Last Updated on June 16, 2022 by Electricalvolt

**Self-induced EMF **is the electromagnetic force induced in the coil when the current in a coil or conductor changes. The changing flux when linked to a coil turn, it produces EMF in the coil. Thus, the induced emf in the coil depends upon the rate of change of linked flux.

We can understand the induced EMF phenomenon using the below diagram.

Let the coil has N number of turns. The current I flows through the coil on closing the switch S. The current I produces flux in a coil and further this flux links to the coil’s own turns. The linking flux produces the EMF in a coil. This EMF is called Self-Induced EMF. The EMF in a coil changes with a change in current flowing through a coil. We can change the magnitude of the current through a coil by changing the resistance of variable resistance(R).

The direction of induced EMF always opposes the very own cause which produces it. The change in the current produce EMF, therefore the induced EMF always opposes the current. This happens because of Lenz’s Law.

**Formula of Self-Induced EMF**

The induced EMF in a coil is proportional to the rate of change of current in a coil.

Here negative sign shows that the induced EMF opposes the rate of change in the current. L is the constant of proportionality. It is the Self Inductance of a coil.

**Solved Problem **

### The inductance of a coil is L =8 H and resistance R = 2 Ω. If the applied voltage of the battery is 8 V and it switches off in 1 ms, find induced emf.

Given Data

L = 8 H

V= 8 Volts

R = 2 Ω

t = 1 ms = 1 X 10^{-3} s

e = ?

The current through the coil

I = V/R = 8/2 = 4 A

The negative sign shows the direction of emf that opposes the change in flux or the flow of current.

**Therefore, e = 32 kV**