When a current in the conducting coil is changed from 5 A in one direction to 5 A in opposite direction in 0.5 second, an average induced e.m.f in the coil is 2 V . The self-inductance of the coil is

The induced eMF in a coil due to self-inductance follows $E = -L \frac{dI}{dt}$, where $L$ represents the self-inductance and $\frac{dI}{dt}$ the rate of current change. Given the initial current $I_1 = 5$ A and final current $I_2 = -5$ A, the current change is $\Delta I = -10$ A over $\Delta t = 0.5$ s, with an average induced eMF magnitude of $2$ V.

The magnitude equation simplifies to $E_{\text{avg}} = L \left| \frac{\Delta I}{\Delta t} \right|$. Substituting values gives $2 = L \left| \frac{-10}{0.5} \right| = L \times 20$, leading to $L = \frac{2}{20} = 0.1$ H.

Converting to millihenrys yields $L = 0.1 \times 1000 = 100$ mH.

The self-inductance is 100 mH, corresponding to option D.