Sol-Ch30 - Exercises 1. I 1 I 2 c/2[1/a1/(a + b)]to the...

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Unformatted text preview: Exercises 1. I 1 I 2 c/2[1/a1/(a + b)]to the right 2. (a) B 1 = (4A)/2(0.08) = 10 T; B 2 = 40 T B T x = B 1 cos53 + B 2 cos37 = 26 T B T y = B 1 sin53 + B 2 sin37 = 32 T (b) 4/d = 12/(0.1 + d), thus d = 5 cm, to the left of I 1 (c) F/L = o I 1 I 2 /2r = 96 N/m, repulsion. 3. (a) B x = (B 1 + B 2 )cos ; B y = (B 1 B 2 )sin , where = 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 = o I /2B = 8 cm, west of the wire. 5. B = o I /2r = 5 10 4 T 6. F = o I 1 I 2 /2d: F 1 = 80 N/m, repulsion; F 2 = 60 N/m, attraction. F x = (F 1 + F 2 )cos60 = 70 N/m; F y = (F 1 F 2 )sin60 = 17.3 N/m 7. B e = 0.5 G due N, B w = o I /2d = 0.06G due W. tan = B w /B e leads to = 6.8 W of N 8. B = o I /2r = 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 = o I /2a at center of loop. B = o I /4a + 2(o I /4a) = ( o I /2a)(1/2 + 1/) = 5.14 10 7 I /a, out 10. From Eq. 30.8, with z = 0, B = o I /2a at center of loop. B = o I /2a + o I /2a = (o I /2a)(1/ + 1) = 8.28 10 7 I /a, out 11. From Eq. 30.8, with z =0, B = o I /2a at center of loop. ( o I /4)(1/a 1/b), into page CHAPTER 30 12. From Example 30.2 : B = (o I /4d) (sin 1 + sin 2 ) Here sin 1 = sin 2 = (L/2) [L 2 /4 + d 2 ] 1/2 Thus B = ( o I /2d)L/(L 2 + 4d 2 ) 1/2 13. From Example 30.2 : B = ( o I /4R)(sin 1 + sin 2 ). R = l /2, and 1 = 2 = 45 , so sin 1 = sin 2 = 1/(2) 1/2 . For four wires, find B = 2(2) 1/2 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...
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Sol-Ch30 - Exercises 1. I 1 I 2 c/2[1/a1/(a + b)]to the...

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