Question

A solenoid inductor (self-inductance of 96 mH) is 0.80 m in length and 0.10 m in diameter. A coil is tightly wound around the solenoid at its center (approximately same diameter). The coil's resistance is 6.0 ohms. The mutual inductance between the coil and solenoid is 31 μH. At a given instant, the current in the solenoid is 540 mA and is decreasing at the rate of 2.5 A/s. At this given instant, what is the magnitude of the induced current in the coil?

Answer 1

Answer: 12.92 μA

Step-by-step explanation:

Given

Coil resistance, R = 6 ohms

The induced emf of the coil is given as

emf = (M) di/dt, where

M = 31*10^-6 H

The rate of change of current in the solenoid is

di/dt = 2.5 A/s

Induced emf of the coil is then

(M) di/dt =

emf = 31*10^-6 * 2.5

emf = 77.5*10^-6 A/s

The induced current in the coil is given as

i = emf / R

i = 77.5*10^-6 / 6

i = 12.92*10^-6 A

Therefore, the magnitude of the induced current in the coil is 12.92*10^-6 A