The inductance of a close-packed coil of $400$ turns is $8 \, mH$. $A$ current of $5 \, mA$ is passed through it. The magnetic flux through each turn of the coil is

An air-cored solenoid with $30 \ cm$ length,area of cross-section $25 \ cm^2$,and $500$ turns carries a current of $2.5 \ A$. The current is suddenly switched off in a brief time of $10^{-3} \ s$. How much is the average back emf induced across the ends of the open switch in the circuit (in $V$)? (Ignore the variation in magnetic field near the ends of the solenoid.)

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:

An average induced emf of $0.5 \,V$ appears in a coil when the current in it is changed from $10 \,A$ in one direction to $10 \,A$ in the opposite direction in $0.8 \,s$. The self-inductance of the coil is: (in $\,mH$)

The current in a coil of self-inductance $2.0 \, H$ is increasing according to $I = 2 \sin(t^2) \, A$. The amount of energy spent during the period when current changes from $0$ to $2 \, A$ is ...... $J$.

$A$ varying current in a coil changes from $10 \,A$ to zero in $1.5 \,s$. If the average emf induced in the coil is $200 \,V$, the self-inductance of the coil is (in $\,H$)