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Fundamentals of Electromagnetics for Teaching and Learning: A Two-Week Intensive Course for Faculty in Electrical-, Electronics-, Communication-, and Computer- Related Engineering Departments in Engineering Colleges in India by Nannapaneni Narayana Rao Edward C. Jordan Professor Emeritus of Electrical and Computer Engineering University of Illinois at Urbana-Champaign, USA Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, India Program for Hyderabad Area and Andhra Pradesh Faculty Sponsored by IEEE Hyderabad Section, IETE Hyderabad Center, and Vasavi College of Engineering IETE Conference Hall, Osmania University Campus Hyderabad, Andhra Pradesh June 3 – June 11, 2009 Workshop for Master Trainer Faculty Sponsored by IUCEE (Indo-US Coalition for Engineering Education) Infosys Campus, Mysore, Karnataka June 22 – July 3, 2009 1-2 Maxwell’s Equations d E dl = – B dS C dt S Electric field intensity V m S C m3 Wb m2 A m Current density A m2 Displacement flux density C m2 V r dv Charge density Magnetic flux density d H dl J dS D dS = + C S dt S Magnetic field intensity D dS = S B dS = 0 1-3 Module 1 Vectors and Fields 1.1 Vector algebra 1.2 Cartesian coordinate system 1.3 Cylindrical and spherical coordinate systems 1.4 Scalar and vector fields 1.5 Sinusoidally time-varying fields 1.6 The electric field 1.7 The magnetic field 1.8 Lorentz force equation 1-4 Instructional Objectives 1. 2. 3. 4. 5. 6. Perform vector algebraic operations in Cartesian, cylindrical, and spherical coordinate systems Find the unit normal vector and the differential surface at a point on the surface Find the equation for the direction lines associated with a vector field Identify the polarization of a sinusoidally time-varying vector field Calculate the electric field due to a charge distribution by applying superposition in conjunction with the electric field due to a point charge Calculate the magnetic field due to a current distribution by applying superposition in conjunction with the magnetic field due to a current element 1-5 Instructional Objectives (Continued) 7. Apply Lorentz force equation to find the electric and magnetic fields, for a specified set of forces on a charged particle moving in the field region 1-6 1.1 Vector Algebra (EEE, Sec. 1.1; FEME, Sec. 1.1) In this series of PowerPoint presentations, EEE refers to “Elements of Engineering Electromagnetics, 6th Edition,” Indian Edition (2006), and FEME refers to “Fundamentals of Electromagnetics for Engineering,” Indian Edition (2009). Also, all “D” Problems and “P” Problems are from EEE. 1-7 (1) Vectors (A) vs. Magnitude and direction Ex: Velocity, Force Scalars (A) Magnitude only Ex: Mass, Charge 1-8 (2) Unit Vectors have magnitude unity, denoted by symbol a with subscript. We shall use the right-handed system throughout. Useful for expressing vectors in terms of their components. A A1a1 + A2 a2 + A3a3 aA = = A A12 + A22 + A32 1-9 (3) Dot Product is a scalar A A A • B = AB cos a a B B Useful for finding angle between two vectors. A • B cos a = AB = A= B= A1a1 + A2a 2 + A3a3 B1a1 + B2a 2 + B3a3 A1 B1 + A2 B2 + A3 B3 A12 + A22 + A32 B12 + B22 + B32 1-10 (4) Cross Product is a vector right hand screw A A A B = AB sin a an B a B an is perpendicular to both A and B. Useful for finding unit vector perpendicular to two vectors. AB AB an = = AB sin a A B 1-11 where a1 a 2 A B = A1 A2 B1 B2 a3 A3 B3 (5) Triple Cross Product A (B C) is a vector A (B C) B (C A) C (A B) in general. 1-12 (6) Scalar Triple Product A•B C= B• C A = C• A B A1 A2 A3 = B1 B2 B3 C1 C2 C3 is a scalar. 1-13 Volume of the parallelepiped = Area of base Height = A × B C a n = A × B C = C A×B = A B×C A×B A×B B an C A 1-14 D1.2 (EEE) A = 3a1 + 2a2 + a3 B = a1 + a2 – a 3 C = a1 + 2a2 + 3a3 (a) A + B – 4C = (3 + 1 – 4)a1 + (2 + 1 – 8)a2 + (1 – 1 – 12)a3 = – 5a2 – 12a3 A + B – 4C = 25 + 144 = 13 1-15 (b) A + 2B – C = (3 + 2 – 1)a1 + (2 + 2 – 2)a2 + (1 – 2 – 3)a3 = 4a1 + 2a2 – 4a3 Unit Vector 4a1 + 2a 2 – 4a 3 = 4a1 + 2a 2 – 4a 3 1 = (2a1 + a 2 – 2a 3 ) 3 1-16 (c) A • C = 3 1 + 2 2 + 1 3 = 10 a1 a 2 (d) B C = 1 1 1 2 a3 –1 3 = (3 + 2)a1 + (–1 – 3)a2 + (2 –1)a3 = 5a1 – 4a2 + a3 1-17 3 2 (e) 1 A • B C = 1 1 –1 1 2 3 = 15 – 8 + 1 = 8 Same as A • (B C) = (3a1 + 2a2 + a3) • (5a1 – 4a2 + a3) = 3 5 + 2 (–4) + 1 1 = 15 – 8 + 1 = 8 1-18 P1.5 (EEE) E D B A C Common Point D= B–A ( A + D = B) E= C–B ( B + E = C) D and E lie along a straight line. 1-19 D×E = 0 B A × C B = 0 B×C A×C B×B + A×B = 0 A×B + B×C+C×A = 0 What is the geometric interpretation of this result? 1-20 E1.1 Another Example Given a1 × A = a 2 + 2a3 a2 × A = a1 2a3 (1) (2) Find A. A =C a 2 + 2a3 × a1 2a3 a1 a 2 a3 = C 0 1 2 = C 2a1 + 2a 2 + a3 1 0 2 1-21 To find C, use (1) or (2). a1 × C 2a1 + 2a2 + a3 = a2 + 2a3 C 2a3 a2 = a2 + 2a3 C =1 A = 2a1 + 2a2 + a3 1-22 Review Questions 1.1. Give some examples of scalars. 1.2. Give some examples of vectors. 1.3. Is it necessary for the reference vectors a1, a2, and a3 to be an orthogonal set? 1.4. State whether a1, a2, and a3 directed westward, northward, and downward, respectively, is a righthanded or a left-handed set. 1.5. State all conditions for which A • B is zero. 1.6. State all conditions for which A × B is zero. 1.7. What is the significance of A • B × C = 0? 1.8. What is the significance of A × (B × C) = 0? 1-23 Problem S1.1. Performing several vector algebraic manipulations 1-24 Problem S1.1. Performing several vector algebraic manipulations (continued) 1-25 1.2 Cartesian Coordinate System (EEE, Sec. 1.2; FEME, Sec. 1.2) 1-26 Cartesian Coordinate System 1-27 Cartesian Coordinate System z az O ay ax y x az ax ay z x y 1-28 Right-handed system a x ay = az ay az = a x xyz xy… az a x = a y ax, ay, az are uniform unit vectors, that is, the direction of each unit vector is same everywhere in space. 1-29 Vector drawn from one point to another: From P1(x1, y1, z1) to P2(x2, y2, z2) z r1 + R12 = r2 P1 R12 = r2 r1 r2 r1 O = x2ax + y2a y + z2az x1ax + y1a y + z1az P2 R12 x = x2 x1 ax + y2 y1 a y + z2 z1 az y 1-30 z R12 P1 (x 2 – x1)ax r1 O x1 x2 x r2 z1 (y2 – y1)ay P2 (z2 – z1)az y1 y2 z2 y 1-31 P1.8 A(12, 0, 0), B(0, 15, 0), C(0, 0, –20). (a) Distance from B to C = (0 – 0)a x + (0 – 15)a y + (–20 – 0)a z = 152 + 20 2 = 25 (b) Component of vector from A to C along vector from B to C = Vector from A to C • Unit vector along vector from B to C 1-32 = 12a x 20a z 15a y 20a z 15a y 20a z 400 = = 16 25 (c) Perpendicular distance from A to the line through B and C = (Vector from A to C) (Vector from B to C) BC = 12ax 20az × 15ay 20az 25 1-33 180az – 240a y – 300a x = 25 = 12 2 (2) Differential Length Vector (dl) az dl Q x + dx, y + dy, z + dz P x, y , z dz dx dy ax ay dl = dx a x + dy a y + dz a z 1-34 dl dx y = f(x) z = constant plane dy = f (x) dx dz = 0 dl = dx ax + dy ay = dx ax + f (x) dx ay Unit vector normal to a surface dl1 dl 2 an = dl1 dl 2 an dl2 dl1 Curve 2 Curve 1 1-35 D1.5 Find dl along the line and having the projection dz on the z-axis. (a) x = 3, y = –4 dx = 0, dy = 0 dl = dz a z (b) x + y = 0, y + z = 1 dx + dy = 0, dy + dz = 0 dy = – dz, dx = – dy = dz d l = dz ax dz a y + dz az = ax a y + az dz 1-36 (c) Line passing through (0, 2, 0) and (0, 0, 1). dy dz x = 0, = 0 – 2 1 –0 dx = 0, dy = – 2 dz d l = 2 dz a y + dz az = 2a y + az dz 1-37 (3) Differential Surface Vector (dS) dS = dl1 dl2 sin a = d l1 × d l2 an a dl2 dS dl1 Orientation of the surface is defined uniquely by the normal ± an to the surface. dS = dS a n = dl1 dl 2 a n = dl1 dl 2 For example, in Cartesian coordinates, dS in any plane parallel to the xy plane is az dx dy a z = dx a x dy a y dy dx dS x y 1-38 (4) Differential Volume (dv) dv = dl1 • dl 2 dl3 dl2 dl3 dv dl1 In Cartesian coordinates, dv = dx a x • dy a y dz a z = dx dy dz dz z dy dx y x 1-39 Review Questions 1.9. What is the particular advantageous characteristic associated with unit vectors in the Cartesian coordinate system? 1.10. What is the position vector? 1.11. What is the total distance around the circumference of a circle of radius 1 m? What is the total vector distance around the circle? 1.12. Discuss the application of differential length vectors to find a unit vector normal to a surface at a point on the surface. 1.13. Discuss the concept of a differential surface vector. 1.14. What is the total surface area of a cube of sides 1 m? Assuming the normals to the surfaces to be directed outward of the cubical volume, what is the total vector surface area of the cube? 1-40 Problem S1.2. Finding the unit vector normal to a surface and the differential surface vector, at a point on it 1-41 1.3 Cylindrical and Spherical Coordinate Systems (EEE, Sec. 1.3; FEME, Appendix A) 1-42 Cylindrical Coordinate System 1-43 Spherical Coordinate System 1-44 Cylindrical and Spherical Coordinate Systems Cylindrical (r, f, z) z Spherical (r, q, f) z az af 90Þ 90 z x ar af q r ar x f r y Only az is uniform. ar and af are nonuniform. y f x All three unit vectors are nonuniform. 90Þ 90 aq y 1-45 x = r cos f y = r sin f z =z x = r sin q cos f y = r sin q sin f z = r cos q D1.7 (a) (2, 5p/6, 3) in cylindrical coordinates z 3 2 x 55p/6 p 6 x = 2 cos 5p 6 = – 3 y = 2 sin 5p 6 = 1 z=3 y 3+1 = 2 1-46 (b) (4, 4p 3, –1) in cylindrical coordinates z 1 4 44p/3 p 3 y x x = 4 cos 4p 3 = – 2 4 + 12 = 4 y = 4 sin 4p 3 = – 2 3 z = –1 1-47 (c) (4, 2p 3, p 6) in spherical coordinates z 2p 3 y p x 6 4 2p p cos = 3 3 6 2p p y = 4 sin sin = 3 9 + 3 + 4 = 4 3 6 p z = 4 cos = – 2 3 x = 4 sin (d) 1-48 8, p 4, p 3 in spherical coordinates. z p 4 8 y p 3 x p p = 1 4 3 p p y = 8 sin sin = 3 1 + 3 + 4 = 8 4 3 p z = 8 cos = 2 4 x = 8 sin cos 1-49 Conversion of vectors between coordinate systems a rc af = a z cos f sin f 0 a x –sin f cos f 0 a y 0 1 0 a z a rs aq = af sin q cos f sin q sin f cos q cos f cos q sin f cos f – sin f af az ay f ax arc az af q ars arc aq cos q a x – sin q a y 0 a z 1-50 P1.18 A = ar at (2, p/6, p/2) B = aq at (1, p/3, 0) C = af at (3, p/4, 3p/2) z A A = sin p 6 x p 2 y p 6 a y + cos 1 3 = ay + az 2 2 1 3 + =1 4 4 p 6 az 1-51 z B = sin p 3 x C B 3 1 y 6 a x – cos 1 3 = ax – az 2 2 z 1 3 + =1 4 4 p /4 3p /2 x p y C = ax p 6 az 1-52 1 3 (a) A B = a y + az 2 2 3 = 4 1 3 (b) A C = a y + az 2 2 =0 1 3 az a x 2 2 ax 1-53 (c) (d) 1 3 B C = ax az ax 2 2 1 = 2 AB•C=C•A B 1 = 0 1 2 0 1 2 0 3 =– 4 0 3 2 3 – 2 1-54 Differential length vectors: Cylindrical Coordinates: dl = dr ar + r df af+ dz az Spherical Coordinates: dl = dr ar + r dq aq + r sin q df af 1-55 Review Questions 1.15. Describe the three orthogonal surfaces that define the cylindrical coordinates of a point. 1.16. Which of the unit vectors in the cylindrical coordinate system are not uniform? Explain. 1.17. Discuss the conversion from the cylindrical coordinates of a point to its Cartesian coordinates. 1.18. Describe the three orthogonal surfaces that define the spherical coordinates of a point. 1.19. Discuss the nonuniformity of the unit vectors in the spherical coordinate system. 1.20. Discuss the conversion from the cylindrical coordinates of a point to its Cartesian coordinates. 1-56 Problem S1.3. Determination of the equality of vectors specified in cylindrical and spherical coordinates 1-57 Problem S1.4. Finding the unit vector tangential to a curve, at a point on it, in spherical coordinates 1-58 1.4 Scalar and Vector Fields (EEE, Sec. 1.4; FEME, Sec. 1.3) 1-59 FIELD is a description of how a physical quantity varies from one point to another in the region of the field (and with time). (a) Scalar fields Ex: Depth of a lake, d(x, y) Temperature in a room, T(x, y, z) Depicted graphically by constant magnitude contours or surfaces. y d3 d1 d2 x 1-60 (b) Vector Fields Ex: Velocity of points on a rotating disk v(x, y) = vx(x, y)ax + vy(x, y)ay Force field in three dimensions F(x, y, z) = Fx(x, y, z)ax + Fy(x, y, z)ay + Fz(x, y, z)az Depicted graphically by constant magnitude contours or surfaces, and direction lines (or stream lines). 1-61 Example: Linear velocity vector field of points on a rotating disk 1-62 (c) Static Fields Fields not varying with time. (d) Dynamic Fields Fields varying with time. Ex: Temperature in a room, T(x, y, z; t) 1-63 D1.10 T(x, y, z, t) = T0 x 1 + sin p t + 2 y 1 cos p t + 4 z 2 (a) 2 2 T x, y, z, 0 = T0 x 1 + 0 + 2 y 1 1 + 4 z 2 2 = T0 x 2 + 4 z 2 Constant temperature surfaces are elliptic cylinders, 2 2 x + 4 z = const. 2 1-64 (b) T x, y, z, 0.5 = T0 x 1 + 1 + 2 y 1 0 + 4 z 2 2 2 = T0 4 x 2 + 4 y 2 + 4 z 2 Constant temperature surfaces are spheres x 2 + y 2 + z 2 = const. (c) T x, y, z, 1 = T0 x 1 + 0 + 2 y 1 + 1 + 4 z 2 2 = T0 x 2 + 16 y 2 + 4 z 2 Constant temperature surfaces are ellipsoids, 2 2 2 x + 16 y + 4 z = const. 2 1-65 Procedure for finding the equation for the direction lines of a vector field The field F is tangential to the direction line at all points on a direction line. F F dl F dl ax ay dl F = dx dy Fx Fy az dz = 0 Fz dx dy dz = = Fx Fy Fz 1-66 Similarly dr r df dz = = Fr Ff Fz cylindrical dr r dq r sin q df = = Fr Fq Ff spherical 1-67 P1.26 (b) xa x + ya y + za z (Position vector) dx dy dz = = x y z ln x = ln y + ln C1 = ln z + ln C2 ln x = ln C1y = ln C2 z x = C1y = C2 z 1-68 Direction lines are straight lines emanating radially from the origin. For the line passing through (1, 2, 3), 1 = C1(2) = C2 (3) 1 1 C1 = , C2 = 2 3 y z x= = 2 3 or, 6 x = 3y = 2z 1-69 Review Questions 1.21. Discuss briefly your concept of a scalar field and illustrate with an example. 1.22. Discuss briefly your concept of a vector field and illustrate with an example. 1.23. How do you depict pictorially the gravitational field of the earth? 1.24. Discuss the procedure for obtaining the equations for the direction lines of a vector field. 1-70 Problem S1.5. Finding the equation for direction line of a vector field, specified in spherical coordinates 1-71 1.5 Sinusoidally Time-Varying Fields (EEE, Sec. 3.6; FEME, Sec. 1.4) 1-72 Sinusoidal function of time 1-73 Polarization is the characteristic which describes how the position of the tip of the vector varies with time. Linear Polarization: Tip of the vector describes a line. Circular Polarization: Tip of the vector describes a circle. 1-74 Elliptical Polarization: Tip of the vector describes an ellipse. (i) Linear Polarization F1 = F1 cos (t + f ) a x Magnitude varies sinusoidally with time Direction remains along the x axis Linearly polarized in the x direction. 1-75 Linear polarization 1-76 F2 = F2 cos (t + q ) a y Magnitude varies sinusoidally with time Direction remains along the y axis Linearly polarized in the y direction. If two (or more) component linearly polarized vectors are in phase, (or in phase opposition), then their sum vector is also linearly polarized. Ex: F = F1 cos (t + f ) a x + F2 cos (t + f ) a y 1-77 Sum of two linearly polarized vectors in phase (or in phase opposition) is a linearly polarized vector 1-78 y F2 a F a F1 x F2 cos (t + f ) –1 = tan F1 cos (t + f ) F2 –1 = tan F1 = constant (ii) Circular Polarization If two component linearly polarized vectors are (a) equal in amplitude (b) differ in direction by 90˚ (c) differ in phase by 90˚, then their sum vector is circularly polarized. 1-79 Circular Polarization 1-80 Example: F = F1 cos t ax + F1 sin t a y F = F1 cos t + F1 sin t 2 2 = F1 , constant a = tan = tan 1 1 F1 sin t F1 cos t tan t = t y F2 F a F1 x 1-81 (iii) Elliptical Polarization In the general case in which either (i) or (ii) is not satisfied, then the sum of the two component linearly polarized vectors is an elliptically polarized vector. Example: F = F1 cos t a x + F2 sin t a y y F2 F F1 x 1-82 Example: F = F0 cos t ax + F0 cos t + p 4 ay y F0 F2 F p 4 F1 F0 x –F0 –F0 1-83 D3.17 F1 = F0 cos 2p 108t 2p z ax F2 = F0 cos 2p 108t 3p z a y F1 and F2 are equal in amplitude (= F0) and differ in direction by 90˚. The phase difference (say f) depends on z in the manner –2pz – (–3pz) = pz. (a) At (3, 4, 0), f = p(0) = 0. F1 + F2 is linearly polarized. (b) At (3, –2, 0.5), f = p(0.5) = 0.5 p. F1 + F2 is circularly polarized. 1-84 (c) At (–2, 1, 1), f = p(1) = p. F1 + F2 is linearly polarized. (d) At (–1, –3, 0.2) = f = p(0.2) = 0.2p. F1 + F2 is elliptically polarized. 1-85 Review Questions 1.25. A sinusoidally time-varying vector is expressed in terms of its components along the x-, y-, and z- axes. What is the polarization of each of the components? 1.26. What are the conditions for the sum of two linearly polarized sinusoidally time-varying vectors to be circularly polarized? 1.27. What is the polarization for the general case of the sum of two sinusoidally time-varying linearly polarized vectors having arbitrary amplitudes, phase angles, and directions? 1.28. Considering the seconds hand on your analog watch to be a vector, state its polarization. What is the frequency? 1-86 Problem S1.6. Finding the polarization of the sum of two sinusoidally time-varying vector fields 1-87 1.6 The Electric Field (EEE, Sec. 1.5; FEME, Sec. 1.5) 1-88 The Electric Field is a force field acting on charges by virtue of the property of charge. Q1Q2 Coulomb’s Law F1 = a 21 F2 2 a12 4 p R 0 Q 2 R F1 a21 Q1 Q2 Q1 F2 = a12 2 4p 0 R 0 = permittivity of free space 10 –9 F/m 36p 1-89 D1.13(b) Q 2a a Q Q2/4p0(2a2) Q2/4p0(4a2) 2/4p (2a2) Q 0 Q Q = 4p 0 Q From the construction, it is evident that the resultant force is directed away from the center of the square. The magnitude of this resultant force is given by 1-90 2 Q 2 4p 0 2a 2 1 1 = + 2 2 2a 4a 0.957 = 2 N a cos 45 + Q 2 4p 0 4a2 1-91 Electric Field Intensity, E is defined as the force per unit charge experienced by a small test charge when placed in the region of the field. F E = Lim q0 q Thus E qE q –q –qE Fe = qE Units: N = N m = V C Cm m Sources: Charges; Time-varying magnetic field 1-92 Electric Field of a Point Charge (Coulomb’s Law) Qq F= aR 2 4p 0 R q Q = q aR 2 4p 0 R = q E due to Q E due to Q = R Q Q 4p 0 R 2 aR aR 1-93 Constant magnitude surfaces are spheres centered at Q. Direction lines are radial lines emanating from Q. E aR R Q E due to charge distributions (a) Collection of point charges Q1 E= n Qj 2 4 p R 0 j j =1 a Rj aRn R1 R2 R3 Q2 Q3 Qn Rn aR3 aR2 aR1 1-94 E1.2 z Q (> 0) d d2 + x2 a e d Q (> 0) y d2 + x2 x Electron (charge e and mass m) is displaced from the origin by D (<< d) in the +x-direction and released from rest at t = 0. We wish to obtain differential equation for the motion of the electron and its solution. 1-95 For any displacement x, F = 2 = Qe 4p 0 d + x 2 2 Qex 2p 0 d 2 + x 2 32 is directed toward the origin, and x D d. Qe x F– a x 3 2p 0 d cos a ax ax 1-96 The differential equation for the motion of the electron is d2 x Qe x m 2 =– dt 2p 0 d 3 d2 x Qe + x=0 2 3 dt 2pm 0 d Solution is given by x = A cos Qe 2p m 0 d 3 t + B sin Qe 2p m 0 d 3 t 1-97 dx Using initial conditions x = D and = 0 at t = 0, dt we obtain x = D cos Qe t 3 2pm 0 d which represents simple harmonic motion about the origin with period 2p Qe 2p m 0 d 3 1-98 (b) Line Charges Line charge density, rL (C/m) dl P dS (c) Surface Charges Surface charge density, rS (C/m2) (d) Volume Charges Volume charge density, r (C/m3) dv 1-99 E1.3 Finitely-Long Line Charge z a rL = rL 0 = 4p0 C m dz z f x –a r2 + z2 y r a E ar dE = 2 E= 1-100 rL dz 4p 0 r + z 2 2 4p 0 2 cos a ar rL 0 r dz a z =0 r 2 +z 2 32 ar a rL 0 r z = 2 2 2 ar 2p 0 r r + z z =0 rL 0 a 2a = 2p 0 r r + a 2 2 ar = r r +a 2 rL 0 For a , E ar 2p0r 2 ar 1-101 Infinite Plane Sheet of Charge of Uniform Surface Charge Density z a y2 + z 2 z rS 0 x dy y y 1-102 dEz = 2 rS 0 dy 2p0 y2 + z2 rS 0 z dy = p0 y2 + z2 rS 0 z Ez = p0 dy y2 + z 2 y =0 rS 0 z 1 = p0 z rS 0 = 20 p 2 a =0 da cos a 1-103 rS 0 E= az for z 20 rS 0 = an 20 rS 0 az 20 z<0 rS 0 az 20 0 rS 0 + z>0 + rS 0 a 20 z + + + z=0 z 1-104 D1.16 rS1 z=0 Given E(3,5,1) = 0 V m E(1, – 2,3) = 6az V m E(3, 4,5) = 4az V m rS 2 z=2 rS 3 z=4 1-105 1 rS 1 rS 2 rS 3 = 0 2 0 1 rS 1 + rS 2 rS 3 = 6 2 0 1 rS 1 + rS 2 + rS 3 = 4 2 0 Solving, we obtain 2 r = 4 C m (a) S 1 0 2 r = 6 C m (b) S2 0 2 r = 2 C m (c) S 3 0 (d) E 2,1, 6 = 4az V m 1-106 Review Questions 1.29. State Coulomb’s law. To what law in mechanics is Coulomb’s law analogous? 1.30. What is the value of the permittivity of free space? What are its units? 1.31. What is the definition of electric field intensity? What are its units? 1.32. Describe the electric field due to a point charge. 1.33. Discuss the different types of charge distributions. How do you determine the electric field due to a charge distribution? 1.34. Describe the electric field due to an infinitely long line charge of uniform density. 1.35. Describe the electric field due to an infinite plane sheet of uniform surface charge density. 1-107 Problem S1.7. Determination of conditions for three point charges on a circle to be in equilibrium 1-108 Problem S1.8. Finding the electric field due to an infinite plane slab charge of specified charge density 1-109 1.7 The Magnetic Field (EEE, Sec. 1.6; FEME, Sec. 1.6) 1-110 The Magnetic Field acts to exert force on charge when it is in motion. Fm = qv B B Fm v q B = Magnetic flux density vector Alternatively, since charge in motion constitutes current, magnetic field exerts forces on current elements. dFm B d Fm = I d l B I dl 1-111 N N–m Units of B: = 2 Am Am Wb = 2 =T m Sources: Currents; Time-varying electric field 1-112 Ampère’s Law of Force I1 a21 dl2 I2 a dl1 12 R 0 I 2 d l 2 × a 21 d F1 = I1d l1 × 2 4 p R = I1d l1 × B 2 0 I1 d l1 × a12 d F2 = I 2 d l 2 × 2 4p R = I 2 d l 2 × B1 1-113 Magnetic field due to a current element (Biot-Savart Law) 0 I dl a R B= 4p R2 a Note B sin a I dl aR R P B 1 B 2 R B right-circular to the axis of the current element 0 = Permeability of free space = 4p 10–7 H m 1-114 E1.4 I d l = I dx a x + a y A situated at 1, 2, 2 . Find B at 2, 1, 3 . B= 0 I d l × a R 0 I d l × R = 2 4p R 4p R3 R since a = R R R = 2 1 a x + 1 + 2 a y + 3 2 a z = ax + a y + az 1-115 0 I dx a x + a y × a x + a y + a z B= 3 4p 3 0 I dx = ax a y 12 3p 1-116 Current Distributions (a) Filamentary Current I (A) (b) Surface Current Surface current density, JS (A/m) JS w I JS = w max 1-117 (c) Volume Current Density, J (A/m2) area A J I J = A max 1-118 P1.44 z a2 P(r, f, z) a2 a dz aR z – z z I r y a1 x a1 r dB = = 0 dz a z × a R 2 2 4p r + z z 0 I dz sin a af 2 2 4p r + z z 1-119 B= z = a2 z = a1 0 I = 4p r 2 z z cot a = r dB d z sin a af a2 a = a1 1 + z z r 2 0 I a cosec2a sin a da = af 2 4p r a = a cosec a 0 I a = cos a a = a af 4p r 0 I = cos a1 cos a 2 af 4p r 2 1 2 1 1-120 For infinitely long wire, a1 – , a2 , a1 0 , a 2 p 0 I B af 2pr 1-121 Magnetic Field Due to an Infinite Plane Sheet of Uniform Surface Current Density This can be found by dividing the sheet into infinitely long strips parallel to the current density and using superposition, as in the case of finding the electric field due to an infinite plane sheet of uniform surface charge density. Instead of going through this procedure, let us use analogy. To do this, we first note the following: 1-122 (a) Point Charge E= Q 4p0 R 2 aR Current Element 0 I d l ×aR B= 4p R2 1-123 (b) Line Charge z rL 0 r Line Current z I P r ar E r=0 rL 0 E= ar 2p0r P B ar r=0 0 I B= af 2p r 0I = az ar 2p r 1-124 Then, (c) Sheet Charge rS 0 Sheet Current JS P P an E rS 0 E= an 20 B an B= 0 2 JS an 1-125 Review Questions 1.36. How is magnetic flux density defined in terms of force on a moving charge? Compare the magnetic force on a moving charge with electric force on a charge. 1.37. How is magnetic flux density defined in terms of force on a current element? 1.38. What are the units of magnetic flux density? 1.39. State Ampere’s force law as applied to current elements. Why is it not necessary for Newton’s third law to hold for current elements? 1.40. Describe the magnetic field due to a current element. 1.41. What is the value of the permeability of free space? What are its units? 1-126 Review Questions (continued) 1.42. Discuss the different types of current distributions. How do you determine the magnetic flux density due to a current distribution? 1.43. Describe the magnetic field due to an infinitely long, straight, wire of current. 1.44. Discuss the analogies between the electric field due to charge distributions and the magnetic field due to current distributions. 1-127 Problem S1.9. Finding parameters of an infinitesimal current element that produces a specified magnetic field 1-128 Problem S1.10. Finding the magnetic field due to a specified current distribution within an infinite plane slab 1-129 1.8 Lorentz Force Equation (EEE, Sec. 1.7; FEME, Sec. 1.6) 1-130 Lorentz Force Equation Fe = qE Fm = qv × B Fm F = Fe + Fm B F = q E + v × B q For a given B, to find E, F E= –vB q One force is sufficient. E v Fe 1-131 D1.21 B0 B= ax + 2a y 2az 3 Find E for which acceleration experienced by q is zero, for a given v. F = q E + v × B = 0 E = v × B (a) v = v0 ax a y + az v0 B0 E= ax a y + az × ax + 2a y 2az 3 = v0 B0 a y + az 1-132 (b) v = v0 2ax + a y + 2az v0 B0 E= 2ax + a y + 2az × ax + 2a y 2az 3 = v0 B0 2ax 2a y az (c) v = v0 along y = z = 2 x v0 = ax 2a y + 2az 3 v0 B0 E= ax + 2a y + 2az × ax + 2a y 2az 3 =0 1-133 For a given E, to find B, F vB= –E q One force not sufficient. Two forces are needed. F1 v1 B = – E = C1 q F2 v2 B = – E = C2 q C1 × C2 = v1 × B × v2 × B = v1 × B B v2 v1 × B v2 B = C1 v2 B 1-134 C2 C1 B = C1 • v 2 provided C1 v 2 0, which means v2 and v1 should not be collinear. 1-135 P1.54 For v = v1 or v = v2, test charge moves with constant velocity equal to the initial value. It is to be shown that for mv1 + nv 2 v= , where m + n 0, m+ n the same holds. qE + qv1 B = 0 (1) qE + qv2 B = 0 (2) qE + qv B = 0 (3) 1 3 v1 v × B = 0 2 3 v2 v × B = 0 1-136 Both v1 v and v2 v are collinear to B. v1 v = k v2 v v1 – kv 2 v= 1– k mv1 + nv 2 = m+ n n for k = – m Alternatively, m n (1) + (2) m+n m+n 1-137 mv1 + nv 2 qE + q m+ n mv1 + nv 2 v = m+n B=0 1-138 Review Questions 1.45. State Lorentz force equation. 1.46. If it is assumed that there is no electric field, the magnetic field at a point can be found from the knowledge of forces exerted on a moving test charge for two noncollinear velocities. Explain. 1.47. Discuss the determination of E and B at a point from the knowledge of forces experienced by a test charge at that point for several velocities. What is the minimum number of required forces? Explain. 1-139 Problem S1.11. Finding the electric and magnetic fields from three forces experienced by a test charge 1-140 The End