A 12 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 ms. The emf induced across the coil is 20 V. The inductance of the coil is :

When the switch is closed, the steady current through the coil:

$$I = \frac{V}{R} = \frac{12}{6} = 2 \text{ A}$$

When the switch is opened, the current drops from 2 A to 0 in $$\Delta t = 1$$ ms.

The induced emf:

$$\varepsilon = L\frac{\Delta I}{\Delta t} = L \times \frac{2}{1 \times 10^{-3}}$$

Given $$\varepsilon = 20$$ V:

$$20 = L \times 2000$$

$$L = \frac{20}{2000} = 0.01 \text{ H} = 10 \text{ mH}$$

The inductance is $$\mathbf{10}$$ mH.