The current in a coil changes from $2 \,A$ to $5 \,A$ in $0.3 \,s$. The magnitude of emf induced in the coil is $1.0 \,V$. The value of self-inductance of the coil is (in $\,mH$)

$A$ 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 $mH$)

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 current through a coil varies according to the following graph. Plot the graph for induced $emf$ versus time for the coil.

$A$ graph of magnetic flux ( $\phi$ ) versus current $(I)$ is shown for four inductors $A, B, C, D$. The smaller value of self-inductance is for inductor:

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?