A solenoidal coil with 25 turns of wire is wound tightly around another coil wit

Question

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 22.0 cm long and has a diameter of 2.40 cm . At a certain time, the current in the inner solenoid is 0.150 A and is increasing at a rate of 1800 A/s .

A) For this time, calculate the average magnetic flux through each turn of the inner solenoid. ( I got 1.06*10^-7 and its incorrect)

B) For this time, calculate the mutual inductance of the two solenoids;

C) For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

Explanation / Answer

a)  the average magnetic flux,phiavg/turn = mu0 I * [pi d22 / 4] * N2 / l2

= (4 pi * 10-7 * 0.150 * pi * 0.0242 * 330) / (4 * 0.22)

= 1.28 * 10-7 Wb

b) M = N1 phi / i1 = (25 * 1.28 * 10-7) / (0.150)

= 2.13 * 10-5 H

c) E = -M * di2/dt = -2.13 * 10-5 * 1800

=-0.0384 V

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