An air-cored coil has a self-inductance of $0.1 \ H$. $A$ soft iron core of relative permeability $1000$ is introduced,and the number of turns is reduced to $\frac{1}{10}$ of the original. The new value of self-inductance is:

When the current changes from $+2 \, A$ to $-2 \, A$ in $0.05 \, s$,an $e.m.f.$ of $8 \, V$ is induced in a coil. The coefficient of self-induction of the coil is ... $H$.

The inductance of a solenoid $0.5 \, m$ long of cross-sectional area $20 \, cm^2$ and with $500$ turns is......$mH$

$A$ magnetic flux of $5 \mu Wb$ is linked with a coil when a current of $1 \ mA$ flows through it. Calculate the self inductance of the coil.

In the given circuit, if $dI/dt = -1 \text{ A/s}$, then the value of $V_{AB}$ at this instant will be: (in $\text{ V}$)

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