The self-inductance of a coil is $L$. Keeping the length and area the same,the number of turns in the coil is increased to four times. The self-inductance of the coil will now be

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:

$A$ long solenoid has $1000$ turns. When a current of $4\, A$ flows through it,the magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3}\, Wb$. The self-inductance of the solenoid is......$H$

The self-inductance $L$ of a solenoid of length $\lambda$ and area of cross-section $A$ with a fixed number of turns $N$ increases as

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: