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Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition

Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering, 1st Edition offers a comprehensive guide to solving every question in your textbook, helping you master the material.

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    Solution Manual for Fundamentals of Electromagnetics for Electrical and Computer Engineering 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 (Amrita University), India

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    CONTENTS CRaPLEr 1 covvn iii eee] CRAPLET 2 Levies een 1S Chapter 3... eee 29) Chapter 4 Lo... eee A Chapter 5 .......oiiiiiiii ie DO) CRAPLEr 6 Levee en 82 Chapter 7 ........oiiiiiiiin iin 104 CHAPLET 8 vivitar. 134 CRAPLEL 9 vise eee eccrine 153 Chapter 10 ......ouviiiiiiiiiiiiiie cee LTO APPENAIX A Louies eee 183) Appendix B o.oo. UBT

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    CHAPTER 1 1.1 (a) Total distance = 1 + ++ Lod =om wr T2727: 1-45" (b) Distance north = [—+ += —=—L_ =0.8m Tae 1+y A 11 1 1, 1.1 Distance east = a (ae =04m .~. Final position is (0.8, 0.4) (c) Straight line distance = +/(0.8)2 + (0.4)? = 0.8944 m 1.2. A+B+C=2a;+3a+2a3 — (1) 2A +B-C=a; + 3a, —(2) A-2B+3C=4a +50, +a; —(3) (1) +2) — 3A +2B=3a; + 16a, + 2a; — (4) 2)x3+(3)— TA +B =Ta, + lday +a; — (5) [G)X2-(@)] +11 >A=a,+2a, — (6) 5)- (6) x7 — B=a; —( H=-©)-N— C=a +a +a; — (8) 13. (A+B)*(A-B)=A*A-A+*B+B*A-B+B=A*—B (A+B)Xx(A-B)=AXA-AXB+BXA-BXxB=2BxA For A = 3a; — 5a; + 4a; and B = a; + a, — 2as, A +B =4a; — 4a; + 2a;, A — B = 2a; — 6a; + 6as, A’=9+25+16=50, and B*=1+1+4=6 A+B)» (A-B)=8+24+12=44=4>-pB 1 — StudyXY

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    a, a, a, (A+B)x(A-B)=|4 -4 2|=-12a,-20a,-16a, 2 6 6 a, a, a, =2|1 1 -2=2BxA 3 5 4 14. Bx C=-4a,+2a,+ 8a, Ax (BxC)=8a,+16a, CxA=-a,-2a,+7a, Bx(CxA)=-12a,-8a,—4a, A xB =a, + 2a, + 3a,, C x (A x B) = 4a, — 8a, + 4a, AxBXC)+Bx(CxA)+Cx(AxB)=0 In fact, this quantity is zero for any A, B, and C. 15. Arca= SABsinc = Lia XB| 2 "2 For the points (1, 2, 1), (3,4, 5), . 8 and (2, -1,-3), 7 B sind A =4a, + 6a, — 4a, B = 5a, + 3a, — 8a, A) A xB =-36a, + 12a, - 18a, ‘ 2 Area = ze? +(12)*+(~18)* =21 units. 1.6. Area of the base = [B Xx | ; Height of parallelepiped = Projection Es = of A onto the normal to the base A [E] ZA. BxC B [BxC]| +. Volume of parallelepiped = Area of base x height =A « Bx C For A =4a,, B=2a, +a, + 3a, and C=2a,+ 6a, A*BxC=0. Hence, volume of the parallelepiped is zero. The three vectors lie in a plane. 2 — StudyXY

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    1.7. The vector A must be perpendicular to both (a, + 2a,) and (a, — 2a,). Hence A = C(-a, + 2a,) X (a, —2a;) = C(2a, + 2a, + a,) where C is a constant. To find C, we note that a, Xx A = a, X C(2a, + 2a, +a,) = 2a, — a, C=1and A =2a, + 2a, +a, Verification: a, x A = a, X (2a, + 2a; + a) = a, — 2a. 1.8. Vector from A(S, 0, 3) to B(3, 3, 2) = 2a, + 3a, — a, Vector from C(6, 2, 4) to D(3, 3, 6) = 3a, + a, + 2a, CD 6+3-2 Component of AB along CD = AB + —— = ————= = 1.8708 P € CD J9+1+4 1.9. Writing the equation for the plane as = 5 tog =1, we find the intercepts on the x, y, and z-axes to be at 15, -12, and 20, respectively. Thus 3 Rup =-15a,— 12a, : foo Rac =-15a, + 20a, ) Rac x Rap = 240a, — 300a, + 180a, < 2 , a= Ryo xR yp _da 5, +3a, As Rac XR gp 572 x . Distance from origin to the plane = 15a, * a, = 2 . 1.10. For y = 2x, z = 4y, we have dy = 2 dx, dz =4 dy = 8 dx. sod =dea+dya,+dza;=dra,+2dxa,+8dxa, = (a, + 2a, + 8a,) dx, independent of the point. 1.11. For x = y = 7%, we have dx = dy = 27 dz. At the point (4, 4,2), dx=dy=4dz sod =dva,+dyay+dza;=4dza, +4 dzay+dza, = (4a, + 4a, + a,) dz 3 — StudyXY

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    1.12. Differential length vector having projection dy a, = dy a, dy . . . i. Differential length vector having xe yer / ik 3x : projection dx a, is J Sai pat 1 [dx % [07 dxay+dza;=doac— 5 da, t L rey a = (a “3a J dx, 4 since forx +2z=2,dz=— : dx, independent of the point. . _ 1 1 sds = a, ~5a; dx Xdya, = Fata, dx dy. 1.13. One vector tangential to the ¥ surface is dz a,. Another tangential vector is given by dl =dxa,+dya, 14 2,4, = dra, +2xdxa, @*) = (a, + 4a,) dx ld .~. Vector normal to the plane = (a, + 4a,) dx X dz a, = (4a, —a,) dx dz Unit vect 1 to the pl fa, nit vector normal to the plane = ——==. ? 7 1.14. Denoting K(x, y) to be the height field, we have Pry +r =4, 7+) <4 or, h= \J4=x2~y2 FP +y <4. 1.15. The number field is x +y + z. .~. Constant magnitude surfaces are the planes x + y + z = constant. 4 — StudyXY

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    1.16. d(x, y, 2) = xa, + ya, + za, Constant magnitude surfaces are x” + y* + z* = constant, and hence are spherical surfaces centered at the corner. Direction lines are radial lines emanating from the corner. 1.17. v = —rwsin ga, + rwcos ga, = aX-ya, + xa,) Y N 9 1.18. f(z,1) =10cos (27 x10't ~ 0.1 77) 1 CUNY aro! t=0 kzLix0 E: id t= Fa t=gxto (® : — Sar Ve SSCS AAV AN a SCO > 320 322-5 3=5 3=1S 310 ’ wi os HZ WT SON . 4 Te >> <> f(z, 1) represents a traveling wave progressing with time in the positive z-direction. 5 + StudyXY

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    1.19. f(z,1) =10cos (27 x10" +0.1 72) 1 1 1 -1 felix k=2x6 bly EzixkG t=0 § z 3 € 3 © ~ RY MN VO) Ll BRN SN . CONN - ARR E TOS << >> : zlo 27.6 =5 22.5 3=0 w 3 PVN VL TY OY TION TSS IN COON B= <> - f(z, ©) represents a traveling wave progressing with time in the negative z-direction. 1.20. f(z, £) =10cos 272 x 10'¢ cos 0.1 72 t=0 = : , exe” “ fi SEIN 0 Bs — — > 0 Thm =7.01 Tez2 xd” | “te Seas xe . 5 - ® Cem 7.01 Pa Se TN + % -1.07 {yztbm “10 Kaz iom A f(z, 1) represents a standing wave. 6

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    1.21. (a) The two components are in phase; hence, linear polarization. (b) The two components are perpendicular in direction, differ in phase by 90° and equal in amplitude; hence, circular polarization. (c) The two components are perpendicular in direction, differ in phase by 90° but unequal in amplitude; hence elliptical polarization. 1.22. F; and F; differ in phase by 90°. 1.3 . . | =+/3+1 coswt = 2cos ar; [Fy = FREE sin wt = 2sin wt . i .. Fy and F; are equal in amplitude. FF, = BLL =0. .. Fy is perpendicular to Fs. Thus Fy + F; is circularly polarized. 1.23. \ [ ~ | H EN H Ly SE Co The polarization is elliptical with major axis in the y-direction, minor axis in the x-direction, and eccentricity equal to V2. 1.24. 10 cos (ax — 30°) + 10 cos (ax + 210°) Iw | l 10 7” + 10M” =10 30° 500” Re .~. The sum is 10 cos (ax — 90°) = 10 sin ar. ~~. ad 1.25. 3 cos (ax + 60°) — 4 cos (ax + 150°) Yor be Im J 3e = de SEAN og J60°-53.13) _ £ , j68T 360 60° s =5e =5e Fa Re — 5 cos (ax + 6.87°). £803 7 — StudyXY

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    1.26. Replacing z by 710°7 i by I, and 13 cos 10% by 13¢/, we have 51070 x j10°T +127 =13¢/" GT =13,T=0 1B mer or, (12+ j5)I =13,1 = 25 WET =le Thus i= 1 cos (10% - 22.62°) =1cos (10% — 0.1267) 1.27. From the construction shown, 0 —= mg =tan45°=1 [3 © El o& 2oN — FR, or, Q =1 87 £91 mg 470,20 Vo VEL Q m3 1.28. (a) At the point (0, 0, 100), I x 476499) 47g, (101) Q . 1 __0 107-99" iN 7 47g) 99% 1012 ° x 0 (100+1)%-(100- 1)? Q 400 YT arrvey seppeey Ll ps pL 475 (100-1)% x (100 +1 Eo (100% —1) 400 EEL By, 47€) 100 100% 78, (b) At the point (100, 0, 0) 20 1 E=————————>a 478, (1007 +12)¥2 °F A 2760(100%) 8 — StudyXY

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    129. E= Q |a,+2a,+2a, LAA, +2, LAAs, ta, 47g, 93/2 932 932 a +a, +2a, 2a,+a, +a, a,+2a,+a, tear Tar TT en Laat, 2a, Lata, ta, 1232 332 =< S42 Lo (a, +a, +a;) dm 3 (or Why WBC ——£ (0.18519+0.27217+0.04811+0.19245)(a +a, +a,) Xx y Zz 47g, 0.0555 Q = ata +a,)N/C. 1.30. For the ith segment, . -3 kd z= 2-1 and charge = EU C. if 100 50 __L FR" 50 Cpe 2 N07 1 ¥ - =a 2% PREY y Eh (2 +1) 50 107° a. a TZ - “Ta 2.10 (2i-1 +1] a, i yy _2i-1 os 2i-1 1.31. For the ith segment, z = 100 , charge density = 10 100 C/m, . 6 and charge = 107° Kami BX = 10 @i-y C. 50 6. Cpl 2 10°°2i-1) 1 SB) 5 2 372 i=1 (2 +1) 7 50 107’ . Ly a TY? = 2,010 @i-1) +1] a, [= 9 — StudyXY

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    1.32. Dividing the circular ring into n segments and using the symmetry of the field about the z-axis, we obtain _ ¥ x 22(2)%107° 1 B=) im a p (2 +1 ) 1 _4mx107 008944x107 Came & ‘ 4 =1.012x10"a, NIC. Xx 1.33. For the (i/)th area, C2i-1 _2j-1 ¥ *= Tao 7 100M charge = —— x10 =4x10°77C J 10000 Wa ITLL: y 4 S& axa” ii - Ee NN Ea, [ATT 77TT AE Led Led (i 4 2 4 1)Y? 3 i=l j=l x _7 50 50 _ 4x10” RTA I i “mm 2210 @i-D*+107*@2j-1) +1] a, i=l j= 2i-1 2j—- 1.34. For the (i/)th area, x= Soy = us, 2-1) 2j-1Y charge density = 107 (G35) =107°(2i-1)2j-1)* ¢/m? _ 4 TT I “130: yin 1\2 charge = Joes X10” 2i-1)(2j 1)? =4x107°@i- D2) -1) C ped Shyer ner, 47g pg +? +1)? : i. 50 50 _4x107? Qi-12j-1)? AT pe [10 Qi-12 +1072) -1) +1] 10 — StudyXY

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    Ne? . 1.35. (a) J = Nev= — Ey sin ax a, maw 12 2.10738 = MOT XLOREXI0 10 sin 2m x10’ a, 9.1083x1077" x 27 x10 =0.4485%x 10 sin 27x 107 a, A/m? (b) AI =J + AS = 0.4485 x 10 sin 27 x10" a, +0.01 (a, +a) =0.4485x 108 sin 27 x 107 A 1.36. Denoting x to be the displacement, we write the equation of motion to be 2 mE = mg kx + gE cos ax l dr ym 2 E, cos wt dx on m=z + kx = mg + qEo cos at 9 t The steady state solution consists of two parts. One is x = = due to mg. To find the second part x, we write (je) mx, + ky = qEge’®, or, (k —a&’m)%, = gE, = = 15 mg 4k X) =——5—. Thus x = x +x, = —>+—"—>— cos wx ? k-a’m TR k-a’m . dx qEyw Velocity = — = ————— sin wr Yd k-a'm I, dya,X(-a 1.37. dF, =I; dxa, x Hela da, x(Ca,) 47(l) ¥ =0 I,dysy I dxa_ xa, . 1 : Fy =I, ya, x| LL 1xEy 47(l) * - Ho Ho =1, dy a,x dxa, =z la dxdya, 11 — StudyXY

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    -a, +a, +a 1.38. (2) For 0, 1, 1), ag = ——2—"% R=4/3, and 3 pote! dx (a, +2a, t2) (a, +a, +a,) ar 3 Ne to I dx =——=—(-a,+a,) of Ch (b) For (2, 2,2) acta, +24, R=3,and or (2,2,2),ag= ——7=——,R=3,an * Nd pot! dx@,+2a,+2,) (a,+2,+2%,) 4 9 3 =0 1.39. (a) At (0,0, 1), the components of B M perpendicular to the z-axis cancel, 1 whereas the z components add. Thus Be1 Hg 0012, X (0.00%, +a) a TY 4 (140.0052)? a “ 5%107 - [2 000005, £0012, a, “22H, (b) At the point (0, 1, 0), poto|__ 001 a+ 0.01 a, 47] (1-0.005) (140.005) N 0.0la, X(=0.005a, +a) B 0.0la, x (0.005a, +a.) (1+0.005%)? (1+0.005%)7? Ho | 0.01x4x1x0.005 2x0.01x0.005 7 EP I Sa Sno I TE Az (1-0.005%) (1+0.0052) 107%, TTT 12 — StudyXY

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    1 1.40. For the ith element, dl = 30% N .\2 kd 2i-1 R= (5%) a" ~ R _1 2i-1 i ag =| a, ~| Tog Jaz |» an I tl 1 (EA), | L x AB =r 505% Too J | 7 Al 2007R 50 io] 50 J" = —_ —4 [ — B=2)" dB 20> [1410 @i-1) ] a, i=l i=1 1.41. Dividing the loop into n segments and using the symmetry of the field 3 about the z-axis, we obtain ol 222) 2 g B= —— a : 32 “2 Dir n (2422) y _ 8mul _ * = Ins =0.179l a, 1.42. Equating the magnetic force to the centripetal force, we have 2 . my my evBy = — on r= By PR e room v . _ oo _ eB Orbital frequency = 5 = om For By =5x107, 13 — StudyXY

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