Question

A solenoidal coil with 30 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.20 cm . At a certain time, the current in the inner solenoid is 0.150 A and is increasing at a rate of 1700 A/s .

PART A

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

PART B

For this time, calculate the mutual inductance of the two solenoids;

PART C

For this time, calculate the emf induced in the outer solenoid by the changing current in the inner solenoid.

Answer 1

A] Current at the given time = 0.15 A

so, the magnetic field through the inner solenoid will be:

the average magnetic flux will then be:

B] Mutual inductance of the two solenoids = M = magnetic flux x number of loops in bigger solenoid / current

=> L = 2.388 x 10^-7 x 30 / 0.15 = 4.777 x 10^-5 H

C] magnitude of EMF induced = mutual inductance x rate of change of current = L(di/dt)

=> E = 4.777 x 10^-5 x 1700 = 0.0812 Volts.