Current in a circuit falls from $5.0 \ A$ to $0.00 \ A$ in $0.1 \ s$. If an average emf of $200 \ V$ is induced,then the self-inductance of the circuit is . . . . . . $H$.

In the given circuit,the bulb will become suddenly bright if:

$A$ current $I$ produces a magnetic flux $\phi$ per turn in a coil of $n$ turns. The self-inductance of the coil is $L$. The relation between them is:

An emf of $2.8 \ mV$ is induced in a rectangular loop of area $150 \ cm^2$ when the current in the loop changes from $3 \ A$ to $8 \ A$ in a time of $0.2 \ s$. Then the self-inductance of the loop is (in $\mu H$)

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

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?