A solenoidal coil with 25 turns of wire is wound tightly around another coil with 320 turns. The inner solenoid is 23.0cm long and has a diameter of 2.00cm . At a certain time, the current in the inner solenoid is 0.150A and is increasing at a rate of 1600A/sPart AFor this time, calculate the average magnetic flux through each turn of the inner solenoid. Part BFor this time, calculate the mutual inductance of the two solenoids;Part CFor this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

see the answers below please

Explanation:

We have that the current in the inner solenoid is:

A) the magnetic flux is given by

B is obtained by computing for a solenoid:

hence, we have

B)

the mutual inductance is obtained by using:

C)

hope this helps!!

Part A, magnetic flux = 9.02 × 10^-8 Wb

Part B, mutual inductance, M = 1.50 × 10^-5H

Part C, emf induced in the outer solenoid = -0.024V

Explanation: Please see the attachments below

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i believe the answer is a hope it and is correct