Question

Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of self inductance L is given by `1/2 LI^2`

Answer 1

Self inductance is the inherent inductance of a circuit, given by the ratio of the electromotive force produced in the circuit by self-induction to the rate of change of current producing it. It is also called coefficient of self-induction.

Self-inductance is also called the inertia of electricity.

Suppose I = Current flowing in the coil at any time

Φ = Amount of magnetic flux linked

It is found that Φ ∝ I

`phi = LI`

Where,

L is the constant of proportionality and is called coefficient of self induction

SI unit of self-inductance is Henry.

Let at t = 0 the current in the inductor is zero. So at any instant t later, the current in the inductor is I and rate of growth of I is dI/dt.

Then, the induced emf is e = L×dI / dt

If the source is sending a constant current I through the inductor for a small time dt, then small amount of work done by the source is given by

dW = eI dt =(LdI/dt)I dt = LIdI

The total amount of work done by the source of e.m.f., till the current increases from its initial value I = 0 to its final value I is given by

`W =int_0^1LIdI = Lint_0^1IdI =L|I^2/2|_0^1 `

This work done by the source of emf is used in building up current from zero to I₀ is stored in the inductor in energy form. Therefore, energy stored in the inductor is

`U = 1/2 LI^2`