Electromagnetics for Engineers

Electromagnetics for Engineers

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Unformatted text preview: CHAPTER 6 219 Shaptet' 6 Sections 6-1 to 6-6: Faraday’s Law and its Applications Problem 6.1 The switch in the bottom 100}: of Fig. 6-17 (Pol) is closed at r = 0 and then opened at a. later time :1. What is the direction of the current I in the top loop (clockwise or counterclockwise) at each of these two times? Figure P6.1: Loops of Problem 6.1. Solution: The magnetic coupling will be strongest at the point where the wires of the two 100ps come closest. When the switch is closed the current in the bottom 100p 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 counterciockwise Problem 6.2 The loop in Fig- 6-18 (136-2) is in the x—y plane and B = $50 sine): with Bo positive. What is the direction of I (h or 453) at (a) r = o, (b) {or = 1t/4, and (c) cur : 112/2? Solution: I = Vemf/R. Since the single-turn loop is not moving or changing shape With time, Vg‘flfi = O V and Vemf 2 ngf. Therefore, from Eq. {6.8}, if we take the surface normal to he +i, then the right hand rule gives positive flowing current to be in the +4: direction. —A 3 -AB (0 I = ———30 sine)! = 0 RBI cos cur (A), .. matin..:..:-m:. - 220 CHAS-HER 6 Figure P62: Loop of Problem 6.2. where A is the area of the loop. (a) A, to and R are positive quantifies. At t = 0, cosmr = 1 so I < 0 and the current is flowing in the me direction (so as to produce an induced magnetic field that opposes B). (b) At tot = T's/4, cos (or = x/i/Z so I < 0 and the current is still flowing in the 4?: direction. {c} At a): = 75/2, cos {or = 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 die open-circuited ends of the coil if the magnetic field is given by (a) B = 2 my?“ (T), (b) B = 210cm (£6th ('1‘), (c) B = a 10cosx sin2y cos 1031' (T). Solufion: Since the coil is not moving or changing shape, Ve‘g‘f = 0V and Vunf = mef. From Eq. (6.6), d d 0.125 0.125 rim: FN— feds = —N—j[ e-(adxdy), dr 5 dr -0125 —o.125 where N = 100 and the surface normal was chosen to be in the +3 direction- (a) For B = 3108—2: (T), d and: —100-&;(10e_2‘(0.25)2) = 125x” (V)- CHAPTER 6 221 (b) For n = tiOcosxcos 10% (T), d 0.125 0.:25 Vemf = — 300-— (10c051032‘f ff cosxdxdy) = 62.3 sin 103: (kV)- dt m—mzs y=—0.125 (c) For B = 210 cosxsin2ycos 103: (T), d 0.125 0.125 Vemfz —100— (10c03103tf / cosxsinlydxdy) :- 0. dz m—mzs y=—0.125 Problem 6.4 A stationary conducting 100p with internal resistance of 0.5 .Q is placed in a time-varying magnetic field. When the loop is closed, a current of 2.5 A flows through it. What will the current be if the loop is opened to create a small gap and a 242 resistor is connected across its open ends? Sointion: Vemf is independent of the resistance which is in the 100p- Therefore, when the loop is intact and the internal resistance is only 0-5 9, Vm=2.5Ax0.5n= 1.25 v. When the Small gap is created, the total resistance in the loop is infinite and. the current flow is zero. With a 2-9 resistor in the gap, 1 = rem/(znws n)=1.25 W25 n = 0.5 (A). Problem 6.5 A circular—loop TV antenna with 0.01 1112 area is in the presence of a nnifonn—arnplitude 300-MI-Iz signal. When oriented for maximum response, the ioop develops an emf with a peak value of 20 (mV). What is the peak magnitude of B of the incident wave? Soinfion: TV loop antennas have one turn. At maximum orientation, Eq. (6.5) evaluates to CD = f B - d5 = :IzBA for a ioop of area A and a nnifonn magnetic fieid with magnitude .8 = |B|. Since we know the frequency of the field is f = 300 MHZ, we can express 8 as B = Bocos(tot+txg) with o} = 211: X 300 ><106 raids and {10 an arbitrary reference phase. From Eq. {6.6), Vemf = "—N? = —A%[B{JCOS{£DI + = Aflowsinfim + me is maximum when sin((m‘ + 01.0} = 1. Hence, 20 x10‘3 = Aing : 10-2 x an x 6:: x108, .—_-,:':ZWzM<-“""'- 222 MR 6 which yields Bo = 1-06 (nAlm). Mblem 6.6 The square loop shown in Fig. 619 (P66) is coplanar with a long, Straight wire carrying a current £(r) = 2.503521: x 104: (A). (3) Determine the emf induced across a small gap created in the loop. (b) Detemfine the direction and magnimde of the cunent that would flow through a 441 resistor connected across the gap. The 100p has an internal resistance of 1 9. Figure P6.6: LOOP coplanar with long wire (Problem 6.6). Solution: {3) The magnetic field due to the wire is _ ~ 1101 _ A #01 B _ 21w ' X 2ng where in the plane of the 100p, @ z: —§; and r : y. The flux passing fimmgh the loop CHAPTER 6 223 is 15 cm «in: [345:] (+sfi‘1{)[—s10(cm)]dy 5 5 2101 cm I 10‘! 5 =M—D x 1n~1— 21: 5 _ 4:: x104 x 2-5cos(2n x104r)x10'1 X1 1 s 21: . : 0.55 x 10-7cos(2n x 104;) (Wb). do . 4 -7 11m: —3 =0.55x21t><1045m(21:x10 z) x 10 = 3.45 x 10-3 5111(21: x 104:) (V). (b) V _ —-3 11m = m — w sin(21: x 104:) = O.69sin(21: x1041) (mA). 4+i_ 5 At 1‘ = 0, B is a madmum, it points in —i-direction, and since it varies as cos(2:n: x 104:), it is decreasing. Hence, the induced current has to be CCW when looking down on the loop, as shown in the figure. Problem 6.”? The rectangular conducting loop shown in Fig. 6-20 (P6,?) rotates at 6,000 revolutions per minute in a uniform magnetic flux density given by B = 3750 (1111'). Determine the current induced in the loop if its internal resistance is 0.5 Q. Soiufioti: <12 = feds = 550 x10‘3-§z(2 x 3 x10“)cos¢(t): 3 X10'5cos¢(t), S ¢(r) = (02’ = W? = 2001:: (mils), at: = 3 X 10—5cos(200m) (Wb), Vemfz 4:? z 3 x 10-5 x 2001asin(200m) = 18.85 x 10-33in(200m) CV), and ~— Vi‘“ = 37.75in(200m) (mA). 224 CHAMBER 6 Figure 136.7: Rotating loop in a mgnetio field (Problem 6.7). The direction of the current is CW (if looking at it along —%—directiorl) when the loop is in the first quadrant (0 g Q) g n/Z). The current reverses direction in the second quadrant, and reverses again every quadrant. Problem 6.8 A rectangular conducting loop 5 cm x10 cm with a small air gap in (me of its sides is spinning at 7200 Evolutions per minute. If the field B is normal to the loop axis and its magnitude is 5 X 10‘6 '1', what is the peak voltage induced across the air gap? Solution: 27: radicycle x 7200 cyclesfmin 60 slmin A z 5 cm x10 era/(100 onv‘n'l)2 = 5.0 x10'3 1112. From Eqs. (6.36) or (6.38), Vemf = AcoBosinmr; it can be seen that the peak voltage is VP“ =AcoBo = 5.0 x10‘3 x 240:: x 5 x10'6 =1s.ss (m7)- (1): = 2401': radfs, E’E'ohlem 6.9 A 50-m-long metal rod rotates about the z—axis at 180 revolutions per minute, with end 1 fixed at the origin as shown in Fig. 6-21 (136.9). Determine the induced emf V12 EB = 23 X 10—4 T. Solution: Since B is constant. Vang : :31? The velocity u for any point on the bar is given by e = era), where 21': radlcycle x (180 cycles’min) (60 51min) a): = 61tradls. cm}? 6 225 Figure P69: Rotating rod of Problem 6.9. From Eq. (6.24), 1 o A Vn=vggfiej£ (uxB)-dl=jos(¢61trxi3><l0“)-f'dr 0 = 18?: x10“‘[ rdr n05 0 = 911: X 10—431 0.5 = —9n: x 10'4 x 0.25 = 407 (w). Problem 6.18 The loop shown in Fig. 6-22 $6.10) moves away from a wire c3113ng :1 current I; = 10 (A) at a constant velocity at = $15 (mfs). HR = 10 fl and the direction of 12 is as defined in the figure, find I: as a function of yo, the distance between the wire and the loop- Ignore the internal resistance of the loop. Solution: Assume that the wire carrying current I; is in the same plane as the loop. The mo identical resistors are in series, so 12 = Vemf/ 2R, where the induced voltage is due to motion of the loop and is given by Eq. (6.26): Vemfz Vergf= j£(flXB)'£fl. C The magnetic field E is created by the wire carrying 31 . Choosing 2 to coincide with the direction of 11, Eq. (5.30) gives the external magnetic field of along wire to be A1101 1 .: .' unimaginsw ' - _ ' axe-t 3% '3‘? ".‘zJ..;'. - 226 CHAPTER 6 Figure P610: Moving 100p of Problem 6.10. For positive values of yo in the 31-: plane, 37 : i", so . A “HOII #0115! B: uxB: ox —= . . ax sll rs 1 tam 2.2m Integrating around the four sides of the 100p with dé = i dz and the limits of integration chosen in accordance with the assumed direction of 12, and recognizing that only the two sides without the resistors contribute to V3139 we have 0-2 I u 0 polls: Vfl‘] = f “‘10 1 ‘ A d A emf o (1 27C? mm (2 z)+ 02 z 2753' 475X10‘7X10X5x0.2(1 1 j 21: yo yo+0-1 1 1 =2 10‘6 — ———— , X (Yo yo—i-OJ) CV) - (fidz) T=Yo+0J and therefore 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 Bzi'fi (T). CHAPTER 6 227 Sliding contact V Figure P613: 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. Soiution: The surface of the cylinder has velocity u given by 1,200 60 u = $0):- = $27: X X 5 ><10'2 = $21: (rials), 0-1 L H 1112:] (uxB)-di= (eznxrs).edz=_3_77 (V)- 0 0 Probiem 6.12 The electromagnetic generator shown in Fig. 6-12 is connected to an electric bulb with a resistance of 100 Q- If the loop area is 0.1 m3 and it rotates at 3,600 revolutions per minute in a uniform magnetic flux density 39 = 0.2 T, determine the amplitude of the current generated in the light bulb. Soiution: Front Eq. (6.38), the sinusoidal voltage generated by the a—c generator is chf = AmBo sin(tuz + Co) = V0 sin(tut + Co). Hence, V0=AmBO=O.1x%§(—)Ex0-2=ls4 (V), v0 7.54 =—=——=7_4 mA. I R 100 5 ( ) Probietn 6.13 The circular disk shown in Fig. 6-24 (P658) lies in the xH-y plane and rotates with uniform angular veiocity to about the z-axis- The disk is of radius a and is present in a uniform magnetic flux density B = 2280. Obtain an expression for the emf induced at the rim relative to the center of the disk. é ..v.‘mz" Main. Va _.. 4mmm.-...Mm.. \. Likwé..w-n.... .'..tt;,z.-.m-.stx;;,.a- .. . 223 CHAPTER 6 Figure P613: (3.) Velocity vector u. Solufien: At a radial distance r, the velocity is I! = $ cur where 4) is the angle in the x—y plane show in the figure. The induced voltage is (I {I A V =] (ex B)-dl :/ [(mr) x230] -i“dr. o o é x i is along En Hence, a V=C030f rdr: . 0 Section 6-7: Displacement Cum-em Problem 6.14- The plates of a parallel-plate capacitor have areas 10 cm2 each and are sepaxa‘ied by 1 cm. The capacitor is filled with a dielectric maierial with CHAPTER 6 229 E = 450, and the voltage across it is given by VG) = ZUcos 21': x 106: (V). Find the displacement current. Selufion: Since the voltage is of the form given by Eq. (6.46) with V0 z 20 V and co : 211: x 106 radls, the displacement current is given by Eq. (6.49): EA L; = —fd—V9msin 031' 4 8‘ —12 -4 x 354 x10 x10 x10 X 20 X 2R X 105 Sine“ x 106:) 1 x 10*2 —445 sin(21tx 106:) 01A). Problem 6.15 A coaxial capacitor of length 3 = 6 cm uses an insulating dielectric material with g = 9. The radii of the cylindrical conductors are 0-5 cm and 1 cm. If the voltage applied across the capacitor is VU) = 1005in(120m) W); What is the displacement current? Figure P615: Solution: To find the displacement cunem, we need to know E in the dielectric space between the cyiindiical conductors. From Eqs. (4-114) and (4.115), _ A Q E 3 21cm” 21:3! 5: Hence, ' 44. E 3: 4L :- _§-1005m(120m) : 41 3 sin(120m‘) (Vim), r1211 (5) r1112 3“ {it-11%.: . 230 cma 6 D=EE = arsoE 144.3 3" = —r9 x 8.85 x 10-12 x sin(1207tr) 1.15 10-8 =—i‘—:——sin(120m) (0:112). The displacement current flows between the conductors through an imaginary cylindrical surface of length I and radius r. The current flewing from the outer conductor to the inner conductor along hf crosses surface S where S = —i’ 21ml. Hence, rd——-S=—r-— — a: a: = 1.15 x10"3 x 120:: x 2micos(120m‘) = 1.63 cos( 120m) 01A)- F -3 BB a(—1'15X10 sin(120mt))-(—i'21tri) Alternatively, since the coaxial capacitor is iossless, 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 21:81 C=ln( hlfi' The current is dV _ 21ml, d! W103?) [1201: x lOOcos(1201tt)] = 1-63cos(1207u) (as), which is the same answer we obtained before- E’rohiem 6.16 The parallel-plate capacitor shown in Fig. 6-25 (P6.16) is filled with a 105531 dielectric material of relative permittivity e, and conductivity 0'. The separation between the plates is d anci each plate is of area A. The capacitor is connected to a time—varying voltage source V0)- (a) Obtain an expression for IC, the conductiOn current flowing between the plates inside the capacitor, in terms of the given quantities. CHAPTER 6 231 Figure P616: Parailel-plate capacitor containing a lossy dielectric material (Problem 6. 1 6). {b} Obtain an expression for lg, 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 forA = 2 c1113, «1‘ 2 0.5 cm, Er = 4, 0' = 2.5 (Sim), and V0) 2 10.20581: x 103:)01). Solution: (3) d V VA 6 R 5A” ‘ R d (b) V BE BE EABV *x— —:——_ E d” d a: may are: (c) The conduction current is directly proportional to V, as characteristic of a resistor, whereas the displacement current varies as 317/32, which is characteiisfic of a capacitor. Hence, d EA R = — = — . 6A and C d {d} 0.5 x 10-2 Rzmflm 232 Actual Circuit Equivalent Circuit Figure P6.16: (a) Equivalent circuit. 4 X 8.85 x 10-“2 x 2 x10—4 _ _ +12 C_ O'Sxma _1.42x10 1:. ?roblem 6.17 An electromagnetic wave propagating in seawater has an electric field with a time variation given by E = iEocosmt. If the permittivity of water is 8180 audits conductivity is 4 (Sim), 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 Nil-11,63 1 GHz, (:1) 100 GHZ. Sointion: From Eq. (6.44), the displacement current density is given by a a = "D = s—E 3“ a: a: and, from Eq. (4.67), the conduction current is IE = (3E. Converting to phasors and taking the ratio of the magnimdes, i lid (a)Atf=1kI-Iz,m=21tx103radls,and @— — 4 z 888 x 103 “jd ‘ 231:)(103 x 81 X 8.854. ><10"12 ' The displacement current is negligible. (b)Atf=1MI-Iz,m=2nx105 radfs,and 3 4 ~— = = gd 21EX106 X 81x 8-854 x10—12 CIMPTER 6 233 The displacement current is practically negligible. (c) Atf = 1 GHZ, fl) = 27tx109 malls, and 3 4 _ ‘ 21:: x109 >< 81 x 3.354 XIO'"12 ‘ 0338' 3d Neither the displacement current nor the conduction current are negligible. (d) Atf: 100 GHz, to: 212x 10“ radfs, and i is 4 3 = — = .88 10‘ . 21: X 1011 x 81 x 3-854- x 10—12”1 8 X The conduction current is practically negligible. Sections 6-9 anfi 6—10: Cont?an Equation anti Charge Dissipation fioblem 6.18 At t = 0, charge density pvo was introduced into the interior of a material with a. relative permittivity er :- 480. If at z = 1 ,us the charge density has dissipated down to 10-35)“), what is the conductivity of the material? Sciatica: We start by using Eq. (6.61) to find If: M!) = EMF”? 01' 104% = Owe—104m, which gives 10-6 m 10—3 = — , tr 01' 10—6 ;=_ =Lexm4 @. 11110-3 But Ir 2 8/6: 483/6. Hence 4 _ 4 —12 43° — fl = 2.44 x 10"1 (Sim). ‘52 1r _ 1.45x10—7 234- Pmbiem 6.19 If the current density in a. conducting medium is given 3(9513’523): (iz * i3?2 + 2276) COS 931‘, determine the corresponding charge disnibution pv (x, 31,2; r). Soflnfion: Eq. (6.58) is given by 39v V- : - _ 3 8: The divergence of J is V‘J = ($3 +3>3 +fii {fiz—ffiyl +22x)cosmr 8x ay 82 = *353-(yzcos (or) 2 —6ycostet. Using this result in Eq. (E4) and then integrating both sides with respect to; DV = —J€{(V-J)dr = —/ ~6ycoswtdr = gsinmI+Cog_ where C9 is a constant of integration. Fromm 6.28 In a certain medium the direction of current density J- radial direction in cylindricai coordinates and its rangintude is indepen ' and. 2. Determine 3, given that the charge density in the medium is Pv : porcoscor (0:133). Sointfion: Based on the given information, 3 = i‘fr(r)- With £3, = I: 2-— 0, in cylindrical coordinates the divergence is given by V-J— liker _r5‘r From Eq. (6-54), = —% = -—§(porCOSW) = pflmsinw‘ CHAPTER 6 235 Hence 3— $01,) = permsiner, r %(Dfl-) I porzmsinmr, f imam = pomsincm / rzdr, 0 3r 0 r3 :- mug = (pomsin our) «- , 3 o ’) J, = powr sincor, 3 and Problem 621 If we were to characterize how good a material is as an insulator by its resistance to dissipating charge, which of the following two materiais is the better insulator? Dry Soil: 5, 2 2-5, G = 10-4 (Sim) Fresh Watch a, = 80, o = 10-3 (Sim) Sokmjon: Relaxation time constant 'tr = E. G . 2.5 For dry sofl, "Cr — if k 2.5 x104 5. 0 For fresh water, 1:r : 10—4; 2 8 x 104 5. Since it rakes Eonger for charge to dissipate in fresh water, it is a better insulator than dry soil. Sections 6411: Electromagnetic yo’renfiais Pmbflem 6.22 The electric field of an electromagnetic wave progragating in air is given by E(z,r) = s4cos(6 x 108: _ 22:) +§r3sin(6 x 108.: — 2z) (Wm). é '§ 2 '5' 'I .. m... :. :. ... 236 CHAPTER 6 Find the associated magnetic field H(z, I). Solution: Converting to phasor form, the electric field is given by fie) = tale—fl: — mas—m mm. which can be used with Eq. (6.87) to find the magnetic field: 1 him 1 x z y =_— a/Bx 3/3)! 3/33 '3‘” 49*!“22 —j3e‘J?“’- 0 1 -J'£0# Vxfi fie.) = (rims-J?z — wage-131) j = mfifi—wsk'fk = jt8.0e—fiz+5r10.6e-fiz (menu). Converting back to instantaneous values, this is H(I,z) : —§8.0sin(6 x 10% — 22) + §10.6cos(6 x 103: — Zz) (mAJm)- Problem 6.23 The magnetic field in. a dielectric material with E r 430, p: = no, and 5 = 0 is given by my) = 15.20501: x 107: +ky) (Aim). Find 1% and the associated electric field E. Selufion: In phase}.- form. the magnetic field is given by Q = :‘iSefl‘Ev (Aim). From Eq. (6.86). .. 1 .. _ ' E = ,—V><H = 4525.2!” 10:38 fine and, from Eq. (6.87), H 1 ._ .. 'kl . H = we ._ _] teem, —J£0# ‘szfii-i which, together with the original phasor expression for fi, implies that cox/ET. 2nx107¢1 47: “W: c =W=§a Wm)- CW6 inserting this valee in the expression for E above, 4112/30 56j435y/30 = _29¢1ei4’9’/30 (Vim)- Ez—i 21n><10'Jr x 4 X 8.854 x104:a Problem 6.24 Given an electric field E = fiEo sinaycos(mt - kz)1 where £0, a, co, and k are constants, find E. Solution: E = 12.50 smaycoswx — kz), E = fiEosinaye'fl‘z, e = J. w e Jam 1 a - 3 _ _ . = —}7m—# [fa (ED sinay e‘fih) — 23 (Ea smaye flail 50 fl _ n _ _ .kz :: —-— k _ J mu [3! smay z jfl cos afle , H : fidfiej‘m] = Sic {g [fksinay + 2 acosay e‘jnn]e_jhejm} ,u _ Ev - v A i E 1 — a); [yksmaycos(mt—~ kz)+zacosaycos (cor kz 2) £9 [fiksinaycosmx — kz) + 2amsay5in(m£ — kz)]. 03;” Problem 6,25 The electric field radiaied by a short dipole antenna is given in Spherical coordinates by E(R=8;r) = 93%sin8 cos(61t x 103: — 215R) (Vim)- Find E(R,B;it). Solmion: Converting to phasor form, the electric field is given by ~ a - 1 ‘1 . E(R,8) = 859 = 821% smfleflm (Vim), 238 CHAPTER 6 which can be used with Eq- (6.87) to find the magnetic field: ~_1~_1.iaz~:e-1a BULB) — _jmpVXE—— [RRSine FEE-F i 1 A2x10-3 a Z __ b - _ —J?.1tR _jmp¢ R smB BR(e ) n 21: 2x10"2 _ flaw _¢61t>< 108 X41tx 10-7 R mag .53 =¢Esinaefi2fl (pA/m). Converting hack to instantaneous value, this is £02, 63) : a} €53mb5 (6n: x108: — 21:3?) (,uAfm). thlem 6.26 A Hertzian dipole is a short conducting wire carrying an approximately constant current over its length I. If such a dipole is placed along the z-axis with its midpoint at the origin and if the current flowing through it is £(r) = Io cos tot, find (a) the retarded vector potential E(R,6,¢) at an observation point Q(R,B,¢} in a spherical coordinate system, and (s) the magnetic field phasor rim, 8, ¢)_ Assume E to be sufficiently small so that the observation point is approximately equidistant to all points on the dipole; that is, assume that R‘ 2 R- Solution: (3.) In phasor form, the current is given by f: 10. 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, ., U2 " . —ficR‘ A: if / More; 41‘: 432 s R’ where sis the wire cross section. The wire is infinitesimafly 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 = 310 / s, and employing the relation R’ x R, ... U2 -ij‘ £32 —ij I _ Azifljr e dzzfitflf e -Hfo fine 47!: 4/2 R’ 4'3: -1112 R 43R CHAPTER 6 239 In spherical coordinates, i = ficos 8 — 3 sin 6, and therefore .. .. A ' HIDE _ .3 = * _ J _ A (Roosfl 851n9)4 e (13) From Eq. (6-85), e=im=ifi jLi 41': A A 9‘“ij Vx [(RcosB—fisine) R ] _Ioz.1 a _ 4R a rib? —E¢§(3+R(—sm6e 1' )v—a—e(cose R Prob} 6.27 The magnetic field in a given dielectric medium is given by H = ificosksina x 107: — 9.1x) (Aim), where x and z are in meters. Determine: (a) E, (b) the displacement current density id, and (e) the charge density pv. Solution: (3) e = §6c0322sin(2 x107: — 0.1x) = $‘6coschos(2 x107: —~ 0.1x— M2), Eleni = fi6cosZze‘jo'l‘e'J-“f2 2 —§Ij6c052ze'j°'h“1 )- A Z Y . 3/8.: 8/33! 3/32 flog 0 +j6c052z e'jo'lx 0 = J'— {3 ['3(—j6cos22£'j°'”)] +2 [3(rj6coslze*j°'1‘)]} Jim 63 3x _ ' _6 = it («-2 sinks—50'”) +2 (—126 cos2ze*-’D‘”) . From the given expression for E, to : 2 x107 (reefs), . 240 CHAPTER 6 I3 = 0.1 (radfm). Hence, up+§=2x103 (mfs), c 2 3 x108 3 up 2X108 Using the. values for (o and 2, we have: i7: = (—fi305in2z + 2j1.5c052z)x life—JILLr (Vim), E = [—1230 sinchos(2 x 107: — 0.1 x) — z 1.5coszzsin(2 x 107: _ 0.130] (lem). (b) fi = 2% = 2,20%": = (—12% siu22+i 111.03 coszz) x lo-fie-In-lx (Clmz), 3]) Ed — E a 01' 3cl = jmfi = (—filesinZzw £0.6coszz)e*j0.1x: 351 = = {i 12sin225in(2 X 107: — 0-1x) — i0.6coschos(2 X 107t — 0- 1 x)] (Mmz). (c) We can find 1:)v from V - D = [3‘r or from a Pu V- = — . “E 3: Applying Maxwell’s equation, BE BE m V- = v -E _—_',, x _z 9‘? D 8 ax + az ) yields 3 . 7 9V = 2,20 3 [—30sm2zcos(2 x 10 t—O-1x)] +353; [—1.5coslzsi_n(2 x107 —0.1x)]} = 2,20 [—35i112zsin(2 x107r— 0.1x) +3sin2zsin(2 x107r—0-1x)] : 0. ...
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Electromagnetics for Engineers - CHAPTER 6 219 Shaptet' 6...

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