Q.When the current changes from +2A to −2A in 0.05 s, an emf of 8 V is induced in a coil. The co-efficient of self-induction of the coil is

Definition. An electric current i flowing through a coil sets up its own magnetic field, whose flux is therefore linked with that same coil. If this self-linked flux changes (because the current itself changes), an emf is induced right there in the very coil carrying the changing current -- this is self-induction, and the induced emf is the self-induced emf. Because the flux linkage $N\Phi_B\propto i$,

where L, the constant of proportionality, is the coil's self-inductance (coefficient of self-induction), $L=N\Phi_B/i$ -- with $i=1$ A, self-inductance equals the flux linkage produced by unit current.

Self-induced emf. A changing current gives, via Faraday's (Lenz-corrected) law, $\varepsilon=-d(N\Phi_B)/dt=-d(Li)/dt=-L\,di/dt$ (for constant L). The minus sign shows the self-induced emf always opposes the current's own change; setting $di/dt=-1$ A/s gives $L=-\varepsilon$, so L equals the opposing emf induced for a current changing at 1 A/s. …