A solenoidal coil with 25 turns of wire is wound tightly aroundanother coil with 300 turns. The inner solenoid is 25.0 cm long andhas a diameter of 2.00 cm. At a certain time, the current in theinner solenoid is 0.120 A and is increasing at a rate of1.75x103A/s. For this time, calculate;

a) the average magnetic flux through each turn of the innersolenoid;
b) the mutual inductance of the two solenoids;
c) the emf induced in the outer solenoid by the changing current inthe inner solenoid
Thanks for any help you can offer! Explanation would be wonderful!

(a). The average magnetic flux through each turn of the inner solenoid is

(b). The mutual inductance of the two solenoids is

(c). The emf induced in the outer solenoid by the changing current in the inner solenoid is -0.0207 V.

Explanation:

Given that,

Number of turns of coil = 25

Number of turns of another coil = 300

Length = 25.0 cm

Diameter = 2.00 cm

Current = 0.120 A

Rate

(a). We need to calculate the magnetic field due to inner solenoid

Using formula of magnetic field

Put the value into the formula

We need to calculate the average magnetic flux through each turn of the inner solenoid

Using formula of magnetic flux

Put the value into the formula

The average magnetic flux through each turn of the inner solenoid is

(b). We need to calculate the mutual inductance of the two solenoids

Using formula of mutual inductance

Put the value into the formula

The mutual inductance of the two solenoids is

(c).  We need to calculate the emf induced in the outer solenoid by the changing current in the inner solenoid

Using formula of emf

Put the value into the formula

The emf induced in the outer solenoid by the changing current in the inner solenoid is -0.0207 V.

Hence, (a). The average magnetic flux through each turn of the inner solenoid is

(b). The mutual inductance of the two solenoids is

(c). The emf induced in the outer solenoid by the changing current in the inner solenoid is -0.0207 V.