In an induction coil with resistance,the induced emf will be maximum when

The adjoining figure shows two bulbs $B_1$ and $B_2$,a resistor $R$,and an inductor $L$. When the switch $S$ is turned off,what happens to the bulbs?

Two coils $A$ and $B$ have $180$ and $360$ turns respectively. $A$ current of $1 \text{ A}$ flows through both the coils. Due to a current of $1 \text{ A}$ in coil $A$, a flux per turn of $0.8 \times 10^{-3} \text{ Wb}$ is linked with coil $A$. Due to a current of $1 \text{ A}$ in coil $B$, a flux per turn of $1 \times 10^{-3} \text{ Wb}$ is linked with coil $B$. The self-inductance of coil $A$ is $L_A$ and the self-inductance of coil $B$ is $L_B$. The ratio $L_A$ to $L_B$ is:

An air-cored solenoid with length $20 \, cm$,area of cross-section $20 \, cm^2$,and number of turns $400$ carries a current of $2 \, A$. The current is suddenly switched off within $10^{-3} \, s$. The average back $emf$ induced across the ends of the open switch in the circuit is (ignore the variation in magnetic field near the ends of the solenoid) ........ $V$.

$A$ wire in the form of a tightly wound solenoid is connected to a $DC$ source and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil,will the current increase or decrease? Explain.

Define the unit of self-inductance. On which factors does self-inductance depend?