$A$ coil has $N$ turns and the current passing through it is $I$ ampere,resulting in a self-inductance of $L$ Henry. If the current changes to $5I$ $A$,the new self-inductance will be . . . . . . $H$.

Write down the formula for the self-inductance of a very long solenoid.

When a current of $1 \,A$ is passed through a coil of $100$ turns, the flux associated with it is $2.5 \times 10^{-5} \,Wb/\text{turn}$. The self-inductance of the coil in millihenry is:

The length of a wire required to make a solenoid of length $l$ and self-induction $L$ is

If in a coil,the rate of change of area is $5 \, m^2/ms$ and the current changes from $2 \, A$ to $1 \, A$ in $2 \times 10^{-3} \, s$. If the magnitude of the magnetic field is $1 \, T$,then the self-inductance of the coil is ... $H$.

Current in a circuit falls from $5.0 \ A$ to $0.0 \ A$ in $0.1 \ s$. If an average emf of $100 \ V$ is induced,then the value of self-inductance of the circuit is . . . . . . . (in $H$)