P30_057 - = 2 B 1 sin 90 = N 2 i 2 A 2 B 1 = N 2 i 2 r 2 2...

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57. (a) We denote the large loop and small coil with subscripts 1 and 2, respectively. B 1 = µ 0 i 1 2 R 1 = (4 π × 10 7 T · m / A)(15 A) 2(0 . 12 m) =7 . 9 × 10 5 T . (b) The torque has magnitude equal to τ = ¯ ¯ ¯ 2 × ~ B 1 ¯ ¯ ¯
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Unformatted text preview: = 2 B 1 sin 90 = N 2 i 2 A 2 B 1 = N 2 i 2 r 2 2 B 1 = (50)(1 . 3 A)(0 . 82 10 2 m) 2 (7 . 9 10 5 T) = 1 . 1 10 6 N m ....
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