Question

when current in a coil changes from 6A to 2A in 0.1s ,and value of induced emf is 40V .the self inductance of the coil is​

Answer 1

Given:

Current in a coil changes from 6A to 2A in 0.1s ,and value of induced emf is 40V.

To find:

Self inductance of coil.

Calculation:

Let self inductance be denoted as L;

General expression for EMF induced when change in current takes place in a solenoid:

$\boxed{ \sf{E = - L \: \dfrac{di}{dt} }}$

Now , putting available values in SI unit:

$\sf{ = > 40 = - L \: \bigg \{ \dfrac{(2 - 6)}{0.1} \bigg \} }$

$\sf{ = > 40 = - L \: \bigg \{ \dfrac{( - 4)}{0.1} \bigg \} }$

$\sf{ = > 40 = - L \: \bigg \{ - 40 \bigg \} }$

$\sf{ = > 40 = 40 L }$

$\sf{ = > L = 1 \: henry}$

So, final answer is:

$\boxed{ \red{\sf{L = 1 \: henry}}}$