Question

Explain self-induction of a coil. Arrive at an expression for the induced emf in a coil and the rate of change of current in it.

Answer 1

Inductance is the property of a component that opposes the change of current flowing through it and even a straight piece of wire will have some inductance.
Inductors do this by generating a self-induced emf within itself as a result of their changing magnetic field. In an electrical circuit, when the emf is induced in the same circuit in which the current is changing this effect is called Self-induction.
Magnetic field gives rise to magnetic flux $\varphi$ passing through each turn in the coil. Coil having N turns will have flux $N\varphi$ .
Magnetic field is proportional to the current i flowing through the coil and so is the flux.
$N\varphi \propto i$
$N\varphi =Li$
$L$ is the self inductance.
Now lets consider the current to change in the coil with time which changes the flux as well.
So this rate of change of flux induces an emf (e) by Farady's law of electromagnetic induction.
$$e=-\frac{d\left(N\varphi \right)}{dt}=-\frac{d\left(Li\right)}{dt}=-L\frac{di}{dt}$$
If $\frac{di}{dt}=1A/s$ we have, $e=-L$
Thus self inductance is numerically equal to the emf induced in coil when current changes at a rate of 1 Ampere per second.