45791-UlabyISMCh06

45791-UlabyISMCh06 - 286 Chapter 6: Maxwell's Equations for...

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Unformatted text preview: 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 I + R1 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 xy plane and B zB0 sin t ^ ^ with B0 positive. What is the direction of I ( or ) at (a) t 0, (b) t 4, and (c) t 2? Solution: I Vemf R. Since the single-turn loop is not moving or changing shape m tr with time, Vemf 0 V and Vemf Vemf . Therefore, from Eq. (6.8), ^ If we take the surface normal to be z, then the right hand rule gives positive ^ direction. flowing current to be in the I A B0 sin t R t AB0 cos t R I tr Vemf R 1 R B ds S t (A) CHAPTER 6 z 293 R I x y Figure P6.2: Loop of Problem 6.2. where A is the area of the loop. (a) A, and R are positive quantities. At t 0, cos t 1 so I 0 and the ^ current is flowing in the direction (so as to produce an induced magnetic field that opposes B). ^ 2 2 so I 0 and the current is still flowing in the (b) At t 4, cos t direction. (c) At t 2, cos t 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 z 20e 3t (T), Vemf 100 3t d 20e dt 0 25 2 375e 3t (V) where N 100 and the surface normal was chosen to be in the ^ (a) For B z20e 3t (T), S 0 125 0 125 ^ z direction. Vemf N B ds N ^ B z dx dy d dt d dt 0 125 0 125 m Solution: Since the coil is not moving or changing shape, Vemf tr . From Eq. (6.6), Vemf Vemf (c) B (b) B ^ z 20 cos x cos 103 t (T), ^ z 20 cos x sin 2y cos 103 t (T). 0 V and 294 CHAPTER 6 ^ z20 cos x cos 103 t (T), Problem 6.4 A stationary conducting loop with internal resistance of 0.5 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- 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 , When the small gap is created, the total resistance in the loop is infinite and the current flow is zero. With a 2- resistor in the gap, Problem 6.5 A circular-loop TV antenna with 0.02 m 2 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 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 t 0 with 2 300 106 rad/s and 0 an arbitrary reference phase. From Eq. (6.6), 30 10 3 AB0 Vemf is maximum when sin t 0 1. Hence, 0 02 B0 6 108 Vemf N A 0 AB0 sin t d dt d B0 cos t dt 0 I Vemf 2 05 25V 25 Vemf 5A 05 25V x 0 125 y 0 125 1 (A) Vemf 100 20 cos 103 t d dt (c) For B ^ z20 cos x sin 2y cos 103 t (T), 0 125 x 0 125 y 0 125 0 125 cos x sin 2y dx dy 0 Vemf 100 20 cos 103 t d dt (b) For B 0 125 0 125 cos x dx dy 124 6 sin 103 t (kV) CHAPTER 6 295 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 (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- resistor connected across the gap. The loop has an internal resistance of 1 . z 10cm I(t) 10cm 5cm y x Figure P6.6: Loop coplanar with long wire (Problem 6.6). Solution: (a) The magnetic field due to the wire is ^ where in the plane of the loop, ^ x and r y. The flux passing through the loop B ^ 0 I 2r ^ x 0 I 2y I t which yields B0 5 cos 2 104 t (A) 296 is CHAPTER 6 S 5 cm (b) ^ At t 0, B is a maximum, it points in x-direction, and since it varies as 4 t , it is decreasing. Hence, the induced current has to be CCW when cos 2 10 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 Determine the current induced in the loop if its internal resistance is 0 5 . Solution: S Iind Vemf 10 3 sin 200t t t 6 103 t 200t (rad/s) 60 3 10 5 cos 200t (Wb) d 3 10 5 200 sin 200t 18 85 dt Vemf 37 7 sin 200t (mA) 05 2 (V) B dS ^ y 50 10 3 ^ y2 3 10 4 cos t 3 10 5 cos t B ^ y 50 (mT) Iind sin 2 104 t 1 38 sin 2 Vemf 4 1 69 10 5 3 69 10 3 sin 2 104 t (V) 104 t Vemf 11 2 104 sin 2 d dt 104 t 11 10 10 7 0 I 10 2 4 10 B ds 15 cm 0 I ^ x 10 (cm) dy 2y 1 15 ln 5 7 5 cos 2 104 t 10 1 2 7 cos 2 104t (Wb) ^ x 11 (mA) CHAPTER 6 z 297 2c m B 3cm B y (t) x Figure P6.7: Rotating loop in a magnetic field (Problem 6.7). ^ The direction of the current is CW (if looking at it along x-direction) when the loop is in the first quadrant (0 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: 2 rad/cycle 7200 cycles/min 240 rad/s 60 s/min 5 cm 10 cm 100 cm/m 2 5 0 10 3 m2 3 6 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 z 2 10 4 T. 2 rad/cycle 90 cycles/min 60 s/min Solution: Since B is constant, Vemf ^ is given by u r, where m Vemf . The velocity u for any point on the bar 3 rad/s Vemf peak AB0 50 10 240 6 10 22 62 From Eqs. (6.36) or (6.38), Vemf AB0 sin t; it can be seen that the peak voltage is V A 298 z CHAPTER 6 B 1 y 2 x Figure P6.9: Rotating rod of Problem 6.9. From Eq. (6.24), 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 y7 5 (m/s). If R 10 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): C ^ The magnetic field B is created by the wire carrying I1 . Choosing z to coincide with the direction of I1 , Eq. (5.30) gives the external magnetic field of a long wire to be B ^ 0 I1 2r Vemf u B dl m Vemf 3 10 4 0 25 236 3 10 r 05 6 10 4 r 05 0 4 2 2 r 05 0 r dr V12 u B dl ^ z2 10 m Vemf 1 0 ^ 3r 4 ^ r dr V CHAPTER 6 z 10 cm R I1 = 10 A 20 cm I2 u R y0 u 299 Figure P6.10: Moving loop of Problem 6.10. ^ z dz and the limits of Integrating around the four sides of the loop with dl integration chosen in accordance with the assumed direction of I2 , and recognizing m that only the two sides without the resistors contribute to Vemf , we have and therefore 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 ^ r 6 (T) y0 01 I2 m Vemf 2R 150 1 y0 3 10 6 1 (nA) 4 10 7 10 7 5 0 2 1 1 2 y0 y0 0 1 1 1 (V) y0 y0 0 1 0 r y0 02 r y0 0 1 m Vemf ^ z ^ z dz ^ z 02 0 I1 u 2r 0 0 I1 u 2r ^ z dz u B ^ yu B ^ ru ^ For positive values of y0 in the y-z plane, y ^ r, so ^ 0 I1 2r ^ z 0 I1 u 2r 300 z CHAPTER 6 5cm 10cm + - V 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 0 0 Problem 6.12 The electromagnetic generator shown in Fig. 6-12 is connected to an electric bulb with a resistance of 150 . If the loop area is 0.1 m 2 and it rotates at 3,600 revolutions per minute in a uniform magnetic flux density B 0 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 V0 sin t C0 . Hence, Vemf AB0 sin t C0 Problem 6.13 The circular disk shown in Fig. 6-24 (P6.13) lies in the xy plane and rotates with uniform angular velocity about the z-axis. The disk is of radius a ^ and is present in a uniform magnetic flux density B zB0 . Obtain an expression for the emf induced at the rim relative to the center of the disk. I V0 R 15 08 150 01 (A) V0 AB0 01 2 3 600 60 04 15 08 (V) V12 u B dl ^ ^ r 6 z dz L 01 ^ 2 3 77 (V) u ^ r ^ 2 1 200 60 5 10 2 ^ 2 (m/s) CHAPTER 6 z 301 - V + a y x Figure P6.13: Rotating circular disk in a magnetic field (Problem 6.13). x r u y Figure P6.13: (a) Velocity vector u. Solution: At a radial distance r, the velocity is where is the angle in the xy plane shown in the figure. The induced voltage is 0 0 0 Section 6-7: Displacement Current Problem 6.14 The plates of a parallel-plate capacitor have areas 10 cm 2 each and are separated by 2 cm. The capacitor is filled with a dielectric material with V B0 ^ ^ ^ z is along r. Hence, a r dr B0 a2 2 V u B dl ^ ^ z B0 r dr a u ^ r a ^ r 302 CHAPTER 6 Problem 6.15 A coaxial capacitor of length l 6 cm uses an insulating dielectric material with r 9. The radii of the cylindrical conductors are 0.5 cm and 1 cm. If the voltage applied across the capacitor is what is the displacement current? l + r V(t) - 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), Hence, b a E ^ r ^ r ^ r V r ln 50 sin 120t r ln 2 V Q b ln 2l a E ^ r Q 2rl 72 1 sin 120t r V t 50 sin 120t (V) Id 2b (V/m) 30 2 106 sin 2 Id A V0 sin t d 4 8 854 10 12 10 10 2 10 2 0 33 sin 2 106 t (mA) 4 106 t Solution: Since the voltage is of the form given by Eq. (6.46) with V0 2 106 rad/s, the displacement current is given by Eq. (6.49): 40 , and the voltage across it is given by V t displacement current. 30 cos 2 10 6 t (V). Find the 30 V and 2a CHAPTER 6 303 r 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 r crosses surface S where Hence, 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 The current is 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 r and conductivity . 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. I C 50 cos 120t 0 82 cos 120t dV dt 2l 120 ln b a C 2l ln b a 0 82 cos 120t (A) 5 75 10 9 120 2l cos 120t r Id ^ r sin 120t D S t t 5 75 10 9 S ^ r 2rl ^ r sin 120t 5 75 10 9 (C/m2 ) ^ r9 8 85 10 12 D E r 0 E 72 1 sin 120t r ^ r 2rl (A) 304 I A CHAPTER 6 + V(t) - , d 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 cm 2 , d 0 5 cm, r 4, 2 5 (S/m), and V t 10 cos 3 103 t (V). Solution: (a) E V D A V Id A A d t t d 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, d A and C R A d (d) 4 (b) R 0 5 10 2 2 5 4 10 5 E R d A Ic V R V A d CHAPTER 6 I + 305 V(t) - , R Ic Id C + V(t) - Actual Circuit Equivalent Circuit Figure P6.16: (a) Equivalent circuit. Problem 6.17 An electromagnetic wave propagating in seawater has an electric ^ field with a time variation given by E zE0 cos t. If the permittivity of water is 810 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 J 4 The displacement current is negligible. (b) At f 1 MHz, 2 106 rad/s, and J 4 Jd 2 106 81 8 854 10 12 Jd 2 103 81 8 854 10 12 888 (a) At f 2 103 rad/s, and 1 kHz, Jd jr 0 E J E r 0 and, from Eq. (4.67), the conduction current is J taking the ratio of the magnitudes, Jd D t E t E. Converting to phasors and 888 103 C 4 8 85 10 12 4 0 5 10 2 10 4 2 84 10 12 F 306 The displacement current is practically negligible. (c) At f 1 GHz, 2 109 rad/s, and J CHAPTER 6 Neither the displacement current nor the conduction current are negligible. (d) At f 100 GHz, 2 1011 rad/s, and 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 v0 was introduced into the interior of a material with a relative permittivity r 9. If at t 1 s the charge density has dissipated down to 10 3 v0 , what is the conductivity of the material? Solution: We start by using Eq. (6.61) to find r : t r or which gives or But r 90 . Hence 90 r 9 8 854 10 1 45 10 7 12 55 10 4 r 10 6 ln 10 3 1 45 10 ln 10 3 10 6 r 7 10 3 v0 v0 e 10 6 v t v0 e r Jd 2 1011 81 8 854 10 12 (s) (S/m) J 4 Jd 2 109 81 8 854 10 12 8 88 10 3 4 0 888 CHAPTER 6 Problem 6.19 If the current density in a conducting medium is given by 307 determine the corresponding charge distribution v x y z;t . Solution: Eq. (6.58) is given by The divergence of J is Using this result in Eq. (24) and then integrating both sides with respect to t gives 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 and z. Determine J, given that the charge density in the medium is Solution: Based on the given information, From Eq. (6.54), J v t 0 r cos t t J With J Jz 0, in cylindrical coordinates the divergence is given by 1 rJr r r J ^ r Jr r v 0 r cos t C/m3 0 r sin t v J dt 8y cos t dt 8y cos t 8y sin t J ^ x ^ ^ y z x y z 2 4 y cos t y ^ xz2 J v t ^ y4y2 ^ z2x cos t (24) C0 J x y z;t ^ xz2 ^ y4y2 ^ z2x cos t 308 Hence CHAPTER 6 0 0 and 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? 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 given by Ezt The electric field of an electromagnetic wave propagating in air is 108 t ^ x4 cos 6 2z ^ y3 sin 6 For fresh water, r 8 For dry soil, r Solution: Relaxation time constant r . 25 10 4 80 10 3 25 104 s. 104 s. 108 t Dry Soil: Fresh Water: r r 2 5, 80, 10 10 4 3 (S/m) (S/m) 2z J ^ r Jr ^ r 0 r2 sin t 3 Jr 0 r2 sin t 3 rJr r 0 0 sin t r3 3 r 0 sin t 0 r2 sin t r 1 rJr r r rJr r r rJr dr 0 r 0 r sin t r2 dr (A/m2 ) (V/m) CHAPTER 6 Find the associated magnetic field H z t . Solution: Converting to phasor form, the electric field is given by 309 which can be used with Eq. (6.87) to find the magnetic field: Converting back to instantaneous values, this is and, from Eq. (6.87), which, together with the original phasor expression for H, implies that k r c 2 107 4 3 108 4 30 H 1 E j E 1 H j jk ^ z5e jky j jk2 ^ x5e jky j2 (rad/m) Solution: In phasor form, the magnetic field is given by H Eq. (6.86), Find k and the associated electric field E. Hyt ^ x5 cos 2 107 t ky (A/m) ^ x5e jky (A/m). From Problem 6.23 The magnetic field in a dielectric material with 0 is given by 4 0 , 0 , and Ht z ^ x8 0 sin 6 108 t 2z ^ y10 6 cos 6 108 t 2z (mA/m) Hz 1 E j ^ ^ ^ x y z 1 y z x j j3e j2z 4e j2z 0 1 ^ ^ x6e j2z y j8e j2z j j ^ ^ x6 y j8 e j2z 8 6 10 4 10 7 ^ jx8 0e j2z ^ y10 6e j2z Ez ^ x4e j2z ^ jy3e j2z (V/m) (mA/m) 310 Inserting this value in the expression for E above, 2 107 10 12 CHAPTER 6 Problem 6.24 Given an electric field where E0 , a, , and k are constants, find H. Solution: Problem 6.25 The electric field radiated by a short dipole antenna is given in spherical coordinates by Find H R ;t . Solution: Converting to phasor form, the electric field is given by ER ^ E ^2 10 R 2 sin e j2R (V/m) E R ;t sin cos 6 ^2 10 R 2 108t 2R ^ z a cos ay sin t kz (V/m) ^ z a cos ay cos t kz E0 ^ y k sin ay cos t kz E0 ^ y k sin ay cos t kz ^ z a cos ay e j 2 E0 ^ y k sin ay H He jt jkz e jkz jt e ^ z H E ^ x E0 sin ay e jkz 1 E j 1 ^ y E0 sin ay e jkz j z E0 ^ ^ y k sin ay z ja cos ay e E ^ x E0 sin ay cos t kz E0 sin ay e y E ^ xE0 sin ay cos t kz jkz 2 E ^ z 5e j4y 30 ^ z941e j4y 4 30 4 8 854 30 (V/m) CHAPTER 6 which can be used with Eq. (6.87) to find the magnetic field: 311 Converting back to instantaneous value, this is 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 I0 cos t, find it (a) the retarded vector potential A R at an observation point Q R in a spherical coordinate system, and (b) the magnetic field phasor H R . 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, 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 zI0 s, and employing the relation R R, l 2 l 2 R R A ^ z dz ^ z I0 4 l 2 e jkR I0 4 l 2 e jkR dz ^ z I0 l e 4R jkR l 2 A 4 l 2 J Ri e R s jkR ds dz H R ;t ^ 53 sin cos 6 R 108 t 2R (A/m) HR 1 E j 1 1 E ^ 1 ^ R RE j R sin R R 1 ^ 2 10 2 e j2R sin j R R 2 10 2 2 ^ sin e 6 108 4 10 7 R ^ 53 sin e j2R (A/m) R j2R 312 CHAPTER 6 ^ R cos ^ sin , and therefore (b) From Eq. (6.85), R Problem 6.27 The magnetic field in a given dielectric medium is given by where x and z are in meters. Determine: (a) E, (b) the displacement current density J d , and (c) the charge density v . Solution: (a) From the given expression for H, 2 107 (rad/s) ^ x j0 1 x ^ z 12 sin 2z e j0 6 cos 2z e j0 1 x ^ x j6 cos 2z e j0 1 x ^ z 1 j z E ^ z z j0 1 x 0 x H ^ y 6 cos 2z e j0 1 x e j 2 1 H j ^ ^ x y 1 y x j 0 j6 cos 2z e ^ y j6 cos 2z e j0 1 x j6 cos 2z e j0 1 x H ^ y 6 cos 2z sin 2 107 t 0 1x ^ y 6 cos 2z cos 2 107 t H ^ y 6 cos 2z sin 2 107 t 0 1x (A/m) 0 1x cos H 1 A I0 l e jkR ^ ^ R cos sin 4 R I0 l ^ 1 sin e jkR 4 R R jkR 1 ^ I0 l sin e jk 4R R e jkR A ^ R cos ^ In spherical coordinates, z I0 l ^ sin e 4R jkR 2 CHAPTER 6 313 Hence, Using the values for and , we have 7 (b) or (c) We can find v from or from Applying Maxwell's equation, yields r 0 3 sin 2z sin 2 107 t 0 1x 3 sin 2z sin 2 1 5 cos 2z sin 2 107 t 0 1x v r 0 30 sin 2z cos 2 x z 107 t 0 1x v D E r 0 J D v v t Ex x Ez z 107t 0 1x 0 ^ x 12 sin 2z sin 2 107 t 0 1x ^ z 0 6 cos 2z cos 2 Jd Jd e jt Jd jD Jd ^ x j12 sin 2z ^ z 0 6 cos 2z e j0 1 x 107 t 0 1x (A/m2 ) D E D t r 0 E ^ x 0 6 sin 2z ^ z j0 03 cos 2z 10 6 e j0 1 x (C/m2 ) E ^ x 30 sin 2z cos 2 10 t 0 1x ^ z 1 5 cos 2z sin 2 E ^ x 30 sin 2z ^ z j1 5 cos 2z 103 e j0 1 x (V/m) r c up 2 and 3 2 108 108 2 2 25 107t 0 1x (kV/m) up 2 108 (m/s) 01 (rad/m) 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 cos t, coupling magnetic energy to a toroidal coil situated in the xy plane and centered at the origin. The toroidal core uses iron material with relative permeability r , around which 100 turns of a tightly wound coil serves to induce a voltage Vemf , as shown in the figure. z I Vemf + N b a c y x (a) Develop an expression for Vemf . 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 Vemf S B ds I cI b ^ ln x c dr 2r 2 a a d cN b dI N ln dt 2 a dt cNI0 b sin t (V) ln 2 a b ^ x (b) Calculate Vemf for f I0 50 A. 60 Hz, r 4000, a Iron core with r 5 cm, b 6 cm, c 2 cm, and CHAPTER 6 (b) 315 Problem 6.29 In wet soil, characterized by 10 2 (S/m), r 1, and r 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 Problem 6.30 In free space, the magnetic field is given by (a) Determine k. (b) Determine E. (c) Determine Jd . Solution: (a) From the given expression, space, 6 k c 3 109 108 6 109 (rad/s), and since the medium is free 20 (rad/m) H ^ 36 cos 6 r 109 t 2 10 12 kz (mA/m) f 10 2 36 8 85 2 or Jd Jc E E0 e jt E D jE0 e jt t t Jc 1 Jd 2 f 5 106 5 MHz E E0 e jt Vemf 2 5 5 sin 377t (V) 4000 4 10 7 2 10 2 100 2 60 50 ln 6 5 sin 377t 316 (b) Convert H to phasor: CHAPTER 6 (c) Jd 0 E t 13 6 ^ r 0 cos 6 109 t 20z r t 13 60 6 109 ^ r sin 6 109 t 20z r 0 72 ^ sin 6 109 t 20z (A/m2 ) r r E (A/m2 ) E H ^ 36 e jkz (mA/m) r 1 H j0 H 1 1 ^ ^ r rH z j0 z r r ^ z 36 jkz 1 ^ 36e jkz e r j0 z r r r j36k jkz 1 ^ r e j0 r 36k 36 377 jkz 13 6 ^ ^ ^ 10 3 r e jkz r e e r 0 r r r Ee jt 13 6 ^ r cos 6 109 t 20z (V/m) r j20z (V/m) ...
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This note was uploaded on 02/09/2011 for the course EE 172 taught by Professor . during the Spring '10 term at UCLA.

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