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Eletromag/911283_498067750292191_186620398_n.jpg Eletromag/ch01.pdf 1-1 Chapter 1 Problem Solutions 111. . (a) 25 10 250 10 2504 3× = × =Ω Ω Ωk (b) 0 035 10 35 10 3504 2. .× = × =Ω Ω Ω (c) 0 00045F 450 10 4506. = × =− F Fµ (d) 0 003 10 0 3 107 9. .× = × =− −F F 0.3nF (e) 0 005 10 50 10 502 6. × = × =− −H H Hµ 112. . (a) 30 12in 254cm 1 1 48 28miles 5280ft 1mile 1ft 1in m 100cm km 1000m km× × × × × =. . (b) 1 12in 1000mils 12 000ft 1ft 1in mils× × = , (c) 100yds 12in 2 54cm 1 9144× × × × =3ft 1yd 1ft 1in m 100cm m. . (d) 5mm 100cm 254cm 1000 19685mils× × × × =1m 1000mm 1m 1in mils 1in. . (e) 20 100cm 254cm 1000 0 7874µ µ m 1m 10 m 1m 1in mils 1in mils6× × × × =. . (f) 880yds 12in 2 54cm 1 804 67× × × × =3ft 1yd 1ft 1in m 100cm m. . 121. . (a) λ = × =3 10 33333 8 m s 90 Hz km = 2,071.2 miles. (b) λ = × =3 10 300 8 m s 10 Hz km = 186.41 miles3 (c) λ = × × = 3 10 85714 8 m s 350 10 Hz m = 0.533 miles3 . (d) λ = × × = 3 10 2 250 8 m s 1 10 Hz m = 820.2 ft6. (e) λ = × × = 3 10 35 8 57 8 m s 10 Hz m = 28.12 ft6 . 1-2 (f) λ = × × = 3 10 110 2 73 8 m s 10 Hz m = 8.95 ft6 . (g) λ = × × = 3 10 335 89 55 8 m s 10 Hz cm = 2.94 ft6 . (h) λ = × × = 3 10 6 5 8 m s 10 Hz cm = 1.97 in9 (i) λ = × × = 3 10 45 6 67 8 m s 10 Hz mm = 262.5 mils9 . 12 2. . (a) ( ) λ λ λ λ = × = × × × × ∴ = = 3 10 5 10 5 10 0 0161 1 62 8 6 6 m s 60 Hz m 50 miles = 80.46 km 80.46 10 m = ? m3 length 1 244 344 1 24 34 ? . (b) ( ) λ λ λ = × × = × ∴ = 3 10 600 152 4 600 0 254 8 m s 500 10 Hz m 500feet = 152.4 m m = ? m 3 length . ? .124 34 124 34 (c) ( ) λ λ λ = × × = × ∴ = 3 10 2 73 5 137 2 73 05 8 m s 110 10 Hz m 4 feet = 1.37 m m = ? m 6 length . . . . ? .123 124 34 (d) ( ) λ λ λ = × × = × ∴ = 15 10 75 508 75 0 677 8. . . . ? . m s 2 10 Hz cm 2 inches = 5.08 cm cm = ? cm 9 length 1234 124 34 1-3 12 3. . (a) β ωβ λ µ φ β = × = = = × = = = = = −2 2 10 1 2 856 10 285 6 10 5 66 2 8. , . , . . rad m MHz, m s m, s, = rad = 3781.5o f v v f T d v d (b) β ωβ λ φ β = = = = × = = = = = 75 4 3 2 5 10 833 4 0 41 7 66 8. , . , . . . rad m GHz, m s mm, inches = 0.102 m, ns, = rad = 438.92o f v v f T d v d (c) β ωβ λ φ β = = = = × = = = = = 315 150 299 10 1995 20 20 4 19 2 8. , . , . . . rad m MHz, m s m, feet = 6.1m, ns, = rad = 1100.2o f v v f T d v d (d) β ωβ λ φ β = × = = = × = = = = = −0126 10 3 15 10 50 50 0 54 1014 3 8. , . , . . rad m kHz, m s km, miles = 80.5 km, ms, = rad = 580.9o f v v f T d v d Eletromag/ch02.pdf 2-1 Chapter 2 Problem Solutions 211. . F m d1 2 = ω θsin , F mg2 = sin cos θ θ , F F1 2= , ∴ =cosθ ω g d2 , ω π π= × =2 2rpm 60 , ( )cos . . .θ π = × = 9 78 2 05 052 , ∴ =θ 603. o 212. . W 45o× = ×cos sin200 θ , sin cosθ = W 200 45o , ∴ =θ 813. o , GS + Wsin45o = 200cosθ , GS = 169.71 mi hr � � � d m F2F1 mg � W GS 200 mi/hr 45° 2-2 213. . 4 3sinθ = , sinθ = 3 4 , ∴ =θ 48 6. o 214. . F N mv r = =sinθ 2 , − = =W N mgcosθ , ∴ =tanθ v rg 2 , v = × × =60 5280 1 88mi hr ft 1mi hr 3600s ft s , ∴ =tan .θ 016 , ∴ =θ 916. o 2 31. . C A B= + , ( ) ( )∴ = • = + • + = • + • + • = + +C A B AB2 2 22 2C C A B A B A A B B A B cosα But α θ= −180o AB and ∴ = −cos cosα θ AB ∴ = + −C A B AB AB 2 2 2 2 cosθ which is the law of cosines. � 3 mi/hr 4 mi/hr N � �AB C B A N W —W F N � 2-3 2 41. . A B C+ + = 0 , ( )A A B C A A A B A C× + + = = × + × + × =0 0 0 123 ( )A B a a× = − = − −AB ABC n o C nsin sinα θ180 where an is a unit normal into the page. Also A C a a× = =AC ACB n B nsin sinα θ . ∴ =AB ACC Bsin sinθ θ giving the law of sines: B C B Csin sinθ θ = . Similarly for ( )B A B C× + + = 0 and ( )C A B C× + + = 0 . 2 4 2. . (a) ( )A B• gives a scalar which cannot be “crossed with” a vector. (c) ( )A B• gives a scalar which cannot be “dotted with” a vector. (d) ( )B C• gives a scalar which cannot be added to a vector. 2 51. . (a) ( ) ( ) ( )( )A a a a a a a= − + − − + − − = − +3 0 4 2 5 4 3 6 9x y z x y z (b) ( ) ( ) ( )A = + − + =3 6 9 11222 2 2 . m (c) a A a a aA x y zA = = − +0 27 053 0 8. . . 2 5 2. . (a) A B a a a+ = + −3 4 3x y z , (b) B C a a a− = − + −2 2 3x y z , (c) A B C a a a+ 3 - 2 = − + −x y z8 9 , (d) A = + + =2 3 1 374 2 2 2 . , (e) ( )a B a a a a a aB x y z x y zB= = + − = + −16 2 41 0 41 0 82. . . , (f) A B• = 7 , (g) B A• = 7 , (h) B C a a a× = − − −x y z7 4 , (i) C B a a a× = + +x y z7 4 , (j) A B C• × = −19 C C B A B A A �B �B �C �C �B�C 2-4 2 5 3. . (a) A B A B• = − − = − =2 2 3 3 cosθ . Therefore B A B A cos .θ = • = =3 14 0 8 , (b) cos .θ θ= • = − ⇒ =A B A B 3 14 6 1091o , (c) unit vector = × × = − − − + + = − − − A B A B a a a a a ax y z x y z 7 5 1 49 25 0115 0808 0 577. . . . 2 5 4. . A B• = + − =α 2 3 0 ∴ =α 1 2 5 5. . ( ) ( ) ( )A B a a a× = − − + + + − =18 3 3 9 2 0β α β αx y z . Hence α = −3 and β = −6 . 2 5 6. . A B a a a× = − − +14 9x y z gives a vector that is perpendicular to the planes containing both A and B and hence is perpendicular to both A and B. The length of this vector is ( ) ( ) ( )− + − + =14 9 1 16 672 2 2 . . Hence ( )C a a a a a a= − − + = − − +1016 67 14 9 8 4 54 0 6. . . .x y z x y z . Check that A C• = 0 and B C• = 0 . 2 5 7. . ( ) ( ) ( )B C a a a× = − + − + −B C B C B C B C B C B Cy z z y x z x x z y x y y x z so that ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) A B C a a a × × = − − − + − − − + − − − A B C B C A B C B C A B C B C A B C B C A B C B C A B C B C y x y y x z z x x z x z y z z y x x y y x y x z x x z y y z z y z ( ) ( ) ( ) ( ) B A C a a a • = + + + + + + + + B A C A C A C B A C A C A C B A C A C A C x x x y y z z x y x x y y z z y z x x y y z z z ( ) ( ) ( ) ( ) C A B a a a • = + + + + + + + + C A B A B A B C A B A B A B C A B A B A B x x x y y z z x y x x y y z z y z x x y y z z z Matching components we find that ( ) ( ) ( )A B C B A C C A B× × = • − • . 2-5 2 5 8. . ( ) ( ) ( )A B a a a× = − + − + −A B A B A B A B A B A By z z y x z x x z y x y y x z so that ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) A B C a a a × × = − − − + − − − + − − − A B A B C A B A B C A B A B C A B A B C A B A B C A B A B C z x x z z x y y x y x x y y x x y z z y z y y z z y y z x x z x z Comparing terms to the result in the previous problem we see that ( ) ( )A B C A B C× × ≠ × × . 2 5 9. . The distance is ( )( ) ( ) ( )( )D = − − + − − + − − =3 1 1 2 5 4 10 2962 2 2 . . By integration we integrate D dlP P = ∫ 1 2 . A straight line between the two points is governed by y x= − +3 4 5 4 and z x= −9 4 7 4 . Hence dl dx dx= + − + =1 3 4 9 4 2 574 2 2 . and D dx x x = ∫ = =− = 2 574 10 296 1 3 . . . 2 510. . The surface is drawn below and lies in the yz plane at x=1. Hence the surface area is ( )A dydz z dz z z y y z z z = ∫ ∫ = −∫ = = = = = − = = 1 2 1 2 1 1 2 2 2 1 . Directly it is the area of a triangle of height 1 and base 2 or ( )( )A = =1 2 2 1 1. (1, 3, 2)y = 2z – 1 (1, 3, 1)(1, 1, 1) y z x 2-6 2 61. . Drawing the rectangular and cylindrical coordinate system axes as shown below we see that x r= cosφ , y r= sin φ , and z z= . From this we form ( )x y r2 2 2 2 2 1 + = +cos sinφ φ 1 244 344 and hence r x y= +2 2 . Similarly we form y x r r = = sin cos tanφφ φ . 2 6 2. . Draw the coordinate system and use the right-hand rule. 2 6 3. . Drawing the vector in the xy plane as shown below shows that A A Ax r= −cos sinφ φφ and A A Ay r= +sin cosφ φφ . P(r, �, z) � y z z x r r � � Ay Ax Ar A� y x 2-7 2 6 4. . At point P, φ π= = 3 60o . Hence Ax = +2 3cos sinφ φ =3.598 and Ay = −2 3sin cosφ φ =0.232 and Bx = −4 6cos sinφ φ =-3.196 and By = +4 6sin cosφ φ =6.464. Directly in cylindrical coordinates A B• = − − = −8 18 2 12 . In rectangular coordinates ( ) ( )A B• = × − + × − = −3598 3196 0 232 6 464 2 12. . . . . 2 6 5. . From problem 2.6.1 x r= cosφ , y r= sin φ , z=z. At P1 2 2 1, , π , x y z1 1 10 2 1= = =, , and at P2 3 3 2= − , , π , x y z2 2 215 2 598 2= = = −. , . , . Hence the distance between the two points is ( ) ( ) ( )D x x y y z z= − + − + − =2 1 2 2 1 2 2 1 2 3407. . 2 6 6. . The surface is 1/6 of the surface of a cylinder of length 4-1=3 and radius 2. Hence the surface area is ( )S = × × =2 2 3 6 2π π . By direct integration we have ( )S r d dr dsz r = ∫ =∫ = = =1 4 0 3 2 2φ π φ π 1 24 34 . 2 6 7. . The volume is 1/6 of the volume of a cylinder of radius 2 minus the volume of a cylinder of radius 1 or ( ) ( )( )V = × − × =16 2 1 1 1 22 2π π π . By direct integration, V rdrd dz z dvr = ∫ ∫ ∫ = = = =0 1 0 3 1 2 2φ π φ π124 34 . 2 71. . Drawing the rectangular and spherical coordinate system axes as shown below we see that x r= sin cosθ φ , y r= sin sinθ φ , and z r= cosθ . From this we form ( )x y z r r r2 2 2 2 2 2 2 2 2 2 1 + + = + + =sin cos sin cosθ φ φ θ 1 244 344 and hence 2-8 r x y z= + +2 2 2 . Similarly we form y x r r = = sin sin sin cos tanθ φ θ φ φ and x y z r r 2 2+ = = sin cos tanθ θ θ . 2 7 2. . Draw the coordinate system and use the right-hand rule. 2 7 3. . Drawing the vector in the zx or zy plane as shown below shows that A A Az r= −cos sinθ θθ . The components parallel to the xy plane are A Ar sin cosθ θθ+ and the φ component, Aφ . Hence the x and y components of these are ( )A A A Ax r= + −sin cos cos sinθ θ φ φθ φ and ( )A A A Ay r= + +sin cos sin cosθ θ φ φθ φ . � � y z z x r sin � r z y x Ar sin � A� sin � Ar sin � + A� cos � Ax Ay A� cos � A� A� Ar cos � Ar x, y plane � � � 2-9 2 7 4. . At point P, θ π= =2 3 120o and φ π= = 3 60o . Hence Ax = + −2 3sin cos cos cos sinθ φ θ φ φ =-0.75 and Ay = + +2 3sin sin cos sin cosθ φ θ φ φ =0.701 and Az = −2 3cos sinθ θ =-3.598. Similarly, Bx = 383. , By = 0 634. , and Bz = −3732. . Directly in spherical coordinates A B• = + − =8 6 3 11. In rectangular coordinates ( ) ( ) ( )A B• = − × + × + − × − =0 75 383 0 701 0 634 3598 3732 11. . . . . . . 2 7 5. . From problem 2.7.1 x r= sin cosθ φ , y r= sin sinθ φ , and z r= cosθ . At P1 2 2 2 3 , ,π π , x y z1 1 11 1732 0= − = =, . , and at P2 3 3 6 = − , , π π , x y z2 2 22 25 1299 15= = − =. , . , . . Hence the distance between the two points is ( ) ( ) ( )D x x y y z z= − + − + − =2 1 2 2 1 2 2 1 2 4 69. . 2 7 6. . The surface is 1/8 of the surface of a sphere of radius 4. Hence the surface area is ( )( ) S = × = 4 4 8 8 2π π . By direct integration we have ( )S r d d dsr = ∫ =∫ = = =θ π π φ π π θ θ φ π 2 2 2 4 8sin1 2444 3444 . 2 7 7. . The volume is ( )S r d d dsr = ∫ =∫ = = =θ π π φ π θ φ θ 4 3 2 0 2 2 521sin .1 2444 3444 . 2-10 2 81. . F l•∫ = ∫ + ∫ − ∫ =− = =− d xdx dy ydz x y z 2 4 1 2 3 4 2 1 . Third integral with respect to z contains y. So we need to determine the equation of the path as y z= +1 3 11 3 . Hence the line integral becomes F l•∫ = ∫ + ∫ − + ∫ = − =− = =− d xdx dy z dz x y z 2 4 1 3 11 3 7 21 2 3 4 2 1 . 2 8 2. . W d xdx zdy dz P P x y z = •∫ = ∫ + ∫ + ∫ = = = F l 1 2 2 3 4 0 1 0 1 0 2 . But along the path z y= 2 which when substituted into the second integral gives W=1+3+8=12J. 2 8 3. . (a) F l•∫ = ∫ + ∫ − ∫ = = = d xdx xydy ydz P P x y z1 2 0 3 0 2 2 0 2 . Along this path x y= 3 2 and y z= − + 2 . substituting these gives ( )F l•∫ = ∫ + ∫ − − +∫ = = = = d xdx y dy z dz P P x y z1 2 0 3 2 0 2 2 0 3 2 29 2 . (b) The integral is the sum of the integrals along the two paths: ( )F l•∫ = ∫ + ∫ − =∫ + ∫ + = ∫ − ∫ = = = = = = d xdx xydy y dz xdx x y ydy ydz P P x y z x y z1 2 0 0 0 0 2 0 0 3 0 2 0 0 2 0 2 3 2 =0+0+0+ 9/2+8+0=25/2. Along the first segment of this path neither x nor y change and y=0. Hence the integral along this first path is zero. Along the second segment of the path, there is no change in z and we substitute the equation for the path, x y= 3 2 , into the y integration. 2 8 4. . F l•∫ = ∫ + ∫ + ∫d rdr zrd dz P P 1 2 2 4φ . The two paths are sketched below. (a) ( ) ( )F l•∫ = =∫ + =∫ + ∫ = + − = − = = = d r dr z r d dz P P r z1 2 2 0 0 4 0 0 12 12 0 0 0 0 3 0 φ φ (b) F l F l F l F l•∫ = •∫ + •∫ + •∫d d d d P P P P P P P P 1 2 1 3 3 4 4 2 2-11 ( )F l•∫ = ∫ + =∫ + ∫ = + + = = = = d rdr z rd dz P P r z1 3 2 3 4 8 0 0 8 0 8 4 4 3 3 φ φ π π ( ) ( )F l•∫ = =∫ + =∫ + ∫ = + − = − = = = d r dr z r d dz P P r z3 4 2 8 8 4 0 0 12 12 8 8 4 4 3 0 φ φ π π ( )F l•∫ = ∫ + =∫ + ∫ = − + + = − = = = d rdr z rd dz P P r z4 2 2 0 4 8 0 0 8 8 0 4 4 0 0 φ φ π π F l F l F l F l•∫ = •∫ + •∫ + •∫ = − − = −d d d d P P P P P P P P 1 2 1 3 3 4 4 2 8 12 8 12 2 8 5. . F l•∫ = ∫ + ∫ − ∫d rdr rd zdz P P 1 2 2 φ . The two paths are sketched below. (a) F l F l F l•∫ = •∫ + •∫d d d P P P P P P 1 2 1 0 0 2 ( ) ( )F l•∫ = =∫ + =∫ − ∫ = + + = = = = d r dr r d zdz P P r z1 0 0 2 0 0 0 2 2 0 0 0 0 2 0 φ φ ( )F l•∫ = ∫ + ∫ − =∫ = + + = = = = d rdr rd z dz P P r z0 2 0 3 0 0 0 0 2 0 9 2 0 0 9 2 φ φ y z x P3(2, 2, 3) P4(2, 2, 0) P2(0, 0, 0) P1(0, 0, 3) 2-12 F l F l F l•∫ = •∫ + •∫ = + =d d d P P P P P P 1 2 1 0 0 2 2 9 2 13 2 (b) F l F l F l F l•∫ = •∫ + •∫ + •∫d d d d P P P P P P P P 1 2 1 3 3 4 4 2 F l•∫ = ∫ + ∫ − ∫ = + + = = = = d rdr rd dz P P r z1 3 0 3 2 2 2 2 2 4 9 2 0 0 9 2 φ φ π π ( ) ( )F l•∫ = =∫ + =∫ − ∫ = + + = = = = d r dr r d dz P P r z3 4 3 2 3 4 0 0 2 2 3 3 2 2 2 0 φ φ π π ( ) ( )F l•∫ = =∫ + =∫ − ∫ = − + = − = = = d r dr r d dz P P r z4 2 3 2 3 4 0 3 0 3 3 3 2 0 0 0 φ π π φ π F l F l F l F l•∫ = •∫ + •∫ + •∫ = + − = −d d d d P P P P P P P P 1 2 1 3 3 4 4 2 9 2 2 3 13 2 3π π 2 8 6. . F l•∫ = ∫ + ∫ + ∫d rdr rd r d P P 1 2 2 3 2θ θ φsin . The two paths are sketched below. (a) F l F l F l•∫ = •∫ + •∫d d d P P P P P P 1 2 1 0 0 2 y z x P3(0, 3, 2) P4(0, 3, 0) 3 3 2 P2(3, 0, 0) P1(0, 0, 2) 2-13 F l•∫ = ∫ + ∫ + ∫ = − + + = − = = = d rdr rd r d P P r1 0 2 3 2 9 0 0 9 3 0 0 0 0 0 θ θ φ θ φ sin F l•∫ = ∫ + ∫ + ∫ = − + + = = = = d rdr rd r d P P r0 2 2 3 2 9 0 0 9 0 3 2 2 0 0 θ θ φ θ π π φ sin F l F l F l•∫ = •∫ + •∫ = − + =d d d P P P P P P 1 2 1 0 0 2 9 9 0 (b) F l F l F l•∫ = •∫ + •∫d d d P P P P P P 1 2 1 3 3 2 ( ) ( ) ( )F l•∫ = =∫ + =∫ + =∫ = + + = = = = d r dr r d r d P P r1 3 2 3 3 3 2 3 0 9 2 0 9 23 3 0 2 2 2 θ θ φ π π θ π φ π π sin ( ) ( ) ( )F l•∫ = =∫ + =∫ + = ∫ = + − = − = = = d r dr r d r d P P r3 2 2 3 3 3 2 3 2 0 0 3 3 3 3 2 2 2 0 θ π φ π π θ π π φ π sin F l F l F l•∫ = •∫ + •∫ = − =d d d P P P P P P 1 2 1 3 3 2 9 2 3 3 2 π π π y z x P3(0, 3, 0) 3 3 3 P2(3, 0, 0) P(0, 0, 3) P0(0, 0, 0) 2-14 2 8 7. . F l•∫ = ∫ + ∫ + ∫d rdr r d r d P P 1 2 2 32 θ θ φsin . The two paths are sketched below. (a) F l F l F l•∫ = •∫ + •∫d d d P P P P P P 1 0 1 3 3 0 ( ) ( ) ( )F l•∫ = =∫ + =∫ + =∫ = + + = = = = d r dr r d r d P P r1 3 2 2 2 3 2 0 2 0 2 2 2 2 0 4 2 2 θ θ φ π π θ π φ π π sin F l•∫ = ∫ + ∫ + ∫ = − + + = − = = = d rdr r d r d P P r3 0 2 0 2 4 4 2 2 2 3 2 0 0 2θ θ φ θ π π φ π π sin F l F l F l•∫ = •∫ + •∫ = −d d d P P P P P P 1 0 1 3 3 0 2 2π (b) F l•∫ = ∫ + ∫ + ∫ = − + + = − = = = d rdr r d r d P P r1 0 2 0 2 0 0 0 0 2 3 2 0 0 2θ θ φ θ φ sin 2 91. . A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ + ∫∫d xdydz ydxdz zdxdy . Top: ( ) y x z dxdy =− =− ∫ =∫ = 1 1 1 1 1 4 . Bottom: ( )− ∫ = −∫ = =− =−y x z dxdy 1 1 1 1 1 4 . Right: ( ) z x y dxdz =− =− ∫ =∫ = 1 1 1 1 1 4 . Left: ( )− ∫ = −∫ = =− =−z x y dxdz 1 1 1 1 1 4 . Front: ( ) z y x dydz =− =− ∫ =∫ = 1 1 1 1 1 4 . Back: ( )− ∫ = −∫ = =− =−z y x dydz 1 1 1 1 1 4 . Total=4+4+4+4+4+4=24. y z x P0(0, 0, 0) P1(0, 0, 2) 2 45° P3(0, 2 , )2 2-15 2 9 2. . A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ − ∫∫d xydydz yzdxdz xzdxdy . Top: ( )− ∫ =∫ = − = =y x x z dxdy 0 2 0 1 3 3 . Bottom: ( ) y x x z dxdy = = ∫ =∫ = 0 2 0 1 0 0 . Right: ( ) z x y zdxdz = = ∫ =∫ = 0 3 0 1 2 9 . Left: ( )− ∫ =∫ = = =z x y zdxdz 0 3 0 1 0 0 . Front: ( ) z y x ydydz = = ∫ =∫ = 0 3 0 2 1 6 . Back: ( )− ∫ =∫ = = =z y x ydydz 0 3 0 2 0 0 . Total=-3+0+9+0+6+0=12. y z x (–1, 1, 1) (–1, 1, –1) y z x 2 1 3 2-16 2 9 3. . A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ − ∫∫d r d dz drdz rdrd2 3 22 φ φ φ . Top: − ∫ ∫ = − = =φ π φ π 0 2 0 2 2 2rdrd r . Bottom: φ π φ π = = ∫ ∫ = 0 2 0 2 2 2rdrd r . Right: z r drdz = = ∫ = ∫ =0 3 0 2 3 2 9φ π π . Left: ( )− ∫ =∫ = = =z r drdz 0 3 0 2 3 0 0φ . Front: ( ) z r d dz = = ∫ =∫ = 0 3 2 0 2 2 2 12φ π φ π . Total=-2π+2π+9π+0+12π=21π. 2 9 4. . A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ − ∫∫d r d dz drdz zrdrd3 2 φ φ φ . Top: ( )− ∫ =∫ = =− =φ π π φ π 2 2 0 2 4 8z rdrd r . Bottom: ( ) φ π π φ =− = ∫ =∫ = 2 2 0 2 0 0z rdrd r . Right: z r drdz = = ∫ = ∫ =0 4 0 2 2 4φ π π . Left: − ∫ = − ∫ = = =z r drdz 0 4 0 2 2 4φ π π . Front: ( ) z z r d dz = =− ∫ =∫ = 0 4 2 2 2 2 48φ π φ π π . Total=8π+0+4π+4π+48π=64π. z 3 2 y x 2-17 2 9 5. . A sketch of the surface is given below. F s•∫ = ∫∫ + ∫∫ − ∫∫d r d d r drd rdrd3 2sin sinθ φ θ θ φ φ θ . Front: ( ) θ π φ π θ φ θ π = = ∫ =∫ = 0 2 3 0 2 2 4r d dsin . Bottom: φ π π φ π = = ∫ ∫ = 0 2 0 2 2 2 2r drd r sin . Right: − ∫ = ∫ = − = =θ π φ π θ π 0 2 0 2 2 2 2 rdrd r . Left: ( ) θ π φ θ = = ∫ =∫ = 0 2 0 2 0 0rdrd r . Total=4 2 2 0 6 2 2 2 π π π π π+ − + = − . z 4 2 y x 2 y x z 2-18 2 9 6. . A sketch of the surface is given below. F s•∫ = ∫∫ − ∫∫ + ∫∫d r d d r drd rdrd3 2 3sin sinθ φ θ θ φ φ θ . Front: ( ) θ π φ π π θ φ θ π = =− ∫ =∫ = 0 2 3 2 2 3 27r d dsin . Bottom: − ∫ ∫ = − =− =φ π π π φ π 2 2 0 3 2 2 9r drd r sin . Right: θ π φ π θ π = = ∫ = ∫ =0 2 0 3 2 3 2 27 8 rdrd r . Left: − ∫ = − ∫ = = =θ π φ π θ π 0 2 0 3 2 3 2 27 8 rdrd r . Total= 27 9 27 8 27 8 18 27 4 2 2 2 π π π π π π − + + = + . 3 y x z Eletromag/ch03.pdf 3-1 Chapter 3 Problem Solutions 311. . Q dv zrdzd drv r z = ∫∫∫ = ∫ ∫ ∫ = = = = ρ φ π φ π π 0 1 4 2 0 2 2 2 C 312. . The surface is shown below. ( )Q ds xy z dxdys x y x = ∫∫ = ∫ =∫ = = = − + ρ 1 2 1 2 5 3 2 13C . 313. . The problem is sketched below. A vector from the second charge to the first is ( )( ) ( ) ( )( )R a a a a a a21 1 1 1 0 1 2 2 3= − − + − + − − = + +x y z x y z whose length is R21 14= . A unit vector pointing from the second charge to the first is a R 21 21 21 = R . Coulomb’s law yields ( )( ) F a a a a21 9 6 6 21 2 219 10 100 10 50 10 1718 0 859 2577= × × × = + + − − R x y z. . . N . y z x y = –2x + 5 (1, 1, 2) (1, 3, 2) (2, 1, 2) y z x (–1, 0, 2) Q2 = 50 �C Q1 = 100 �C (1, 1, 1) 3-2 314. . (a) F=0, (b) ( ) ( )F a= × × × × − 4 9 10 100 10 2 459 6 2 2 cos o z , (c) ( ) ( )F a a1 9 6 2 29 10 100 10 1 90= × × = − x x , ( ) ( )F a a2 9 6 2 29 10 100 10 1 90= × × = − y y , ( )F a a3 905= +cos sinθ θx y , ( )F a a4 905= +sin cosθ θx y . But cosθ = 25 and sinθ = 1 5 . Hence F F F F F a a= + + + = +1 2 3 4 11415 11415. .x x . 315. . The problem is sketched below. For the forces exerted on Q3 by the other two charges to be equal and oppositely directed, we must have ( )( ) ( )( ) ( )9 10 18 10 8 10 9 10 72 10 8 10 0 03 9 6 6 2 9 6 6 2× × × = × × × − − − − − d d. . Solving for d gives d=1cm. F2 F4 F3 F1 –1 � � � � 1 y x 5 5 1 2 4 3 d 3 cm Q1 = 18 �C Q3 = –8 �C Q2 = 72 �C –1 1 3-3 316. . The charge of an electron is e = − × −16 10 19. C. Placing the positive charge on the left and the negative charge on the right, the force exerted on the electron by the positive charge is directed to the left and is ( )( ) ( )F1 9 6 19 2 2 129 10 35 10 16 10 10 10 504 10= × × × × = × − − − − . . N . The force exerted on the electron by the negative charge is of the same magnitude and in the same direction so that the net force is 1 10 11× − N directed toward the positive charge. 317. . The problem is sketched below. In order that the forces balance, the Coulomb force acting in the horizontal direction is ( )F Q lo = 2 24 2π ε θsin . The component of the restraining force along the string that is horizontally directed is T sinθ and the force of gravity acting downward on the charges is T mgcosθ = . Hence F mg= sin cos θ θ . Equating the two and solving gives ( )Q l mgo2 2 31 16= cos sinθ πε θ . 3 21. . The problem is sketched below. The electric field due to the positive charge is ( )E a a1 9 1 29 10 4 2 8125= × =Q y y, . . The electric field due to the negative charge is ( )E a a2 9 2 29 10 2 22 500= − × = −Q z z, . Hence the total electric field is E E E a a= + = −1 2 2 8125 22 500, . ,y z V m . � 2l sin � QF mg Q T l 3-4 3 2 2. . The problem is sketched below. The angle θ is θ = 60o . The distances from the triangle vertices to the center is, according to the law of cosines, ( )5 2 22 2 2 2= + −d d d cos θ giving d=2.887 m. The vector contributions are ( )E Q d Q do o o = + = 4 2 4 60 1082 2π ε π ε cos kV m . 3 2 3. . The problem is sketched below. (a) First we determine the electric field along the z axis. Superimposing the fields due to the two charges gives E a a a a= − − + = − + = − 1 4 2 1 4 2 2 2 2 2 4 2 2 2 2 2 2 2π ε π ε π ε π εo z o z o z o z Q z l Q z l Ql z z l z l Ql z z l z y (0, 4, 2) Q1 = 5 �C Q2 = –10 �C z = 2 y = 4 E1 E2 E 5 m 5 m 5 m Q —Q —Q d d d � � 3-5 (b) Now we determine the electric field along the y axis. Superposing the fields as shown gives E a a= − + + = − + 2 4 1 4 2 4 4 4 2 2 2 2 1 2 2 2 3 2 Q y l l y l Ql y l o z o zπ ε π ε θsin 1 244 344 . 3.2.4 The problem is sketched below. Divide the charge into chunks of charge, {dQ adl dl = ρ φ . At a distance d from the center and on a line perpendicular to the ring, the horizontal components cancel out leaving only the vertical components so that ( )E a= ∫ = 1 4 20 2 π ε ρ φ α φ π o l z ad R cos where R d a= +2 2 and ( )cos α = d R . Substituting these gives ( ) ( ) E a a = + ∫ = + > = 1 4 1 2 0 2 2 3 20 2 2 2 3 2 π ε ρ φ ε ρ φ π o l z o l z ad d a d ad d a z At a large distance from the center, d a>> , this result reduces to E a a = >> = >> 1 2 2 4 2 2 ε ρ π ρ π ε o l z l o z a d d a a d d a y l 2 � � � E+ E+ E– E– E z z y l 2 3-6 3.2.5 The problem is sketched below. The chunks of charge are dQ dr rds ds = ρ φ123 and again, by symmetry, the horizontal components cancel leaving only the vertical (z- directed) components. Summing these contributions gives ( ) E a a a = ∫ ∫ + + = − + > = >> = =r a s o R z s o z s o z rd dr d r d d r d d a z a d d a 0 0 2 2 2 1 2 2 2 2 2 2 4 1 2 1 0 4 2 ρ φ π ε ρ ε π ρ π ε φ π α 124 34 1 24 34 cos E R z = d a y z x � � �l ad� E R z = d �s rdrd� a r y z x � � 3-7 3.2.6 The problem is sketched in the xy plane below. Using the results of Example 3.3 and superpositioning the fields gives (a) E a a a= − − + = − ρ π ε ρ π ε ρ π ε l o y l o y l o y y l y l l y l 2 1 2 2 1 2 2 4 2 2 . Similarly the fields along the x axis become E a= − + + 2 2 4 2 4 2 2 2 2 ρ π ε θ l o y x l l x l cos 1 24 34 . 3.2.7 The problem is sketched below. Place the strip in the xz plane centered on the origin. Divide the strip into infinite line charges with distribution ρ ρl sdz= C m . Use the result in Example 3.3, equation (3.10). Accounting for symmetry, E a= ∫ = 2 20 2 ρ πε αs oz W y dz R cos ` where R z d= +2 2 and cosα = d R . Hence, using the integral 1 1 2 2 1 d z dz d z d+ ∫ = −tan , ( )E a a= +∫ = = − ρ πε ρ πε s o z W y s o y d d z dz d W d 1 22 20 2 1 ` tan . Er– Er+ Er– Er+l 2 l 2 R R y y x x —� � � � 3-8 3 31. . The electric field intensity vector is E V d = = 105 V m . The polarization vector is P D Eo= − ε . Substituting D Er o= ε ε gives ( ) ( )P Eo r= − = × − × =−ε ε π µ1 1 36 10 54 1 10 3899 5. . C m2 . 3.4.1 The problem is sketched below. (a) the total charge enclosed is Q r drd d kr drd d ka r a v dv r a enc = ∫ ∫ =∫ ∫ ∫ =∫ = = = = = =0 0 2 2 0 0 0 2 3 0 4 φ π θ π φ π θ π ρ θ φ θ θ φ θ πsin sin1 244 344 . (b) Using Gauss’ law, ε ε π πo od E r Q kaE s•∫ = = =4 2 4enc giving E a= ka ro r 4 24ε . (c) The charge enclosed is Q krenc = π 4 . Hence the electric field is E a= kr o r 2 4ε . � � E– E+ z d R R W —z z y W 2 — W 2 � v = kr a r dv y z x 3-9 3 4 2. . No. No closed surface can be found for which the electric field is perpendicular to all sides and hence no simplification of D s•∫ d can be obtained. 3 4 3. . The problem is sketched below. Since D is directed in the z direction, there is no flux through the sides. Hence Gauss’ law gives ( ) ( )Q d d z rrdrd z rrdrd a r a ds r a ds enc top bottom C= •∫ + •∫ = ∫ =∫ − ∫ =∫ = = = = = = D s D s123 123 123 123φ π φ π φ φ π π 0 2 0 0 2 0 3 4 0 8 3 64 3 . 3 4 4. . For r b≥ ( )Q dv kr r drd d k b av r a b dv enc = ∫ = ∫ ∫ ∫ = − = = = ρ θ φ θ π φ π θ π 0 2 2 0 2 22sin1 244 344 . By symmetry, the field is radially directed. Hence D s•∫ =d Qenc so that ( )ε π πo rE r k b a4 22 2 2= − . Thus ( ) E k b a r r o = − 2 2 22ε . For r a≤ , Er = 0 since no charge is enclosed. For a r b≤ ≤ , ( )Q k r aenc = −2 2 2π . Hence ( )ε π πo rE r k r a4 22 2 2= − . Thus E k a r r o = − 2 1 2 2ε . 3.5.1 The problem is sketched below. The work required to move q around the paths is W q d q xdx q ydy q zdz= − •∫ = − ∫ − ∫ − ∫E l . (a) z 4 Dz a y x Dz 3-10 W q zdz q ydy q zdz q ydy z y z y = − ∫ − ∫ − ∫ − ∫ = = = = =0 1 0 2 1 0 2 0 0 . (b) W q zdz q ydy q xdx q xdx z z x x = − ∫ − ∫ − ∫ − ∫ = = = = =0 1 1 0 0 1 1 0 0 . 3.5.2 The problem is sketched below. Applying superposition and equation (3.37) yields V Q o = − =4 1 2 1 3 15 π ε kV . 3.5.3 The problem is sketched below. Applying superposition and equation (3.37) yields ( ) ( ) ( ) ( ) V Q Q o o = + − − + + − − = − − = − 1 2 2 2 2 24 1 5 5 3 1 3 4 1 5 5 2 1 2 13 287 42 1478256 28 π ε π ε , . . ,070V y z x (0, 2 m, 1 m) (0, 2 m, 0) (0, 0, 1 m) x = 1 y = 2 z = 1 (0, 0, 0) (0, 3, 2) – + y = 3 Q = 10 �C V z y 3-11 3.5.4 The problem is sketched below. Applying superposition and equation (3.37) yields V Q Q o o = + − + + − = − − = −1 2 2 2 2 24 1 5 2 1 2 4 1 5 3 1 3 28 287 42 7 28256 35 570 π ε π ε , . , . , V 3.5.5 The problem is sketched below. Applying superposition and equation (3.37) yields V Q Q o o = + − + + − = − + =1 2 2 2 2 24 1 5 3 1 3 4 1 5 2 1 2 7 28256 28 287 42 21 π ε π ε , . , . ,005V . (0, 5, 5) – + y = 2 z = 3 V Q2 = 5 �C Q1 = 10 �C z y (5)2 + (5 – 3 )2 29 = (5 – 2)2 + (5)2 34= (0, 0, 5) – + y = 3y = –2 V Q2 = 5 �CQ1 = 10 �C z y 52 + 22 52 + 32 3-12 3.5.6 The problem is sketched below. Applying superposition and equation (3.38) yields V o o = + + + = − + = ρ π ε ρ π ε 1 2 2 2 2 22 2 5 2 2 3 5 3 89145 119 30 477Vln ln , ,622 , . 3.5.7 The problem is sketched below. By the law of cosines, the distance from each charge to the center of the triangle is related to the side length as ( )l d d d o2 2 2 22 120= + − cos so that d l= 1732. . Applying superposition and equation (3.42) yields V Q l Q l o = = × =3 4 1732 4 677 10 31177V10 π ε . . , . y z x x = 3 Q1 = 5 �C Q2 = –10 �C (0, 5, 0) z = 2 52 + 22 52 + 32 – + y = 2 z = 5 y = –3 V �l = –10 �C/m �1 = 5 �C/m z y 52 + 32 52 + 22 3-13 3.5.8 The problem is sketched below. Using (3.44) gives V ad z a a z a l o l o = + ∫ = += ρ φ π ε ρ εφ π 4 22 20 2 2 2 . The potential is only a function of z and hence the gradient is ( ) ( )E a a a= − = − = − + = + − gradientV V z a z z a a z z a z l o z l o z ∂ ∂ ρ ε ∂ ∂ ρ ε2 2 2 2 1 2 2 2 3 2 which agrees with the results of Problem 3.2.4. 120° 120°120° l ll Q Q Q dd d R a y z x �l ad�C 3-14 3.5.9 The problem is sketched below. Using (3.44) gives V rdrd z r z a z r a s o s o = ∫ + ∫ = + − = =0 2 20 2 2 2 4 2 ρ φ π ε ρ εφ π . The potential is only a function of z and hence the gradient is E a a a= − = − = − + − = − + gradientV V z z z a z z z a z s o z s o z ∂ ∂ ρ ε ∂ ∂ ρ ε2 2 12 2 2 2 which agrees with the results of Problem 3.2.5. 3.5.10 (a) ( ) ( ) ( ) E a a a a a a = − = − − − = + + + + + + + + gradientV V x V y V z x x y z y x y z z x y z x y z x y z ∂ ∂ ∂ ∂ ∂ ∂ 2 2 2 3 2 2 2 2 3 2 2 2 2 3 2 (b) E a a a a a a = − = − − − = − + +− − − gradientV V r r V V z e e re r z z r z z z ∂ ∂ ∂ ∂ φ ∂ ∂ φ φ φ φ φ 1 cos sin cos R �s rdrd� �s a r y z x 3-15 E a a a a a a = − = − − − = − + gradientV V r r V r V r r r r r ∂ ∂ ∂ ∂ θ θ ∂ ∂ φ θ φ θ φ φ θ φ θ φ 1 1 2 3 3 3 sin sin cos cos cos sin 3 61. . E V d = = 10 kV m , D Eo r= =ε ε µ0 478. C m2 , P D Eo= − =ε µ0 389. C m2 , C A d pFr o= =ε ε 4775. . 3 6 2. . ( )C A do o = = = − ε ε π 01 10 277 8 2 3 . . pF . 3 6 3. . There are two capacitors in series: C A dr o1 1 1 = ε ε and C A dr o2 2 2 = ε ε . Capacitors in series add like resistors in parallel so that C C C C C A d d d d o r r r r = + = + 1 2 1 2 1 2 1 2 1 1 2 2 ε ε ε ε ε . The total free charge on the upper (and lower) plate is Q CVf = . Using Gauss’ law and surrounding the upper plate with a closed surface yields DA Q f= . Therefore D Q A CV A f = = . In the upper dielectric, E D CV Ar o r o 1 1 1 = = ε ε ε ε and in the lower dielectric, E D CV Ar o r o 2 2 2 = = ε ε ε ε . Evaluating these gives C = 70 74. pF , E1 4000= V m , E2 2000= V m . 3 6 4. . There are two capacitors in parallel: C A dr o1 1 1 = ε ε and C A dr o2 2 2 = ε ε . Capacitors in parallel add like resistors in series so that ( )C C C A A do r r = + = + 1 2 1 1 2 2ε ε ε . Using Gauss’ law and surrounding each portion of the upper plate with a closed surface yields D A Q f1 1= and D A Q f2 2= . Therefore E D C V Ar o r o 1 1 1 1 1 1 = = ε ε ε ε and (c) 3-16 E D C V Ar o r o 2 2 2 2 2 2 = = ε ε ε ε . Evaluating these gives C1 88 42= . pF , C2 707 4= . pF , C = 7958. pF , E1 5000= V m , E2 5000= V m . Observe that V E d E d= = =1 2 10V and Q Q Q CVf f f= + = =1 2 7 96. nC . 3 6 5. . We observe that there are essentially two spherical capacitors in series. First we obtain the capacitance of a spherical capacitor with a homogeneous dielectric filling the interior. Using Gauss’ law and surrounding the inner sphere with a sphere of radius r gives D Q r f = 4 2π . Hence the electric field is radially directed and is E Q r f = 4 2π ε . The voltage between the spheres is V Q r dr Q a b f b a f = − ∫ = − 4 4 1 1 2π ε π ε . Hence the capacitance is ( )C Q V ab b a f = = − 4π . This result was derived in Exercise Problem 3.8. Hence the capacitances are ( )C ar r a1 1 1 4 = − π and ( )C br b r2 1 1 4 = − π . Since these are in series and capacitors in series add like resistors in parallel we obtain C C C C C r b a r o r r = + = − + − 1 2 1 2 2 1 1 1 4 1 1 1 1 1 1 1 π ε ε ε . Substituting the values gives 0.303pF. 3 6 6. . The per-unit-length capacitance for a coaxial cable filled with a homogeneous dielectric was obtained in Example 3.15 as c b a = 2π ε ln F m . For this problem we observe that there ae two such capacitors in series: c r a o r 1 1 1 2 = π ε ε ln and c b r o r 2 2 1 2 = π ε ε ln . Since capacitors in 3-17 series add like resistors in parallel, c c c c c b r r a o r r r r = + = + 1 2 1 2 1 2 1 1 2 1 2π ε ε ε ε εln ln . Evaluating this for the given dimensions yields 82 06. pF m . 3.7.1 Converting the radius from mils to meters gives 16 254cm 1 4 064 10 4mils 1inch 1000mils 1inch meter 100cm m× × × = × −. . . Hence the resistance is ( )R A= × × = × =− 1000 4 064 10 3323 4 2 m = 5.8 107σ π . . Ω . 3.7.2 The electric field is E V d = . Hence the current density is J E V d = =σ σ . The total current is I JA A V d = = σ . Hence the resistance is R V I d A = = σ Ω . Evaluating this gives 50mΩ. 3.7.3 The electric field intensity at a radius r was determined in Problem 3.6.5 as E Q r f = 4 2π ε . The voltage between the spheres was also determined as V Q r dr Q a b f b a f = − ∫ = − 4 4 1 1 2π ε π ε . The current flowing between the two spheres is I JA EA Q r r Qf f = = = =σ σ π ε π σ ε4 42 2 . Hence the resistance is R V I a b = = − 1 4 1 1 π σ . Evaluating this for the given dimensions yields 106.1Ω. 3.7.4 The voltage between the inner wire and the shield was determined in Example 3.15 in terms of the charge distribution on the inner wire of ρ l C m as V b a l = ρ π ε2 ln . Similarly, the electric field in this region was determined in Example 3.7 to be 3-18 E r l = ρ π ε2 . Hence the current flowing from one cylinder to the other (per-unit-of line length) is I JA r rl l= = =σ ρ π ε π σ ρ ε2 2 . Thus the resistance per unit length is r V I b a = = 1 2π σ ln /Ω m . Evaluating this for the given dimensions yields 01. /Ω m 3.7.5 From the results of Example 3.16, the magnetic flux density vector at a perpendicular distance from the midpoint of the current element is B a= + µ π φ o I r L r L 2 2 4 2 2 . (a) At (0,3m,0) the field is 0.2858 Taφ µ . (b) Off the ends of the curent element d Rl a× is zero and hence the field is zero off the ends. 3.7.6 Utilizing the result obtained in Example 3.16 for an infinite current, the magnetic flux density at a distance r is B I r o = µ π2 Wb m2 . Hence the total magnetic flux penetrating the loop is ψ = •∫ B sd . By the right-hand rule the B field is directed into the page and hence the dot product can be removed. However, the B field depends on distance away from the current and cannot be removed from the integral. Hence ψ µ π µ π = ∫ ∫ = = =z l o r r r oI r drdz Il r r0 2 12 21 2 ln Wb . For the given parameters we obtain ψ µ= 0139. Wb . 3.7.7 The problem is sketched below. Using the results of Example 3.16 we may superimpose the contributions from the four identical sides. Observe that the contributions from opposite sides cause the net B from those two to be directed in the +z direction. similarly the other two sides cause a similar result. Hence B a= + 4 2 2 4 2 2 µ π αo z I R l R l cos where 3-19 R z l= +2 2 4 and cosα = l R 2 . Combining these gives B a= + + 2 4 4 2 2 2 2 2 2 µ π o z I l z l z l . For l = 2m and I = 10A , at the center of the loop, z = 0 , we obtain B a= 566. µ T z . 3.7.8 The problem is sketched below. Using the results of Example 3.16 we may superimpose the contributions from the three identical sides. The radial distance from the center of each side to the center of the triangle is r l= 0 577 2 . . Using (3.60) of Example 3.16 we obtain B I r l r l I l o o = + =3 2 2 4 9 22 2 µ π µ π into the page. For l = 5cm and I = 4A we obtain B = 144µ T . R � �� � R I I I I l 2 ( ,l 2 , 0)l 2 (– ,l 2 , 0)l 2 l 2 y z x 3-20 3.7.9 The problem is sketched below. According to the Biot Savart law, the magnetic field off the ends of a current is zero since d Rl a× is zero. Hence there are no contributions to the magnetic field at point P due to sides DA and BC. For side CD d rdRl a× = θ and is out of the page. Hence the Biot Savart law gives for this contribution ( ) µ π θ µ π θ θ θ o oI r r d I r4 42 2 2 0 2= ∫ = . Similarly the contribution from the segment AB is µ π θo I r4 1 which is into the page. Hence the total is B I r r o = − µ θ π4 1 1 1 2 which is directed into the page since r r2 1> . I 30° l l l r rr I I � I AD C B P I I I r1 r2 3-21 3.9.1 The problem is sketched below. Use the result for the magnetic field from an infinitely long current filament obtained in Example 3.19: H I r = 2π . (a) At the center, the H fields add giving H I d I d = =2 2 2 2 π π . (b) At a distance D from the center and in a plane containing the currents we superimpose the fields to give ( ) ( )H I D d I D d I d D d = − − + = − 2 2 2 2 2 4 2 2π π π . 3.9.2 The problem is sketched below. Place the strip on the z axis centered about the origin. Treat the linear current density as filaments of current Kdz A . Superimpose the magnetic fields using equation (3.81) of Example 3.19. The magnetic field intensity vector is H a= ∫ = 2 20 2 Kdz Rz w zπ αcos where R d z= +2 2 and cosα = d R . Hence ( )H Kd d z dz K z d K w dz w z w = + ∫ = = = − = − π π π 1 22 20 2 1 0 2 1tan tan . For an infinite strip as w → ∞ , H K→ 2 which agrees with the result for an infinite strip in Example 3.21. d 2 d 2 II D 3-22 3.9.3 For (b) a r b< < , H rd NIφ φ π φ = ∫ = 0 2 . Hence H NI rφ π= 2 . For (a) r a< and (c) r b> the net current penetrating this loop of radius r is zero and hence H=0. 3.9.4 The problem is sketched below. Construct a rectangular contour as shown. By symmetry, the magnetic field is directed along the solenoid and is directed right to left. Applying Ampere’s law to this yields H l•∫ = = =d HL I InLenclosed . Hence H nI= and B H nIr o r o= =µ µ µ µ . 3.11.1 The magnetic flux density along the axis of the solenoid was obtained in Problem 3.9.4 as B nIr o= µ µ . Since this is uniform over the cross section of the core and is axially directed, the flux is ψ π= •∫ =B sd B a core cross section 2 . The flux linkages per unit length are � �� z d R R Kdz y z = w 2 z = _ w 2 L C H H II 3-23 Λ = nψ . Hence the per-unit-length self inductance is l I n ar o= = Λ µ µ π2 2 H m . For the given dimensions, ( )l = × × × = −1000 4 10 2000 0 01 1587 2 2π π turns m H m . . . 3.11.2 Treating this as a long parallel wire line of length l and separation w and neglecting the contribution from the end segments gives, using the result of Example 3.25 for an infinite line, L l w a o≅ µ π ln . 3.11.3 The problem is sketched below. Assuming the plate width is much greater than the plate separation we can use the result in Example 3.18 for the magnetic flux density from an infinite plate carrying a linear current density of K A m . The magnetic flux density is parallel to the plate and opposite to the direction of the current: B Ko= µ 2 . Superimposing the fields due to both plates gives B Ko= µ which is constant across the cross section between the plates. The flux penetrating the surface between the plates is ψ µ= •∫ = =B sd Bs l Ks lo∆ ∆ . The total current on each plate is I Kw= . Hence the flux is ψ µ= o I l s w ∆ . Thus the inductance per unit length is l I l s wo = = ψ µ ∆ . s w B 3-24 3.11.4 First we solve the basic problem shown below. The magnetic field intensity due to the infinitely long current is B I r o = µ π2 . The flux through the loop is obtained by integration as ψ µ π µ π = ∫ = = l I r dr Il b a o r a b o 2 2 ln . Applying this to the original problem and superimposing the fluxes due to each current gives ( ) ( ) ( ) ( )ψ µ π µ π2 2 2 2 2 2 2 2 2 2 2 = − + − − − + + + − o o Il D s w D s w I D s w D s w ln ln . Hence the mutual inductance is ( )( ) ( )( )M I l D s w D s w D s w D s w o = = − + + − − − + + ψ µ π 2 2 2 2 2 2 2 2 2 2 ln . 3.12.1 The particle is traveling with constant velocity so that F ma= = 0 . In order for the vertical forces to balance so that the particle passes through the hole we must have the electric force, F qEe = , equal the magnetic force, F qvBm = . Solving gives the critical velocity of the particle as v E B = . Evaluating this for the given conditions yields v = × = × − 2000 1 10 2 103 6 m s . 3.12.2 The magnetic flux density vector is radially directed about each wire and is given by B I r o = µ π2 . The magnetic flux density at wire 2 due to the current of wire 1 is therefore b I a l 3-25 B I s o 21 1 2 = µ π and is perpendicular (into the page) to current I2 . The force exerted on a ∆l section of wire 2 is, according to the Lorentz force equation, F I lB I l I s o 21 2 21 2 1 2 = =∆ ∆ µ π . Hence the force per unit length exerted on wire 2 is f F l I I s o 21 21 1 2 2 = = ∆ µ π N m . 3.12.3 From the previous problem, the force exerted on the left side of the loop is F I I a wo1 1 22 = µ π , and the force exerted on the right side of the loop is F I I b wo3 1 22 = µ π . Along the upper and lower segments of the loop, the B field varies along the wire. Hence the force along the upper segment is F I I r dr I I b a Fo r a b o 2 1 2 1 2 42 2 = ∫ = = = µ π µ π ln . 3.12.4 The sliding bar cuts the magnetic field resulting in a voltage source , Bvw , inserted in it as shown below. Hence the current is I Bvw R = − . 3.12.5 The vertical side of the loop cuts the magnetic field so that a voltage is induced in it. The horizontal sides have no inserted source since v B× is perpendicular to the wire. Shown below is a view in the xy plane. The tangential velocity is v l= ω and ( )v B× = l B tω ωsin assuming that the loop starts at the x axis at t=0. The voltage source is inserted as shown below so that the current through the resistor is ( )I wl B t R = − ω ωsin . I R Bvw+ – 3-26 B B l v �t � x y I R l� B sin �t+ – Eletromag/ch04.pdf 4-1 Chapter 4 Problem Solutions 411. . The flux in the left loop is ψ1 301 1 0 2 10= × × = × −B t. . Wb , and the flux in the right loop is ψ 2 301 0 5 01 10= × × = × −B t. . . Wb . The induced sources are shown below: V d dt1 1 0 2= =ψ . mV and V d dt2 2 01= =ψ . mV . Solving the resulting circuit gives V = − − + = −01 100 100 50 0 2 0 233. . .mV mV mV . 412. . The flux in the inner loop is ψ1 301 0 5 01 10= × × = × −B t. . . Wb , and the flux in the outer loop is ψ 2 30 3 1 0 6 10= × × = × −B t. . Wb . The induced sources are shown below: V d dt1 1 01= =ψ . mV and V d dt2 2 0 6= =ψ . mV . Solving the resulting circuit gives V = − + =0 6 100 100 50 01 0533. . .mV mV mV . – V + 50 100 0.1 mV + – 0.2 mV + – – V + 50 100 0.1 mV 0.6 mV + – + – 4-2 413. . The flux in the left loop is ( )ψ π1 305 0 2 10 2 60= × × = −B t. . sin Wb , and the flux in the right loop is ( )ψ π2 303 0 2 0 6 10 2 60= × × = × −B t. . . sin Wb . The induced sources are shown below: ( )V d dt t1 1 377 2 60 2 60= = ×ψ π πcos1 2444 3444 mV and ( )V d dt t2 2 266 2 60 0 6 2 60= = × ×ψ π π. cos1 24444 34444 mV . Solving the resulting circuit gives ( ) ( ) ( )V t t t= × + + × = ×226 2 60 50 200 50 377 2 60 3016 2 60cos cos . cosπ π πmV mV mV . 414. . (b) and (d) are not correct. In (b) a current must be established that will induce a secondary current I that will oppose the change in the original magnetic field. For this case the induced current must be counterclockwise to keep the field from decreasing. Similarly, in (d) a current must be established that will induce a secondary current I that will oppose the change in the original magnetic field. For this case the induced current must be clockwise to keep the field from increasing. 415. . The flux in the loop is ( ) ( )ψ π π= × = × × × ×B t t t tArea mWb = mWb2 2 60 05 10 10 2 60cos . cos . The induced – V + 200 50 V1 V2 +– +– 4-3 voltage is ( ) ( )[ ]V ddt t t t= = × − × ψ π π10 2 60 3770 2 60cos sin mV . Hence the current is ( ) ( )[ ]I V t t t= = × − ×100 01 2 60 37 7 2 60. cos . sinπ π mA . 416. . The position of the bar is ( )L d tt= ∫ =100 10 10 10 0 cos sin( )τ τ m . The flux in the loop is ( ) ( )ψ = × = × ×B t tArea mWb = 0.05 Wb10 05 10 10 10. sin sin . Hence the induced voltage is ( )V d dt t= =ψ 05 10. cos V . Hence the current is ( ) ( )I t t= =05 10 100 5 10 . cos cos mA . 417. . A view in the xy plane is shown below. The flux through the loop is ( ) ( )ψ ω ω= •∫ = × × =B sd B t tArea mWbcos cos1 . Hence an induced voltage in the loop is ( ) ( )V d dt t t= = −ψ ω ω ωsin sinmV = -5 mV with polarity shown below. Hence the current is ( )I V t= = − 2 2 5. cos ω mA . I V 100 50 cm v + – I 100 50 cm v +– 4-4 418. . The magnetic flux density a distance r from an infinitely long current was obtained in the previous chapter as B t I t r o( ) ( )= µ π2 and is circumferentially directed about the wire. Hence the flux through the loop formed by the household power wiring is ψ µ π = •∫ = ∫ ∫ = × = × = − −B sd I t r drdz I t I t z 0 3 7 8 2 6 10 109 517 10 m o r=1km 1.09km km 1km ( ) ln . ( ) . ( ) . The induced voltage is V d dt dI t dt = = × − ψ 517 10 8. ( ) and is sketched below. Hence V = 2 585, V for 0 1< <t µ s and is V = −287V for 1 10µ µs s< <t . 419. . The magnetic flux density threading the loop is B I r o = µ π2 . Assuming that the loop starts at t=0 barely touching the wire, at some time t the flux through the loop is (downward) I V 2 + – B B �t x y 2585 V –287 V V (t) 1 �s 10 �s t 4-5 ψ µ π µ π = •∫ = ∫ ∫ = + = = + B sd I r drdz Il vt w vtz l o r vt vt w o 0 2 2 ln . The induced voltage in the loop (tending to push current counter clockwise) is ( )V d dt d dt Il vt w vt Il vt vt w w vt Ilw vt w t o o o = = + = + − = − + ψ µ π µ π µ π2 2 22 ln . Hence ( )I V R Ilw R vt w t o = − = + µ π2 . 4110. . A sketch of the problem looking down on the plane of rotation is shown below. The further we go out along the bar length, the greater the velocity cutting the magnetic field lines. The linear velocity of a section of the bar at a radius r is v r= ω . The voltage induced in the bar is ( )v B l× •∫ = ∫ = = d B rdr B l r l ω ω 0 2 2 . By the right-hand rule, the rotating end is the positive end. 4111. . Forming ∇ × = − + − + − E a a a ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ E y E z E z E x E x E y z y x x z y y x z . From the form of E this reduces to ( ) ( )∇ × = − + = − + −E a a a a∂∂ ∂ ∂ β α ω β α α ω β E z E x E x t z E x t zy x y z m x m zsin cos cos sin l B � 4-6 . From Faraday’s law: ∇ × = −E Hµ ∂∂o t we obtain ( ) ( )H a a= − − + −E x t z E x t zm o x m o z β ω µ α ω β α ω µ α ω βsin sin cos cos . 4112. . Forming ∇ × = − + − + − E a a a ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ E y E z E z E x E x E y z y x x z y y x z . From the form of E this reduces to ( )∇ × = =E a a∂∂ β β ω E z E z tx y m ycos cos . From Faraday’s law: ∇ × = −E Hµ ∂∂o t we obtain ( )H a= − E z tm o y β ω µ β ωcos sin . 4 21. . Above a frequency for which σ ω ε εr o = 1 the displacement current dominates the conduction current. This occurs for f > 45kHz . 4 2 2. . Forming ∇ × = − + − + − H a a a ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ H y H z H z H x H x H y z y x x z y y x z . From the form of H this reduces to ( ) ( )∇ × = − = − + −H a a ∂ ∂ ∂ ∂ β α α ω β H z H x H H x t zx z y x z ysin cos . From Ampere’s law: ∇ × =H Eε ∂∂o t we obtain ( ) ( )E a= − + −β α ωε α ω βH H x t zx z o ysin sin 4 2 3. . Forming ∇ × = − + − + − H a a a ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ H y H z H z H x H x H y z y x x z y y x z . From the form of H this reduces to ( )∇ × = − =H a a∂∂ β β ω H z H z ty x m xsin sin . From Ampere’s law: ∇ × =H Eε ∂∂o t we obtain ( )E a= − β ωε β ωH z tm o xsin cos 4-7 Gauss’ law becomes, in rectangular coordinates, ( )∇ • = = + +ε ∂∂ ∂ ∂ ∂ ∂o x y zE x E y E z E 0 . But for the given electric field, ( )E a= −E x t zm ysin sinα ω β , it has only a y component which is independent of y. Hence, the divergence is zero. 4 3 2. . Gauss’ law becomes, in rectangular coordinates, ( )∇ • = = + +ε ∂∂ ∂ ∂ ∂ ∂o x y zE x E y E z E 0 . But for the given electric field, ( )E a= E z tm xsin cosβ ω , it has only an x component which is independent of x. Hence, the divergence is zero. 4 3 3. . Gauss’ law becomes, in rectangular coordinates, ( )∇ • = = + +µ ∂∂ ∂ ∂ ∂ ∂o x y zH x H y H z H 0 . For the given magnetic field, ( ) ( )H a a= − + −H x t z H x t zx x z zsin sin cos cosα ω β α ω β , so that ( ) ( )∂∂ ∂ ∂ α α ω β β α ω β H x H z H x t z H x t zx z x x z z+ = − + −cos sin cos sina a . Hence the divergence is zero and Gauss’ law is satisfied only if α βH Hx y+ = 0 . 4 3 4. . Gauss’ law becomes, in rectangular coordinates, ( )∇ • = = + +µ ∂∂ ∂ ∂ ∂ ∂o x y zH x H y H z H 0 . But for the given magnetic field, ( )H a= H z tm ycos sinβ ω , the y component is independent of y. Hence, the divergence is zero. 4 61. . The Poynting vector is S E H a= × = − − + −715 1 2 2 8 4 1 2 4 8. cos cose t zz zω π π W m2 . The Poynting vector averaged over one cycle is S aaverage 2 W m = − 715 2 4 8. cose z z π . This is perpendicular only to the sides at z=0 and z=1m. Hence the average power exiting the cube is 4 31. . 4-8 ( ) ( ) P S S e e z z z z average average average m m Area+ Area = - = -25.27W = − × × + = = − = − = 0 1 8 0 8 1715 2 4 715 2 4 . cos . cosπ π 4 71. . The components of E that are tangential to the boundary must be continuous. Hence E a a2, tan = +β γy z . The components of D that are normal to the boundary must be continuous. Hence D E a D E1 1 1 1 2 2 2, , , , norm norm norm norm= = = =ε ε α εx so that E a2 1 2 , norm = ε ε α x . Therefore, E E E a a a2 0 2 2 1 2 x x y z= = + = + +, , norm tan ε ε α β γ . 4 7 2. . The components of B that are normal to the boundary must be continuous. Hence B a2, norm = α x . The components of H that are tangential to the boundary must be continuous. Hence H B a a H B1 1 1 1 1 2 2 2 1 1 , , , , tan tan tan tan= = + = =µ β µ γ µ µy z so that B a a2 2 1 2 1 , tan = + µ µ β µ µ γy z . Therefore, B B B a a a2 0 2 2 2 1 2 1 x x y z= = + = + +, , norm tan α µ µ β µ µ γ . 4 7 3. . The components of D that are normal to the boundary must be continuous. Hence D a2, norm = α x . The components of E that are tangential to the boundary must be continuous. Hence E D a a E D1 1 1 1 1 2 2 2 1 1 1 1 , , , , tan tan tan tan= = + = =ε ε β ε γ εy z so that D a a2 2 1 2 1 , tan = + ε ε β ε ε γy z . Therefore, D D D a a a2 0 2 2 2 1 2 1 x x y z= = + = + +,tan ,norm α β ε ε γ ε ε . 4 7 4. . The components of H that are tangential to the boundary must be continuous. Hence H a a2, tan = +β γy z . The components of B that are normal to the boundary must be continuous. Hence B H a B H1 1 1 1 2 2 2, , , , norm norm norm norm= = = =µ µ α µx so that 4-9 B a a2 2 1 2 1 , tan = + µ µ β µ µ γy z . Therefore, B B B a a a2 0 2 2 2 1 2 1 x x y z= = + = + +, , norm tan α µ µ β µ µ γ . 4 7 5. . The tangential components of E must be zero at the surface of a perfect conductor. Hence γ = 0 . Also, the x and y components of E must be equal in order that there be no tangential component from these two components and hence α β= . the normal components of B must be zero at the surface of a perfect conductor. Hence the x and y components must form a resultant that is tangent to the surface so that σ δ= − . 4 81. . The components tangent to the plane are 2 3a ax y− . Reversing these gives − +2 3a ax y . Similarly reversing the z component gives −4a z . Hence the image current is I a a aimage A= − + −2 3 4x y z at (0,0,-2). 4 91. . To convert a phasor quantity to the time domain we simply multiply by e ej t j tω π= ×2 10 6 and take the real part of the result. Hence, (a) ( ) ( )E E a a a a = = − + − = − × + + × + Re $ Re Re cos cos e j e j e t t j t j t x j t y x y ω ω ω π π π π 30 10 30 2 10 2 10 2 10 2 6 6 , (b) ( )H H a a= = = × + Re $ Re cose e e tj t j t j z zω ω π π π10 10 2 10 434 3 6 , (c) ( ) ( ) ( )B B a a= = = × −− − −Re $ Re cose e e e e t zj t z j t j z x z xω ω π π π4 4 2 10 42 4 2 6 , (d) $E a= − −10 3 3e ex j z yπ , (e) ( )$ sinB a= −5 3 6z e j z x . 4-10 The phasor form of the field is $ sinE a= −E xem j z yα β . Faraday’s law in phasor form is ∇ × = −$ $E Hj oω µ . Writing out the curl gives ∇ × = − + − + − $ $ $ $ $ $ $ E a a a∂∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ E y E z E z E x E x E y z y x x z y y x z . From the form of the field this reduces to ∇ × = − + = +− −$ $ $ sin cosE a a a a ∂ ∂ ∂ ∂ β α α α β βE z E x j E xe E xey x y z m j z x m j z z . Hence ( )$ $ $ sin cos sin cos H E E a a a a = − ∇ × = ∇ × = + = − + − − − − 1 1 1 j j j j E xe E xe E xe j E xe o o o m j z x m j z z m o j z x m o j z z ω µ ω µ ω µ β α α α β ω µ α α ω µ α β β β β The time-domain field is ( ) ( ) ( ) ( ) H H a a a a = = − − + − + = − − − − Re $ sin cos cos cos sin cos cos sin e E x t z E x t z E x t z E x t z j t m o x m o z m o x m o z ω β ω µ α ω β α ω µ α ω β π β ω µ α ω β α ω µ α ω β 2 4 9 3. . Consider two field vectors, ( )E Ee= Re $ j tω and ( )H He= Re $ j tω . The instantaneous power density or Poynting vector is S E H= × . Rewriting the field vectors as ( ) ( )E Ee Ee E e= = + ∗ −Re $ $ $j t j t j tω ω ω12 and substituting into S E H= × yields ( ) ( )S E H E H E H E H E H= × = × + × + × + ×∗ ∗ ∗ ∗ −14 14 2 2$ $ $ $ $ $ $ $e ej t j tω ω . But since ( ) ( )$ $ $ $E H E H× = ×∗ ∗ ∗ and ( )$ $ Re $A A A+ =∗ 2 , this may be written as ( ) ( )S E H E H= × + ×∗12 12 2Re $ $ Re $ $ e j tω . The time average of this is S Sav = ∫1 0T dt T and T is the period of the sinusoid. The first term of our result is a constant and the second term is sinusoidal and hence averages to zero giving the desired result. 4 9 2. . 4-11 The magnetic flux density in the core due to a current I is, from Chapter 3, B I r r oφ µ µ π = 2 where the flux is circumferentially directed and r is approximately the mean radius of the core. The total flux through the windings is approximately ψ = BA where A is the area of the core cross section. According to Faraday’s law a voltage will be generated in the turns of wire of V N d dt NA r dI dt r o = = ψ µ µ π2 . But dI dt j I⇔ ω $ . Hence $ $ $Z V I NA rT r o = = ω µ µ π2 . Substituting numerical values gives ( )( ) ( )( ) ( )$ . .Z f N A rT r o = = = = = × = = −ω π µ µ π 2 200 10 4 10 2 0 02 50 3 4 Ω Ω= 34dB 4101. . Eletromag/ch05.pdf 5-1 Chapter 5 Problem Solutions 511. . The time derivative is related to the phasor form by j t ω ∂ ∂⇔ . Hence − ⇔ −j H z H z t ty y ω µ µ ∂ ∂ $ ( ) ( , ) and − ⇔ −j E z E z t tx xω ε ε ∂ ∂ $ ( ) ( , ) . 51 2. . The time-domain equations are ( ) ( )E z t E t z E t zx m m( , ) cos cos= − + ++ −ω β ω β and ( ) ( )H z t E t z E t zy m m( , ) cos cos= − − + + − η ω β η ω β . Substituting into the first equation gives ( ) ( )∂ ∂ β ω β β ω β E z t z E t z E t zx m m ( , ) sin sin= − − ++ − and ( ) ( )− = − − ++ −µ ∂ ∂ µω η ω β µω η ω β H z t t E t z E t zy m m ( , ) sin sin . Matching the corresponding coefficients of the sin terms gives the requirement that β µω η = . But substituting β ω µ ε= and η µ ε = shows this to be true. Similar results are shown for the second equation of Problem 5.1.1. 51 3. . (a) pvc, ( )ε r = 35. β ω µ ε= =vo r r 0392. rad m , η η µ ε = =o r r 202 Ω , v vo r r = = × µ ε 16 108. m s , λ = =v f 16 m (b) Teflon, ( )ε r = 21. β ω µ ε= = vo r r 0304. rad m , η η µ ε = =o r r 260 Ω , v vo r r = = × µ ε 2 07 108. m s , λ = =v f 20 7. m . (c) Mylar, ( )ε r = 5 β ω µ ε= =vo r r 0 468. rad m , η η µ ε = =o r r 169 Ω , v vo r r = = × µ ε 134 108. m s , λ = =v f 134. m . (d) Polyurethane ( )ε r = 7 5-2 β ω µ ε= = vo r r 0554. rad m , η η µ ε = =o r r 142 Ω , v vo r r = = × µ ε 113 108. m s , λ = =v f 113. m . 51 4. . A sketch is shown below. The phase constant is β ω= = vo 0105. rad m and the intrinsic impedance is η η= =o 377 Ω . The electric field intensity vector is given by $ .E a= 10 0105e j y z or ( )E a= × +10 10 10 01056cos .π t y z . In order for the power flow E H× to be in the -y direction, the magnetic field intensity vector must be in the -x direction so that $ . .H a= −0 0265 0105e j y x or ( )H a= − × +0 0265 10 10 01056. cos .π t y x . 515. . A sketch is shown below. Since the wavelength is λ = v f , the frequency of the wave is 800MHz. Also the velocity of propagation is v vo r = ε . Hence ε r = 2 25. . The phase constant is β ω πλ= = =v 2 251. rad m . The electric field is ( )E a= × −100 16 10 2518cos .π t x z . The intrinsic impedance is η η ε = = o r 251Ω . In order that E H× be in the +x direction, the magnetic field must be directed in the
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