The current in self-inductance $L=40\; mH$ is to be increased uniformly from $1\; A$ to $11\; A$ in $4\; ms$. The $e.m.f.$ induced in the inductor during this process is ..... $V$.

Calculate the self-inductance of a very long solenoid.

The length of a wire required to make a solenoid of length $l$ and self-induction $L$ is

The momentum in mechanics is expressed as $m \times v$. The analogous expression in electricity is

$A$ solenoid is tightly wound with wire of diameter $0.10\,cm$,has a cross-sectional area $\frac{1}{\pi}\,cm^2$,and is $40\,cm$ long. If the current through the solenoid decreases uniformly from $10\,A$ to $0\,A$ in $0.10\,s$,then the $emf$ induced inside the solenoid is.....$mV$.

Magnetic flux of $10 \mu Wb$ is linked with a coil,when a current of $2 \ mA$ flows through it. What is the self-inductance of the coil (in $mH$)?