$12 \mathrm{~V}$ 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 :

When the switch is closed, the circuit is a simple DC circuit and the current in the circuit is given by Ohm's law:<br/><br/> $I = \frac{V}{R} = \frac{12\text{V}}{6\Omega} = 2\text{A}$. <br/><br/> When the switch is opened, the current in the circuit drops to zero instantaneously. <br/><br/>However, the magnetic field generated by the current in the coil does not disappear immediately, and it continues to produce a back EMF that opposes the change in current. <br/><br/>This back EMF induces a voltage across the coil that can be calculated using Faraday's law of induction: $\mathcal{E} = -L\frac{\Delta I}{\Delta t}$, where $\mathcal{E}$ is the induced voltage, $L$ is the inductance of the coil, and $\Delta I/\Delta t$ is the rate of change of current in the coil. <br/><br/> In this case, we know that the induced voltage is $20\text{V}$ and the rate of change of current is<br/><br/> $\Delta I/\Delta t = -2\text{A}/(1\text{ms}) = -2\times 10^3\text{A/s}$.<br/><br/> Substituting these values into the equation above, we get: $20\text{V} = -L\times(-2\times 10^3\text{A/s})$. <br/><br/> Solving for $L$, we get: $L = \frac{20\text{V}}{2\times 10^3\text{A/s}} = 0.01\text{H}$. <br/><br/> Therefore, the inductance of the coil is $0.01\text{H}$, or 10 mH.