Question

For this time, calculate the average magnetic flux through each turn of the inner solenoid. A solenoidal coil with 27 turns of wire is wound tightly around another coil with 320 turns. The inner solenold is 22.0 cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is 0.130 A and is increasing at a rate of 1800 A/s Submit RequestAnswer Part B For this time, calculate the mutual inductance of the two solenoids; Submit RequestAnswer Part C For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

Answer 1

Solution,

A)

Flux, phi = BA = (u0 N2 I/L) pi r^2

Phi = (4 pi x 10^-7 x 320 x 0.13/0.22) (pi x 0.01^2)

Phi = 7.46 x 10^-8 Wb

B)

Mutual inductance, M = N1 phi/I = 27 x 7.46 x 10^-8/0.13 = 1.55 x 10^-5 H

C)

Emf, E = - M dI/dt = - 1.55 x 10^-5 x 1800 = - 0.028 V

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