The current in a coil falls from 10.0 A to 0.0 A in 0.1 s. If an average emf of 100 V is induced in the coil, then the self- inductance of the coil is.

1 H

The correct answer is Option (2) → 1 H

Induced emf is given by:

$\mathcal{E} = L \, \frac{\Delta I}{\Delta t}$

Given: $\mathcal{E} = 100 \, \text{V}$, $\Delta I = 10 - 0 = 10 \, \text{A}$, $\Delta t = 0.1 \, \text{s}$

$100 = L \cdot \frac{10}{0.1}$

$100 = L \cdot 100$

$L = 1 \, \text{H}$

Self-inductance L = 1 H