The current in an inductor of self-inductance $L=40 \text{ mH}$ is to be increased uniformly from $2 \text{ A}$ to $12 \text{ A}$ in $8 \text{ ms}$. The emf induced in the inductor during this process is (in $\text{ V}$)

An air-cored solenoid with length $30 \; cm$,area of cross-section $25 \; cm^{2}$ and number of turns $500$,carries a current of $2.5 \; A$. The current is suddenly switched off in a brief time of $10^{-3} \; s$. How much is the average back $emf$ induced (in $V$) across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

$A$ circular coil of radius $5\, cm$ has $500$ turns of a wire. The approximate value of the coefficient of self-induction of the coil will be:

The self-inductance of a coil is $5 \, H$. If a current of $1 \, A$ changes to $2 \, A$ within $5 \, s$ through the coil,the magnitude of the induced $e.m.f.$ will be $... \, V$.

$A$ circular coil consists of $70$ closely wound turns and has a radius of $10 \,cm$. An externally produced magnetic field of magnitude $2 \times 10^{-3} \,T$ is applied perpendicular to the coil. The net flux through the coil is found to vanish when the current in the coil is $2.2 \,A$. The inductance of the coil is: (in $\,mH$)

The dimensional formula of self-inductance is . . . . . . .