$A$ coil of wire of radius $r$ has $600$ turns and self-inductance of $108 \ mH$. The self-inductance of a coil with the same radius and $500$ turns is (in $mH$)

In the circuit shown below,a long time after the key $K$ is closed,the reading in the ammeter is $20 \,mA$. What was the reading (in $mA$) immediately after it was closed?

An inductor coil wound uniformly has self-inductance $L$ and resistance $R$. The coil is broken into two identical parts. The two parts are then connected in parallel across a battery of $E$ volt of negligible internal resistance. The current through the battery at steady state is

$A$ magnetic flux of $5 \mu Wb$ is linked with a coil when a current of $1 \ mA$ flows through it. Calculate the self inductance of the coil.

The self-induced $e.m.f.$ in a $0.1 \, H$ coil when the current in it is changing at the rate of $200 \, A/s$ is......$V$

The self-inductance of a solenoid of length $31.4 \ cm$,area of cross-section $10^{-3} \ m^2$ having a total number of turns $500$ will be nearly $\left[\mu_0 = 4 \pi \times 10^{-7} \ SI \ unit\right]$.