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 $\,V$)

Calculate the self-inductance of a very long solenoid.

For a toroid,$N = 500$,radius $R = 40 \text{ cm}$,and area of cross-section $A = 10 \text{ cm}^2$. Find the inductance in $\mu\text{H}$.

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 an air-cored solenoid of length $40 \ cm$,diameter $7 \ cm$ having $200$ turns is nearly: (in $\mu H$)

In the given circuit, if $dI/dt = -1 \text{ A/s}$, then the value of $V_{AB}$ at this instant will be: (in $\text{ V}$)