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 current in a coil of inductance $0.2 \,H$ changes from $5 \,A$ to $2 \,A$ in $0.5 \,s$. The magnitude of the average induced emf in the coil is (in $\,V$)

In the part of an a.c. circuit as shown,the resistance $R=0.2 \ \Omega$. At a certain instant,$(V_{A}-V_{B})=0.5 \ V$,$I=0.5 \ A$ and $\frac{dI}{dt}=8 \ A/s$. The inductance of the coil is (in $H$)

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

$A$ coil of $Cu$ wire (radius $r$,self-inductance $L$) is bent into two concentric turns,each having a radius of $r/2$. What is the new self-inductance?

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: