When current $I$ passes through an inductor having self-inductance of $4 \ H$. If the current is made double,what will be the new self-inductance of the inductor?

$A$ coil of $Cu$ wire (radius $r$,self-inductance $L$) is bent into two concentric turns,each having a radius of $r/2$. What is the new self-inductance?

$A$ 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:

$A$ circular coil of radius $R$ has $N$ turns of a wire. The coefficient of self-induction of the coil will be $(\mu_0 = \text{permeability of free space})$

$A$ choke coil works on the principle of

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