When the number of turns in a coil is doubled without any change in the length of the coil,its self-inductance becomes

Two coils have self-inductance $L_1$ and $L_2$. The current through them is increasing at a constant rate. If the power dissipated in both the coils is the same, then the ratio of energy stored in the coil having inductance $L_1$ to that in $L_2$ is

An air-cored coil has a self-inductance of $0.1 \ H$. $A$ soft iron core of relative permeability $1000$ is introduced,and the number of turns is reduced to $\frac{1}{10}$ of the original. The new value of self-inductance is:

$A$ magnetic flux of $5 \mu Wb$ is linked with a coil when a current of $1 \ mA$ flows through it. Calculate the self inductance of the coil.

The average $e.m.f.$ induced in a coil in which a current changes from $0$ to $2 \,A$ in $0.05 \,s$ is $8 \,V.$ The self-inductance of the coil is:

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