A solenoidal coil with 20 turns of wire is wound tightly around another coil wit

Question

A solenoidal coil with 20 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.50 cm. At a certain time, the current in the inner solenoid is 0.100 A and is increasing at a rate of 1800 A/s For this time, calculate the average magnetic flux through each turn of the inner solenoid For this time, calculate the mutual inductance of the two solenoids For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

Explanation / Answer

N1 = 20 , N2 = 350

l =25 cm , d = 2.5 cm , I = 0.1 A

dI/dt = 1800 A/s

(a) phi2 = BA = (uoI*N2/l) *(pi*d^2/4)

phi2 = (4*3.14*10^-7*0.1*350/0.25) (3.14*0.025^2/4)

phi2 = 8.63*10^-8 Wb

(b) M = uo*N1N2pi*d^2/4l

M = (4*3.14*10^-7*20*350*3.14*0.025^2/(4*0.25))

M = 1.7*10^-5 H

(c) V = - Mdi/dt

V = -(1.7*10^-5)(1800)

V = -0.031 V

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