Q.When a current changes from +2A to –2A in 0.05 second, an emf of 8 volt is induced in a coil. The coefficient of self-induction (L) of the coil is

Self-inductance measures how strongly a coil's own changing current induces an opposing emf in itself.

When the current $I$ through a coil changes, the magnetic flux $\Phi$ linked with the coil itself changes, inducing an emf in the coil (by Faraday's law) that opposes the change (Lenz's law). This property is called self-inductance. Since flux linkage $N\Phi \propto I$,

$N\Phi = LI \;\Rightarrow\; L = \dfrac{N\Phi}{I}$

…

Using |emf| = L|dI/dt| with ΔI = −4 A over 0.05 s gives L = 0.1 H.

The emf induced by self-induction is

$|\varepsilon| = L\left|\dfrac{dI}{dt}\right|$

Here the current changes from $+2\,A$ to $-2\,A$, so $\Delta I = -2-(+2) = -4\,A$, over $\Delta t = 0.05\,s$: …

Self-induced emf is $-L\,dI/dt$.

Magnitude of self-induced emf: …

Switching off breaks the current very suddenly; by self-induction this abrupt $dI/dt$ generates a large back-emf which ionises the air gap in the switch, producing a spark.

Any electric circuit possesses some self-inductance $L$ (from its wiring, coils, or the filament/choke of the light itself). While current $I$ is steadily flowing, the circuit stores magnetic energy $(1/2)LI^2$.

When the switch is opened, the current is forced to fall to zero almost instantaneously, i.e. $dI/dt$ becomes very large (in magnitude) for a brief instant. By Faraday's law of self-induction,

$\varepsilon = -L\dfrac{dI}{dt}$ …

For inductors in series (no mutual coupling), inductances simply add.

For two inductors in series (ignoring mutual inductance): …

$L = \varepsilon / (dI/dt)$; here $dI/dt = 250$ A/s, giving $L=0.2$ H.

Change in current: $\Delta I = 10-(-10) = 20$ A, in time $\Delta t = 0.08$ s.

$\frac{dI}{dt} = \frac{20}{0.08} = 250\text{ A/s}$ …