Field Theory10ES36 FIELD THEORY Sub Code : 10ES36 IA Marks : 25 Hrs/ Week : 04 Exam Hours : 03 Total Hrs. : 52 Exam Marks : 100 PART: A UNIT 1: ELECTROSTATICS Hrs: 8 Experimental law of coulomb, Electric field intensity, Field due to continuous volume charge distribution, Field of a line charge, Electric flux density, Electric flux density, Gauss law, Divergence, Vector Operator, Divergence theorem UNIT 2: Energy and potential & Conductors, dielectrics and capacitance Energy expended in moving a point charge in an electric field, Line integral, Definition of potential difference and potential, Potential field of a point charge & system of charges. Energy density in an electrostatic field. Potential gradient, Current and current density, Continuity of current, metallic conductors, Conductor properties and boundary conditions for dielectrics, Conductor properties and boundary conditions for capacitance Hrs: 8 UNIT 3: Poisson’s and Laplace’s equations Derivation of Poisson’s equation, Uniqueness theorem, Examples of Laplace Equations, Examples of Poisson’s Equations Hrs: 6 UNIT – 4: MAGNETOSTATIC FIELDS Biot- savart law, Amphere’s circuitary law, curl’s theorem, Stoke’s theorem, Magnetic flux and flux density, Vector magnetic potentials. Hrs: 6 PART - B UNIT -5: MAGNETIC FORCES, MATERIALS AND DEVICES Force on a moving charge and differential current element, Force between differential current elements, Force and Torque on a closed circuit, Magnetic materials and inductance, Magnetization and permeability, Magnetic boundary condition ns, Magnetic circuits, Potential energy and forces on magnetic materials. Hrs: 8 Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Field Theory 10ES36 UNIT -6: TIME VARYING MAGNETIC FIELDS AND MAXWELL’S EQUATIONS Faraday’s law, Displacement current, Maxwell’s equation in point and integral form, Retarded potentials. Hrs: 6 UNIT -7: UNIFORM PLANE WAVES Wave propagation in free space, Wave propagation in dielectrics, Poynting’s theorem and wave power, Propagation in good conductors Hrs: 6 UNIT -8: REFLECTION AND REFARACTION OF PLANE WAVES Reflection of uniform plane waves at normal incidence, SWR, Plane wave propagation in general directions Hrs: 6 Recommended readings: 1. Energy Electromagnetics, William H Hayt Jr. and John A Buck, Tata McGraw-Hill, 7th edition, 2006. 2. Electromagnetics with Applications, John Krauss and Daniel A Fleisch McGraw-Hill, 5th edition, 1999 3. Field and Wave Electromagnetics, David K Cheng, Pearson Education Asia, 2nd edition, - 1989, Indian Reprint – 2001. Dept. Of ECE/SJBIT Field Theory 10ES36 INDEX SHEET SL.NO TOPIC PAGE NO. PART A UNIT 1: ELECTROSTATICS 01 Experimental law of coulomb 1 02 Force on a point charge 1 03 Force due to several charges 2 204 Electric field intensity 2 05 Electric flux 10 06 Gauss law 14 07 Divergence 22 08 Vector Operator 20 09 Divergence theorem 25 UNIT 2: ENERGY & POTENTIAL, CONDUCTORS, DIELECTRICS & CAPACITANCE 01 Energy expended in moving a point charge in an electric field 28 02 Line integral. 30 03 Definition of potential difference and potential 33 04 Potential field of a point charge & system of charges. 35 05 Energy density in an electrostatic field 36 06 Potential gradient 40 07 Current and current density 45 Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Field Theory 10ES36 08 Continuity of current 47 09 Conductor properties and boundary conditions for dielectrics 50 10 Conductor properties and boundary conditions for capacitance 53 UNIT 3: Poisson’s and Laplace’s Equation 01 Derivation of Poisson’s equation 59 02 Uniqueness theorem 60 03 Examples of Laplace Equations 62 04 Example of Poisson’s equatiom 64 UNIT 4:STEADY MAGNETIC FIELD 01 Biot- savart law. 70 02 Amphere’s circuitary law 78 03 curl’s theorem 79 04 Stoke’s theorem 80 05 Magnetic flux and flux density 81 06 Vector magnetic potentials 84 PART B UNIT 5:MAGNETIC FORCES 01 Force on a moving charge and differential current element. 89 02 Force between differential current elements. 91 03 Force and Torque on a closed circuit. 93 04 Magnetic materials and inductance 115 05 Magnetization and permeability. 118 06 Magnetic boundary conditions 120 Dept. Of ECE/SJBIT 198 02 Reflection of uniform plane waves at normal inicidence 200 Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP . 122 UNIT 6: TIME VARYING FIELDS & MAXWELL’S EQUTION 01 Faraday’s law. 133 UNIT 7: UNIFORM PLANE WAVES 01 Wave propagation in free space 156 02 Wave propagation in dielectrics 157 03 Poynting’s theorem and wave power 171 04 Propagation in good conductors 175 UNIT 8:PLANE WAVE AT BOUNDARIES & DISPERSIVE MEDIUM 01 Reflection of uniform plane waves at normal inicidence. 124 02 Displacement current. 129 03 Maxwell’s equation in point and integral form.Field Theory 07 10ES36 Potential energy and forces on magnetic materials. Force on Q1 is given by Dept.. k= . m/F in SI units. In SI unit. and inversely proportional to the square of the distance between them. R^ is the unit vector along the line joining the two charges. and is the force acting on points from charge 2 toward charge 1. . F kq1q2 Or F=(kq1q2)/(r²) N Where k is the constant of proportionality whose value varies with the system of units. Where is called the permittivity of the free space. The unit vector . as per the Coulomb’s Law.834 F/m. Then. It has an assigned value given as =8. and R is the distance between the two charges.Field Theory 10ES36 UNIT 1 ELECTROSTATICS Coulomb’s Law Coulomb’s law states that the electrostatic force F between two point charges q1 and q2 is directly proportional to the product of the magnitude of the charges. Accordingly. and it acts along the line joining the two charges. is the force exerted on . Of ECE/SJBIT Page 1 . FORCE ON A POINT CHARGE The forces of attraction/repulsion between two point charges and (charges whose size is much smaller than the distance between them) are given by Coulomb’s law: where Here. F= q{ }N Electric field intensity Electric field intensity at any point in an electric field is the force experienced by positive unit charge placed at that point. it is required To find the electric field intensity E... At the point B in the electric fields set up by Q. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 2 ...Fn at the charges q1..q3.. The force F1.. Dept......r3.....q2.qn at distances r1..... Hence..Field Theory 10ES36 F1 = Newtons F1 q1 q2 q1 q2 F2 @ 2 Force due to several charges Let there be many point charges q1.. F3......q3.......q2...r2..qn respectively are: q{ rˆ + } F=Fq1+Fq2+Fq3.. Consider a charge Q located at a point A..rn from charge q.. F2....... As per the Coulomb’s Law.Field Theory 10ES36 Let the charge at B be and let the charge Q be fixed at A. the magnitude of the electric field strength is: E=Q/(4r Let r be the unit vector along the line joining A and B.qn at distances r1...q3....q2. Thus... i. Let r be the distance between A and B.......rn be the corresponding unit vectors. the vector relation between E is written as: E=Q/(4 or²) V/m Electric Field intensity due to several charges Let there be many point charges q1.... then by definition.q2...En at the charges q1....r2.. the force between Q and q is given by: rˆ N F= If it is a unit positive charge.. The field E1.e.... Dept... E2. Of ECE/SJBIT E= Page 3 .r3....qn respectively are: rˆ + E=Eq1+Eq2+Eq3... E=F when Therefore... E3.. F in the above equation gives the magnitude of the electric field intensity E..q3....... Hence... Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 4 .Field Theory 10ES36 Electric field intensity at a point due to a infinite sheet of charge Dept. Field Theory Dept. Of ECE/SJBIT 10ES36 Page 5 . Field Theory Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP 10ES36 Page 6 . Field Theory Dept. Of ECE/SJBIT 10ES36 Page 7 . Field Theory Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP 10ES36 Page 8 . So.Field Theory 10ES36 Electric field at a point on the axis of charged circular ring Let ρ be the charge density of the ring. ρ=dq/dl dq=ρdl Electric field due to an infinitely small element=dE dE=dq/4πεo r² rˆ where rˆ is the unit vector along AP. Of ECE/SJBIT Page 9 . Dept. cancels with that due to element A. the equation becomes. Dept. Therefore. If the area is not planar. r=(R²+x²)½ Therefore. The total field at P is the sum of the fields due to all the elements of the ring. If we consider an element of the ring at a point diametrically opposite to A. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 10 . dEx=dEcosθ. ∫dEy=0. dEx= The component dEy is directed downwards. Taking the magnitude of dE from above.Field Theory 10ES36 dE can resolved into two rectangular components. The electric flux through a planar area is defined as the electric field times the component of the area perpendicular to the field. Now. E=∫dE=∫dEx+∫dEy=∫dEx E=∫dEx= = But. then the evaluation of the flux generally requires an area integral since the angle will be continually changing. dEx= cosθ= substituting for dq from above. Electric flux The concept of electric flux is useful in association with Gauss' law. E= ax where a x is the unit vector along the x axis. then its dEy component points upwards and hence. dEx and dEy. we have. The dEx components add up. Of ECE/SJBIT Page 11 . Dept. it is understood that the magnitude of the vector is equal to the area and the direction of the vector is perpendicular to the area.Field Theory 10ES36 When the area A is used in a vector operation like this. Of ECE/SJBIT Page 12 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP .Field Theory 10ES36 Dept. Of ECE/SJBIT Page 13 .Field Theory 10ES36 Dept. Field Theory 10ES36 Gauss law Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 14 Field Theory 10ES36 Dept. Of ECE/SJBIT Page 15 Field Theory 10ES36 Dept. Of ECE/SJBIT Page 16 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Field Theory 10ES36 Dept. Of ECE/SJBIT Page 17 Of ECE/SJBIT Page 18 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP .Field Theory 10ES36 Dept. Field Theory 10ES36 Dept. Of ECE/SJBIT Page 19 . Field Theory 10ES36 Application of Gauss law \ Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 20 . Of ECE/SJBIT Page 21 .Field Theory 10ES36 Dept. Field Theory 10ES36 Divergence Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 22 . Field Theory 10ES36 Maxwell’s First equation From divergence theorem. Of ECE/SJBIT Page 23 . we have Dept. Field Theory 10ES36 Dept. Of ECE/SJBIT Page 24 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP . Field Theory 10ES36 Divergence theorem From Gauss law. we have Dept. Of ECE/SJBIT Page 25 . Field Theory 10ES36 Dept. Of ECE/SJBIT Page 26 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP . Conductor properties and boundary conditions for capacitance Recommended readings: 1. Conductor properties and boundary conditions for dielectrics. . John Krauss and Daniel A Fleisch McGraw-Hill. Definition of potential difference and potential. 1968. Electromagnetics with Applications. Energy density in an electrostatic field. Indian Reprint – 2001. Current and current density. Edward C. Of ECE/SJBIT Page 27 . Energy Electromagnetics. 2. dielectrics and capacitance Energy expended in moving a point charge in an electric field. Electromagnetic Waves And Radiating Systems. Tata McGraw-Hill. metallic conductors. William H Hayt Jr . Line integral. Field and Wave Electromagnetics. Potential gradient.1989. 7th edition. Continuity of current. Jordan and Keith G Balmain. Dept. David K Cheng. Prentice – Hall of India / Pearson Education. 1999 3.Reprint 2002 4. Potential field of a point charge & system of charges. 5th edition.2006. and John A Buck.Field Theory 10ES36 UNIT 2: Energy and potential & Conductors. 2nd edition. 2nd edition. Pearson Education Asia. Field Theory 10ES36 UNIT 2: Energy and potential & Conductors. dielectrics and capacitance Energy expended in moving a point charge in an electric field Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 28 . Field Theory 10ES36 Dept. Of ECE/SJBIT Page 29 . Field Theory 10ES36 Dept. Of ECE/SJBIT Page 30 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP . Of ECE/SJBIT Page 31 .Field Theory 10ES36 Dept. Field Theory 10ES36 Dept. Of ECE/SJBIT Page 32 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP . Of ECE/SJBIT Page 33 .Field Theory 10ES36 Definition of Potential &Potential difference Dept. Field Theory 10ES36 Dept. Of ECE/SJBIT Page 34 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP . Of ECE/SJBIT Page 35 .Field Theory 10ES36 Potential field of a point charge Dept. Field Theory 10ES36 Dept. Of ECE/SJBIT Page 36 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP . Field Theory 10ES36 Potential field of system of charges: Conservative property Dept. Of ECE/SJBIT Page 37 . Of ECE/SJBIT Page 38 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP .Field Theory 10ES36 Dept. Field Theory 10ES36 Potential gradient Dept. Of ECE/SJBIT Page 39 . Field Theory 10ES36 Dept. Of ECE/SJBIT Page 40 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP . Field Theory 10ES36 Dept. Of ECE/SJBIT Page 41 . Of ECE/SJBIT Page 42 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP .Field Theory 10ES36 Dept. Of ECE/SJBIT Page 43 .Field Theory 10ES36 Dept. Of ECE/SJBIT Page 44 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP .Field Theory 10ES36 Dept. Of ECE/SJBIT Page 45 .Field Theory 10ES36 Dipole Dept. Of ECE/SJBIT Page 46 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP .Field Theory 10ES36 Dept. Field Theory 10ES36 Energy density in an electrostatic field Dept. Of ECE/SJBIT Page 47 . Field Theory 10ES36 Dept. Of ECE/SJBIT Page 48 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP . Of ECE/SJBIT Page 49 .Field Theory 10ES36 Dept. Field Theory 10ES36 Capacitance Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 50 . Field Theory 10ES36 Dept. Of ECE/SJBIT Page 51 . Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 52 .Field Theory 10ES36 Problem Dept. Field Theory 10ES36 Dept. Of ECE/SJBIT Page 53 . Field Theory 10ES36 Boundary condition for perfect dielectric Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 54 . Field Theory 10ES36 Dept. Of ECE/SJBIT Page 55 . Field Theory 10ES36 Dept. Of ECE/SJBIT Page 56 GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP . Of ECE/SJBIT Page 57 .Field Theory 10ES36 Dept. 2. Edward C. Field and Wave Electromagnetics. William H Hayt Jr . 7th edition.Field Theory 10ES36 UNIT 3: Poisson’s and Laplace’s equations Derivation of Poisson’s equation. 1968. Electromagnetic Waves And Radiating Systems. Pearson Education Asia.1989. Electromagnetics with Applications. David K Cheng. and John A Buck. Indian Reprint – 2001. Tata McGraw-Hill. 2nd edition. Prentice – Hall of India / Pearson Education.Reprint 2002 4. Examples of Poisson’s Equations Recommended readings: 1. Uniqueness theorem. 2nd edition.2006. Energy Electromagnetics. 5th edition. Dept. Examples of Laplace Equations. Jordan and Keith G Balmain. . John Krauss and Daniel A Fleisch McGraw-Hill. 1999 3. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 58 . Field Theory 10ES36 UNIT 3: Poisson’s and Laplace’s equations Laplace’s & Poisson’s equation Dept. Of ECE/SJBIT Page 59 . Field Theory 10ES36 Uniqueness theorem Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 60 . Field Theory 10ES36 Dept. Of ECE/SJBIT Page 61 . Field Theory 10ES36 Example of laplace’s equation Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 62 . Field Theory 10ES36 Dept. Of ECE/SJBIT Page 63 . Field Theory 10ES36 Example 2: Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 64 . Of ECE/SJBIT Page 65 .Field Theory 10ES36 Example 3: Dept. Field Theory 10ES36 Example 4: Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 66 . Field Theory 10ES36 Dept. Of ECE/SJBIT Page 67 . Dept.1989. Recommended readings: 1. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 68 . Pearson Education Asia. Magnetic flux and flux density. Energy Electromagnetics. curl’s theorem. 1999 3. Electromagnetics with Applications.savart law. 1968.2006. 2. Indian Reprint – 2001.Field Theory 10ES36 UNIT – 4: MAGNETOSTATIC FIELDS Biot. and John A Buck. Prentice – Hall of India / Pearson Education. William H Hayt Jr . Jordan and Keith G Balmain. Electromagnetic Waves And Radiating Systems. 7th edition. 2nd edition. Field and Wave Electromagnetics. John Krauss and Daniel A Fleisch McGraw-Hill. Edward C. 5th edition. 2nd edition. Vector magnetic potentials.Reprint 2002 4. Tata McGraw-Hill. Stoke’s theorem. . David K Cheng. Amphere’s circuitary law. A magnetostatic field is produced by a constant current flow (or direct current). and so on. magnetic separators. microphones. The development of the motors. If the charges are moving with constant velocity. advertising displays. Static magnetic fields. This current flow may be due to magnetization currents as in permanent magnets. An electrostatic field is produced by static or stationary charges. television focusing controls. memory stores. Like Coulomb's law. involve magnetic phenomena and play an important role in our everyday life. a static magnetic (or magnetostatic) field is produced. H and B are related according to B = H. transformers. telephone bell ringers. Just as Gauss's law is a special case of Coulomb's law. are characterized by H or B. or conduction currents as in current-carrying wires. compasses. Ampere's law is a special case of BiotDept. and (2) Ampere's circuit law. electron-beam currents as in vacuum tubes. Of ECE/SJBIT Page 69 . Biot-Savart's law is the general law of magnetostatics.4 MAGNETOSTATIC FIELDS Static electric fields are characterized by E or D. There are similarities and dissimilarities between electric and magnetic fields. There are two major laws governing magnetostatic fields: (1) Biot-Savart's law. A definite link between electric and magnetic fields was established by Oersted in 1820. magnetically levitated high speed vehicles.Field Theory 10ES36 UNIT . As E and D are related according to D = E for linear material space. k is the constant of proportionality.2) becomes dH I dl sin 4R 2 (1. k = 1/4. (1.1. In SI units. dH I dl sin R2 (1. BIOT SAVART's LAW Biot-Savart's law states that the magnetic field intensity dH produced at a point P. eq. by the differential current element I dl is proportional to the product I dl and the sine of the angle between the element and the line joining P to the element and is inversely proportional to the square of the distance R between P and the element. as shown in Figure 1. So.1) or dH KI dl sin R2 (1. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 70 .Field Theory 10ES36 Savart's law and is easily applied in problems involving symmetrical current distribution.2) where.3) Dept. That is. 3) is better put in vector form as dH Idl a R 4R 2 Idl R 4R 3 (1. Figure 1. Thus. Dept. one can use the right-handed screw rule to determine the direction of dH: with the screw placed along the wire and pointed in the direction of current flow. it is easy to notice that eq. the direction of dH can be determined by the right-hand rule with the right-hand thumb pointing in the direction of the current. Alternatively.Field Theory 10ES36 From the definition of cross product equation A x B = AB Sin AB an. Of ECE/SJBIT Page 71 .1: Magnetic field dH at P due to current element I dl. the right-hand fingers encircling the wire in the direction of dH as shown in Figure 1. (1.2(b).4) where R in the denominator is |R| and aR = (vector R/|R|}.2(a). the direction of advance of the screw is the direction of dH as in Figure 1. Field Theory 10ES36 Figure 1. one can have different current distributions: line current. or into. It is customary to represent the direction of the magnetic field intensity H (or current I) by a small circle with a dot or cross sign depending on whether H (or I) is out of. the page as illustrated in Figure 1. surface current and volume current as shown in Figure 1. the source elements are related as I dl K dS J dv (1. As like different charge configurations.3. If we define K as the surface current density (in amperes/meter) and J as the volume current density (in amperes/meter square).5) Dept. or (b) the righthanded screw rule.4. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 72 .2: Determining the direction of dH using (a) the right-hand rule. 8) As an example. We assume that the conductor is along the z-axis with its upper and lower ends respectively subtending angles Figure 1.5.4) becomes H L Idl a R H S KdS a R V Jdv a R 4R 2 4R 2 (Line current) (1.Field Theory 10ES36 Thus.6) (Surface current) (1. Biot-Savart law as in eq. Of ECE/SJBIT Page 73 . let us apply eq.7) H 4R 2 (Volume current) (1. Dept. (1. in terms of the distributed current sources. (1.3: Conventional representation of H (or I) (a) out of the page and (b) into the page.6) to determine the field due to a straight current carrying filamentary conductor of finite length AB as in Figure 1. 9) But dl = dz az and R = a . dH Idl R 4R 3 (1. 2 and 1 at P. as the formula to be derived will have to be applied accordingly.zaz .10) Dept. If we consider the contribution dH at P due to an element dl at (0. so dl x R = dz a (1.Field Theory 10ES36 Figure 1. the point at which H is to be determined.4: Current distributions: (a) line current (b) surface current (c) volume current. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 74 . 0. z). Particular note should be taken of this assumption. dz = - cosec2 d. Of ECE/SJBIT 2 a sin d 1 Page 75 . equation (1.11) becomes I H 4 I 4 2 2 cos ec 2 d 1 3 cos ec 3 a Dept. Letting z = cot .11) Figure 1.Field Theory 10ES36 Hence.5: Field at point P due to a straight filamentary conductor. H I dz 4 z 2 2 3 a 2 (1. 2 = 0. (1.14) is not always easy.Field Theory 10ES36 Or H I 4 cos 2 cos 1 a (1. point A is at (0. . 0) while B is at (0.12) The equation (1.15) Dept. eq. and eq. ). (1. 2 = 0.. (1. ). Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 76 . 0. Note from eq. along concentric circular paths) irrespective of the length of the wire or the point of interest P. As a special case.12) reduces to H I 2 a (1.12) is generally applicable for any straight filamentary conductor of finite length. 0. For this case.14) To find unit vector a in equations (1. A simple approach is to determine a from a a a (1.12) becomes H I 4 a (1.) while B is at (0. So. so that point A is now at O(0.12) that H is always along the unit vector a (i.e.13) Another special case is when the conductor is infinite in length. 0. 1 = 180.12) to (1. 1 = 90. when the conductor is semi-infinite (with respect to P). 0. The key point to be kept in mind in applying eq. Find H at (0. 2.Field Theory 10ES36 where al is a unit vector along the line current and a is a unit vector along the perpendicular line from the line current to the field point. 5) due to side 1 of the loop in Figure 1.6(a) carries a current of 10 A. consider Figure Dept. To find H at (0. Solution: This example illustrates how eq. current-carrying conductor. 0.12) is figuring out 1. Of ECE/SJBIT Page 77 . and a.6(a). 0. (1. 5) due to side 1 of the loop.12) is applied to any straight. (1. thin. Illustration: The conducting triangular loop in Figure 1. so a = ax x az = -ay Hence. According to eq.12). (1. Notice that we join the Point of interest (0. 0.12) is based. 2 and are assigned in the same manner as in Figure 1. Dept. 29 To determine a is often the hardest part of applying eq. 1. where side 1 is treated as a straight conductor. (1. 2 =5 . cos 2 cos 1 = cos 90 = 0.5 on which eq.6: (a) conducting triangular loop (b) side 1 of the loop. H1 1 4 cos 2 cos 1 a 10 2 0 (a y ) 4 (5) 29 = -59.Field Theory 10ES36 Figure 1.1 ay mA/m AMPERE'S CIRCUIT LAW Ampere's circuit law states that the line integral of the tangential components of H around a closed path is the same as the net current Ienc enclosed by the path In other words. Observe that 1.6(b). Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 78 .15). (1. al = ax and a = az. 5) to the beginning and end of the line current. that is. the circulation of H equals Ienc . Field Theory 10ES36 H dl I enc (1. It should be noted that eq.17) and (7. Of ECE/SJBIT Page 79 . (7. (1.18) Comparing the surface integrals in eqs.16).16) Ampere's law is similar to Gauss's law and it is easily applied to determine H when the current distribution is symmetrical.17) But S I enc J dS (1.19) Dept. we obtain L S I enc H dl ( H ) dS (1. Ampere's law is a special case of Biot-Savart's law.18) clearly reveals that xH=J (1. the former may be derived from the latter. By applying Stoke's theorem to the left-hand side of eq. (1.16) always holds whether the current distribution is symmetrical or not but we can only use the equation to determine H when symmetrical current distribution exists. which Ampere's law is to be applied. This path on. (1. Dept. that is.Field Theory 10ES36 This is the third Maxwell's equation to be derived. magnetostatic field is not conservative. We choose a concentric circle as the Amperian path in view of eq. To determine H at an observation point P.7: Ampere's law applied to an infinite filamentary. it is essentially Ampere's law in differential (or point) form whereas eq. which shows that H is constant provided p is constant.14). we should observe that X H = J 0. line current. (1. we allow a closed path pass through P.19). according to Ampere's law I H a d a H d H 2 Figure 1. 7. APPLICATIONS OF AMPERE'S LAW Infinite Line Current Consider an infinitely long filamentary current I along the z-axis as in Figure 1. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 80 . is known as an Amperian path (analogous to the term Gaussian surface).16) is the integral form. Since this path encloses the whole current I. (1. From eq. in terms of the magnetic force. The constant is in henrys/meter (H/m) and has the value of 0 = 4 x 10-7 H/m (1.Field Theory 10ES36 Or H 1 2 a (1. can be discussed later. 0 is a constant known as the permeability of free space.22) The precise definition of the magnetic field B. MAGNETIC FLUX DENSITY The magnetic flux density B is similar to the electric flux density D. As D = 0E in free space.21) where. the magnetic flux density B is related to the magnetic field intensity H according to B = 0 H (1.14). Of ECE/SJBIT Page 81 .20) As expected from eq. (1. Dept. Field Theory 10ES36 Figure 1.24) This equation is referred to as the law of conservation of magnetic flux or Gauss'’s law for magnetostatic fields just as D. Dept. An isolated magnetic charge does not exit. B dS 0 (1.23) Where the magnetic flux is in webers (Wb) and the magnetic flux density is a webers/square meter (Wb/m2) or teslas. that is. magnetic flux is conserved. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 82 . Total flux through a closed surface in a magnetic field must be zero. Although the magnetostatic field is not conservative.8: Magnetic flux lines due to a straight wire with current coming out of the page The magnetic flux through a surface S is given by B dS S (1. dS = Q is Gauss's law for electrostatic fields. Equation (1. TABLE 1.B=0 xE=0 xH=J SB dS 0 Nonexistence of magnetic monopole LE dl 0 Conservativeness of electrostatic field L H dl J dS Ampere's law s The Table 1.2: Maxwell's Equations for Static EM Fields Differential (or Point) Integral Form Remarks Form .Field Theory 10ES36 By applying the divergence theorem to eq.2 gives the information related to Maxwell's Equations for Static Electromagnetic Fields. we obtain S B dS B dv 0 v Or .25) This equation is the fourth Maxwell's equation to be derived.25) suggests that magnetic field lines are always continuous. (1. Dept.25) shows that magnetostatic fields have no sources or sinks. Of ECE/SJBIT Page 83 .24). Equation (1.24) or (1. D = v S D dS v dv Gauss's law v .B=0 (1. Just as E = -V. ( x A) = 0 (1. (1. Combining eq.29) Page 84 . Vm if J = 0 (1. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP (1.28) The condition attached to this equation is important and will be explained. we can define a potential associated with magnetostatic field B. we define the magnetic scalar potential Vm (in amperes) as related to H according to H = .26) . (1.27) which must always hold for any scalar field V and vector field A.28) and eq. Similarly.Field Theory 10ES36 MAGNETIC SCALAR AND VECTOR POTENTIALS We recall that some electrostatic field problems were simplified by relating the electric Potential V to the electric field intensity E (E = -V). To define Vm and A involves two important identities: x (V) = 0 (1. the magnetic potential could be scalar V m vector A.19) gives J = x H = - x (- Vm) = 0 Dept. In fact. 27) simultaneously. We should also note that V m satisfies Laplace's equation just as V does for electrostatic fields.30) We know that for a magnetostatic field. (1. Of ECE/SJBIT L S 0 I dl 4R 0 K dS 4R for line current (1. Thus the magnetic scalar potential V m is only defined in a region where J = 0 as in eq.26). must satisfy the condition in eq. (1. 2 Vm = 0.33) for surface current (1. we can define the vector magnetic potential A (in Wb/m) such that B=xA (1. (J = 0) (1.28). (1.Field Theory 10ES36 since Vm. x B = 0 as stated in eq.25) and (1. In order to satisfy eqs.25).32) We can define A A Dept. hence. (1.31) Just as we defined V dQ 4 0 r (1.34) Page 85 . 75 Wb Illustration 2: Identify the configuration in figure 1. 0 z 5m. 2 1 2 5 2 z0 1 d dS d dz a dz 1 2 15 (5) 4 4 = 3. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 86 . Solution: B A B dS Az a a .35) Illustration 1: Given the magnetic vector potential A = -2/4 az Wb/m.Field Theory 10ES36 A v 0 J dv for volume current 4R (1. 1 2m. calculate the total magnetic flux crossing the surface = /2.9 that is not a correct representation of I and H. Dept. Of ECE/SJBIT Page 87 . Dept. The direction of H field should have been outwards for the given I direction.9 (c) is not a correct representation.Field Theory 10ES36 Figure 1.9: Different I and H representations (related to Illustration 2) Solution: Figure 1. 5th edition. Jordan and Keith G Balmain. Magnetic circuits. Field and Wave Electromagnetics. David K Cheng. Force between differential current elements. 2. Recommended readings: 1. . Potential energy and forces on magnetic materials. 1968. John Krauss and Daniel A Fleisch McGraw-Hill. Energy Electromagnetics.Reprint 2002 4. Prentice – Hall of India / Pearson Education. 2nd edition. Magnetic boundary conditions. Magnetic materials and inductance. Electromagnetics with Applications. Pearson Education Asia. Indian Reprint – 2001. and John A Buck. Edward C. William H Hayt Jr .Field Theory 10ES36 PART .2006. MATERIALS AND DEVICES Force on a moving charge and differential current element. Tata McGraw-Hill. Magnetization and permeability. Force and Torque on a closed circuit. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 88 . Dept. 2nd edition. Electromagnetic Waves And Radiating Systems. 1999 3. 7th edition.1989.B UNIT -5: MAGNETIC FORCES. the electric force Fe.1) This shows that if Q is Positive. From experiments.2) This clearly shows that Fm is perpendicular to both u and B. Fm depends on the charge velocity and Page 89 . (2. Fe is independent of the velocity of the charge and can perform work on the charge and change Dept. A magnetic field can exert force only on a moving charge. From eqs. Of ECE/SJBIT its kinetic energy. it is found that the magnetic force Fm experienced by a charge Q moving with a velocity u in a magnetic field B is Fm = Qu x B (2. Unlike Fe.Field Theory 10ES36 PART -B UNIT -5 MAGNETIC FORCES.1) and (2.2). Fe and E have the same direction. on a stationary or moving electric charge Q in an electric field is given by Coulornb's experimental law and is related to the electric field intensity E as Fe = QE (2. a comparison between the electric force Fe and the magnetic force Fm can be made. MATERIALS AND DEVICES Force on a Charged Particle According to earlier information. 4) The solution to this equation is important in determining the motion of charged particles in E and B fields. The magnitude of Fm is generally small compared to Fe except at high velocities. A summary on the force exerted on a charged particle is given in table 2. the total force on the charge is given by F = Fe + Fm or F = Q (E + u x B) (2. If the mass of the charged Particle moving in E and B fields is m. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 90 . We should bear in mind that in such fields. It relates mechanical force to electrical force.dl = 0). of the charge (Fm. For a moving charge Q in the Presence of both electric and magnetic fields. it does not cause an increase in kinetic energy of the charge.1.Field Theory 10ES36 is normal to it. Dept. energy transfer can be only by means of the electric field.3) This is known as the Lorentz force equation. by Newton's second law of motion. F m du Q E u B dt (2. Fm cannot perform work because it is at right angles to the direction of motion. 1. As per equation dF = I dl x B2 d(dF1) = I1 dl1 x dB2 Dept.2) as the vector which satisfies F m / q = u x B just as we defined electric field E as the force per unit charge. (2. both current elements produce magnetic fields. Of ECE/SJBIT (2. Fe / q.Field Theory 10ES36 TABLE 2. According to BiotSavart's law. B may be defined from eq. So we may find the force d(dF1) on element I1 dl1 due to the field dB2 produced by element I2 dl2 as shown in Figure 2.5) Page 91 . Force between Two Current Elements Let us now consider the force between two elements I1 dl1 and I2 dl2.1: Force on a Charged Particle State of Particle E Field B Field Combined E and B Fields Stationary QE - QE Moving QE Qu x B Q(E + u x B) The magnetic field B is defined as the force per unit current element Alternatively. This equation is essentially the law of force between two current elements and is analogous to Coulomb's law.7) Figure 2. which expresses the force between two stationary charges. (2.6) Hence. we obtain the total force F1 on current loop 1 due to current loop 2 shown Figure 2.1: Force between two current loops. d (dF1 ) 0 I1dl1 I 2 dl 2 a R 21 2 4R21 (2.1 as Dept. dB2 0 I 2 dl 2 a R 21 4R 2 21 (2.7).Field Theory 10ES36 But from Biot-Savart's law. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 92 . From eq. It is eq. it experiences a force that tends to rotate it.8) was experimentally established by Qersted and Ampete.8) Although this equation appears complicated. Dept. That is.c. It can be shown that F2 = . If the loop is placed parallel to a magnetic field.8) by interchanging subscripts 1 and 2. (2. d. vector product of the force F and the moment arm r. The torque T (or mechanical moment of force) on the loop is the. (8. The force F2 on loop 2 due to the magnetic field B1 from loop 1 is obtained from eq. thus F1 and F2 obey Newton's third law that action and reaction are equal and opposite.F1. we can determine the torque on it. (2. The concept of a current loop experiencing a torque in a magnetic field is of paramount importance in understanding the behavior of orbiting charged particles. (2. MAGNETIC TORQUE AND MOMENT Now that we have considered the force on a current loop in a magnetic field. 10) that is of fundamental importance. and generators. Biot and Savart (Ampere's colleagues) actually based their law on it.5). It is worthwhile to mention that eq.Field Theory 10ES36 F1 0 I1 I 2 4 L1 dl1 dl 2 a R21 L2 2 R21 (2. motors. we should remember that it is based on eq. Of ECE/SJBIT Page 93 . From this figure. Let us apply this to a rectangular loop of length l and width w placed in a uniform magnetic field B as shown in Figure 8.2: Rectangular planar loop in a uniform magnetic field.Field Theory 10ES36 T=rxF (2.5(a).9) and its units are Newton-meters. we notice that dl is parallel to B along sides 12 and 34 of the loop and no force is exerted on those sides. Thus 3 1 2 4 F I dl B I dl B l 0 2 l I dz a z B I dz a z B Figure 2. or Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 94 . F0 and –F0 act at different points on the loop. T = BIS sin (2. no force is exerted on the loop as a whole. Thus. However. (2. an is a unit normal vector to the plane of the loop and its direction is determined by the right-hand rule: fingers in the direction of current Hand thumb along an.10) Where.11) But lw = S. the area of the loop. In eq. thereby creating a couple.12) We define the quantity m = I S an (2. as shown in the cross-sectional view of Figure 2. Hence.2(b).13).13) as the magnetic dipole moment (in A/M2) of the loop. If the normal to the plane of the loop makes an angle with B. Dept. the torque on the loop is |T| = |F0| w sin or T = B I l w sin (2. Of ECE/SJBIT Page 95 .Field Theory 10ES36 F= F0 – F0 = 0 (2. |F0| = I Bl because B is uniform. This force is given by Dept. B (=H). 3.2) If a charge. (2. V is placed in a magnetic field.1) FORCE ON A MOVING CHARGE DUE TO ELECTRIC AND MAGNETIC FIELDS If there is a charge or a moving charge. Q in an electric field. Q moving with a velocity.13) in eq.0 STOKE'S THEOREM Stoke's Theorem relates a line integral to the surface integral and vice-versa. we obtain T=mxB (2. Introducing eq. that is C H dL ( H ) dS S (3. its reaction is normal to the loop.14) 3. This force is given by FE = QE (3. E. then there exists a force on the charge (Fig.Field Theory 10ES36 The magnetic dipole moment is the product of current and area of the loop. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 96 . there exists a force on the charge.1). (2.12). (wb/m2) V = velocity of the charge. Q is placed in both electric and magnetic fields. velocity and force If the charge.3) B = magnetic flux density. Problem 1: A charge of 12 C has velocity of 5ax + 2ay . then the force on the charge is F =Q (E + V x B) (3.+5ay +10az V/m Dept.1: Direction of field.3az m/s. Determine F on the charge in the field of (a) E=18ax.Field Theory 10ES36 FH = Q(V x B) (3. 3.4) This equation is known as Lorentz force equation. m/s Fig. Of ECE/SJBIT Page 97 . 3az m/s B = 4 ax + 4 ay + 3 az wb / m2 F = 12 [18ax . Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 98 . F=Q |E|= 12 18 2 5 2 10 2 F = 254. Solution: (a) The force.27 N (b) The force F on the charge due to B is F = Q[V x B) Here V = 5ax + 2ay .Field Theory 10ES36 (b) B = 4ax + 4ay + 3az wb/m2.27ay + 12az] Dept. Q due to E is F = QE = 12 (18ax + 5ay + 10az) = 216ax + 60ay + 120az or. F on the charge. 6) where is the angle between the direction of the current element and the direction of magnetic flux density B = magnetic flux density. F= 12 (324 729 144) F = 415. Then the differential force on the charge is given by dF = dQ (V x B) (3. wb/m2 IL = current element.5) or. dQ to be moving with a velocity.Field Theory 10ES36 or.7) Dept.17 N FORCE ON A CURRENT ELEMENT IN A MAGNETIC FIELD The force on a current element when placed in a magnetic field. V in a magnetic field. Amp-m Proof: Consider a differential charge. Of ECE/SJBIT Page 99 . B is F = IL x B (3. H = (B/). F = I L B Sin Newton (3. dF =IdL x B or. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 100 . Newton (3. Determine the force on the current element due to the magnetic field if the magnetic field H = (5ax/) A/m.8) Problem 2: A current element 4 cm long is along y-axis with a current of 10 mA flowing in y-direction. F = IL x B. Solution: Dept.Field Theory 10ES36 But dQ = d dF = d (V x B) = ( V x B) d But V = J dF = J d x B Jd is nothing but IdL. 04ay = 4 x l0-4 ay H = (5ax/) A/m B = 5ax wb/m2 F = 4 x l0-4 ay x 5ax F = (0. IL = 10 x 10-3 x 0. H is continuous across any boundary except at the surface of a perfect conductor.Field Theory 10ES36 The force on a current element under the influence of magnetic field is F = IL x B Here. Of ECE/SJBIT Page 101 .0az mN BOUNDARY CONDITIONS ON H AND B 1.4ay x 5ax) x 10-3 or F = -2. Dept. The tangential component of magnetic field. that is. 2 in which a differential rectangular loop across a boundary separating medium 1 and medium 2 are shown. The normal component of magnetic flux density. Fig. Js = 0. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 102 .2: A rectangular loop across a boundary From Ampere's circuit law. Bnl = Bn2 (3.Htan2 = Js (3. we have Dept.10) Proof: Consider Fig.9) At non-conducting boundaries. B is continuous across any discontinuity. 3. 3. that is. 2.Field Theory 10ES36 Htanl . 11) Here.12) Now consider a cylinder shown in Fig. we get H dL H x1 x H x 2 x I or. So.Field Theory 10ES36 H dL 50 H y4 01 12 23 34 45 y y H y3 H x1x H y1 2 2 y y H y2 H x 2 x I 2 2 As y 0. 3. Of ECE/SJBIT Page 103 . Hxl and Hx2 are tangential components in medium 1 and 2. Htan1 – Htan2 = Js (3.3. H x1 H x 2 I Js x (3. respectively. Dept. 14) Bn1 S . Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 104 .3: A differential cylinder across the boundary Gauss's law for magnetic fields is B dS 0 s (3. Bnl = Bn2 (3.Field Theory 10ES36 Fig. 3.15) Dept.13) In this case.Bn2 S = 0 that is. Therefore. for y 0 B dS B s n1 a y dS a y B n 2 a y dS (a y ) s s (3. 400ay + 250az A/m.4. Of ECE/SJBIT Page 105 .400ay +250az A/m Consider Fig. Fig. 3. x < 0 describes medium 1 and x > 0 describes medium 2. The magnetic field in medium 1 is 150ax . linear and isotropic media have an interface at x = 0. Determine: (a) Magnetic field in medium 2 (b) Magnetic flux density in medium 1 (c) Magnetic flux density in medium 2.4: Illustrative figure Dept. Solution: The magnetic field in medium 1 is H1 = 150ax . 3. r1 = 2 and r2 = 5.Field Theory 10ES36 Problem 3: Two homogeneous. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 106 .Field Theory (a) 10ES36 H1 = Htan1 + Hn1 Htanl = -400ay + 250az A/m Hn1 = 150ax The boundary condition is Htanl = Htan2 Htan2 = .400ay + 50az A/m The boundary condition on B is Bn1 = Bn2 that is. 1 Hn1 = 1 Hn2 H n2 1 H n1 2 2 150a x 5 = 60ax Dept. 1004ay + 627.400ay + 250az) = (376.2513.98ax . Of ECE/SJBIT Page 107 .400ay + 250az) = (376.2ay +1570.5ax .75az)wb/m2 SCALAR MAGNETIC POTENTIAL Dept.Field Theory 10ES36 H2 = Htan2 + Hn2 (b) B1= 1 H1 = 0 r H1 =4 x 10-7 x 2(150ax .5az) wb/m2 (c) B2 = 2 H2 = 4 x 10-7 x 5 (60ax . scalar magnetic potential exists in a region where J = 0.18) But. XH=J J=0 In other words. that is. (3. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 108 .17) But curl of the gradient of any scalar is always zero.16) Vm = scalar magnetic potential (Amp) Taking curl on both sides. by Ampere's circuit law or. Dept.Field Theory 10ES36 Like scalar electrostatic potential. it is possible to have scalar magnetic potential. It is defined in such a way that its negative gradient gives the magnetic field. we get x H = - x Vm (3. x H =0 So. H = Vm (3. It is defined in such a way that its curl gives the magnetic flux density. Of ECE/SJBIT Page 109 .B = 0 . It is directly defined as B Vm H dL A 5. that is. It has the unit of Ampere. Dept. It satisfies Laplac’s equation. It exists where J = 0 3. VECTOR MAGNETIC POTENTIAL Vector magnetic potential exists in regions where J is present.20) Characteristics of Scalar Magnetic Potential (Vm) 1. The negative gradient of Vm gives H.19) The scalar potential satisfies Laplace's equation.H = 0 = m (-Vm) = 0 or. 2 Vm = 0 (J = 0) (3.Field Theory 10ES36 H = -Vm (J=0) (3. that is. 4. we have . or H = -Vm 2. Field Theory 10ES36 BxA (3.22) A or. 4R (3. It exists even when J is present. It is defined in two ways Dept. s A or. 4R (K = current sheet) (3. It is also defined as A 0 IdL Henry Amp 4R m (3.23) 0 Jdv .21) where A = vector magnetic potential (wb/m). Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 110 . 2.24) Characteristics of Vector Magnetic Potential 1. v 0 Kds . Field Theory 10ES36 BxA v and 0 Jd 4R 3. Problem 4: The vector magnetic potential. 6. 2A = 0 j 4. Vector magnetic potential. 3). apertures and also to obtain radiation leakage from transmission lines. A has applications to obtain radiation characteristics of antennas. waveguides and microwave ovens. A is used to find near and far-fields of antennas. 2A = 0 if J = 0 5. Solution: A = (x2 +y2) az wb/m2 We have B = x A 10 6 Dept. Of ECE/SJBIT ax x 0 ay y 0 az z 2 x y 2 Page 111 . 2. A due to a direct current in a conductor in free space is given by A = (X2 + Y2) az wb /m2. Determine the magnetic field produced by the current element at (1. 3) 1 4ax 2 4a y 106 5a x 6a y 10 6 H 1 0 5a x 6a y 10 6 1 5a x 6a y 10 6 7 4 10 H = (3. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 112 .5 in which a rectangular loop is placed under a uniform magnetic flux density. Dept.774ay).Field Theory 10ES36 2 x y 2 a x x 2 y 2 a y 10 6 x x x 2 2 y a x 2 x y 2 a y 10 6 B / at (1.978ax – 4. B.2. A/m FORCE AND TORQUE ON A LOOP OR COIL Consider Fig. 3. the force.5. the force on QR due to B is F1 =IL x B =-ILaz x Bax (3.26) that is. F1 on QR moves it downwards. Of ECE/SJBIT Page 113 .5: Rectangular conductor loop in x-z plane From Fig. 3.28) Dept. 3. Now the force on PS is F2 = IL x B = -ILaz x Bax (3.Field Theory 10ES36 Fig.ILBay (3.27) F2 = .25) F1 = -ILBay (3. The torque on the loop is defined as the vector product of moment arm and force. N-m where (3. T r x F.Field Theory 10ES36 Force. that is. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 114 . The forces on QR and PS exert a torque.29) r = moment arm F = force Applying this definition to the loop considered above. This torque tends to rotate the coil about its axis. The torque. It may be noted that the sides PQ and SR will not experience force as they are parallel to the field. T is nothing but a mechanical moment of force.31) = -BILwaz Dept. the expression for torque is given by T = r1 x F1 + r2 x F2 (3.30) w w a x ( ILBa y ) a x ( ILBa y ) 2 2 (3. B. F2 on PS moves it upwards. Find the torque about the z-axis. It carries a current of 1 A.34) where m = I l w ay = I S ay Problem 5: A rectangular coil is placed in a field of B = (2ax + ay) wb/m2. The coil is in y-z plane and has dimensions of 2 m x 2 m.Field Theory or. Solution: m=IS an = 1 x 4ax T = m x B = 4ax x (2ax + ay) T = 4az. 10ES36 T = -BISaz (3. Of ECE/SJBIT Page 115 .32) where S = wL = area of the loop The torque in terms of magnetic dipole moment. m is T = m x B. N-m (3. N-m Dept. Field Theory 10ES36 MATERIALS IN MAGNETIC FIELDS A material. Dept. Examples are copper. m < 0 and r 1. r = 1. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 116 . compass needle. r = 1 A material is said to be non-magnetic if m = 0. diamond and bismuth. Magnetic properties are described in terms of magnetic susceptibility and relative permeability of the materials. The term 'Magnetism' is commonly discussed in terms of magnets with basic examples like north pole. Diamagnetic materials 2. Ferromagnetic materials Diamagnetic Materials A material is said to diamagnetic if its susceptibility. is said to be magnetic if m 0.0. Magnetic materials are classified into 1. silicon. Paramagnetic materials 3. lead. horse shoe magnets and so on. Magnetic susceptibility m is (-)ve. r = 1 B=0 Most of the materials exhibit diamagnetism. Examples are air. Of ECE/SJBIT Page 117 . Dept. potassium and platinum. tungsten. Magnetic fields due to orbiting and spinning electrons do not cancel each other. Permanent magnetic moment of each atom is zero.Field Theory 10ES36 Characteristics of diamagnetic materials Magnetic fields due to the motion of orbiting electrons and spinning electrons cancel each other. These materials are widely affected by magnetic field. Diamagnetism is not temperature dependent. Characteristics of paramagnetic materials They have non-zero permanent magnetic moment. Paramagnetism is temperature dependent. m lies between 10-5 and 10-3. These materials acquire magnetisation opposite to H and hence they are called diamagnetic materials. They are linear magnetic materials. Paramagnetic Materials A material for which m > 0 and r 1 is said to be paramagnetic. They lose their ferromagnetic properties when the temperature is raised. m >> 0 r >> l They are strongly magnetised by magnetic field. Dept. r >> 1 is said to be ferromagnetic. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 118 . cobalt and their alloys. Ferromagnetic Materials A material for which m >> 0. it loses its magnetisation completely. nickel. Characteristics of ferromagnetic materials They exhibit large permanent dipole moment. Examples are iron. They are non-linear magnetic materials. m > 0 r 1 They are linear magnetic materials. magnetisation is not determined by the field present. 770C. B = H does not hold good as depends on B. In these materials. If a permanent magnet made of iron is heated above its curie temperature. These materials acquire magnetisation parallel to H and hence they are called paramagnetic materials. It depends on the magnetic history of the object.Field Theory 10ES36 These are used in MASERS. They retain magnetism even if the magnetic field is removed. 35) Here N = number of turns = flux produced I = current in the coil 1 Henry l wb/Amp L is also defined as (2WH/I2). that is. L N (Henry) I (3. L of a conductor system is defined as the ratio of magnetic flux linkage to the current producing the flux. It also opposes a sudden change in current that has been established. Inductance.Field Theory 10ES36 INDUCTANCE Inductor is a coil of wire wound according to various designs with or without a core of magnetic material to concentrate the magnetic field. an inductance is the inertial property caused by an induced reverse voltage that opposes the flow of current when a voltage is applied. device or circuit. L In a conductor. L (Henry): The inductance. Of ECE/SJBIT Page 119 . Definition of Inductance. or Dept. One side of each turn of the coil is threaded through the ring to form a Toroid (Fig. 3.6). a straight conductor carrying current has the property of inductance. Aircore coils are wound to provide a few pico henries to a few micro henries. The requirements of such coils are: Stability of inductance under all operating conditions High ratio inductive reactance to effective loss resistance at the operating frequency Low self capacitance Small size and low cost Low temperature coefficient STANDARD INDUCTANCE CONFIGURATIONS Toroid It consists of a coil wound on annular core.Field Theory 10ES36 L 2WH I2 (3. interstage coupling coils and so on. WH = energy in H produced by I.36) where. Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 120 . These are used at IF and RF frequencies in tuning coils. In fact. 7. 3. A solenoid is also used to demonstrate electromagnetic induction.37) N = number of turns r = average radius S = cross-sectional area Magnetic field in a Toroid. A bar of iron. Dept. 3. is usually provided for this purpose.Field Theory 10ES36 Fig. It is used to produce a known magnetic flux density along its axis. A typical solenoid is shown in Fig. Solenoid It is a coil of wire which has a long axial length relative to its diameter. H NI 2r (3. Of ECE/SJBIT Page 121 .38) I is the current in the coil. The coil is tubular in form. which is free to move along the axis of the coil. L 2r Here (3.6: Toroid 0 N 2S Inductance of Toroid. Field Theory 10ES36 Fig.oo0oo ----------Dept. L of a solenoid is L 0 N 2 S l (3.7: Solenoid The inductance.40) I is the current ----------. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 122 .39) l = length of solenoid S = cross-sectional area N = Number of turns The magnetic field in a solenoid is H NI l (3. 3. Discuss briefly about magnetic materials.Field Theory 10ES36 Recommended Questions: 1. 5. Derive for force on a moving charge. Derive for Lorentz force equation. 6. Discuss briefly about magnetic boundary conditions. 2. 3. Derive for force & torque on a closed circuit. 4. Dept. Of ECE/SJBIT Page 123 . Derive for force between differential current elements. Energy Electromagnetics.1989. Dept. Prentice – Hall of India / Pearson Education. Pearson Education Asia. David K Cheng. William H Hayt Jr. Tata McGraw-Hill. Indian Reprint – 2001. Edward C. 2nd edition. Field and Wave Electromagnetics. 2nd edition. Electromagnetic Waves And Radiating Systems. Retarded potentials. 5th edition. Recommended readings: 1. 1999 3. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 124 . Maxwell’s equation in point and integral form.Reprint 2002 4. and John A Buck. 1968. 2006. Electromagnetics with Applications.Field Theory 10ES36 UNIT -6: TIME VARYING MAGNETIC FIELDS AND MAXWELL’S EQUATIONS Faraday’s law. Displacement current. . 2. 7th edition. Jordan and Keith G Balmain. John Krauss and Daniel A Fleisch McGraw-Hill. 1 where N is the number of turns in the circuit and is the flux through each turn. Of ECE/SJBIT Page 125 . in any closed circuit is equal to the time rate of change of the magnetic flux linkage by the circuit This is called Faraday’s Law. Vemf (in volts). The negative sign shows that the induced voltage acts in such a way as to oppose the flux producing it. This is known as Lenz’s Law. and it can be expressed as Vemf d d N dt dt 1.Field Theory 10ES36 UNIT -6 TIME VARYING MAGNETIC FIELDS AND MAXWELL’S EQUATIONS Introduction Electrostatic fields are usually produced by static electric charges whereas magnetostatic fields are due to motion of electric charges with uniform velocity (direct current) or static magnetic charges (magnetic poles). Stationary charges Electrostatic fields Steady current Magnetostatic fields Time-varying current Electromagnetic fields (or waves) Faraday discovered that the induced emf. and it emphasizes the fact that the direction of current flow in the Dept. time-varying fields or waves are usually due to accelerated charges or time-varying current. Field Theory 10ES36 circuit is such that the induced magnetic filed produced by the induced current will oppose the original magnetic field. It is clear from eq. (1.3) are in Dept. (1. For a circuit with a single (N = 1). eq.3 where.1) becomes Vemf N d dt 1. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 126 . eq. both electric and magnetic fields are present and are interrelated.2 In terms of E and B. (1. 1 A circuit showing emf-producing field Ef and electrostatic field Ee TRANSFORMER AND MOTIONAL EMFS Having considered the connection between emf and electric field. Fig. has been replaced by B dS and S is the surface area of the circuit bounded by the S closed path L.2) can be written as Vemf E dl L d B dS dt S 1.3) that in a time-varying situation. (1. we may examine how Faraday's law links electric and magnetic fields. Note that dl and dS in eq. (1. Dept. A.3) may be caused in three ways: 1. 2: Induced emf due to a stationary loop in a time varying B field.4 Fig. Of ECE/SJBIT Page 127 . By having a time-varying loop area in a static B field 3. This should be observed in Figure 2. Equation (1.Field Theory 10ES36 accordance with the right-hand rule as well as Stokes's theorem.3) becomes Vemf E dl L S B dS t 1. (1. STATIONARY LOOP IN TIME-VARYING B FIELD (TRANSFORMER EMF) This is the case portrayed in Figure 2 where a stationary conducting loop is in a time varying magnetic B field. The variation of flux with time as in eq. By having a stationary loop in a time-varying B field 2.1) or eq. By having a time-varying loop area in a time-varying B field. 7) that the force on a charge moving with uniform velocity u in a magnetic field B is Fm = Qu x B Dept. We recall from eq. The work done in taking a charge about a closed path in a timevarying electric field. E B t 1.6 This is one of the Maxwell's equations for time-varying fields. we obtain B E dS t dS S S 1. This does not imply that the principles of energy conservation are violated. that is. It shows that the time varying E field is not conservative ( x E 0). (1. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP 1. MOVING LOOP IN STATIC B FIELD (MOTIONAL EMF) When a conducting loop is moving in a static B field. B. for example.Field Theory 10ES36 This emf induced by the time-varying current (producing the time-varying B field) in a stationary loop is often referred to as transformer emf in power analysis since it is due to transformer action.7 Page 128 .5 For the two integrals to be equal. an emf is induced in the loop. their integrands must be equal. (1.4). is due to the energy from the time-varying magnetic field. By applying Stokes's theorem to the middle term in eq. generators.9 gives the total emf as Dept.9 This type of emf is called motional emf or flux-cutting emf because it is due to motional action.4 and 1. and alternators.Field Theory 10ES36 We define the motional electric field Em as Em Fm uB Q 1. moving with uniform velocity u as consisting of a large number of free electrons. C. MOVING LOOP IN TIME-VARYING FIELD This is the general case in which a moving conducting loop is in a time-varying magnetic field.8 If we consider a conducting loop. It is the kind of emf found in electrical machines such as motors. Combining equation 1. Of ECE/SJBIT Page 129 . the emf induced in the loop is Vemf Em dl u B dl L L 1. Both transformer emf and motional emf are present. ( x H) = 0 = . Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP 1.12 DISPLACEMENT CURRENT For static EM fields. .10 Em u B 1.13 But the divergence of the curl of any vector field is identically zero. E B u B t 1. Hence.Field Theory 10ES36 B dS u B dl t S L Vemf E dl L 1.14 Page 130 . J Dept.11.11 or from equations 1.6 and 1. we recall that xH=J 1. 1. 1. so that it becomes x H = J + Jd 1. ( x H) = 0 = . 1.17 to agree with eq. 1. To do this.17 In order for eq.13. 1. the divergence of the curl of any vector is zero. we add a term to eq. J + .15 Thus eqs.15.15 are obviously incompatible for time-varying conditions.18 or Dept. Of ECE/SJBIT Page 131 .Field Theory 10ES36 The continuity of current requires that J v 0 t 1.16 where Jd is to be determined and defined.13 to agree with eq.15. Again. We must modify eq. J d J v D D t t t 1. Jd 1.14 and 1. Hence: . 1. 19 into eq. the two terms are comparable.Field Theory Jd 10ES36 D t 1. 3 Two surfaces of integration showing the need for Jd in Ampere’s circuit law The insertion of Jd into eq. At the time of Maxwell. Fig. at radio frequencies. 1. 1.20 could not be verified experimentally. we define the displacement current as Dept. Without the term Jd. Jd is usually neglected compared with J.15 results in H J D t 1. 1. 1. high-frequency sources were not available and eq. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 132 .19 Substituting eq. Based on displacement current density.13 was one of the major contribution of Maxwell. however. The term Jd = D/t is known as displacement current density and J is the conduction current density (J = E)3.20 This is Maxwell's equation (based on Ampere's circuit law) for a time-varying field. electromagnetic wave propagation (radio or TV waves. for example) would be impossible. At low frequencies. Of ECE/SJBIT Page 133 . A typical example of such current is that through a capacitor when an alternating voltage source is applied to its plates. Calculate the displacement current assuming = 2 0. given by Dept.Field Theory 10ES36 I d J d dS D dS t 1.21 We must bear in mind that displacement current is a result of time-varying electric field. PROBLEM: A parallel-plate capacitor with plate area of 5 cm2 and plate separation of 3 mm has a voltage 50 sin 103 t V applied to its plates. Solution: D E Jd V d D dV t d dt Hence. Id Jd S S dV d dt C dV dt which is the same as the conduction current. 4 cos 103 t nA Dept.Field Theory 10ES36 Ic d dQ dD dE S dV dV S s S S C dt dt dt dt d dt dt Id 2 10 9 5 10 4 10 3 50 cos 10 3 t 3 36 3 10 = 147. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 134 . v v t = vector differential operator (1/m) ax ay az x y z Proof: Consider a closed surface enclosing a charge Q.Field Theory 10ES36 EQUATION OF CONTINUITY FOR TIME VARYING FIELDS Equation of continuity in point form is . Dept. Of ECE/SJBIT Page 135 . J = conduction current density (A/M2) P = volume charge density (C/M3). There exists an outward flow of current given by I J dS S This is equation of continuity in integral form. J = -v where. J d By definition. J d . we have I J dS S dQ dt From the divergence theorem. = volume charge density (C/m3) So. we have I J dS J d S v Thus. d d t t Dept. where . Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 136 . Q d dQ dt where .Field Theory 10ES36 From the principle of conservation of charge. Of ECE/SJBIT Page 137 . J = . .B=0 B t xH=J+ monopole E dl t B dS L Faraday’s Law s D t L H dl J dS Ampere's circuit law s MAXWELL’S EQUATIONS FOR TIME VARYING FIELDS These are basically four in number. . Maxwell's equations in differential form are given by xH= D +J t Dept. Thus.Field Theory 10ES36 The volume integrals are equal only if their integrands are equal. MAXWELL'S EQUATIONS FOR STATIC EM FIELDS Differential (or Integral Form Point) Form S D dS v dv . D = v Remarks Gauss's law v x E =- Nonexistence of magnetic SB dS 0 . D = .B = 0 Here. H dL L S D J dS E dL B dS L S D dS d S B dS 0 S Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 138 .Field Theory xE=- 10ES36 B t . (C/m2) (D/t) = displacement electric current density (A/m2) J = conduction current density (A/m 2) E = electric field (V/m) B = magnetic flux density wb/m2 or Tesla (B/t) = time-derivative of magnetic flux density (wb/m2 -sec) B is called magnetic current density (V/m2) or Tesla/sec P = volume charge density (C/m3) Maxwell's equations for time varying fields in integral form are given by . H = magnetic field strength (A/m) D = electric flux density. Of ECE/SJBIT 10ES36 Page 139 .Field Theory Dept. The fourth law states that the total magnetic flux passing through any closed surface is zero. 4. The third law states that the total electric displacement flux passing through a closed surface (Gaussian surface) is equal to the total charge inside the surface. The first Maxwell's equation states that the magnetomotive force around a closed path is equal to the sum of electric displacement and. 3.Field Theory 10ES36 DEFINATIONS OF MAXWELL'S EQUATIONS 1. conduction currents through any surface bounded by the path. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 140 . 2. The second law states that the electromotive force around a closed path is equal to the inflow of magnetic current through any surface bounded by the path. MAXWELL’S EQUATIONS FOR STATIC FIELDS Maxwell’s Equations for static fields are: H J H dL J dS L S E 0 E dL 0 L Dept. xH=. 0. RHS = . that is.Field Theory 10ES36 D D dS d S B 0 B dS 0 S As the fields are static. Of ECE/SJBIT Page 141 . all the field terms which have time derivatives are zero. t PROOF OF MAXWELLS EQUATIONS 1. D = t B = 0. From Ampere's circuital law. we have xH=J Take dot product on both sides . J = 0 Dept.J As the divergence of curl of a vector is zero. F + (-) = 0 Dept.J from the equation of continuity in the above expression. Hence we can write xH=J+F Take dot product on both sides . .xH=.J+. As the equation of continuity is more fundamental.Field Theory 10ES36 But the equation of continuity in point form is J t This means that if x H = J is true.xH=0=. we get .F that is.F Substituting the value of . Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 142 . Ampere's circuital law should be modified.J+. it is resulting in . J = 0. . 2.D The divergence of two vectors are equal only if the vectors are identical.F=. D = or. D = From the above expressions.Field Theory or. F = D So. xH=D+J Hence proved. Of ECE/SJBIT Page 143 . According to Faraday's law. we get . Dept. 10ES36 . that is. F = - The point form of Gauss's law is . B S B t Applying Stoke's theorem to LHS. (wb) and by definition. we get E dL E dS L S Dept.Field Theory 10ES36 emf d dt = magnetic flux. emf E dL L E dL L But d dt B dS S B E dL t dS L S B dS . Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 144 . 3. Of ECE/SJBIT Page 145 . we have D dS Q d S Applying divergence theorem to LHS. D = Dept. x E = .B Hence proved. that is. . that is. From Gauss's law in electric field.Field Theory 10ES36 E dS B dS S S Two surface integrals are equal only if their integrands are equal. we get D dS Dd d S Two volume integrals are equal if their integrands are equal. we get B d 0 or. in free space. determine D. We have Gauss's law for magnetic fields as B dS 0 S RHS is zero as there are no isolated magnetic charges and the magnetic flux lines are closed loops.y) ay V/m.y) ay. B and H. Solution: E = 10 sin (t . Applying divergence theorem to LHS. PROBLEM 1: Given E = 10 sin (t . V/m Dept. 4.Field Theory 10ES36 Hence proved. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 146 . .B=0 Hence proved. C/m2 Second Maxwell’s equation is x E = -B ax That is. x E becomes Dept.y) ay. Of ECE/SJBIT Page 147 .z) V/m E y x 0 Now.Field Theory 10ES36 D = 0 E.854 x 10-12 F/m D = 100 sin (t . 0 = 8. As ay y Ey az z 0 E a x E y 0 a z E y z x Ey = 10 sin (t . E x 0 or. wb / m 2 10 0 sin t z a z . 0 We have B/t = - x E Dept. find the magnetic field.Field Theory 10ES36 E E y z ax = 10 cos (t . E of an electromagnetic wave in free space is given by E = 2 cos t Solution: z a y V/m.z) ax B t B 10 cost z dt a x or B and H 10 B 0 sin t z a z . Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 148 . H. A / m PROBLEM 2: If the electric field strength. H ax 2 0 z cos t a x 0 0 2 B 0 z cos t a x 0 0 0 2 z cos t a x 0 0 2 Dept. H Thus.Field Theory 10ES36 ax x 0 ay y Ey az z 0 a x E y a y (0) a z E y x z E y z z sin t a x 0 0 2 B or. B or. Of ECE/SJBIT z dt a x 0 sin t 0 0 0 0 120 0 0 1 Page 149 . 0 cos (t . V/m.0 x 103 (mho)/cm. V/m The displacement current density Jd D D t Dept.y) az. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 150 . If the same field exists in a medium whose conductivity is given by 2. find the conduction current density.0 cos (t . determine the displacement current density. V/m Electric flux density D = 0 E = 5 0 cos (t .Field Theory 10ES36 H 1 z cos t a x A / m 60 0 PROBLEM 3: If the electric field strength of a radio broadcast signal at a TV receiver is given by E = 5. Solution: E at a TV receiver in free space = 5.y) az.y) az. y) az Jc = 106 cos (t . Of ECE/SJBIT Page 151 .Field Theory 10ES36 5 0 cost y a z t Jd = -50 sin (t .y) az V/m2 Dept. Jc = E = 2.y) az.0 x 103 (mho) /cm = 2 x 105 mho /m Jc = 2 x 105 x 5 cos (t . V/m2 The conduction current density. 2. Poynting’s theorem and wave power. 2nd edition. and John A Buck. Electromagnetics with Applications. David K Cheng. Tata McGraw-Hill. Field and Wave Electromagnetics.Reprint 2002 4. 1999 3. Dept. 1968.Field Theory 10ES36 UNIT -7: UNIFORM PLANE WAVES Wave propagation in free space. Edward C. Prentice – Hall of India / Pearson Education. Electromagnetic Waves And Radiating Systems. Pearson Education Asia. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 152 . 2nd edition.1989. Indian Reprint – 2001. Jordan and Keith G Balmain. Energy Electromagnetics. Propagation in good conductors Recommended readings: 1. John Krauss and Daniel A Fleisch McGraw-Hill. William H Hayt Jr . 5th edition.2006. Wave propagation in dielectrics. . 7th edition. E = Em cos t E = Em sin t Here. y. t Re E r e jt ________ (11) The symbol ‘tilda’ placed above the E vector represents that E is time – varying quantity. Therefore we consider fields having sinusoidal time variations. t . in terms of corresponding phasor quantity E (r) as E r . the electric field E can be represented as E x. Of ECE/SJBIT Page 153 . any periodic variation can be represented as a weight sum of fundamental and harmonic frequencies. z . w = 2f. Therefore every field or field component varies sinusoidally. for example. y. most generators produce voltage and currents and hence electric and magnetic fields which vary sinusoidally with time. z Where E is the time varying field. Representing sinusoidal variation. f = frequency of the variation.. mathematically by an additional term. r x. The time varying electric field can be equivalently represented.Field Theory 10ES36 UNIT -7 UNIFORM PLANE WAVES Sinusoidal Time Variations: In practice. For example. Further. t as ie. The phasor notation: Dept. E r . Field Theory 10ES36 We consider only one component at a time, say Ex. The phasor Ex is defined by r, t R E r e jt ________ (12) E x e x | Ex | | Ex | t E x Ex r denotes Ex as a function of space (x,y,z). In general Ex r is complex and hence can be represented as a point in a complex and hence can be represented as a point in a complex plane. (see fig) Multiplication by e jwt results in a rotation through an angle wt measured from the angle . At t increases, the point Ex e jwt traces out a circle with center at the origin. Its projection on the real axis varies sinusoidally with (varying sinusoidally time & we get the time-harmonically varying electric field Ex with time). We note that the phase of the sinusoid is determined by , the argument of the complex number Ex. Therefore the time varying quantity may be expressed as E x Re Ex e j e jt ________ (13) Ex cos( t ) ________ (14) Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 154 Field Theory 10ES36 Maxwell’s eqn. in phasor notation: In time – harmonic form, the Maxwell’s first curl eqn. is: D H J _______ (15) t using phasor notation, this eqn. becomes, Re He jt Re De jt Re Je jt ________ (16) t The diff. Operator & Re part operator may be interchanged to get, j t j t Re He j t Re De Re Je t j t j t Re j D e Re Je Re H j D J e j t 0 This relation is valid for all t. Thus we get H J j D ________ (17) This phasor form can be obtained from time-varying form by replacing each time derivative by jw ie., is to be replaced by t For the sinusoidal time variations, the Maxwell’s equation may be expressed in phasor form as: Dept. Of ECE/SJBIT Page 155 Field Theory 10ES36 (17) H J j D (18) E j B (19) D (20) B 0 L H dL J j D ds S E dl j B ds D ds B ds 0 L S S V dV V S The continuity eqn., contained within these is, J j S J ds j dv _______ (21) vol The constitutive eqn. retain their forms: D E B H J E ____ (22) For sinusoidal time variations, the wave equations become E 2 E 2 H 2 H 2 ( for electric field ) ( for electric field ) _________ (23) Vector Helmholtz eqn. In a conducting medium, these become 2 E 2 j E 0 ________ (24) 2 H 2 j H 0 Wave propagation in a loss less medium: In phasor form, the wave eqn. for VPW is Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 156 Field Theory 10ES36 2 E 2 2 E Ey 2 2 E y _______ (25) x ; 2 x 2 E E y C1 e j x C2 e j x _______ (26) C1 & C2 are arbitrary constants. The corresponding time varying field is E y x, t Re E y x e j t Re C1 e j t z C2 e j t z ______ (27) C1 cos t z C2 cos t z ______ (28) When C1 and C2 are real. Therefore we note that, in a homogeneous, lossless medium, the assumption of sinusoidal time variations results in a space variation which is also sinusoidal. Eqn. (27) and (28) represent sum of two waves traveling in opposite directions. If C1 = C2 , the two traveling waves combine to form a simple standing wave which does not progress. If we rewrite eqn. (28) with Ey as a fn of (x-t), we get = Let us identify some point in the waveform and observe its velocity; this point is t x a constant Then dx dt ' a ' t x t This velocity is called phase velocity, the velocity of a phase point in the wave. Dept. Of ECE/SJBIT Page 157 Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 158 .. Wavelength: These distance over which the sinusoidal waveform passes through a full cycle of 2 radians ie.Field Theory 10ES36 is called the phase shift constant of the wave. 2 E 2 E 0 Where 2 2 j j j is called the propagation constant is. Of ECE/SJBIT Page 159 .Field Theory 10ES36 2 2 2 or But or f . in general. for UPW of electric field strength is 2 E 2 E x 2 One possible solution is E x E0e x Therefore in time varying form. we get Dept. Therefore. complex. : 2 f f in H Z 1 0 Wave propagation in a conducting medium We have. The eqn. = + j = Attenuation constant = phase shift constant. H E j E J c J disp where J c E conduction current density ( A/m2 ) J disp j E displacement current density ( A/m2 ) J cond J disp We can choose a demarcation between dielectrics and conductors. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 160 . Dept. The phase shift factor 2 and velocity f = Real part of = RP = j jt 2 1 2 2 1 2 2 1 2 2 1 2 Conductors and dielectrics: We have the phasor form of the 1st Maxwell’s curl eqn. shown that a up wave traveling in the +x direction and attenuated by a factor e x . t Re E e x e jt e x Re E0e jwt This eqn.Field Theory 10ES36 E x. 0002 @ audio and RF * For good conductors. Of ECE/SJBIT Page 161 .5*108 @ 30 GHz * 1 is dielectric. & are function of freq.Field Theory 10ES36 1 * 1 is conductor.1 a) Express E y 100 cos 2 108 t 0.5 z 300 v / m as a phasor E y Re 100 e j 2 10 6 t 0. PF = sin = tan-1D PF & D difference by <1% when their values are less than 0. & are independent of freq. * For most dialectics. Mica: 0.5 z 300 Drop Re and suppress ejwt term to get phasor Dept. is relatively constant over frequency range of interest * Therefore dielectric “ constant “ * dissipation factor D if D is small.15. dissipation factor is practically as the power factor of the dielectric. Example 11. Cu: 3. 5 z 30 10ES36 0 Whereas Ey is real. Eys is in general complex. the procedure is still effective.2 Given ˆ 20 500 ay ˆ 40 2100 az ˆ .5z is in radians. Example 11.Field Theory Therefore phasor form of Eys = 100e0. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 162 .y or z. Even if so. b) Consider Dept. Note: 0.V / m Es 100e j 30 ax E Re Es e j t j t 300 j t 500 j t 2100 Re 100e 20e 40e V /m E 100 cos t 300 20 cos t 500 40 cos t 2100 V / m None of the amplitudes or phase angles in this are expressed as a function of x. 300 in degrees.V / m Es 100 300 ax find its time varying form representation Let us rewrite Es as 0 0 0 ˆ 20e j 50 ay ˆ 40e j 210 az ˆ . 2 at t 20 ns. a) From given data. 4. Of ECE/SJBIT Page 163 . y . y .1 at t 10 ns. 3.Field Theory 10ES36 ˆ A/ m H s 20e 0. z Note : consider Ex Re E x x.4 x V / m E0 s 500 400 ay Find a b E at 2.1 j 20 z ax 0. Dept.1 at t 0 c E at 2. d E at 3.1 j 20 z ˆ H t Re 20e ax e j t ˆ A/ m 20e 0.1z cos t 20 z ax E x E x x. Example Given ˆ 200 j 600 az ˆ e j 0. z t t Re j Ex e j t e j t Therefore taking the partial derivative of any field quantity wrt time is equivalent to multiplying the corresponding phasor by j . 3. 076 az ˆ V /m 36.4 x 71.4 x 71.565 az ˆ 500 cos t 0.4 x ay 0 500e j 0.4 2 71. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 164 .4 0 0 0.565 az ˆ E at 2.456 cos t 0.5650 ˆ 632.4 x 400 ay ˆ 632.4 2 400 ay ˆ 632.456e j 71.565 ay ˆ 632.4 x 71.5650 az ˆ 417.456 0.1 t 0 500 cos 0. 3.565 e j 0.4 x Es 500 400 ay ˆ 632.456e ˆ az 0 j 0.1 ˆ 500 cos 120 106 10 109 0.823 ay d) Dept.Field Theory 10ES36 0.4 x 400 j 0.4 x az ˆ 500e j 40 e j 0.3.4 x 400 j 0.4 x 71.4 3 108 4 107 10 9 36 9 120 106 f 19.473 az ˆ V /m 477.456 cos 120 106 10 109 0.297 ay c) E at t 10 ns at 2.456 e j t e ˆ E t 500 Re e j t e ay az ˆ 632.4 x 400 ay ˆ 291.1 106 Hz b) Given ˆ 200 j 600 az ˆ e j 0. 736 ay D 11.2. Of ECE/SJBIT Page 165 .07 z Given H s 2 400 ax A/ m for a UPW traveling in free space.07 0.3. (c) H at t=0 at the origin. Find (a) (b) Hx at p(1.07 3 108 21.644 az ˆ V /m 438.1 ˆ 631.0 106 rad / sec 21.2: ˆ 320 ay ˆ e j 0.3) at t = 31 ns.Field Theory 10ES36 at t = 20 ns. E at 2. (a) we have p = 0.07 (e j z term) 0.07 0.0 106 rad / sec (b) Dept. t Ey we have Hx Hx 120 Ey 120 ˆ sin t z ay 120 120 1 sin t z H z.07 z 400 ax H x (t ) 2 cos t 0. 2.07 z 400 H x (t ) at p 1.3 ay ˆ H t 2 cos 0.21 400 At t 31n sec. t 120 sin t z ay find H V /m z.20666 A / m In free space. 2 cos 2.9333 A/ m (c) ˆ 3cos 0.07 z ax ˆ 3 cos t 0.35 ay ˆ H t at t 0 2 cos 0.7 ax ˆ 2.07 z 200 ay ˆ 2 cos t 0.7 ax ˆ 3cos 0.07 z ay ˆ e j t H t Re 2 e j 40 e j 0.Field Theory 10ES36 0 0 ˆ 3 e j 20 e j 0.21 400 2 cos 651103 0.53ax 3.82ay ˆ 1. ˆ E z . 3 2 cos 2.1106 t 0. t 1 ˆ sin t z ax Dept.1106 3110 9 0.21 400 1. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 166 .07 z 0.7 z 0. Field Theory 10ES36 Problem 3. Show that the function F e z sin x t satisfies the wave equation 2 F 1 2 F c 2 t 2 provided the wave velocity is given by 2c 2 e 1 2 Ans: From the given eqn. 2 F 2 F F x 2 y 2 F e z cos x t x 2 F 2 e z z e F x t sin 2 x 2 F e z sin x t z 2 F 2 z e sin x t 2 F 2 z 2 Dept. J&B Non uniform plans waves also can exist under special conditions. Of ECE/SJBIT Page 167 . for F. we note that F is a function of x and z. Dept.Field Theory 10ES36 2 F 2 2 F dF e z cos x t dt d 2F e z sin x t 2 dt 2 2 F The given wave equation is 1 2 F c 2 t 2 2 1 2 F 2 c 2 F 2 F 2 2 2 2 c2 2 2 2 2 2 c 2 2 2 2 2 2 c 2c 2 2c 2 2 c2 2 or 1 2c 2 2 c 2c 2 1 2 Example The electric field intensity of a uniform plane wave in air has a magnitude of 754 V/m and is in the z direction. If the wave has a wave length = 2m and propagating in the y direction. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 168 . 14 rad / m 2m Ez 754 cos 2 150 106 t y e (ii) For a wave propagating in the +y direction. In air or free space. Of ECE/SJBIT Page 169 . E Ez x Hz Hz For the given wave. c 3 108 m / sec (i) 3 108 f m / sec 1. Dept.5 108 Hz 150 MHz 2m 2 2 3.Field Theory 10ES36 Find (i) (ii) Frequency and when the field has the form A cos t z . Find an expression for H . 3MHz. the wave equation in phasor form may be expressed as 2 E 2 E Z 2 0r Dept.3459 10 3 m 50 66 103 (ii ) 3. the uniform plane wave is defined as one for which the equiphase surface is also an equiamplitude surface.8105 10 7 m 6 3 10 (i ) Wave Propagation in a loss less medium: Definition of uniform plane wave in Phasor form: In phasor form.8*107 (/m) at 50Hz. it is a uniform plane wave.8105 10 5 m 3 106 66 103 (iii ) 3.2 2 f f 66 10 3 9.Field Theory 10ES36 Ez 754 V / m. For a uniform plane wave having no variations in x and y directions. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP 2 E 2 E ________ (i ) Z 2 Page 170 .8 10 f 1 1 4 2 5.8 f 1 66 10 3 23. 30GHz. 2 1 1 f 1 1 1 7 7 4 10 5. Ex 0 754 754 H x 754 A/ m 120 377 ˆ H 2 cos 2 150 106 t y ax A/ m Example find for copper having = 5. If C1 = C2. The corresponding time varying form of E y is E y z . Let us consider eqn. the two wave combine to form a standing wave which does not progress. Phase velocity and wavelength: The wave velocity can easily obtained when we rewrite Ey as a function and z t . t C1 cos t z C2 cos t z _______ (4) From (3) we note that. the Ey component. (4). we get 2 Ey Z 2 2 Ey E y has a solution of the form.(i) for. sinusoidal time variation results in space variations which is also sinusoidal. the result of real part extraction operation is. Equations (3) and (4) represent sum of two waves traveling in opposite directions. Ey C1e j z C2e j z ________ (2) Where C1 and C2 are arbitrary complex constants. t Re E y z e j t Re C1 e j z C2 e j z e jt _______ (3) If C1 and C2 are real.Field Theory 10ES36 where . Ey z. as in eqn. Of ECE/SJBIT Page 171 . in a homogeneous lossless medium. This shows that Dept. this point is given by t z a constant. 2 2 2 2 1 . written in the form of Helmholtz eqn. is Dept. dz . 1 0 _______(9) C 3 108 m / sec Wave propagation in conducting medium: The wave eqn. identifying a some reference point on the waveform and observing its velocity may obtain the same result. f in Hz ________(8) For the value of given in eqn. For a wave traveling in the +Z direction. as in eqn.. 0 C .Field Theory 10ES36 ________(5) In phasor form. (1). Wavelength: Wavelength is defined as that distance over which the sinusoidal waveform passes through a full cycle of 2 radius. is called the phase-shift constant and is a measure of phase shift in radians per unit length. the phase velocity is. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 172 . ie. ________(7) 2 f f f . (5) dt This velocity of some point on the sinusoidal waveform is called the phase velocity. Field Theory 10ES36 2 E 2 E 0 _______(10) where 2 2 j j j _______(11) . Of ECE/SJBIT j j ________(11) Page 173 . t Re E0 = e z Re E0 e Z e jt _______(15) e j t z ________(16) This is the equation of a wave traveling in the +Z direction and attenuated by a factor e Z . for the uniform plane wave traveling in the z direction. Dept. are given by 2 f The propagation constant We have. the propagation constant is complex = + j _________(12) We have. as in the lossless case. the electric field E must satisfy 2 E 2 E _______(13) Z 2 This equation has a possible solution E Z E0e Z _______(14) In time varying form this is becomes E z. . velocity. The phase shift factor and the wavelength phase. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 174 .. Dept. 2 2 2 ________(18) ________(19) 2 Therefore (19) in (18) gives: 2 2 4 4 4 4 2 2 2 2 2 0 2 4 2 2 2 2 2 2 4 0 2 4 2 2 2 2 2 2 2 2 1 2 2 2 2 2 1 1 2 2 2 2 2 2 1 2 2 2 1 _________(20) and 2 2 1 2 2 1 ___________(21) We choose some reference point on the wave. the cosine function. the cosine is an integer multiple of 2 at erest.(say a rest).Field Theory 10ES36 2 j 2 2 j 2 j 2 ________(17) 2 2 2 2 . The value of the wave ie. Let us further consider only +ve z traveling wave: We have iˆ 0 x Ex Dept. nothing that the entire cosine argument is the same multiple of 2 for all time in order to keep track of the point. ie.. Now let us fix our position on the wave as this mth erest and observe time variation at this position. ( ) having a cosine argument t 0 z describes a wave that moves in the negative direction (as + increases z must decrease to keep the argument constant).Field Theory 10ES36 k0 z 2m at mth erest. Of ECE/SJBIT ˆj 0 y Ey kˆ z 0 Page 175 . position z must also increase to satisfy eqn. Thus the wave erest (and the entire wave moves in a +ve direction) with a speed given by the above eqn. Similarly. t k0 0 z 2m t z / c Thus at t increases. ( ). These two waves are called the traveling waves. eqn. Therefore also called a TEM wave. ˆ direction is 250 V/m. E and H are perpendicular to the direction of propagation.1. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 176 . both lie in a plane that is transverse to the direction of propagation. The UPW is called so because is uniform thought any plane Z = constant. Dept. If E 11. Energy flow is in +Z direction. t Ex 0 Ex Hy 0 .Field Theory 10ES36 Es j H s E y Ex i kˆ0 j iH 0 x j by j z z Exs j H 0 y z 0 1 H oy Ez 0 e jk0 z E x 0 e j 0 z j 0 0 H y z . find (i) f (ii) (iii) period (iv) amplitude of H = Ex ax . cos t 0 z 0 377 120 Ey and Hx are in phase in time and space. The electric field amplitude of a UPW in the az ˆ and = 1m rad/sec. 2.155 KHz 2 2 2 C 1.Field Theory 10ES36 2 f 106 f 159. the wave equation becomes 2 Es k 2 s k k0 r r 0 r r For Ex component We have d 2 Exs k 2 Exs 2 dz for Ex comp.2.3) at t = 31ns and (iii) H at t = 0 at the orign. Of ECE/SJBIT Page 177 . Wave propagation in dielectrics: For an isotopic and homogeneous medium. Find (i).283 s f E amplitude of H y x 120 Hy Hy Ex 250 0. k can be complex one of the solutions of this eqn.6631 A / m 120 120 11. (ii)Hx at p(1. Of electric field wave traveling in Z – direction.88495 km f 1 period 6. Dept. Given ˆ 3200 ay ˆ e j 0. is.07 z A / m H s 2 400 ax for a certain UPW traveling in free space. Exs Ex 0e z cos t z This is UPW that propagates in the +Z direction with phase constant but losing its amplitude with increasing Z e z . becomes for sinusoidal time variations: 2 E 2 j E 0 Problem: Dept. the wave eqn.Field Theory 10ES36 jk j Exs Ex 0 e z e j z Therefore its time varying part becomes. Wave propagation in a conducting medium for medium for time-harmonic fields: (Fields with sinusoidal time variations) For sinusoidal time variations. Thus the general effect of a complex valued k is to yield a traveling wave that changes its amplitude with distance. If is +ve = attenuation coefficient if is +ve wave decays If is -ve = gain coefficient In passive media. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 178 . is +ve wave grows is measured in repers per meter In amplifiers (lasers) is –ve. the electric field for lossless medium ( = 0) becomes 2 E 2 E In a conducting medium. Of ECE/SJBIT Page 179 . & are constants and of finite values and 0 D 0 Dept.D 0 in a conductor if ohm’s law and sinusoidal time variations are assumed. (1) show that . we get. the first Maxwell’s curl equation is H E j E Taking divergence on both sides. When ohm’s law and sinusoidal time variations are assumed. H E j E 0 E j 0 or D j 0 .Field Theory 10ES36 Using Maxwell’s eqn. it is a uniform plane wave. the uniform plane wave is defined as one for which the equiphase surface is also an equiamplitude surface. the Ey component. sinusoidal time variation results in space variations which is also sinusoidal. Dept. The corresponding time varying form of E y is E y z .(i) for. Ey z. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 180 . the wave equation in phasor form may be expressed as 2 E 2 E Z 2 2 E 2 E ________ (i ) Z 2 0r where . we get 2 Ey Z 2 2 Ey E y has a solution of the form. Let us consider eqn.Field Theory 10ES36 Wave Propagation in a loss less medium: Definition of uniform plane wave in Phasor form: In phasor form. in a homogeneous lossless medium. For a uniform plane wave having no variations in x and y directions. the result of real part extraction operation is. t C1 cos t z C2 cos t z _______ (4) From (3) we note that. Ey C1e j z C2e j z ________ (2) Where C1 and C2 are arbitrary complex constants. t Re E y z e j t Re C1 e j z C2 e j z e jt _______ (3) If C1 and C2 are real. 2 2 2 2 1 . (5) dt This velocity of some point on the sinusoidal waveform is called the phase velocity. as in eqn. Phase velocity and wavelength: The wave velocity can easily obtained when we rewrite Ey as a function and z t .. as in eqn. identifying a some reference point on the waveform and observing its velocity may obtain the same result.Field Theory 10ES36 Equations (3) and (4) represent sum of two waves traveling in opposite directions. Of ECE/SJBIT Page 181 . This shows that ________(5) In phasor form. the two wave combine to form a standing wave which does not progress. For a wave traveling in the +Z direction. f in Hz ________(8) Dept. ________(7) 2 f f f . ie. (4). Wavelength: Wavelength is defined as that distance over which the sinusoidal waveform passes through a full cycle of 2 radius. dz . this point is given by t z a constant. is called the phase-shift constant and is a measure of phase shift in radians per unit length. If C1 = C2. the electric field E must satisfy 2 E 2 E _______(13) Z 2 This equation has a possible solution E Z E0e Z _______(14) In time varying form this is becomes E z. 0 C .Field Theory 10ES36 For the value of given in eqn. 1 0 _______(9) C 3 108 m / sec Wave propagation in conducting medium: The wave eqn. t Re E0 = e z Re E0 e Z e jt _______(15) e j t z ________(16) Dept. the phase velocity is. the propagation constant is complex = + j _________(12) We have. is 2 E 2 E 0 _______(10) where 2 2 j j j _______(11) . Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 182 . . (1). written in the form of Helmholtz eqn. for the uniform plane wave traveling in the z direction. as in the lossless case.Field Theory 10ES36 This is the equation of a wave traveling in the +Z direction and attenuated by a factor e Z . Of ECE/SJBIT Page 183 . 2 2 2 ________(18) ________(19) 2 Therefore (19) in (18) gives: Dept. j j ________(11) 2 j 2 2 j 2 j 2 ________(17) 2 2 2 2 . velocity. The phase shift factor and the wavelength phase. are given by 2 f The propagation constant We have. Dept. the cosine is an integer multiple of 2 at erest. ie.. the cosine function.Field Theory 10ES36 2 2 4 4 4 4 2 2 2 2 2 0 2 4 2 2 2 2 2 2 4 0 2 4 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 1 1 2 2 2 2 1 1 _________(20) 2 2 2 and 2 1 2 2 1 ___________(21) 2 We choose some reference point on the wave. k0 z 2m at mth erest. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 184 . ( ).(say a rest). The value of the wave ie. nothing that the entire cosine argument is the same multiple of 2 for all time in order to keep track of the point.. t k0 0 z 2m t z / c Thus at t increases. Now let us fix our position on the wave as this mth erest and observe time variation at this position. Thus the wave erest (and the entire wave moves in a +ve direction) with a speed given by the above eqn. position z must also increase to satisfy eqn. Field Theory 10ES36 Similarly. Of ECE/SJBIT 0 . Let us further consider only +ve z traveling wave: We have iˆ 0 x Ex kˆ z 0 ˆj 0 y Ey Es j H s E y Ex i kˆ0 j iH 0 x j by j z z Exs j H 0 y z 0 1 H oy Ez 0 e jk0 z E x 0 e j 0 z j 0 0 H y z . ( ) having a cosine argument t 0 z describes a wave that moves in the negative direction (as + increases z must decrease to keep the argument constant). t Ex 0 Ex Hy Dept. cos t 0 z 0 377 120 Page 185 . These two waves are called the traveling waves. eqn. 283 s f E amplitude of H y x 120 Hy Hy Ex 250 0. ˆ direction is 250 V/m.07 z A / m H s 2 400 ax for a certain UPW traveling in free space. Find (i). find (i) f (ii) (iii) period (iv) amplitude of H = Ex ax .155 KHz 2 2 2 C 1. The UPW is called so because is uniform thought any plane Z = constant. Wave propagation in dielectrics: For an isotopic and homogeneous medium.1. If E 11.6631 A / m 120 120 11. The electric field amplitude of a UPW in the az ˆ and = 1m rad/sec. Energy flow is in +Z direction.3) at t = 31ns and (iii) H at t = 0 at the orign. both lie in a plane that is transverse to the direction of propagation. Therefore also called a TEM wave. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 186 . E and H are perpendicular to the direction of propagation.3.Field Theory 10ES36 Ey and Hx are in phase in time and space.2. the wave equation becomes Dept. (ii)Hx at p(1. 2 f 106 f 159. Given ˆ 3200 ay ˆ e j 0.88495 km f 1 period 6. jk j Exs Ex 0 e z e j z Therefore its time varying part becomes. is. If is +ve = attenuation coefficient if is +ve wave decays If is -ve = gain coefficient In passive media. Exs Ex 0e z cos t z This is UPW that propagates in the +Z direction with phase constant but losing its amplitude with increasing Z e z . Wave propagation in a conducting medium for medium for time-harmonic fields: Dept. Thus the general effect of a complex valued k is to yield a traveling wave that changes its amplitude with distance. Of electric field wave traveling in Z – direction. k can be complex one of the solutions of this eqn. Of ECE/SJBIT Page 187 . is +ve wave grows is measured in repers per meter In amplifiers (lasers) is –ve.Field Theory 10ES36 2 Es k 2 s k k0 r r 0 r r For Ex component We have d 2 Exs k 2 Exs 2 dz for Ex comp. (1) show that . H E j E 0 E j 0 or D j 0 . the first Maxwell’s curl equation is H E j E Taking divergence on both sides. the wave eqn. & are constants and of finite values and 0 D 0 POLARISATION: Dept. the electric field for lossless medium ( = 0) becomes 2 E E 2 In a conducting medium. becomes for sinusoidal time variations: 2 E 2 j E 0 Problem: Using Maxwell’s eqn. When ohm’s law and sinusoidal time variations are assumed.D 0 in a conductor if ohm’s law and sinusoidal time variations are assumed.Field Theory 10ES36 (Fields with sinusoidal time variations) For sinusoidal time variations. we get. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 188 . have equal magnitudes and a 900 phase and Ey In the particular case where Ex difference.. then the direction of the resultant electric vector will vary with time. they reach their maxima at different instances of and Ey If Ex time. Consider of a UPW traveling along Z direction with E and H vectors lying in the xy plane. are not in phase ie. and Ey field E has a direction that depends on the relative magnitudes of Ex The angle which this resultant direction makes with the x axis is tan-1 Ey . the locus of the resultant E is a circle and the wave is circularly polarized. In this case it can be shown that the locus of the end point of the resultant E will be an ellipse and the wave is said to be elliptically polarized. is present. then the resultant electric and Ey If both Ex . the wave is said to be polarized in the xIf Ey direction. the wave is said to be polarized in the y = 0 and only Ey If Ex direction. 0 and only Ex is present. Of ECE/SJBIT Page 189 . Linear Polarisation: Dept. and this Ex angle will be constant with time. Therefore the direction of E is the direction of polarization are present and are in phase. the direction of the resultant vector is constant with time and the wave is said to be linearly polarized.Field Theory 10ES36 It shows the time varying behavior of the electric field strength vector at some point in space. (a) Linear polarization: In all the above three cases. Dept. we have.. in general. In general. E ax The corresponding time varying version is. Therefore E z lies in the x-y plane. E0 Er jE0i Where E0 and E0i are real vectors having. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 190 .. If Ey leads Ex by 900 and Ex and Ey have the same amplitudes. At some point in space.Field Theory 10ES36 Consider the phasor form of the electric field of a UPW traveling in the Z-direction: E Z E0 e j z Its time varying or instanious time form is E Z . Circular Polarisation: Here the x and y components of the electric field vector are equal in magnitude. a vector whose components are complex numbers. Therefore we can write E0 as. different directions. E0 is a complex vector ie. ˆ j ay ˆ E0 Ie. Ex E y . t Re E0e j z e jt The wave is traveling in Z-direction. t Re E0 r j E0i e j t E0 r cos t E0i sin t Therefore E not only changes its magnitude but also changes its direction as time varies. (say z = 0) the resultant time varying electric field is E 0. t traces a circle of radius E0 as time progresses. Then.Field Theory 10ES36 ˆ cos t ay ˆ sin t E0 E 0. ˆ A j ay ˆ B E0 ax Where A and B are +ve real constants. Of ECE/SJBIT Page 191 . Its time varying form is Dept. Therefore the wave is said to the circularly polarized. t ax Ex E0 cos t and E y E0 sin t Ex2 E y2 E02 Which shows that the end point of E 0 0. in the direction of propagation). Assume that Ey leads Ex by 900. Elliptical Polarisation: Here x and y components of the electric field differ in amplitudes E x E y . Similar remarks hold for a right-circularly polarized wave represented by the complex vector. Then this wave is said to be left circularly polarized. Further we see that the sense or direction of rotation is that of a left handed screw advancing in the Z-direction ( ie. ˆ j ay ˆ E0 E ax It is apparent that a reversal of the sense of rotation may be obtained by a 180 0 phase shift applied either to the x component of the electric field.. SWR. Jordan and Keith G Balmain. Dept. and John A Buck. Plane wave propagation in general directions Recommended readings: 1.2006.1989. Prentice – Hall of India / Pearson Education. Field and Wave Electromagnetics. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 192 . David K Cheng. Pearson Education Asia. Tata McGraw-Hill. Electromagnetics with Applications. t axA E A cos t x E y B sin t E x2 A2 E y 2 B2 1 Thus the end point of the E 0. Indian Reprint – 2001. 1999 3. John Krauss and Daniel A Fleisch McGraw-Hill. The polarization is completely specified by the orientation and axial ratio of the polarization ellipse and y the sense in which the end point of the electric field moves around the ellipse. 2nd edition. 7th edition. t vector traces out an ellipse and the wave is elliptically polarized. 2. .Reprint 2002 4. 1968. William H Hayt Jr . 5th edition. Electromagnetic Waves And Radiating Systems. 2nd edition. UNIT -8: REFLECTION AND REFARACTION OF PLANE WAVES Reflection of uniform plane waves at normal incidence. Elliptical polarization is a more general form of polarization. the sense of polarization is left-handed. Energy Electromagnetics. Reflection of uniform plane waves at normal incidence.Field Theory 10ES36 ˆ cos t ayB ˆ sin t E 0. Edward C. at x = 0. The reflected wave is Er e j x __________(2) Er must be determined from the boundary conditions.Field Theory 10ES36 UNIT -8 REFLECTION AND REFARACTION OF PLANE WAVES REFLECTION BY A PERFECT CONDUCTOR: NORMAL INCIDENCE: When an em wave traveling in one medium impinges upon a second medium having a different . the amplitudes of E and H in the reflected wave are the same as in the incident wave. ie. Let Ei e j x ________(1) be the incident wave. the surface of the perfect conductor be at x = 0. When a plane wave in air is incident normally on the surface of a perfect conductor the wave is for fields that vary with time. (i) (ii) Etan is continuous across the boundary E is zero within the conductor. then the wave will be partially transmitted. Of ECE/SJBIT Page 193 .. As there can be no loss within a perfect conductor. therefore none of the energy is obsorbed. This requires that. Dept. neither E nor H can exist within a conductor. With respect to.. Therefore at the boundary. the sum of the electric field strengths in the initial and reflected waves add to give zero resultant field strength in the plane x = 0. Let the boundary. and partially reflected. therefore no energy of the incident wave is transmitted. Therefore. the electric field is zero. the only difference is in the direction of power flow. or . Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 194 . Dept. It is zero at the surface and at multiples of half wave lengths from the surface. given by ET x Ei e j x Er e j x 2 jEi e j x e j x 2 jEi sin x _______ 4 Its time varying version is ET x . The resultant electric field strength at any point at any point a distance –x from the x = 0 plane is the sum of the field strengths of the incident and reflected wave at that point. (3) shows that (1) the incident and reflected waves combine to produce a standing wave. 3. t Re 2 jEi sin x e j t 2 Ei sin x sin t . if Ei real _______ 5 1. which does not progress. The magnitude of the electric field varies sinusoidally with distance from the reflecting plane. 4.Field Theory 10ES36 Er Ei _______(3) The amplitude of the reflected electric field strength is equal to that of the incident electric field strength but its phase has been reversed on reflection. 2. Eqn. It has a maximum value of twice the electric field strength of the incident wave at distances from the surface that are odd multiples of a quarter wavelength. which is required in this case). (4) and (6) Dept. From the boundary conditions for H its follows that there must be a surface current of Js amperes per such that JS = HT (at x = 0). where as the zero points occur at odd multiples of a quarter wavelength from the surface. H T x. Therefore if follows that H must be reflected without phase reversal. This SWD has maximum value at the surface of the conductor and at multiples of a half from the surface. This is the case when the energy transmitted in the forward direction is equaled by that reflected back. (otherwise if both are reversed. for it indicates no average flow of power. This is as it should be. t Re HT x e jt 2 H i cos x cos t ______ 7 The resultant magnetic field strength H also has a standing was distribution. Of ECE/SJBIT Page 195 .Field Theory 10ES36 In as much as the BCs require that the electric field is reversed in phase on reflection to produce zero resultant field at the boundary surface. on reversal of direction of energy propagation). eqns. (6) and (7) show that ET and HT are 90 0out of time phase because of the factor j in eqn. Since Ei and Hi were in phase in the incident plane wave. Let us rewrite eqns. (4). Therefore the phase of the mag field strength is the same as that of the incident mag field strength Hi at the surface of reflection. HT x H i e j x H r e j x 2 H i e j x e j x 2 H i cos x _______ 6 Hi is real since it is in phase with Ei Further. Field Theory 10ES36 ET x, t Re 2 Ei sin x e j / 2 e jt 2 Ei sin x cos t / 2 _______ 8 HT x, t 2 H i cos x cos t _______ 9 Eqns. (8) and (9) show that ET and HT differ in time phase by 900. REFLECTION BY A PERFECT CONDUCTOR – OBLIQUE INCIDENCE: TWO SPACIAL CASES: 1. Horizontal Polarisation: (also called perpendicular polarization) Here the electric field vector is parallel to the boundary surface, or perpendicular to the plane of incidence. ( Transverse electric TE) 2. Vertical Polarisation: (also called parallel polarization) Here the magnetic field vector is parallel to the boundary surface, and the electric field vector is parallel to the plane of incidence. (Transverse magnetic TM) TE or TM are used to indicate that the electric or magnetic vector respectively is parallel to the boundary surface/plane. When a wave is incident on a perfect conductor, the wave is totally reflected with the angle of incidence equal to the angle of reflection. Case 1: E perpendicular to the plane of incidence: (perpendicular Polarisation) The incident and reflected waves have equal wavelengths and opposite directions along the Z axis, the incident and reflected waves form a standing wave distribution pattern along this axis. Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 196 Field Theory 10ES36 In the y direction, both the incident and reflected waves progress to the right (+y direction) with the same velocity and wavelength and so there will be a traveling wave along the +y direction. The expression for reflected wave, using the above fig, is Erefelected Er e j y sin z cos ______ 8 and Eincident Ei e j y sin z cos ______ 9 j y sin z cos j y sin z cos E Ei e e 2 jEi sin z cos e j y sin 2 jEi sin z z e j y y _______ 10 From the BCs we have, Er = - Ei Therefore total electric field strength E is given by Erefelected Er e j y sin z cos ______ 8 and Eincident Ei e j y sin z cos ______ 9 j y sin z cos j y sin z cos E Ei e e 2 jEi sin z cos e j y sin 2 jEi sin z z e j y y _______ 10 Where, Dept. Of ECE/SJBIT Page 197 Field Theory 10ES36 2 Phase shift constant of the incident wave, z cos = Phase shift constant in the Z direction. y sin = Phase shift constant in the y direction. z 2 2 : wavelength: distance twice between modal points of cos cos the standing wave distribution. The planes of zero electric field strength occur at multiples of z from the reflecting 2 surface. The planes of max electric field strength occurs at odd multiples of z from the 4 surface. The whole standing wave distribution of electric field strength is seen from eqn. (10) above to be traveling in the y direction with a velocity, y y sin sin This is the velocity with which a erest of the incident wave moves along the y axis. The wavelength in this direction is, g sin Case 2: E parallel to the plane of incidence: (parallel polarization) Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 198 Field Theory 10ES36 Here, Ei and Er will have the instantaneous directions shown above, because the components parallel to the perfectly conducting boundary must be equal and opposite. The magnetic field strength vector H will be reflected without phase reversal. The magnitudes of E and H are related by Ei E r Hi H r For the incident wave, the wave expression for the magnetic field strength would be H incident H i e j y sin z cos and for the reflected wave, H reflected H r e j y sin z cos Therefore Hi = Hr The total magnetic field H is, H 2 H i cos z z e where z cos j y y and y sin The magnetic field strength has a standing wave distribution in the Z-direction with the planes of maximum H located at the conducting surface and at multiples of z 2 from the surface. The planes of zero magnetic field strength occur at odd multiples of z from the surface. 4 For the incident wave, Dept. Of ECE/SJBIT Page 199 Field Theory 10ES36 Ei Hi , Ez sin Hi ; Ey cos Hi For the reflected wave, H r Hi , Ez sin H r ; Ey cos H r The total z component of the electric field strength is, Ex 2 sin H i cos z z e j y y The total y component of the electric field strength is, E y 2 j cos H i sin z z e j y y Both Ey and Ez have a standing wave distribution above the reflecting surface. However, for the normal or z components of E , the maxima occur at the plane and multiples of z from the plane, whereas for the component E parallel to the 2 reflecting surface the minima occur at the plane and at multiples of z 2 from the plane. REFLECTION BY PERFECT DIELECTRICS Normal incidence: In this case part of the energy is transmitted and part of the energy is reflected. Perfect dielectric: = 0. no absorption or loss of power in propagation through the dielectric. Boundary is parallel to the x = 0 plane. Plane wave traveling in +x direction is incident on it. Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 200 Field Theory 10ES36 Ei 1 H r We have, Er 1 H r Et 2 H t BC: Tang comp. Of E or H is continuous across the boundary. ie., Hi H r H z Ei Er Et Hi H r 1 Ei Er H z 1 2 Ei Er 1 Ei Er 1 2 Ei Er Ei 2 1 Er 2 1 Er 2 1 Ei 2 1 Also, Et E Er E 21 i 1 r c Ei Ei 1 2 Further , Hr E 2 r 1 Ht E 1 2 H t 1 Et 21 H i 2 Ei 1 2 Dept. Of ECE/SJBIT Page 201 2. Of wave in medium (2) Then from figure. Dept. reflected wave and the incident wave. we get CB 1 AD 2 Now CB = AB sin1 and AD = AB sin2.Field Theory 10ES36 The permeabilities of all known insulators do not differ appreciably from that of free space. Of wave in medium (1) 2 ______ vel. There is a transmitted wave. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 202 . 1 2 Er Ei Et Ei Hr Hi Ht Hi 0 / 2 0 / 2 0 / 1 0 / 1 1 2 1 2 2 1 1 2 2 1 2 1 2 2 1 2 REFLECTION BY PERFECT DIELECTRIC: OBLIQUE INCIDENCE: 1. The transmitted wave is refracted 9direction of propagation is altered) 1 ______ vel. so that. Therefore by conservation of energy we get 1 1 Et2 cos 1 1 1 Et2 cos 1 1 2 Et2 cos 2 Er2 1 Et2 cos 2 1 Et2 2 Ei2 cos 1 1 2 1 Dept. AE = CB sin1 = sin3 or 1= 3 The power transmitted = E2 E and H are perpendicular to each other. Incident power striking AB Reflected power leaving AB The power transmitted = 1 2 1 1 1 2 E12 cos 1 E22 cos 1 Et2 cos 2 .Field Theory 10ES36 2 2 2 1 1 1 sin 1 1 sin 2 2 In addition. Of ECE/SJBIT Et2 cos 2 Ei2 cos 1 Page 203 . According to BCs. Across the boundary. E r2 1 E i2 2 E t2 cos 2 2 1 E i cos 1 E r2 1 E i2 2 1 E 1 r Ei 2 cos 2 cos 1 2 cos 2 cos 1 E r2 1 2 Ei 2 1 E 1 r Ei 1 Er Ei 2 1 E cos 2 1 r E i cos 1 Er Ei 1 cos 1 2 cos 2 1 cos 1 2 cos 2 But we have. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 204 . Ei E r Et Et E 1 r Ei Ei But we have.Field Theory 10ES36 Case 1: Perpendicular polarization (HP): ( E is perpendicular to the plane of incidence parallel to the reflecting surface) Let Ei propagate along +x direction. so as the direction of Er and Et. Etan and Htan are continuous. Dept. we get Ei Er cos 1 Et cos 2 Et E cos 1 1 r Ei Ei cos 2 Dept. . The BCs on tangential components give Htan = Etan is continuous across the boundary. H is parallel to the reflecting surface.Field Theory sin 1 2 sin 2 1 10ES36 2 cos 2 2 1 sin 2 2 2 1 sin 2 1 1 cos 1 2 1 sin 2 1 Er Ei 1 cos 1 2 1 sin 2 1 cos 1 2 sin 2 1 1 cos 1 2 sin 2 1 1 This equation gives the ratio of the reflected to incident electric field strength for the case of a perpendicular polarized wave. Therefore this BC when applied. Of ECE/SJBIT Page 205 . Case II: Parallel Polarisation: Here E is parallel to the plane of incidence. the ratio of reflected to incident electric field strength when E is parallel to the plane of incidence..Field Theory 10ES36 But we already have 2 E t2 cos 2 E r2 1 2 E i2 1 E i cos 1 2 E r2 E r2 cos 2 2 cos 2 1 1 E i2 Ei2 cos 2 1 cos 1 1 E r2 1 2 Ei 2 E r2 cos 1 1 2 Ei cos 2 1 1 Er Ei 2 E cos 1 1 r E i cos 2 1 Er Ei 2 cos 1 1 1 cos 2 Er Ei 2 cos 1 1 1 cos 2 2 cos 1 1 cos 2 2 cos 1 1 cos 2 But from Snell’s law we get 1 sin 2 cos 1 1 1 sin 2 2 2 cos 1 1 2 2 sin 2 2 1 / 2 sin 2 1 Therefore we get Er Ei 2 / 1 cos 1 2 sin 2 1 1 2 / 1 cos 1 2 sin 2 1 1 This equation gives the reflection coefficient for parallel or vertical polarization. Dept. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 206 . ie. but the reflected wave is entirely of perpendicular (or horizontal) polarization. we have. which is called the Bresoster angle.Field Theory 10ES36 BRESNSTER ANGLE: We have Er Ei 2 / 1 cos 1 2 sin 2 1 1 2 / 1 cos 1 2 sin 2 1 1 When Nr = 0. there will he some reflection. Therefore no reflection at all. there is no reflected wave when the incident wave is parallel (or vertically) polarized. Er = 0. 2 cos 1 1 2 sin 2 1 1 22 2 2 cos sin 2 1 1 2 1 1 22 22 2 2 sin sin 2 1 1 2 2 1 1 1 22 22 sin 2 1 12 12 sin 2 1 2 1 1 22 sin 2 1 2 1 2 2 1 2 sin 2 1 2 1 2 sin 2 1 2 1 2 cos 2 1 1 1 2 tan 1 2 1 At this angle. Dept. Therefore for zero reflection condition. Of ECE/SJBIT Page 207 . If the incident wave is not entirely parallel polarized. Dept.. then. We can show that Er tan 1 2 Ei tan 1 2 and for perpendicular polarization.Field Theory 10ES36 Note:1 For perpendicular paolarisation. we get cos 1 2 / 1 sin 2 1 cos 2 1 2 / 1 sin 2 1 or 2 1 ie. we have E Ei cos 1 2 / 1 sin 2 1 cos 1 2 / 1 sin 2 1 putting N r 0. we can show that. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 208 . both the reflection coefficients given by equations. there is no corresponding Bresvster angle for this polarization. Note 2: For parallel polarization. Er sin 2 1 Ei sin 2 1 TOTAL INTERNAL REFLECTION: If 1 2 . Of ECE/SJBIT 1 2 2 sin 2 1 1 sin 2 1 1 Page 209 . e j 2 Z cos2 e j 2 Z e e 1sin 2 j 2 Z j 1 2 2Z Dept. (2) but total reflection does not imply that there is no field in medium (2). In medium (2). the fields have the form. In other Both coefficients take the form a jb words. e j 2 y 1 /2 And the Z variation as. e j 2 y sin 2 Z cos 2 Snell’s law gives the y variation as.Field Theory Er Ei 10ES36 2 sin 2 1 1 cos 1 ( perpendicular polarization ) 2 cos 1 sin 2 1 1 and Er Ei 2 sin 2 1 1 2 / 1 cos 1 ( parallel polarization ) 2 2 / 1 cos 1 sin 2 1 1 become complex numbers when. sin 1 2 1 a jb and thus have a unit magnitude. the reflection is total provided that 1 is great enough and also provided that medium (1) is denser than medium. under the conditions of TIR is less than the phase velocity of a UPW in 2 medium (2). the non-uniform plane wave in medium (2) is a slow wave. Consequently. it is called a surface wave. 2 1 sin 1 2 Which. this field has a phase progression along the boundary and decreases exponentially away from it. Dept. However. cos 2 j 1 2 sin 2 1 1 j 1 2 sin 2 1 2 2 Therefore under the condition of TIR.Field Theory 10ES36 In the above expression. If is thus the example of a non-uniform plane wave. Of ECE/SJBIT GET ALL VTU NOTES AND QUESTION PAPERS! DOWNLOAD VTUCAMPUS APP Page 210 . ie.. a field does exist in the rarer medium. the lower sign must be chosen such that the fields decrease exponentially as Z becomes increasingly negative. The phase velocity along the interface is given by . since some kind of a surface between two media is necessary to support this wave. Also.