What will be the self-inductance of a coil of $100$ turns if a current of $5 \ A$ produces a magnetic flux of $5 \times 10^{-5} \ Wb$?

When the current in a coil changes from $8 \ A$ to $2 \ A$ in $3 \times 10^{-3} \ s$,the induced $e.m.f.$ in the coil is $2 \ V$. The self-inductance of the coil in millihenry $(mH)$ is:

The self-inductance of an air-core inductor (solenoid) is $0.03 \ mH$. By introducing an iron core into the inductor, the self-inductance increases to $30 \ mH$. The relative permeability of the core used is:

Calculate the self-inductance of a very long solenoid.

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 magnetic field is $1 \, T$,then the self-inductance of the coil is......$H$

When the number of turns in a coil is made $3$ times without any change in the length of the coil,its self-inductance becomes