$A$ $10\, V$ battery connected to $5\,\Omega$ resistance coil having inductance $10\, H$ through a switch drives a constant current in the circuit. The switch is suddenly opened and the time taken to open it is $2\, ms$. The average $emf$ induced across the coil is

The self-inductance of a solenoid is:

The self-inductance of a coil is $50 mH$. When a current of $1 A$ passing through the coil reduces to zero at a steady rate in $0.1 s$,find the self-induced emf. (in $V$)

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 $\,V$)

To manufacture a solenoid of length $1 \,m$ and inductance $1 \,mH$, the length of thin wire required is (cross-sectional diameter of a solenoid is considerably less than the length).

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: