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Homework+13 (1) - 30.2 If we have a current I2 going...

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30.2 If we have a current I 2 going through the inner solenoid, it produces a field: B = μ 0 n 2 I 2 The flux through the outer solenoid is φ 12 = BA = ( μ 0 n 2 I 2 )( πr 2 2 ) Therefore the mutual inductance is M = N 1 φ 12 I 2 = N 1 μ 0 n 2 I 2 πr 2 2 I 2 = N 1 μ 0 n 2 πr 2 2 Finally, we want the mutual inductance per unit length so we divide by the length. We notice that N 1 /l = n 1 (total number of turns divided by the length is number of turns per unit length). So our final answer is: m = N 1 μ 0 n 2 πr 2 2 /l = μ 0 n 1 n 2 πr 2 2 30.4 The magnetic field due to a long wire is B = μ 0 I 2 πr So the flux through the rectangular loop is: φ = Z BdA = w Z l 2 l 1 μ 0 I 2 πr dr = w μ 0 I 2 π ln l 2 l 1 The mutual inductance is M = φ I = μ 0 w 2 π ln l 2 l 1 30.9 E = - L dI dt = - (440 mH )(3 . 60 A/s ) = - 1 . 584 V 1
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30.10 The inductance per unit length for a coaxial cable is (see example 30-5 for the derivation of the formula): L l = μ 0 2 π ln r 2 r 1 Plug in r 2 = 3 . 0mm and L/l = 55nH, we will be able to find r 1 .
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