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

Explanation:

Data given

temperature =700c

Density=10.35g/cm^3

but

Nv is the number of vacant site and N is the number of lattice site.

Since the number of lattice site can also b computed as

if we substitute the value of the number of lattice into the first equation, we arrive at