Question

Two coils of wire are placed close together. Initially, a current of 3.04 A exists in one of the coils, but there is no current in the other. The current is then switched off in a time of 1.75 x 10-2 s. During this time, the average emf induced in the other coil is 4.29 V. What is the mutual inductance of the two-coil system

Answer 1

Answer:

$M=0.0247H$

Explanation:

Given data

$V_{voltage}=4.29V\\I_{current}=3.04A\\t_{time}=1.75*10^{-2}s$

To find

Mutual inductance of the two-coil system

Solution

The mutual inductance given as:

M= (-VΔt)/ΔI

Substitute the given values

So

$M=-\frac{4.29V*1.75*10^{-2}s}{(0-3.04A)}\\ M=0.0247H$