In a coil,current falls from $5\, A$ to $0\, A$ in $0.2\, s$. If an average $emf$ of $150\, V$ is induced,then the self-inductance of the coil is.....$H$

Why does the current not rise immediately in a circuit containing inductance?

$A$ coil has $N$ turns and the current passing through it is $I$ ampere, resulting in a self-inductance of $L$ Henry. If the current is doubled, the new self-inductance will be . . . . . . $H$.

If $N$ is the number of turns in a coil,the value of self-inductance varies as

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

$A$ circular coil consists of $70$ closely wound turns and has a radius of $10 \,cm$. An externally produced magnetic field of magnitude $2 \times 10^{-3} \,T$ is applied perpendicular to the coil. The net flux through the coil is found to vanish when the current in the coil is $2.2 \,A$. The inductance of the coil is: (in $\,mH$)