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Sol-Ch30 - CHAPTER 30 E xercises 1 0 I 1 I 2 c/2[1/a1(a...

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Exercises 1. µ 0 I 1 I 2 c/2π[1/a–1/(a + b)]to the right 2. (a) B 1 = µ 0 (4A)/2π(0.08) = 10 µT; B 2 = 40 µT B Tx = –B 1 cos53 ° + B 2 cos37 ° = 26 µT B T y = B 1 sin53 ° + B 2 sin37 ° = 32 µT (b) 4/d = 12/(0.1 + d), thus d = 5 cm, to the left of I 1 (c) F/L = µ o I 1 I 2 /2πr = 96 µN/m, repulsion. 3. (a) B x = (B 1 + B 2 )cos θ ; B y = (B 1 – B 2 )sin θ , where θ = 33.7 ° . B = (6.92i – 1.54 j   ) × 10 –5 T; (b) F = I L × B = (4.6i + 20.8 j   ) × 10 –5 N 4. r = µ o I /2πB = 8 cm, west of the wire. 5. B = µ o I /2πr = 5 × 10 –4 T 6. F = µ o I 1 I 2 /2πd: F 1 = 80 µN/m, repulsion; F 2 = 60 µN/m, attraction. F x = (F 1 + F 2 )cos60 ° = 70 µN/m; F y = (F 1 – F 2 )sin60 ° = 17.3 µN/m 7. B e = 0.5 G due N, B w = µ o I /2πd = 0.06G due W. tanθ = B w /B e leads to θ = 6.8 ° W of N 8. B = µ o I /2πr = 5 × 10 –5 T in the –z direction. Let F = evB = 8 × 10 – 18 N and note that charge is negative. (a) –F j   ; (b) + Fi ; (c) zero 9. From Eq. 30.8, with z = 0, B = µ o I /2a at center of loop. B = µ o I /4a + 2(µo I /4πa) = (µ o I /2a)(1/2 + 1/π) = 5.14 × 10 –7 I /a, out 10. From Eq. 30.8, with z = 0, B = μ o I /2a at center of loop. B = µ o I /2πa + µ o I /2a = (µo I /2a)(1/π + 1) = 8.28 × 10 –7 I /a, out 11. From Eq. 30.8, with z =0, B = µ o I /2a at center of loop. o I /4)(1/a – 1/b), into page CHAPTER 30
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12. From Example 30.2 : B = (µo I /4πd) (sin α 1 + sin α 2 ) Here sin α 1 = sin α 2 = (L/2) [L 2 /4 + d 2 ] 1/2 Thus B = (µ o I /2πd)L/(L 2 + 4d 2 ) 1/2 13. From Example 30.2 : B = (µ o I /4πR)(sin α 1 + sin α 2 ). R = l /2, and α 1 = α 2 = 45 ° , so sin α 1 = sin α 2 = 1/(2) 1/2 . For four wires, find B = 2(2) 1/2 µ o I . 14. d l = 10 –3 k m. Note that the unit vector is r/r (a) r = L(i + k ); dB = 8.84 × 10 –7 j   T (b) r = L(i + j   + k ); dB = 4.81(–i + j   )   × 10 –7 T (c) r = L( j   + k ); dB = –8.84 × 10 –7 i T (d) r = L (i + j   ); dB = 8.84 (–i + j   ) × 10 –7 T (e) r = L j   ; dB = –25 × 10 –7 i T 15. (a) Plot B = (µ o I a 2 )/2(a 2 + z 2 ) 3/2 .
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