The $SI$ unit of inductance,the henry,can be written as

If in a coil the rate of change of area is $5 \, m^2/ms$ and the current changes from $2 \, A$ to $1 \, A$ in $2 \times 10^{-3} \, s$. If the magnetic field is $1 \, T$,then the self-inductance of the coil is......$H$

The self-inductance of a solenoid of length $31.4 \ cm$,area of cross-section $10^{-3} \ m^2$ having a total number of turns $500$ will be nearly $\left[\mu_0 = 4 \pi \times 10^{-7} \ SI \ unit\right]$.

The inductance of a coil is $60\,\mu H$. $A$ current in this coil increases from $1.0\,A$ to $1.5\,A$ in $0.1\,s$. The magnitude of the induced $e.m.f.$ is

$A$ long solenoid has $500$ turns. When a current of $2 \,A$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \,Wb$. The self-inductance of the solenoid is: (in $\,H$)

$A$ long solenoid has $1500$ turns. When a current of $3.5 \, A$ flows through it, the magnetic flux linked with each turn of the solenoid is $2.8 \times 10^{-3} \, Wb$. The self-inductance of the solenoid is (in $ \, H$)