single dev vice called a multimeter that can be switched int any of to evices above ementioned. the three de DEVICE ER AMMETE VOLTMET TER OHMMETER Yo will need t refer to the appendix to this hando ou to e t out! HO TO CONS OW STRUCT See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 39 6. RC CIR RCUITS RC Circuits Con ntain a resisto and a capacitor or Curr rent I, flows in a single dire n ection, but its magnitude varies with tiime s App plications lie iin the behav vior of the RC Circuit, tha is, RC Circ C at cuit has charg ging and discharging abilities When we sa charging, we put the maximum amount of ch ay a harge possib for that capacitor ble c over a speciific time cons stant When we say discharg ging, we rem move all the charge in the capac e citor until its value is negligible. DISCHAR RGING A CAP PACITOR Discharge happens because when t the switch is closed at t = 0, there is a potential drop acros the resisto meaning s ss or, there is curre in it. ent The current is due to the flow of ch harge from the positive t of acitor through the resis stor to the conductor o the capa negative conductor of th capacitor he r. time, the ch harge on the capacitor is reduced, e After some t hence the current is also reduced! (W is this hap Why ppening?) ns d arge This happen again and again, until at some time, the cha and the curr rent are both negligible h h hence “discharge” 1. Q0 is known as th initial cha he arge contain ned in the capacitor be c efore discharged. It has a va alue of CV0, where V0 is the potential being d differenc between the plates of a capacitor ce f 2. The ch harge in the capacitor “d decays exponentially “ 3. The tiime constan τ is the tim it would take the ca nt me apacitor to fully discharg if at a con ge nstant rate CHARGING A CAPAC CITOR We assume t that the cap pacitor is initia uncharge ally ed. Charging ha appens because when the switch is closed at time t=0, s charge imm mediately be egins to flow through the resistor o w t onto the positive plate of the capacitor. e Charge will increase in th capacitor however, current decre he r, c eases. Charge in th capacito at some time later, will reach its m he or maximum value of Q = Cξ when the current I eq e quals zero. 1. Qf is th maximum charge that can be stor he m red in the ca apacitor, it is dictated by the emf source and t the capacita ance of the c capacitor. 2. When the capacit is fully cha tor arged, no mo current will flow in it! ore w 3. Charg will increase logarithmiically and tends to appro ge oach a satura ation value. See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 40 See mo @ teodorickbarry ore d y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 41 UNIT FOUR: THE MAGNETIC FIELD AND ITS SOURCES OUTLINE 1. THE FORCE EXERTED BY A MAGNETIC FIELD 2. MOTION OF A POINT CHARGE IN A MAGNETIC FIELD 3.THE MAGNETIC FIELD OF MOVING POINT CHARGES 4. THE MAGNETIC FIELD OF CURRENTS 5. GAUSS’S LAW FOR MAGNETISM 6. AMPERE’S LAW: LIMIT AND CORRECTION 7. MAGNETISM IN MATTER OBJECTIVES At the end of this chapter you must be able to: 1. Calculate the force exerted by a magnetic field; 2. Calculate the magnetic field from various field-source configurations; 3. Calculate parameters from velocityselector applications; 4. Define the Ampere; 5. Apply Gauss’s Law for Magnetism; and 6. Apply Ampere’s Law; PART 1: THE MAGNETIC FIELD BRIEF HISTORY Ancient Greeks (around 2000 years ago) were aware that magnetite attracts pieces of iron. There are written references to the use of magnets for navigation during the 12th century. In 1269, Pierre de Maricourt discovered using simple observations, the existence of magnetic poles. Note that like poles repel and unlike poles attract. In 1600, William Gilbert discovered that the earth itself is a natural magnet. Although electric charges and magnetic poles are similar in many respects, there is an important difference: Magnetic Poles always exist as pairs. No isolated magnetic poles were ever observed. 1. THE FORCE EXERTED BY A MAGNETIC FIELD In this course, we will examine the force exerted by a magnetic field on a › Moving point charge › Current-Carrying Wire Before we proceed, there are something we need to convene with F is force, q is charge, v is velocity, B is the magnetic field F is force, I is current, ∆L is the length vector, B is the magnetic field We will also, exhaustively discuss, the right-hand rule, to determine the direction of the vector that results from a cross product. Magnetic Force on a Moving Point Charge Experimental observations reveal that magnetic force on a moving point charge › Is proportional to q and v, and to the sine of the angle between v and B. › Surprisingly perpendicular to both the velocity and the field. The abovementioned observations are summarized as the equation below This is the force exerted on a point charge moving with a velocity v in a magnetic field Since F is perpendicular to both v and B, it is perpendicular to the plane defined by this two vectors. The direction of F is given by the right-hand rule as v is rotated into B, as illustrated below. See more @ teodorickbarry.multiply.com PHYSICS DIVISION P H Y S I C S 1 3 L E C T U R E N O T E S | 42 Magneti Force on a Current-Car ic rrying Wire Since the Ele ectric curren is basically moving charges, the w nt y wire that con ntains the current, will then experie ence magnet forces, on tic nce it is subjected to a ma agnetic field. This magnet force is p tic proportional to the current, the lengt of the wir segment, and the th re magnetic fie eld. This is summa arized in the e equation below Where ∆L is called the length vect tor, whose magnitude is the length of the wire and its m s h e direction is th direction of the curren he nt GNETIC FIELD 2. UNITS OF THE MAG The equation for the ma agnetic force on a movin point char e ng rge allows us to define th unit of s he the magnetiic field. The SI unit of the magnet field is the TESLA (T). f tic e A charge o 1C moving with a ve of g elocity of 1m perpendicular to a magnetic fie of 1T m/s eld experiences a force of 1N N The Tesla, ho owever, is a very large q quantity; we need to de efine the tesla in terms of a more popular unit– the gauss. – See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 43 EXAMPLE ES 1. A pro oton is movin with a ve ng elocity of 10 0Mm/s. It exp periences a magnetic fiield of 0.6G which is directed downward and northw d ward, making an angle of 70o with the horizonta Find the magnetic g o al. m force on the proton. n e mm rries a current of 3A in the x-direction. It lies in a m e . magnetic field of 0.02T d 2. A wire segment 3m long car that is in the xy-plan and make an angle of 30o with the x-axis, as shown in th figure. Wh is the n ne es t s he hat magnetiic force exer rted on the w segment. wire 3. MAGN NETIC FIELD LI INES Just as the E Electric Field E can be rep presented by Electric Fielld Lines, the Magnetic Fie B can y eld be represent ted by Magn netic Field Lin nes. In both case es › (1) D Direction of th field is ind he dicated by the direction o the lines of › (2) M Magnitude of the field is in f ndicated by the density o the lines of There are ho owever, two important di ifferences be etween elect tric field line and magn es netic field lines: (1) E Electric field lines are in th direction of the electr force on a positive charge, the he ric › mag gnetic field lin are perp nes pendicular to the magneti force on a moving cha ic arge › (2) Electric field lines begin on positiv charges and end on negative charges; d ve mag gnetic field l lines form cl losed loops! Note: Magn netic field lin nes emerge from the north pole and e enters the sou pole! uth See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 44 ON NT TIC 4. MOTIO OF A POIN CHARGE IN A MAGNET FIELD CASE 1: CYCLOTRON N The magnet force on a charged particle mov tic ving through a h magnetic fie is always perpendicular to the velocity of t eld the particle. The magnet force, thu changes th direction of the veloc tic us he city but not its magnitude. Therefore, m magnetic field do no wo on partic ds ork cles and do n not change their kinetic energy. r In the spec cial case, w where the velocity of a particle is perpendicular to a unif form field as shown in the figure, t the particle undergoes uniform circular m motion (UCM) ). In this spec case, the magnetic force cial e provides th centripeta force nece he al essary for the c centripetal acceleratio on in circular mo otion. ewton’s Seco ond Law to r relate We use Ne the quantit ties. are known a the cycl as lotron T and f a period and frequency r d respectively. uency The cyclotron period and frequ on rge-to-mass ratio depend o the char q/m but are independ dent of r and v of d e! the particle ATH CASE 2: THE HELIX PA Suppose tha a charged particle en at d nters a unifor magnetic field with a velocity no entirely rm c ot perpendicular to the field B. d olved into tw componen wo nts The velocity vector is reso › (1) the v| experie ences magne force, an is thus acc etic nd celerated. (2) the v|| experiiences no ma agnetic force and remains constant. e, › The trajector is then callled a helix, w ry which is illustra ated below. TIC CASE 3: THE MAGNET BOTTLE The motion o particles in non-uniform magnetic fi of n m ields can be quite compllex. The figure b below shows a magnetic bottle. This interesting c c configuration happens when the n w field is weak at the cente and strong at both end er g ds. See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 45 This configur ration is used to trap dens beams of plasmas. se A similar phe enomenon is the oscillation of ions back and fort between t s th the earth’s magnetic m poles in the V Van Allen Be elts 5. THE VE ELOCITY SELECTOR AND A APPLICATIONS S agnetic force on a charg e ged particle moving in a uniform magnetic field c can be bala anced by “The ma an elect force if th magnitude and direct tric he es tions of the tw fields are chosen prop wo perly.” Since the electric force is in the dire ection of the electric field (for positiv particles) and the e ve orce is perpen ndicular to th magnetic field. he magnetic fo The electric and magne etic fields in the region th hrough whic the particle is moving must be ch perpendicular to each o other if the for rces are to balance! b Such regions is said to ha crossed f s ave fields. Th Figure shows a reg he gion of space between the plates of a capacitor b e f where there is an electr field and a w ric d perpendicula magnetic field into t p ar the plane of the p p paper. Consider a pa C article of cha arge q enteriing th space fro the left, the net for his om rce (t termed Loren force) on the particle is ntz n e given by: g f ectric force of If q is positive, the ele magnitude qE is dow m wn and t the magnetic forc of magnit m ce tude qvB is up p. he es f Th two force balances if qE = qvB or For given ma agnitudes of the electric and magnet fields, the forces balance only for particles tic e with the spee v = E/B. ed The arrangement of the f fields gives us the velocity selector v = E/B. y Any particle with this sp e peed, regard dless of its mass or charg will trave m ge erse the cros ssed-field space undef flected. › A p particle with greater spe eed than the velocity selector will be deflecte in the e ed direc ction of the m magnetic fiel ld. › A pa article with le esser speed t than the velo ocity selector will be defle r ected in the direction of th electric fie he eld. APPLICA ATIONS r ds pplications that were dis scovered The velocity selector for crossed-field has very important ap during the la 19th and e ate early 20th cen ntury. In this course three appliications will b discussed: e, be See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 46 1. Thoms son’s Measur rement of q/m for Electron m ns JJ Thomson, in 1897, illustrated the technique fo measuring the , or g q/m of elect trons. In his experim ment, he sho owed that the rays of a cathode-ray tube c can be def flected by E and B field thus they must consiist of ds, y charged particles. By measuring the deflect g tions of these particles, Th e homson, show wed that all the particle have the same q/m. Thomson als showed th particles with this q/m can be o so hat m obtained usin any mate ng erial for a source, whic only mea ch ans that the ese particles (now called electrons) are a fund d damental constituent o all matters. of vo is the velo ocity selector (E/B) r q/m can be determined from the equation below w. eld roduced at th entrance. he The magnetic fie is only intr EXAMPLE E Electrons pass undef flected throu ugh the plate of Thomso es on’s apparat when the electric fiel is 3000 tus e ld V/m and there is a crossed magn d netic field of 1.40G. If the pla ates are 4cm long and the ends of the plates are 30cm from th screen. e 3 he Find the deflection o the screen when the m on n magnetic field is turned off. d 2. The Mass Spectrom meter The mass sp pectrometer, first design ned by Fran ncis William Aston in 1919, was d n developed as a means a of meas suring the ma asses of isoto opes. Such mea asurements are an ant way of determining both the g importa presenc of isotope and their abundance ce es a in natur re. For examplle, natural magnesium m en 7 has bee found to consist of 78.7% 24Mg, 5 10.1% 25Mg, and 11.2% 26Mg. pes asses in the These isotop have ma approxiimate ratio o 24:25:26. of Ions ejected from the so d ource move in a semi-circ cular orbit and strike a photographic film a P2. at MASS SP PECTROMETER EQUATION S R SET See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 47 EXAMPLE E 26 A 58Ni io of charge +e and mass 9.62 x 10-2 kg is accelerated throu on ugh a potential differenc of 3kV ce and deflected in a m magnetic field of 0.12T. d (a) Find the radius of curvature of the orbit of the ion. f f (b) Find the differenc in the rad of curvatu of 58Ni ion and 60Ni io ce dii ure ns ons. (Assume that the ma ratio is ass 58/60.) 3. The Cy yclotron The cyclo otron was invented by E.O. Lawren nce and M.S. Livings ston in 19 934 to accele erate particle such as protons or deuterons es to high kinetic ener h rgies*. OTRON EQUAT TION SET CYCLO EXAMP PLE A cyclotron for ac ccelerating protons has a magnetic field of 1.5 and a p c 5T maxim mum radius of 0.5m f (a) Wh is the cyclotron freque hat ency? (b) Wh is the kine energy of the protons when they e hat etic s emerge? PART 2: SOURCES OF THE MA : AGNETIC FIELD STORY BRIEF HIS Permanent M Magnets wer the earliest known sour re rces of magn netism. Oersted ann nounced his d discovery tha a compass needle is de at s eflected by a electric current. an Jean Baptiste Biot and Fe Savart an elix nnounced the results of th measurem e heir ments of the force on ear urrent-carryin wire and analyzed resu in terms o the magne field. ng a ults of etic a magnet ne a long cu Andre-Marie Ampere ex e xtended the ese experime ents and sho owed that c current elements also experience a force in the presence of a magnetic field and that two currents exert forces on f each other. 1. THE MAGNETIC FIEL OF MOVIN POINT CH LD NG HARGES When a poin charge q moves with a velocity v, it nt produces a magnetic fie B in space given by: eld e (For Left) Wh here r is calle the positio vector tha points from the charge to the field point P. ed on at m e d μ0 is a constant of propo ortionality ca alled the perm meability of free space, w which has the value e (For Right) θ is the angle between r and v. See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 48 EXAMPLE E A point charge of m magnitude q = 4.5 nC is m moving with speed v = 3 x 107 m/s parallel to the x-axis 3.6 s t along th line y = 3m he m. Find the magnetic fie elds produce by this cha ed arge when th charge is a the point (x = -4m, y = 3m) he at (1) at the origin e (2) at the point (0,3m e m) (3) at the point (0, 6m e m) Ans: (1) 3.89 x 10-10 T in the paper (2) 0 (Why? (3) 3.89 x 10-10 T out the the paper r. ?). 1 e LD AVART LAW 2. THE MAGNETIC FIEL OF CURRENTS: BIOT-SA The calc culation of th magnetic field caused by a point charge he c can be extended to calculate t o the magnetic field cause by an c ed electric current in a w wire. The equation below is known as t the Biot-Sava Law. art ELECTRIC C-MAGNETIC ANALOGY C The two equ uations used t calculate the magneti field are analogous to Coulomb’s Law. to ic L However, the is a distinct difference in the direc ere e ctional aspec cts. E is in the dire ection of the force. e B is in the direction perpe endicular to t the force. B DUE TO DIFFERE CON O ENT NFIGURA ATIONS— —See APP PENDIX A S’S 3. GAUSS LAW FOR MAGNETISM We know tha magnetic field lines dif at ffer from elec ctric field lines. Magnetic fie lines form closed loops eld s. The magnetic equivalen of the elec nt ctric charge is called a m magnetic pole. w sm as: Gauss’s Law for Magnetis is stated a That is, no magnetic mon nopoles! RE’S LAW 4. AMPER Ampere’s La is very ana aw alogous to G Gauss’s Law fo Electricity. or It relates the magnetic field to the c e current enclosed by an im maginary loop (called Amperian Lo oop). Ampere’s La works for configuratio that hav a high de aw r ons ve egree of symmetry. It is stated as s: See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 49 EXAMPLES: 1. Find the magnetic field caused by a wire that carries a current I at a point P which is at a perpendicular distance R from the wire. 2. A long straight wire of radius R carries a current I that is distributed uniformly over he cross-sectional area of the wire. Find the magnetic field both inside and outside the wire. LIMITATIONS OF AMPERE’S LAW Ampere’s Law will only work if and only if the following statements hold: 1. The configuration has a very high level of symmetry 2. The current is continuous everywhere in space. Therefore, there are only three cases where Ampere’s Law can be used: 1. Long straight lines 2. Long, tightly wound solenoids 3. Toroids 5. MAGNETISM IN MATTER Unlike the E field and the dipole moment p, magnetic moments inside all materials tend to increase the magnetic field during alignment. Materials fall into three categories: (1) Paramagnetic (2) Diamagnetic (3) Ferromagnetic PARAMAGNETISM Paramagnetism arises from partial alignment of the electron spins (in metals) or of atomic or molecular magnetic moments by an applied magnetic field in the direction of the field. In paramagnetic materials, the magnetic dipoles do not interact strongly with each other and are randomly oriented. In the presence of an external magnetic field, the dipoles are partially aligned in the direction of the field, thereby increasing the field. However, in external magnetic fields of ordinary strength at ordinary temperatures, only a small fraction of the molecules are aligned. The total increase in the field is therefore small. DIAMAGNETISM Diamagnetism arises from the orbital magnetic dipole moments induced by an applied magnetic field. These magnetic moments are opposite the direction of the applied magnetic field so they decrease the total magnetic field B This effect actually happens to all material, but because of the induced magnetic moments are very small compared to the permanent magnetic moments, diamagnetism is masked by paramagnetic or ferromagnetic moments. Diamagnetism is thus only observed in materials that have no permanent magnetic moments. See more @ teodorickbarry.multiply.com PHYSICS DIVISION P H Y S I C S 1 3 L E C T U R E N O T E S | 50 FERROMAGNETISM Ferromagnetism is much more complicated than paramagnetism because of a strong interaction between neighboring magnetic dipoles. A high degree of alignment occurs even in weak external magnetic fields, thus causing a great increase in the total field. Even when there is no external field, ferromagnetic materials may have its dipoles aligned and have its own magnetic field just like a permanent magnet. See more @ teodorickbarry.multiply.com PHYSICS DIVISION P H Y S I C S 1 3 L E C T U R E N O T E S | 51 UNIT FIVE: M MAGNE ETIC IND DUCTION OUTLINE 1. Magnetic Flux c 2. Induced EMF and Far raday’s Law 3. Lenz’s La aw 4. Motional EMF l 5. Eddy Currents nce: Self-Indu uctance and 6. Inductan Mutual Indu uctance 7. RL Circuits OBJ JECTIVES he you must be At th end of thiis chapter, y able to: 1. Define and c D compute for magnetic flux; 2. Define and u utilize Farada ay’s Law to the induced emf for d calculate for t eral ations; seve configura 3. Define and utilize Lenz D z’s Law to calculate for t the directio on of the ced current iin Faraday’s Law; induc 4. Ch haracterize a and enumera ways to ate reduce Eddy Cur rrents; 5. Co ompute Inductances; 6. Le earn means and ways in storing s s mag gnetic energy and y; 7. Compute for circuital par rameters of RL Circuits. INTRODU UCTION day (Englan nd) and Joseph Henry (USA) 1830’s – Michael Farad independen discovere that chan ntly ed nging magne etic field indu uces a current in the wire. e The emfs an currents caused by changing magnetic fields are nd m called induc ced emfs and induced cu d urrents. The process itself, is referr red to as mag gnetic induc ction. pull the plug of an elec g ctric cord fro its socke you om et, When you p sometimes o observed a sm spark. Th phenomen mall his non is explain ned by magn netic inductio on! 1. MAGN NETIC FLUX The flux of a magnetic fiield through a surface is defined similarly to the flux of an electric field. n The magnetiic flux Φm is d defined as The unit of f flux is that of a magnetic field times area, tesla-m f c meter square which is called a ed, weber (Wb) 1 Wb = 1 T•m2 m e: volt. Exercise Show that a weber per second is a v We are oft ten intereste in the flux through a coil ed containing several turns o wire. of ontains N tur rns, the flux through the coil is N c If the coil co times the flux through each turn. x See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 52 EXERCISE Find the magnetic fllux through a solenoid th is 40cm lo hat ong, has a r radius of 2.5c cm, as 600 tu urns, and carries a current of 7.5A 2. INDUC CED EMF AND FARADAY’S LAW D Experiments by Faraday, Henry and o others showed that: If the magnetic f e flux through a area bounded by a c an circuit changed by any means, an m emf equal in ma agnitude to th rate of ch he hange of the flux is induce in the circuit! ed We usually d detect the em by: mf Observing a cur rrent in the c circuit, but it’s present eve when the circuit is inc s en e complete current. and there is no c BEFORE: were localize in a spec ed cific part of the circuit, such as f We considered emfs that w betw ween the terminals of the battery e HOWEVER: Indu uced emfs ca be consid an dered to be distributed thr d rough out the circuit e How to chan nge magnetic flux??? c The current p producing th magnetic field may be increased o decreased he e or d Permanent M Magnets may be moved toward the circuit or awa from it y c ay Circuit itself m be mov may ved toward o away from the source o the flux or of The orientation of the circ may be changed cuit the circuit in a fixed magnetic field may be increa ased or decre eased. The area of t 1. 2. 3. 4. 5. In every case, an emf is induce in the circ y ed cuit that is equal in mag gnitude to th rate of ch he hange of magnetiic flux. Figure at the right shows a single loop of wire in a magnetic fie e p eld. If the flux thro ough the loo is changing, an emf is induced in th loop. op i he Since emf is W/q, there m must be force exerted on the charge associated w e n with the emf. The F/q is the E, which in this case is in e nduced by th changing flux. he E fields that resulted from static elect m tric charges are conserva ative! (Mean ning, work done a across a close curve is ze ed ero) E fields tha resulted f at from changi ing magnetic flux is n nonconservat tive! (Meaning, w work done ac cross a closed curve is NO ZERO!) d OT These finding are summa gs arized as Fara aday’s Law, which is give below: w en The negativ sign in Fa ve araday’s Law has to do with the d w o direction of the induced emf, which we w discuss sh will hortly! EXAMPLES: 1. A uniform magnetic field makes an a d angle of 30o with the axis of a circular coil of 300 turns and w r t a radius of 4 cm. The field changes a a rate of 85T/s. Find the magnitude of the indu at 8 e e uced emf in the coil. 2. An 80-turn coil has a radius of 5.0cm and a resistance of 30Ω. At what rate must a t e perpendicular magnetic field to prod duce a current of 4.0A in t the coil? 3. A solenoid o length 25 c and radiius 0.8cm wit 400 turns is in an exter of cm th rnal magnetic field of 600 G that m makes an an ngle of 50o with the axis of the solenoid. (a) Find the magni s d itude flux through the solenoid. (b) Find the ma ) agnitude of the emf induced in the so t olenoid if the external e magnetic fie is reduced to zero in 1.4s. eld d See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 53 3. LENZ’S LAW S The negative sign in Fara e aday’s law ha to do with the directio of the indu as h on uced emf, which can w be found fro a generall physical prin om nciple known as Lenz’s La n aw “The induced em and induc e mf ced current are in such a direction so as to opp pose the cha ange the prod duces them.” ” Note: We diidn’t specify just what kind of chang causes th induced e ge he emf and cur rrent. The statement w left vague to cover a variety of co was e onditions we w now illustrate. will There is a alternative statement to Lenz’s Law to make it more operat an w tional! “For a change in magne etic flux, a c counter flux is produced so that there will be “no overall” e o change in th flux!” he This “counter flux” will the give the d en direction of th induced c he current or em mf! ENZ’S LAW IN EACH N APPLY LE See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 54 EXAMPLE A rectangular coil of 80 turns, 20 cm wide and 3 cm long, is located 0 m 30 s in a magn netic field B = 0.8T direc cted into the page, wit only a th portion of the coil in th region of t he the magnetic field. The re c esistance of the coil is 30Ω. Find the m magnitude an direction o the induce current if the coil is nd of ed t moved wit a speed of 2m/s (a) to the right, (b) up, and (c) down. th ) 4. EDDY CURRENTS uced by cha anging flux we set up in definite circu ere uits. Previously, currents produ Often a cha anging flux se up circula ets ating current called Edd currents in conductor as they ts, dy n rs move throug a region o changing m gh of magnetic flux. The heat pro oduced by su uch current c constitutes a power loss in the conduc n ctor and the system. s e duced? How are Eddies prod ts y because pow is lost in th form of he generate by the wer he eat ed Eddy current are usually unwanted b current, and that heat its must be d self dissipated. he e f Power loss is reduced by increasing th resistance on the possible paths of the eddies. Eddy current are not alw ts ways undesira able. Eddies are o often used to lessen unwa anted oscillations in severa applications al Eddies are a also used in th magnetic breaking sys he stem of magnetic transit t trains. 5. INDUC CTANCE 5.1 SELF-INDUCTANCE E The magnet flux throug a circuit iis related to the current in that circuit and the currents in tic gh other, nearb circuits* by The current produces a magnetic field B that varies from point to po oint, but B is always i proportional to I at every point. y See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 55 The magnetiic flux throug the coil is t gh therefore also proportiona to I, hence the definitio of selfo al e on inductance! o Where L is a constant called the self-inductance of the coil. The self-inductance depe ends on the geometric sh hape of the c coil. SI UNIT O INDUCTAN OF NCE From the eq quation that d defines self-in nductance, we see that the unit of in nductance is the unit s of flux divide by the unit of current. ed 1 H = 1 Wb/A = 1 Tm2/A, (H) is called t A the henry. After Joseph Henry, wh also disc ho covered and studied th phenome d he enon of ind ductance th thoroughly d during the 19t century. CALCULA ATING SELF-IN NDUCTANCE EXAMPLE ES: 1. Find th self-inductance of a so he olenoid of len ngth 10 cm, area 5 cm2, and 100 turn a ns. 2. At wh rate must the current in the solenoid in the example abo hat t t e ove change to induce an emf of a 20V? The Equa ation that rela ates Faraday Law with Inductance y’s See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 56 5.2 MUTU INDUCTA UAL ANCE When two o more circu are close to each oth or uits e her, as in the figure abov the magnetic flux e ve, through one circuit does not depend only to its ow but the o e d wn, other’s contribution as we ell. The flux, for example through circuit 2, is due to it’s self-inductance and current I2, and the t o mutual induc ctance M(2,1) and current I. t The mutual in nductance M21 = M12, we drop the subscript and c it M. e call Mutual Induc ctances, dep pend on the geometric arrangement of circuits! a SEE DERIVATION ON T BOARD O HOW TO C THE ON CALCULATE FO MUTUAL INDUCTANCE OR ES: NOTES HERE: NETIC ENERGY AND THE IN Y NDUCTOR 6. MAGN An inductor stores magnetic energy through the current build ding up tores electric energy. cal in it, just as a capacitor st Consider the circuit at the right. e The energy s stored in an inductor carr rying a curren I is given b nt by: The magnetiic energy de ensity is given by: See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 57 7. RL CIR RCUITS circuit containing a resisto and an inductor such as or RL Circuit- c the one in th right. he For all RL Circuits in Physic 13, we ca apply Kirchhoff’s Rules to cs an solve for the circuital par rameters. Current I, flo ows in a sing direction but it’s valu is changing gle ue with time. s Circuits, howe ever, we do on’t RL Circuits is very similar with RC C charge/discharge RL circ cuits. t e f t We just want to know the behavior of currents in them. THE GRO OWTH OF I IN RL CIRCUITS witch is clos sed, current does not When the sw build up insta antaneously. . It grows expo onentially via the equatio above, a on until it reaches the final c current If = ξ0/ /R. τ is called tim constant, τ = L/R, whic is the time it takes the c me ch circuit to reac maximum current. ch m When curren reaches its maximum, t nt s the inductor acts as a “sh a hort” or just a wire. EXAMPLE E: A coil o self-inducta of ance 5.0mH and a resist tance of 150 Ω is place across the terminals of a 12-V 0 ed e o battery o negligible internal resistance. of (A) What is the final c current? (B) What is the curren after 100μs t nt s? (C) How much energ is stored in this inductor when the final current h been atta w gy n has ained? THE DECA OF I IN RL CIRCUITS AY L The circuit above is very similar to the circuit for the “growth e of I”. e ditional switc ches in order to remove However, we place add the battery a and, R1 to pr rotect the ba attery from su urge and short. Here, we let the circuit at ttain If, then we proceed with the dec caying proce of I. ess Initially the c current I0 = ξ0/ then it stea /R adily decays until it is neg s gligible. See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 58 UNIT SIX ALTER U X: RNATING CURRENT CIR G RCUITS OUTLINE 1. AC GENE ERATORS ATING CURREN NT: 2. ALTERNA RESISTORS & RMS VALUE ES, INDUCTOR AND CAPA RS ACITORS S 3. PHASORS CIRCUITS 4. LC, RLC C LC and RLC Without a G C Generator Series RLC W a Gener With rator Parallel RLC With a Generator C ORMERS 5. TRANSFO OBJ JECTIVES At th end of thiis chapter, y he you must be able to: 1. Understand h how an AC Generator k pute for the maximum e work and comp EMF; C d 2. Comprehend the behavior of an Alter rnating cur rrent in a Resistor, Capacitor, and an Indu d uctor and cuital parame eters; calculate for circ Define Ph hasors and d identify 3. ween circuital potential relationships betw C; differences in AC ompute for A AC-circuital parameters 4. Co involved in an LC and RLC Cir C rcuits driven (or not) by an AC generator; C ompute for th Transforme he er’s 5.Co char racteristics. INTRODU UCTION ectrical energ used toda is produce gy ay ed More than 99% of the ele by electricall generators iin the form o alternating current (ac). of AC’s advan ntage over D because electrical energy can b DC e be transported over long distances at v very high voltage and lo ow currents to re educe energ losses due to Joule hea gy at! AC can the be transfo en ormed, with almost no energy loss, to e lower and safer voltage and correspondingly higher curren es h nts for everyday use! y ANGE IN WAV FUNCTIONS VE S THE CHA Here are som of the bas formulas f obtaining changes in wave functio me sic for g ons: Constants:ω is angular fre equency, δ is phase difference s These formullas are very im mportant in o obtaining the value of some AC circu e uital paramet ters. ENERATOR an the GENER nd RATION OF ALTERNATING CURRENT 1. AC GE Figure below shows a sim w mple AC gene erator. It consists of a coil of ar f rea A and N turns rotatin (with freq ng quency ω) in a uniform magnetic n m field. the coil are c connected to rings (called slip rings) th rotate wiith the coil. o hat The ends of t They make e electrical con ntact through stationary conducting b h c brushes in co ontact with th rings. he See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 59 The emf in th coil will the be: he en Or Where s sinusoidal em in a coil by rotating it w constant angular velocity in a mf y with t We can thus produce a s magnetic fie eld. As we all kno with an in ow, nduced emf, there is also an induced current. , o With an Alternating EMF, there is also an alternatin current! ng E: EXAMPLE A 250-turn coil has an area of 3cm2. If it rotate in a magne field of 0 n m es etic 0.4T at 60Hz, w what is ξmax? See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 60 2. ALTERN NATING CURRENT IN CIRC CUITAL ELEMENTS 2.1 RESIS STORS IN AC THE POW DISSIPATE IN A RESIST WER ED TOR RMS VAL LUES voltmeters ar designed to measure r re root-mean-square (rms) values of v Most ac ammeters and v current and voltage rather than the m maximum or peak values s! RULE: The RM value of a MS any quantity that varies sinusoidally e equals the ma aximum valu of that ue quantity divided by √2. rrent equals t the steady d current that would pro dc oduce the sa ame Joule he eating as *The rms cur the actual a current. ac Example: The rms value o a current, Irms is given by: e of b See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 61 RMS EXE ERCISE: 1. Find Pav in terms of Irms and R f 2. Find Pav in terms of ξmax and Ima f ax 3. Find Pav in terms of ξrms and Irms f 4. Find Irm in terms of ξrms and R ms 5. A 12-Ω resistor is c Ω connected a across a sinus soidal emf th has a pe hat eak value of 48V. Find (a) the rms current, (b) the avera age power, ( the maxim (c) mum power. hat of tor esistor, the vo oltage drop across a Note: In a circuit, th consists o more than a generat and a re resistor is not usually equal to the generator v s e voltage, so we write volta w age drop across a resistor in terms of VR,rms! RNATING CURRENT IN IND DUCTORS AND CAPACITOR D RS 2.2 ALTER Alternating c current behav differentl than direct current in in ves ly t nductors and capacitors. When a cap pacitor becomes fully cha arged in a dc circuit, it st c tops the curre ent, that is, it acts like t an open circ cuit. But if the cu urrent alterna ates, charge continually flows onto o off the pla or ates of the capacitor c and at highe frequencie the capac er es, citor, will hard impede c dly current at all, which means, it acts , like a short c circuit! Conversely, an inductor coil usually h a very sm resistanc and is esse has mall ce entially a sho circuit ort for dc. But when th current is alternating, a back emf is generate in an ind he ed ductor, and at higher a frequencies, the back em is so large, the inducto acts like an open circuit! mf , or n ORS IRCUITS INDUCTO IN AC CI VL Leads I by 90o s In the previo set of slide we see th functional difference o VL and I. ous es, he of This functional difference is due to the current I’s phase differe e e p ence with voltage VL. hase” with VL, more preci isely VL leads current by 9 o. 90 We say that I is “out of ph This is illustrat ted by the plot at the righ ht. As with earlie techniques we can transform the equation to p er prove this “lea ading” pheno omenon EXAMPLE E A 40mH inductor is placed across an ac gene s erator that ha a maximum emf of 120 Find the inductive as 0V. i reactance and the m maximum current when th frequency is he y a) 60 Hz See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 62 b) 2000 H Hz What ca you conclude about th relation of inductive re an he f eactance an current? nd CAPACIT TORS IN AC C CIRCUITS VC Lags I by 90o ous es, he of In the previo set of slide we see th functional difference o VC and I. This functional difference is due to the current I’s phase differe e e p ence with voltage VC. We say that I is “out of ph hase” with VC, more prec cisely VC lags current by 90o. 0 This is illustrat ted by the plot at the righ ht. As with earlie techniques we can transform the equation to p er prove this “lag gging” pheno omenon EXAMPLE E: A 20-μF capacitor is placed acro a genera oss ator that has a maximum emf of 100V Find the ca V. apacitive reactance and the m maximum current when th frequency is he y A) 60 Hz B) 5000 H Hz What ca you conclude about th relation of capacitive reactance a an he f and current? 3. PHASO ORS relations bet tween the c current and the voltage drop in a resistor, cap pacitor or The phase r inductor can be represen n nted by two dimensional vectors calle phasors. ed Phasors rot tates counterclockwise (since in e ncreasing a angular degrees are moving counterclockwise in the C CCS). al nected toget ther in series circuit, their voltages ad when r dd, When severa components are conn they are con nnected in parallel, their c current add. Meaning, c complications in the co s omputation of circuital para ameters can be simplified using vector n d addition, usin phasors! ng Consider a circuit containing an inductor L, a C, capacitor C and a resistor R, all connected in series. are arry the sam me Since they a in series, they all ca current, whic is represe ch ented as the x-component of the curren phasor I. nt The voltages are ob btained usin ng the prior definitions w which include the resistive, capacitive, es and inductiv reactance ve es. See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 63 4. RLC C CIRCUITS 4.1 LC CIRCUITS WITH HOUT A GENER RATOR Figure to the right shows an LC Circuit e t. In an LC Circ cuit, we assum that the c me capacitor ca arries an initia charge Q0. al When the sw witch is closed charge be d, egins to flow through the iinductor. t The effect is that, the ca apacitor is be eing discharg ged by the lo of charge then it incr oss e, rease the current in the inductor, w which in turn, re-charges the capacit and deca the curre in the tor ays ent inductor. This circuit is very similar to a mass atta o ached to a spring. s out ator: There are important parameters iinvolving LC Circuits witho a genera Angular “Na atural” Freque ency: Current: EXAMPLE: A 2-μF c capacitor is charged to 20V and is then connec cted across a 6-μH indu uctor. (a) Wh is the hat frequenc of oscillatiion? (b) Wha is the maxim cy at mum value of the current o t? 4.2 RLC C CIRCUITS WIT THOUT A GENERATOR If we includ a resistor in series w de with a inductor, we have an RLC Circuit. e C It is basically the same as an LC c circuit, scharging d does not ha appen charging/dis It is a spr ring system that enco ounters forces! capac citor and howev ver, the “forev ver” frictional See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 64 ES R 4.3 SERIE RLC WITH A GENERATOR RESONANCE IN SERIE RLC ES Resonance iin circuit is wh hen the impe edance is at its smallest, a and the curre is at its greatest. ent There condit tions for reson nance are gi iven in the rig ght. At resonance, the power factor is 1. r EXAMPLE ES: 1. A serie RLC Circuiit with L = 2H, C = 2μF, and R = 20Ω is driven by a g es , d generator wit a maximum emf of th 100 V an a variable frequency Find (a) the resonance frequency (f0), (b) the maximum current at nd y. e c resonanc (c) the phase angle δ (d) the pow factor, and (e) the av ce, δ, wer a verage powe delivered. er 2. A serie RLC Circuiit with L = 2H, C = 2μF, and R = 20Ω is driven by a g es , d generator wit a maximum emf of th 100 V an a variable frequency Find the m nd y. maximum volt tage across the resistor, t the inductor and the r capacito or. 3. A resiistor R and c capacitor C are in series with a gen s nerator with peak vo oltage of 220 60Hz, as shown in the Figure. If R=20 Ω and 0V, e R C=14.7μF, Find Vout,rm . ms See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 65 4.4 PARA ALLEL RLC WIT A GENERA TH ATOR RESONANCE IN PARA ALLEL RLC Conditions a basically the same, ho are owever, we note some im n mportant fea atures of reso onance in parallel RLC: : m, 1) Impedance is a maximum current is a minimum ctor and cap pacitor are equal, but 2) The currents in the induc they are opp posite, so the cancel, th total current is just ey he the current in the resistor. n . SFORMERS 5. TRANS A transformer is a device u used to raise or lower the voltage in v out a circuit witho an appreciable loss of power A simple transformer con nsisting of two wire coils around a o a common iron core. n The coil carry ying the input power is calle the primar ed ry. The coil carry ying the outpu power is ca ut alled the seco ondary. The transform operates o the princip that an alternating cur mer on ple rrent in one circuit induce an alterna es ating emf in a nearby cir rcuit due to t the mutual inductances of the two cir rcuits. The iron core increases the magnetic fie fir a given current and guides it so e eld n all etic flux throu ugh one coil goes through the other h that nearly a the magne coil. THE TRAN NSFORMER EQ QUATIONS For a transf former with N1 turns in the primary and N2 tu y urns in the secondary, the voltage across the secondary coil is related to the e generator em across the primary coiil by: mf e See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N P H Y S I C S 1 3 L E C T U R E N O T E S | 66 If there are n losses, due to Joule He no e eating (which is due to ne h egligible resist tance in the coils), EXAMPLE E: A doorb requires 0 bell 0.4A at 6V. It is connecte to a transformer whose primary co t ed ontaining 200 00turns, is connect ted to a 120-V ac line. (a How many turns should there be in the second a) dary? (b) Wh is the hat current in the primary y? _____ ___________ ____end___ ___________ ______ See mo @ teod ore dorickbarry y.multiply.c com PHYSICS DIVISION S N
Report "Digested Handout in Electricity and Magnetism"