A $$10$$ $$\Omega$$, $$20$$ mH coil carrying constant current is connected to a battery of $$20$$ V through a switch. Now after switch is opened current becomes zero in $$100$$ $$\mu$$s. The average e.m.f. induced in the coil is ______ V.

A coil with resistance $$R = 10$$ $$\Omega$$ and inductance $$L = 20$$ mH carries a constant current from a 20 V battery. After the switch is opened, the current drops to zero in $$100$$ $$\mu$$s. We need to find the average induced EMF.

Since the coil carries a steady current before the switch is opened, there is no self-induced EMF. By Ohm’s law, the initial current is

$$I = \frac{V}{R} = \frac{20}{10} = 2\text{ A}$$

When the current changes, the average EMF induced in the coil is given by Faraday’s law:

$$|\varepsilon| = L \left|\frac{\Delta I}{\Delta t}\right|$$

The current decreases from 2 A to 0 A in $$100$$ $$\mu$$s, so

$$\left|\frac{\Delta I}{\Delta t}\right| = \frac{2 - 0}{100 \times 10^{-6}} = \frac{2}{10^{-4}} = 2 \times 10^4\text{ A/s}$$

Substituting into the expression for emf gives

$$|\varepsilon| = L \times \left|\frac{\Delta I}{\Delta t}\right| = 20 \times 10^{-3} \times 2 \times 10^4$$

$$|\varepsilon| = 400\text{ V}$$

Therefore, the average EMF induced in the coil is 400 V.