The current in a coil decreases from $1 \, A$ to $0.2 \, A$ in $10 \, s$. Calculate the coefficient of self-inductance if the induced $e.m.f.$ is $0.4 \, V$. (in $, H$)

The current passing through a coil of $120$ turns and inductance $40 \text{ mH}$ is $30 \text{ mA}$. The magnetic flux linked with the coil is:

$A$ coil has $N$ turns and the current passing through it is $I$ ampere, resulting in a self-inductance of $L$ Henry. If the current is doubled, the new self-inductance will be . . . . . . $H$.

$A$ 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 total number of turns and cross-section area in a solenoid are fixed. However,its length $L$ is varied by adjusting the separation between windings. The inductance of the solenoid will be proportional to

$A$ coil of wire of a certain radius has $600$ turns and a self-inductance of $108 \, mH$. The self-inductance of a $2^{nd}$ similar coil of $500$ turns will be.........$mH$.