Question

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.

Answer 1

Answer: 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

Answer 2

see the answers below please

Step-by-step explanation:

We have that the current in the inner solenoid is:

`i(t)=0.150+1600t`

A) the magnetic flux is given by

`\Phi_B=BS=\pi r^2 B`

B is obtained by computing for a solenoid:

`B=(\mu_0n_1i(t))/(L)`

hence, we have

`\Phi_B=(\pi r^2 \mu_0 n_1)/(L)(0.150+1600t)`

B)

the mutual inductance is obtained by using:

`M_(12)=M_(21)=M=\mu_0n_1n_2\pi r_(1)^2i(t)=\pi \mu_0 r^2 (25)(320)(0.150A+1600t)`

C)

`emf=-(d\Phi_B)/(dt)=-(\pi r_2^2\mu_0n_2)/(L)(1600)`

hope this helps!!