A $12 \mathrm{V}$ battery connected to a coil of resistance $6 \Omega$ through a switch, drives a constant current in the circuit. The switch is opened in $1 \mathrm{ms}$. The emf induced across the coil is $20 \mathrm{V}$. The inductance of the coil is :

The current $\left(i\right)$ through the circuit is given by

$\begin{array}{rcl}i& =& \frac{V}{R}\\ & =& \frac{12 \mathrm{V}}{6 \mathrm{\Omega }}\\ & =& 2 \mathrm{A}\end{array}$

The formula to calculate the induced emf in the coil can be written as

$\epsilon =L\frac{di}{dt} ...\left(1\right)$

Substitute the values of the known parameters into equation (1) and solve to calculate the inductance of the coil.

$\begin{array}{rcl}20 \mathrm{V}& =& L\times \frac{2 \mathrm{A}}{1\times {10}^{-3} \mathrm{s}}\\ & \Rightarrow & L= \frac{20 \mathrm{V}\times 1\times {10}^{-3} \mathrm{s}}{2 \mathrm{A}}\\ & =& 10 \mathrm{mH}\end{array}$