Question

A current of 2.5 A flowing in a solenoid having
self-inductance 3.8 H is reduced steadily to
zero in 1.0 ms. Calculate the magnitude of the
back emf of the solenoid while the current is
being switched off.

Answer 1

Solution. The magnitude of the induced emf is equal to

$$\Epsilon=\frac {\Delta \Phi} {\Delta t}$$

where

$$\Delta \Phi$$

 is change in flux; t is time.

On the other hand the magnitude of the change in flux can be represented as

$$\Delta \Phi=L \Delta I$$

where L is inductance; I is current. Combining the two formulas we get

$$\Epsilon=\frac {L \Delta I} {\Delta t}.$$

According to the data in the condition of the problem we get

$$\Epsilon=\frac {3.8*2.5} {10^{-3}}=9500V$$

Answer. 9500V