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?

The adjoining figure shows two bulbs $B_1$ and $B_2$,a resistor $R$,and an inductor $L$. When the switch $S$ is turned off,what happens to the bulbs?

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

$A$ long solenoid has $500$ turns. When a current of $2 \,A$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \,Wb$. The self-inductance of the solenoid is: (in $\,H$)

Two coils have self-inductance $L_1 = 4 \, mH$ and $L_2 = 1 \, mH$ respectively. The currents in the two coils are increased at the same rate. At a certain instant of time,both coils are given the same power. If $I_1$ and $I_2$ are the currents in the two coils at that instant of time respectively,then the value of $\frac{I_1}{I_2}$ is:

The self-inductance of a long solenoid of cross-sectional area $A$,length $l$ and $n$ turns per unit length is given by