In an inductor of self-inductance $L = 2 \, mH$,current changes with time according to the relation $i = t^2 e^{-t}$. At what time is the $emf$ zero? (in seconds)

$A$ toroid is a long coil of wire ($N$ turns) wound over a circular core. The coefficient of self-induction of the toroid is [The magnetic field in it is uniform and $R >> r$,where $r=$ radius of wire,$R=$ radius of coil] ($\mu_0=$ permeability of free space).

Two coils have self-inductance $L_1 = 4 \, mH$ and $L_2 = 1 \, mH$ respectively. The currents in the two coils are increased at the same rate. At a certain instant of time,both coils are given the same power. If $I_1$ and $I_2$ are the currents in the two coils at that instant of time respectively,then the value of $\frac{I_1}{I_2}$ is:

The back $e.m.f.$ induced in a coil,when current changes from $1 \ A$ to $0 \ A$ in $1 \ ms$,is $4 \ V$. The self-inductance of the coil is:

$A$ closely wound coil of $100$ turns and of cross-section $1 \,cm^2$ has a coefficient of self-inductance $1 \,mH$. The magnetic induction at the centre of the core of the coil when a current of $2 \,A$ flows in it will be (in $Wb/m^2$):

In a circuit,the current falls from $14 \ A$ to $4 \ A$ in a time $0.2 \ ms$. If the induced emf is $150 \ V$,then the self-inductance of the circuit is: