EECE 380 Problems Solution Manual, Fall 2011

286 Chapter 6: Maxwell’s Equations for Time-Varying Fields Lesson #37 Chapter — Section: 6-1, 6-2 Topics: Faraday’s law, stationary loop in changing magnetic field Highlights: • Faraday’s law • EMF Special Illustrations: • Example 6-1 • Example 6-2 • CD-ROM Demo 6.1 • CD-ROM Modules 6.1 and 6.2 287 Lesson #38 Chapter — Section: 6-3, 6-4 Topics: Ideal transformer, moving conductor Highlights: • Transformer voltage and current relations • EMF for moving conductor Special Illustrations: • CD-ROM Modules 6.3 and 6.4 • CD-ROM Demo 6.2 288 Lesson #39 Chapter — Section: 6-5, 6-6 Topics: EM Generator, moving conductor in changing field Highlights: • Motor and generator reciprocity • EMF for combination of motional and transformer Special Illustrations: • Technology Brief on “EMF Sensors” (CD-ROM) EMF Sensors An electromotive force (emf) sensor is a device that can generate an induced voltage in response to an external stimulus. Three types of emf sensors are profiled in this Technical Brief: the piezoelectric transducer, the Faraday magnetic flux sensor, and the thermocouple. Piezoelectric Transducers Piezoelectricity refers to the property of certain crystals, such as quartz, to become electrically polarized when the crystal is subjected to mechanical pressure, thereby exhibiting a voltage across it. The crystal consists of polar domains represented by equivalent dipoles (A). Under the absence of an external force, the polar domains are randomly oriented throughout the material (A1), but when compressive or tensile (stretching) stress is applied to the crystal, the polar domains align themselves along one of the principal axes of the crystal, leading to a net polarization (electric charge) at the crystal surfaces (A2 and A3). Compression and stretching generate voltages of opposite polarity. The piezoelectric effect (piezein means to press or squeeze in Greek) was discovered by the Curie brothers, Pierre and Paul-Jacques, in 1880, and a year later Lippmann predicted the converse property, namely that if subjected to an electric field, the crystal would change in shape. Thus, the piezoelectric effect is a reversible (bidirectional) electro-mechanical process. 289 Lesson #40 Chapter — Section: 6-7, 6-8 Topics: Displacement current, boundary conditions Highlights: • Concept of “displacement current” • Boundary conditions for the dynamic case Special Illustrations: • Example 6-7 290 Lesson #41 Chapter — Section: 6-9, 6-10 Topics: Charge-current continuity, charge dissipation Highlights: • Continuity equation • Relaxation time constant Special Illustrations: 291 Lesson #42 Chapter — Section: 6-11 Topics: EM potentials Highlights: • Retarded potential • Relation of potentials to fields in the dynamic case Special Illustrations: Example 6-8 292 CHAPTER 6 Chapter 6 Sections 6-1 to 6-6: Faraday’s Law and its Applications Problem 6.1 The switch in the bottom loop of Fig. 6-17 (P6.1) is closed at t � 0 and then opened at a later time t1. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times? R2 R1 I + - Figure P6.1: Loops of Problem 6.1. Solution: The magnetic coupling will be strongest at the point where the wires of the two loops come closest. When the switch is closed the current in the bottom loop will start to flow clockwise, which is from left to right in the top portion of the bottom loop. To oppose this change, a current will momentarily flow in the bottom of the top loop from right to left. Thus the current in the top loop is momentarily clockwise when the switch is closed. Similarly, when the switch is opened, the current in the top loop is momentarily counterclockwise. Problem 6.2 The loop in Fig. 6-18 (P6.2) is in the x–y plane and B � ˆzB0 sinωt with B0 positive. What is the direction of I (ˆφφφ or � ˆφφφ) at (a) t � 0, (b) ωt � π 4, and (c) ωt � π 2? Solution: I � Vemf R. Since the single-turn loop is not moving or changing shape with time, V m emf � 0 V and Vemf � V tr emf. Therefore, from Eq. (6.8), I � V tr emf R � � 1 R � S ∂B ∂t � ds If we take the surface normal to be � ˆz, then the right hand rule gives positive flowing current to be in the � ˆφφφ direction. I � � A R ∂ ∂t B0 sinωt � � AB0ω R cosωt (A) � CHAPTER 6 293 R z y x I Figure P6.2: Loop of Problem 6.2. where A is the area of the loop. (a) A, ω and R are positive quantities. At t � 0, cosωt � 1 so I � 0 and the current is flowing in the � ˆφφφ direction (so as to produce an induced magnetic field that opposes B). (b) At ωt � π 4, cosωt � � 2 2 so I � 0 and the current is still flowing in the � ˆφφφ direction. (c) At ωt � π 2, cosωt � 0 so I � 0. There is no current flowing in either direction. Problem 6.3 A coil consists of 100 turns of wire wrapped around a square frame of sides 0.25 m. The coil is centered at the origin with each of its sides parallel to the x- or y-axis. Find the induced emf across the open-circuited ends of the coil if the magnetic field is given by (a) B � ˆz20e  3t (T), (b) B � ˆz20cosx cos103t (T), (c) B � ˆz20cosx sin2y cos103t (T). Solution: Since the coil is not moving or changing shape, V m emf � 0 V and Vemf � V tr emf. From Eq. (6.6), Vemf � � N d dt � S B � ds � � N d dt � 0 � 125  0 � 125 � 0 � 125  0 � 125 B � � ˆzdxdy � � where N � 100 and the surface normal was chosen to be in the � ˆz direction. (a) For B � ˆz20e  3t (T), Vemf � � 100 d dt � 20e  3t � 0 25 � 2 �� 375e  3t (V) 294 CHAPTER 6 (b) For B � ˆz20cosxcos103t (T), Vemf � � 100 d dt � 20cos103t � 0 � 125 x �  0 � 125 � 0 � 125 y �  0 � 125 cosxdxdy � 124 6sin103t (kV) (c) For B � ˆz20cosxsin2ycos103t (T), Vemf � � 100 d dt � 20cos103t � 0 � 125 x �  0 � 125 � 0 � 125 y �  0 � 125 cosxsin2ydxdy � 0 Problem 6.4 A stationary conducting loop with internal resistance of 0.5 Ω is placed in a time-varying magnetic field. When the loop is closed, a current of 5 A flows through it. What will the current be if the loop is opened to create a small gap and a 2-Ω resistor is connected across its open ends? Solution: Vemf is independent of the resistance which is in the loop. Therefore, when the loop is intact and the internal resistance is only 0 5 Ω, Vemf � 5 A 0 5 Ω � 2 5 V When the small gap is created, the total resistance in the loop is infinite and the current flow is zero. With a 2-Ω resistor in the gap, I � Vemf � 2 Ω � 0 5 Ω �� 2 5 V 2 5 Ω � 1 (A) Problem 6.5 A circular-loop TV antenna with 0.02 m2 area is in the presence of a uniform-amplitude 300-MHz signal. When oriented for maximum response, the loop develops an emf with a peak value of 30 (mV). What is the peak magnitude of B of the incident wave? Solution: TV loop antennas have one turn. At maximum orientation, Eq. (6.5) evaluates to Φ � � B � ds � � BA for a loop of area A and a uniform magnetic field with magnitude B � � B � . Since we know the frequency of the field is f � 300 MHz, we can express B as B � B0 cos � ωt � α0 � with ω � 2π 300 106 rad/s and α0 an arbitrary reference phase. From Eq. (6.6), Vemf � � N dΦ dt � � A d dt � B0 cos � ωt � α0 �� � AB0ωsin � ωt � α0 � Vemf is maximum when sin � ωt � α0 �� 1. Hence, 30 10  3 � AB0ω � 0 02 B0 6π 108 � CHAPTER 6 295 which yields B0 � 0 8 (nA/m). Problem 6.6 The square loop shown in Fig. 6-19 (P6.6) is coplanar with a long, straight wire carrying a current I � t �� 5cos2π 104t (A) (a) Determine the emf induced across a small gap created in the loop. (b) Determine the direction and magnitude of the current that would flow through a 4-Ω resistor connected across the gap. The loop has an internal resistance of 1 Ω. y x z 5cm I(t) 10cm 10cm Figure P6.6: Loop coplanar with long wire (Problem 6.6). Solution: (a) The magnetic field due to the wire is B � ˆφφφ µ0I 2πr � � ˆx µ0I 2πy � where in the plane of the loop, ˆφφφ � � ˆx and r � y. The flux passing through the loop 296 CHAPTER 6 is Φ � � S B � ds � � 15 cm 5 cm � � ˆx µ0I 2πy � � � ˆx10 (cm) � dy � µ0I 10  1 2π ln 15 5 � 4π 10  7 5cos � 2π 104t �� 10  1 2π 1 1 � 1 1 10  7 cos � 2π 104t � (Wb) Vemf � � dΦ dt � 1 1 2π 104 sin � 2π 104t �� 10  7 � 6 9 10  3 sin � 2π 104t � (V) (b) Iind � Vemf 4 � 1 � 6 9 10  3 5 sin � 2π 104t �� 1 38sin � 2π 104t � (mA) At t � 0, B is a maximum, it points in � ˆx-direction, and since it varies as cos � 2π 104t � , it is decreasing. Hence, the induced current has to be CCW when looking down on the loop, as shown in the figure. Problem 6.7 The rectangular conducting loop shown in Fig. 6-20 (P6.7) rotates at 6,000 revolutions per minute in a uniform magnetic flux density given by B � ˆy50 (mT) Determine the current induced in the loop if its internal resistance is 0 5 Ω. Solution: Φ � � S B � dS � ˆy50 10  3 � ˆy � 2 3 10  4 � cosφ � t �� 3 10  5 cosφ � t � � φ � t �� ωt � 2π 6 103 60 t � 200πt (rad/s) � Φ � 3 10  5 cos � 200πt � (Wb) � Vemf � � dΦ dt � 3 10  5 200πsin � 200πt �� 18 85 10  3 sin � 200πt � (V) � Iind � Vemf 0 5 � 37 7sin � 200πt � (mA) CHAPTER 6 297 y x z B B ω φ(t) 3cm 2cm Figure P6.7: Rotating loop in a magnetic field (Problem 6.7). The direction of the current is CW (if looking at it along � ˆx-direction) when the loop is in the first quadrant (0 � φ � π 2). The current reverses direction in the second quadrant, and reverses again every quadrant. Problem 6.8 A rectangular conducting loop 5 cm 10 cm with a small air gap in one of its sides is spinning at 7200 revolutions per minute. If the field B is normal to the loop axis and its magnitude is 6 10  6 T, what is the peak voltage induced across the air gap? Solution: ω � 2π rad/cycle 7200 cycles/min 60 s/min � 240π rad/s � A � 5 cm 10 cm � 100 cm/m � 2 � 5 0 10  3 m2 From Eqs. (6.36) or (6.38), Vemf � AωB0sinωt; it can be seen that the peak voltage is V peak emf � AωB0 � 5 0 10  3 240π 6 10  6 � 22 62 � µV � Problem 6.9 A 50-cm-long metal rod rotates about the z-axis at 90 revolutions per minute, with end 1 fixed at the origin as shown in Fig. 6-21 (P6.9). Determine the induced emf V12 if B � ˆz2 10  4 T. Solution: Since B is constant, Vemf � V m emf. The velocity u for any point on the bar is given by u � ˆφφφrω, where ω � 2π rad/cycle � 90 cycles/min � � 60 s/min � � 3π rad/s 298 CHAPTER 6 ω 1 2 x y z B Figure P6.9: Rotating rod of Problem 6.9. From Eq. (6.24), V12 � V m emf � � 1 2 � u B � � dl � � 0 r � 0 � 5 � ˆφφφ3πr ˆz2 10  4 � � ˆrdr � 6π 10  4 � 0 r � 0 � 5 r dr � 3π 10  4r2 � � � 0 0 � 5 � � 3π 10  4 0 25 � � 236 � µV � Problem 6.10 The loop shown in Fig. 6-22 (P6.10) moves away from a wire carrying a current I1 � 10 (A) at a constant velocity u � ˆy7 5 (m/s). If R � 10 Ω and the direction of I2 is as defined in the figure, find I2 as a function of y0, the distance between the wire and the loop. Ignore the internal resistance of the loop. Solution: Assume that the wire carrying current I1 is in the same plane as the loop. The two identical resistors are in series, so I2 � Vemf 2R, where the induced voltage is due to motion of the loop and is given by Eq. (6.26): Vemf � V m emf � � � C � u B � � dl The magnetic field B is created by the wire carrying I1. Choosing ˆz to coincide with the direction of I1, Eq. (5.30) gives the external magnetic field of a long wire to be B � ˆφφφµ0I1 2πr CHAPTER 6 299 u u I1 = 10 A I2 20 cm R z R y0 10 cm Figure P6.10: Moving loop of Problem 6.10. For positive values of y0 in the y-z plane, ˆy � ˆr, so u B � ˆy � u � B � ˆr � u � ˆφφφµ0I1 2πr � ˆzµ0I1u 2πr Integrating around the four sides of the loop with dl � ˆz dz and the limits of integration chosen in accordance with the assumed direction of I2, and recognizing that only the two sides without the resistors contribute to V m emf, we have V m emf � � 0 � 2 0 � ˆzµ0I1u 2πr � � � � r � y0 � � ˆzdz � � � 0 0 � 2 � ˆzµ0I1u 2πr � � � � r � y0 � 0 � 1 � � ˆzdz � � 4π 10  7 10 7 5 0 2 2π � 1 y0 � 1 y0 � 0 1 � 3 10  6 � 1 y0 � 1 y0 � 0 1 (V) � and therefore I2 � V m emf 2R � 150 � 1 y0 � 1 y0 � 0 1 (nA) Problem 6.11 The conducting cylinder shown in Fig. 6-23 (P6.11) rotates about its axis at 1,200 revolutions per minute in a radial field given by B � ˆr6 (T) 300 CHAPTER 6 z 10cm + - V 5cm ω Sliding contact Figure P6.11: Rotating cylinder in a magnetic field (Problem 6.11). The cylinder, whose radius is 5 cm and height 10 cm, has sliding contacts at its top and bottom connected to a voltmeter. Determine the induced voltage. Solution: The surface of the cylinder has velocity u given by u � ˆφφφωr � ˆφφφ2π 1 � 200 60 5 10  2 � ˆφφφ2π (m/s) � V12 � � L 0 � u B � � dl � � 0 � 1 0 � ˆφφφ2π ˆr6 � � ˆz dz � � 3 77 (V) Problem 6.12 The electromagnetic generator shown in Fig. 6-12 is connected to an electric bulb with a resistance of 150 Ω. If the loop area is 0.1 m2 and it rotates at 3,600 revolutions per minute in a uniform magnetic flux density B0 � 0 4 T, determine the amplitude of the current generated in the light bulb. Solution: From Eq. (6.38), the sinusoidal voltage generated by the a-c generator is Vemf � AωB0 sin � ωt � C0 �� V0 sin � ωt � C0 � . Hence, V0 � AωB0 � 0 1 2π 3 � 600 60 0 4 � 15 08 (V) � I � V0 R � 15 08 150 � 0 1 (A) Problem 6.13 The circular disk shown in Fig. 6-24 (P6.13) lies in the x–y plane and rotates with uniform angular velocity ω about the z-axis. The disk is of radius a and is present in a uniform magnetic flux density B � ˆzB0. Obtain an expression for the emf induced at the rim relative to the center of the disk. CHAPTER 6 301 z y x a + - V ω Figure P6.13: Rotating circular disk in a magnetic field (Problem 6.13). y x φ u r Figure P6.13: (a) Velocity vector u. Solution: At a radial distance r, the velocity is u � ˆφφφωr where φ is the angle in the x–y plane shown in the figure. The induced voltage is V � � a 0 � u B � � dl � � a 0 � � ˆφφφωr � ˆzB0 � � ˆr dr ˆφφφ ˆz is along ˆr. Hence, V � ωB0 � a 0 r dr � ωB0a2 2 Section 6-7: Displacement Current Problem 6.14 The plates of a parallel-plate capacitor have areas 10 cm2 each and are separated by 2 cm. The capacitor is filled with a dielectric material with 302 CHAPTER 6 ε � 4ε0, and the voltage across it is given by V � t � � 30cos2π 106t (V). Find the displacement current. Solution: Since the voltage is of the form given by Eq. (6.46) with V0 � 30 V and ω � 2π 106 rad/s, the displacement current is given by Eq. (6.49): Id � � εA d V0ωsinωt � � 4 8 854 10  12 10 10  4 2 10  2 30 2π 106 sin � 2π 106t � � � 0 33sin � 2π 106t � (mA) Problem 6.15 A coaxial capacitor of length l � 6 cm uses an insulating dielectric material with εr � 9. The radii of the cylindrical conductors are 0.5 cm and 1 cm. If the voltage applied across the capacitor is V � t �� 50sin � 120πt � (V) � what is the displacement current? l r Id + - V(t) 2a 2b Figure P6.15: Solution: To find the displacement current, we need to know E in the dielectric space between the cylindrical conductors. From Eqs. (4.114) and (4.115), E � � ˆr Q 2πεrl � V � Q 2πεl ln � b a Hence, E � � ˆr V rln � b a � � � ˆr 50sin � 120πt � rln2 � � ˆr 72 1 r sin � 120πt � (V/m) � CHAPTER 6 303 D � εE � εrε0E � � ˆr9 8 85 10  12 72 1 r sin � 120πt � � � ˆr 5 75 10  9 r sin � 120πt � (C/m2) The displacement current flows between the conductors through an imaginary cylindrical surface of length l and radius r. The current flowing from the outer conductor to the inner conductor along � ˆr crosses surface S where S � � ˆr2πrl Hence, Id � ∂D ∂t � S � � ˆr ∂ ∂t � 5 75 10  9 r sin � 120πt � � � � ˆr2πrl � � 5 75 10  9 120π 2πl cos � 120πt � � 0 82cos � 120πt � (µA) Alternatively, since the coaxial capacitor is lossless, its displacement current has to be equal to the conduction current flowing through the wires connected to the voltage sources. The capacitance of a coaxial capacitor is given by (4.116) as C � 2πεl ln � b a � The current is I � C dV dt � 2πεl ln � b a � � 120π 50cos � 120πt �� � 0 82cos � 120πt � (µA) � which is the same answer we obtained before. Problem 6.16 The parallel-plate capacitor shown in Fig. 6-25 (P6.16) is filled with a lossy dielectric material of relative permittivity εr and conductivity σ. The separation between the plates is d and each plate is of area A. The capacitor is connected to a time-varying voltage source V � t � . (a) Obtain an expression for Ic, the conduction current flowing between the plates inside the capacitor, in terms of the given quantities. 304 CHAPTER 6 + - V(t) I A d ε, σ Figure P6.16: Parallel-plate capacitor containing a lossy dielectric material (Problem 6.16). (b) Obtain an expression for Id, the displacement current flowing inside the capacitor. (c) Based on your expression for parts (a) and (b), give an equivalent-circuit representation for the capacitor. (d) Evaluate the values of the circuit elements for A � 4 cm2, d � 0 5 cm, εr � 4, σ � 2 5 (S/m), and V � t �� 10cos � 3π 103t � (V). Solution: (a) R � d σA � Ic � V R � VσA d (b) E � V d � Id � ∂D ∂t � A � εA ∂E ∂t � εA d ∂V ∂t (c) The conduction current is directly proportional to V, as characteristic of a resistor, whereas the displacement current varies as ∂V ∂t, which is characteristic of a capacitor. Hence, R � d σA and C � εA d (d) R � 0 5 10  2 2 5 4 10  4 � 5 Ω � CHAPTER 6 305 + - V(t) I Id C Ic R V(t) + - ε, σ Actual Circuit Equivalent Circuit Figure P6.16: (a) Equivalent circuit. C � 4 8 85 10  12 4 10  4 0 5 10  2 � 2 84 10  12 F Problem 6.17 An electromagnetic wave propagating in seawater has an electric field with a time variation given by E � ˆzE0 cosωt. If the permittivity of water is 81ε0 and its conductivity is 4 (S/m), find the ratio of the magnitudes of the conduction current density to displacement current density at each of the following frequencies: (a) 1 kHz, (b) 1 MHz, (c) 1 GHz, (d) 100 GHz. Solution: From Eq. (6.44), the displacement current density is given by Jd � ∂ ∂t D � ε ∂ ∂t E and, from Eq. (4.67), the conduction current is J � σE. Converting to phasors and taking the ratio of the magnitudes, � � � � � � J � Jd � � � � � � � � � � � σ � E jωεrε0 � E � � � � � � σ ωεrε0 (a) At f � 1 kHz, ω � 2π 103 rad/s, and � � � � � � J � Jd � � � � � � 4 2π 103 81 8 854 10  12 � 888 103 The displacement current is negligible. (b) At f � 1 MHz, ω � 2π 106 rad/s, and � � � � � � J � Jd � � � � � � 4 2π 106 81 8 854 10  12 � 888 306 CHAPTER 6 The displacement current is practically negligible. (c) At f � 1 GHz, ω � 2π 109 rad/s, and � � � � � � J � Jd � � � � � � 4 2π 109 81 8 854 10  12 � 0 888 Neither the displacement current nor the conduction current are negligible. (d) At f � 100 GHz, ω � 2π 1011 rad/s, and � � � � � � J � Jd � � � � � � 4 2π 1011 81 8 854 10  12 � 8 88 10  3 The conduction current is practically negligible. Sections 6-9 and 6-10: Continuity Equation and Charge Dissipation Problem 6.18 At t � 0, charge density ρv0 was introduced into the interior of a material with a relative permittivity εr � 9. If at t � 1 µs the charge density has dissipated down to 10  3ρv0, what is the conductivity of the material? Solution: We start by using Eq. (6.61) to find τr: ρv � t �� ρv0e  t � τr � or 10  3ρv0 � ρv0e  10 � 6 � τr � which gives ln10  3 � � 10  6 τr � or τr � � 10  6 ln10  3 � 1 45 10  7 (s) But τr � ε σ � 9ε0 σ. Hence σ � 9ε0 τr � 9 8 854 10  12 1 45 10  7 � 5 5 10  4 (S/m) CHAPTER 6 307 Problem 6.19 If the current density in a conducting medium is given by J � x � y � z;t � � � ˆxz2 � ˆy4y2 � ˆz2x � cosωt � determine the corresponding charge distribution ρv � x � y � z;t � . Solution: Eq. (6.58) is given by ∇ � J � � ∂ρv ∂t (24) The divergence of J is ∇ � J � � ˆx ∂ ∂x � ˆy ∂ ∂y � ˆz ∂ ∂z � � ˆxz2 � ˆy4y2 � ˆz2x � cosωt � � 4 ∂ ∂y � y2 cosωt �� � 8ycosωt Using this result in Eq. (24) and then integrating both sides with respect to t gives ρv � � � � ∇ � J � dt � � � � 8ycosωt dt � 8y ω sinωt � C0 � where C0 is a constant of integration. Problem 6.20 In a certain medium, the direction of current density J points in the radial direction in cylindrical coordinates and its magnitude is independent of both φ and z. Determine J, given that the charge density in the medium is ρv � ρ0rcosωt � C/m3 � Solution: Based on the given information, J � ˆrJr � r � With Jφ � Jz � 0, in cylindrical coordinates the divergence is given by ∇ � J � 1 r ∂ ∂r � rJr � From Eq. (6.54), ∇ � J � � ∂ρv ∂t � � ∂ ∂t � ρ0rcosωt �� ρ0rωsinωt 308 CHAPTER 6 Hence 1 r ∂ ∂r � rJr �� ρ0rωsinωt � ∂ ∂r � rJr �� ρ0r2ωsinωt � � r 0 ∂ ∂r � rJr � dr � ρ0ωsinωt � r 0 r2 dr � rJr � r 0 � � ρ0ωsinωt � r3 3 � � � � r 0 � Jr � ρ0ωr2 3 sinωt � and J � ˆrJr � ˆr ρ0ωr2 3 sinωt (A/m2) Problem 6.21 If we were to characterize how good a material is as an insulator by its resistance to dissipating charge, which of the following two materials is the better insulator? Dry Soil: εr � 2 5, σ � 10  4 (S/m) Fresh Water: εr � 80, σ � 10  3 (S/m) Solution: Relaxation time constant τr � ε σ . For dry soil, τr � 2 5 10  4 � 2 5 104 s. For fresh water, τr � 80 10  3 � 8 104 s. Since it takes longer for charge to dissipate in fresh water, it is a better insulator than dry soil. Sections 6-11: Electromagnetic Potentials Problem 6.22 The electric field of an electromagnetic wave propagating in air is given by E � z � t �� ˆx4cos � 6 108t � 2z � � ˆy3sin � 6 108t � 2z � (V/m) CHAPTER 6 309 Find the associated magnetic field H � z � t � . Solution: Converting to phasor form, the electric field is given by � E � z �� ˆx4e  j2z � jˆy3e  j2z (V/m) � which can be used with Eq. (6.87) to find the magnetic field: � H � z �� 1 � jωµ∇ � E � 1 � jωµ � � � � � � ˆx ˆy ˆz ∂ ∂x ∂ ∂y ∂ ∂z 4e  j2z � j3e  j2z 0 � � � � � � � 1 � jωµ � ˆx6e  j2z � ˆyj8e  j2z � � j 6 108 4π 10  7 � ˆx6 � ˆyj8 � e  j2z � jˆx8 0e  j2z � ˆy10 6e  j2z (mA/m) Converting back to instantaneous values, this is H � t � z �� � ˆx8 0sin � 6 108t � 2z � � ˆy10 6cos � 6 108t � 2z � (mA/m) Problem 6.23 The magnetic field in a dielectric material with ε � 4ε0, µ � µ0, and σ � 0 is given by H � y � t �� ˆx5cos � 2π 107t � ky � (A/m) Find k and the associated electric field E. Solution: In phasor form, the magnetic field is given by � H � ˆx5ejky (A/m). From Eq. (6.86), � E � 1 jωε∇ � H � � jk jωε ˆz5ejky and, from Eq. (6.87), � H � 1 � jωµ∇ � E � � jk2 � jω2εµ ˆx5ejky � which, together with the original phasor expression for � H, implies that k � ω � εµ � ω � εr c � 2π 107 � 4 3 108 � 4π 30 (rad/m) 310 CHAPTER 6 Inserting this value in the expression for � E above, � E � � ˆz 4π 30 2π 107 4 8 854 10  12 5ej4πy � 30 � � ˆz941e j4πy � 30 (V/m) Problem 6.24 Given an electric field E � ˆxE0 sinaycos � ωt � kz � � where E0, a, ω, and k are constants, find H. Solution: E � ˆxE0 sinaycos � ωt � kz � � � E � ˆxE0 sinaye  jkz � � H � � 1 jωµ ∇ � E � � 1 jωµ � ˆy ∂ ∂z � E0 sinaye  jkz � � ˆz ∂ ∂y � E0 sinaye  jkz � � � E0 ωµ � ˆyksinay � ˆz jacosay � e  jkz � H � �� � � Hejωt � � �� � E0 ωµ � ˆyksinay � ˆzacosaye  jπ � 2 � e  jkzejωt � � E0 ωµ � ˆyksinaycos � ωt � kz � � ˆzacosaycos  ωt � kz � π 2  � � E0 ωµ � ˆyksinaycos � ωt � kz � � ˆzacosaysin � ωt � kz �� Problem 6.25 The electric field radiated by a short dipole antenna is given in spherical coordinates by E � R � θ;t �� ˆθθθ2 10  2 R sinθ cos � 6π 108t � 2πR � (V/m) Find H � R � θ;t � . Solution: Converting to phasor form, the electric field is given by � E � R � θ �� ˆθθθEθ � ˆθθθ2 10  2 R sinθe  j2πR (V/m) � CHAPTER 6 311 which can be used with Eq. (6.87) to find the magnetic field: � H � R � θ �� 1 � jωµ∇ � E � 1 � jωµ � ˆR 1 Rsinθ ∂Eθ ∂φ � ˆφφφ1 R ∂ ∂R � REθ � � � 1 � jωµ ˆφφφ 2 10  2 R sinθ ∂ ∂R � e  j2πR � � ˆφφφ 2π 6π 108 4π 10  7 2 10  2 R sinθe  j2πR � ˆφφφ 53 R sinθe  j2πR (µA/m) Converting back to instantaneous value, this is H � R � θ;t �� ˆφφφ 53 R sinθcos � 6π 108t � 2πR � (µA/m) Problem 6.26 A Hertzian dipole is a short conducting wire carrying an approximately constant current over its length l. If such a dipole is placed along the z-axis with its midpoint at the origin and if the current flowing through it is i � t �� I0 cosωt, find (a) the retarded vector potential � A � R � θ � φ � at an observation point Q � R � θ � φ � in a spherical coordinate system, and (b) the magnetic field phasor � H � R � θ � φ � . Assume l to be sufficiently small so that the observation point is approximately equidistant to all points on the dipole; that is, assume that R � � R. Solution: (a) In phasor form, the current is given by � I � I0. Explicitly writing the volume integral in Eq. (6.84) as a double integral over the wire cross section and a single integral over its length, � A � µ 4π � l � 2  l � 2 � � s � J � Ri � e  jkR � R � dsdz � where s is the wire cross section. The wire is infinitesimally thin, so that R � is not a function of x or y and the integration over the cross section of the wire applies only to the current density. Recognizing that � J � ˆzI0 s, and employing the relation R � � R, � A � ˆzµI0 4π � l � 2  l � 2 e  jkR � R � dz � ˆzµI0 4π � l � 2  l � 2 e  jkR R dz � ˆz µI0l 4πRe  jkR 312 CHAPTER 6 In spherical coordinates, ˆz � ˆRcosθ � ˆθθθsinθ, and therefore � A � � ˆRcosθ � ˆθθθsinθ � µI0l 4πRe  jkR (b) From Eq. (6.85), � H � 1 µ∇ � A � I0l 4π∇ � � ˆRcosθ � ˆθθθsinθ � e  jkR R � � I0l 4π ˆφφφ 1 R � ∂ ∂R  � sinθe  jkR  � ∂ ∂θ � cosθe  jkR R � ˆφφφ I0l sinθe  jkR 4πR � jk � 1 R Problem 6.27 The magnetic field in a given dielectric medium is given by H � ˆy6cos2zsin � 2 107t � 0 1x � (A/m) � where x and z are in meters. Determine: (a) E, (b) the displacement current density Jd, and (c) the charge density ρv. Solution: (a) H � ˆy6cos2zsin � 2 107t � 0 1x �� ˆy6cos2zcos � 2 107t � 0 1x � π 2 � � � H � ˆy6cos2ze  j0 � 1xe  jπ � 2 � � ˆy j6cos2ze  j0 � 1x � � E � 1 jωε ∇ � H � 1 jωε � � � � � � ˆx ˆy ˆz ∂ ∂x ∂ ∂y ∂ ∂z 0 � j6cos2ze  j0 � 1x 0 � � � � � � � 1 jωε � ˆx � � ∂ ∂z � � j6cos2ze  j0 � 1x � � � ˆz � ∂ ∂x � � j6cos2ze  j0 � 1x � � � � ˆx � � 12 ωε sin2ze  j0 � 1x � ˆz � j0 6 ωε cos2ze  j0 � 1x From the given expression for H, ω � 2 107 (rad/s) � CHAPTER 6 313 β � 0 1 (rad/m) Hence, up � ω β � 2 108 (m/s) � and εr � � c up 2 � � 3 108 2 108 2 � 2 25 Using the values for ω and ε, we have � E � � � ˆx30sin2z � ˆz j1 5cos2z �� 103e  j0 � 1x (V/m) � E � � � ˆx30sin2zcos � 2 107t � 0 1x � � ˆz1 5cos2zsin � 2 107t � 0 1x � � (kV/m) (b) � D � ε � E � εrε0 � E � � � ˆx0 6sin2z � ˆz j0 03cos2z �� 10  6e  j0 � 1x (C/m2) � Jd � ∂D ∂t � or � Jd � jω � D � � � ˆx j12sin2z � ˆz0 6cos2z � e  j0 � 1x � Jd � �� � � Jdejωt � � � ˆx12sin2zsin � 2 107t � 0 1x � � ˆz0 6cos2zcos � 2 107t � 0 1x � � (A/m2) (c) We can find ρv from ∇ � D � ρv or from ∇ � J � � ∂ρv ∂t Applying Maxwell’s equation, ρv � ∇ � D � ε∇ � E � εrε0 � ∂Ex ∂x � ∂Ez ∂z yields ρv � εrε0 � ∂ ∂x � � 30sin2zcos � 2 107t � 0 1x �� � ∂ ∂z � � 1 5cos2zsin � 2 107t � 0 1x � � � � εrε0 � � 3sin2zsin � 2 107t � 0 1x � � 3sin2zsin � 2 107t � 0 1x � � � 0 314 CHAPTER 6 Problem 6.28 The transformer shown in the figure consists of a long wire coincident with the z-axis carrying a current I � I0 cosωt, coupling magnetic energy to a toroidal coil situated in the x–y plane and centered at the origin. The toroidal core uses iron material with relative permeability µr, around which 100 turns of a tightly wound coil serves to induce a voltage Vemf, as shown in the figure. a b x y z I Vemf c N + - Iron core with µr (a) Develop an expression for Vemf. (b) Calculate Vemf for f � 60 Hz, µr � 4000, a � 5 cm, b � 6 cm, c � 2 cm, and I0 � 50 A. Solution: (a) We start by calculating the magnetic flux through the coil, noting that r, the distance from the wire varies from a to b Φ � � S B � ds � � b a ˆx µI 2πr � ˆxc dr � µcI 2π ln � b a Vemf � � N dΦ dt � � µcN 2π ln � b a dI dt � µcNωI0 2π ln � b a sinωt (V) CHAPTER 6 315 (b) Vemf � 4000 4π 10  7 2 10  2 100 2π 60 50ln � 6 5 � 2π sin377t � 5 5sin377t (V) Problem 6.29 In wet soil, characterized by σ � 10  2 (S/m), µr � 1, and εr � 36, at what frequency is the conduction current density equal in magnitude to the displacement current density? Solution: For sinusoidal wave variation, the phasor electric field is E � E0ejωt Jc � σE � σE0ejωt Jd � ∂D ∂t � ε∂E ∂t � jωεE0ejωt � � � � Jc Jd � � � � � 1 � σ ωε � σ 2πε f or f � σ 2πε � 10  2 2π 36 8 85 10  12 � 5 106 � 5 MHz Problem 6.30 In free space, the magnetic field is given by H � ˆφφφ 36 r cos � 6 109t � kz � (mA/m) (a) Determine k. (b) Determine E. (c) Determine Jd. Solution: (a) From the given expression, ω � 6 109 (rad/s), and since the medium is free space, k � ω c � 6 109 3 108 � 20 (rad/m) 316 CHAPTER 6 (b) Convert H to phasor: � H � ˆφφφ 36 r e  jkz (mA/m) � E � 1 jωε0 ∇ � H � 1 jωε0 � � ˆr ∂Hφ ∂z � ˆz 1 r ∂ ∂r � rHφ � � � 1 jωε0 � � ˆr ∂ ∂z � 36 r e  jkz � ˆz r ∂ ∂r � 36e  jkz � � � 1 jωε0 � ˆr j36k r e  jkz � � ˆr 36k ωε0r e  jkz � ˆr 36 377 r e  jkz 10  3 � ˆr 13 6 r e  j20z (V/m) E � �� � � Eejωt � � ˆr 13 6 r cos � 6 109t � 20z � (V/m) (c) Jd � ε0 ∂E ∂t � ˆr 13 6 r ε0 ∂ ∂t � cos � 6 109t � 20z � � � � ˆr 13 6ε0 6 109 r sin � 6 109t � 20z � (A/m2) � � ˆr 0 72 r sin � 6 109t � 20z � (A/m2)