A solenoidal coil with 23 turns of wire is wound tightly around another coil wit

Question

A solenoidal coil with 23 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 21.0 cm long and has a diameter of 2.50 cm . At a certain time, the current in the inner solenoid is 0.150 A and is increasing at a rate of 1500 A/s .

A.)For this time, calculate the average magnetic flux through each turn of the inner solenoid.

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

1.

Average flux through each turn in the inner coil:

phi = BA

magnetic field due to a solenoid is

B = u0*N*i

N = n/l

B = 4*3.14*10^-7*300*0.15/0.21

B = 2.69*10^-4

A = pi*r^2 = 3.14*0.0125^2

phi = B*A = 2.69*10^-4*3.14*0.0125^2

phi = 1.32*10^-7 Wb

2.

The flux is same through each of the solenoid

so mutual inductance will be

M = N2*phi/i1

M = 1.32*10^-7*23/0.150

M = 2.024*10^-5 Amp.

3.

EMF = -M*di/dt

EMF = -2.024*10^-5*1500

EMF = -3.036*10^-2 V

Let me know if you have any doubt.

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