In a coil of self inductance 10 H, the rate of change of current is $4\, A s^{-1}$. The electromotive force (emf) induced in the coil will be

-40 V

The correct answer is Option (4) → -40 V

Given: Self-inductance of the coil $L = 10 \ \text{H}$, rate of change of current $\frac{dI}{dt} = 4 \ \text{A/s}$.

Electromotive force (emf) induced in a coil is given by Faraday’s law for self-inductance:

$\mathcal{E} = - L \frac{dI}{dt}$

Substitute the given values:

$\mathcal{E} = - 10 \times 4 = -40 \ \text{V}$

The negative sign indicates that the induced emf opposes the increase in current (Lenz's law).

Induced emf = -40 V