When current in a coil changes from $5\,A$ to $2\,A$ in $0.1\,s,$ an average voltage of $50\,V$ is produced. The self-inductance of the coil is.......$H$.

$A$ $30\text{ cm}$ 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 . . . . . . .

The self-inductance $L$ of a solenoid of length $l$ and area of cross-section $A$,with a fixed number of turns $N$,increases as:

When a current of $2 \ A$ is passed through a coil of $100$ turns,the magnetic flux associated with it is $5 \times 10^{-5} \ Wb$. Find the self-inductance of the coil.

$A$ graph of magnetic flux ( $\phi$ ) versus current $(I)$ is shown for four inductors $A, B, C, D$. The smaller value of self-inductance is for inductor:

When a current $i$ through a solenoid is increasing at a constant rate, then the induced current is