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# Sol-Ch34 - CHAPTER 34 E xercises 1 a o = C 2/N m 2 0 = N/A...

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Exercises 1. (a) [ ϵ o ] = C 2 /N m 2 ; [µ 0 ] = N/A 2 ; so [ ϵ o µ o ] = (s/m 2 ) (b) [E] = N/C; [B] = N/A.m; [1/µ o ] = A 2 /N [EB/µ] = (N.m)/m 2 s = W/m 2 2. (a) [E] = N/C; [cB] = (m/s)(N/A.m) = N/C (b) [ ϵ o d(EA)/dt] = (C 2 /N m 2 )(N/C)(m 2 /s) = C/s (c) From 1b, [S] = [EB/µ o ] = W/m 2 , so [S/c] = W s/m 3 = N/m 2 3. dE/dt = (1/d) dV/dt, Ф = EA; so I D = ( ϵ o A/d) dV/dt = 2.9 ×   10 –7 A 4. (a) 3A; (b) dV/dt = I D /C = I D d/ ϵ o A = 3.78 ×   10 11 V/s 5. I D = ( ϵ o A/d)(dV/dt) = C dV/dt 6. I D = ( ϵ o A/d)(dV/dt) = 92.7 ×   10 – 10 A 7. From Example 34.1:B = ( ϵ o µ o /2d)(dV/dt)r; Peak dV/dt = ωV o . Find B = 6.28 ×   10 –13 T 8. (a) Ampere’s law: (2πr)B = µ o I D (b) I (r) = I D (r/R) 2 , so B = µ o I /2πr = µ o I D r/2πR 2 9. From Example 34.1 and In-chapter exercise 1: (a) B = µ o ϵ o (dE/dt)r/2 = µ o I r/2A = µ o I r/2πR 2 = 5 ×   10 –8 T; (b) B = µ o ϵ o (dE/dt)R 2 /2r = µ o I R 2 /2Ar = µ o I /2πr = 8 ×   10 –8 T 10. B = (E/c) j   = 7.0 ×   10 –8j T 11. (a) 1.26 cm, 23.9 GHz; (b) E z = 60sin(500x + 1.5 ×   10 11 t) V/m 12. B x = –E o /c sin(ky + ωt); B y = B z = 0 13. u av = ϵ o E 2 /2 = B o 2 /2µ o . (a) 150 V/m; (b) 0.5 µT 14. (a) u av = ϵ o E o 2 /2 = 1.11 ×   10 –8 J/m 3 CHAPTER 34

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(b) B o = E o /c = 1.67 ×
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Sol-Ch34 - CHAPTER 34 E xercises 1 a o = C 2/N m 2 0 = N/A...

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