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

$A$ long solenoid has $1000$ turns. When a current of $4\, A$ flows through it,the magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3}\, Wb$. The self-inductance of the solenoid is......$H$

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:

In a coil,the current changes from $-2 \,A$ to $+2 \,A$ in $0.2 \,s$ and induces an emf of $0.1 \,V$. The self-inductance of the coil is: (in $\,mH$)

An air-cored solenoid with $30 \ cm$ length,area of cross-section $25 \ cm^2$,and $500$ turns carries a current of $2.5 \ A$. The current is suddenly switched off in a brief time of $10^{-3} \ s$. How much is the average back emf induced across the ends of the open switch in the circuit (in $V$)? (Ignore the variation in magnetic field near the ends of the solenoid.)

Physically,the self-inductance plays the role of