Question

A solenoidal coil with 27 turns of wire is wound tightly around another coil with 320 turns. The inner solenoid is 21.0 cm long and has a diameter of 2.40 cm. At a certain time, the current in the inner solenoid is 0.130 A and is increasing at a rate of 1800 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

Part A: Average magnetic flux = 1.13 × 10^-7 Wb

Part B: The mutual inductance of the two solenoids = 2.35× 10^-5H

Part C: The emf induced in the outer solenoid = -0.0423V

Step-by-step explanation: