What is the self-inductance of a coil which produces a self-induced $EMF$ of $5 \ V$ when the current changes from $3 \ A$ to $2 \ A$ in $1 \ ms$?

$A$ 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 $\,H$)

$A$ varying current in a coil changes from $10 \, A$ to $0 \, A$ in $0.5 \, s$. If the average $EMF$ induced in the coil is $220 \, V$,the self-inductance of the coil is ... $H$.

$A$ branch of a circuit is shown in the figure. If the current is decreasing at the rate of $10^3 \ A \ s^{-1}$, then the potential difference between $A$ and $B$ is (in $V$)

$A$ $30\text{ cm}$ long solenoid has $10$ turns per cm and area of $5\text{ cm}^2$. The current through the solenoid coil varies from $2\text{ A}$ to $4\text{ A}$ in $3.14\text{ s}$. The e.m.f. induced in the coil is $\alpha \times 10^{-5}\text{ V}$. The value $\alpha$ is . . . . . . .

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