Induced Electric Fields and Energy Stored in an Inductor

Induced Electric Fields

When a conductor moves in a magnetic field, emf is induced. Let us consider a long, thin solenoid of cross-sectional area (A), n number of turns per unit length encircled at its centre by a circular conducting loop. The galvanometer G is used to measure current in the loop. When current I passes through the solenoid, magnetic field is produced that is \(B = \mu_0 nI\), when \(\mu_0\) is the permeability of free space. If we neglect the small field outside the solenoid and take area vector \(\vec A\) to point in the same direction as be \(\vec B\)m then magnetic flux through the loop is given by,

$$\phi = BA = \mu_0 nIA$$

From Faraday's law, the induced emf is given by

$$\begin{align*} \epsilon &= -\frac {d\phi }{dt} = -\frac {d}{dt}(\mu_0\: nIA)\dots (i) \\ \text {or,} \: \epsilon &= -\mu _0nA\frac {dI}{dt} \dots (ii) \\ \end{align*}$$

If \(\epsilon \) be the induced emf, R be the total resistance of the loop, the induced current is

$$\text {i.e.} \: \:\: \text {Induced current} = \frac {\epsilon }{R} $$

The change in magnetic flux is a source of electric field. When a charge goes once around the loop, the total work done on it by the electric field must be equal to q times the emf \(\epsilon \). We can conclude that the electric field in the loop is not conservative because line integral of \(\vec E\) around a closed path is not zero. This line integral representing the work done by the induced \(\vec E\) field per unit charge is equal to the induced emf \(\epsilon \).

$$\begin{align*} \text {i.e.} \:\oint \vec E.\vec {dl} &= \epsilon \dots (iii) \\ \text {From equations} \:(i) \: \text {and} \: (iii),\: \text {we get} \\ \oint \vec E.\vec {dl} &= -\frac {d\phi }{dt} \dots (iv) \\ \end{align*}$$

Equation (iv) is true only when the path around which we integrate is stationary.

Let us consider the stationary circular loop of radius r. The electric field \(\vec E\) has the same magnitude at every point on the circle and is tangent to it at each point due to cylindrical symmetry. The line integral of equation (iv) becomes simply the magnitude E times the circumference \(2\pi r\) of the loop.

$$\begin{align*} \text {i.e.}\: \oint \vec E.\vec {dl} &= 2\pi rE \\ \text {or,} \: -\frac {d\phi }{dt} &= -2\pi r\:E \\ \therefore E &= \frac {1}{2\pi r} \frac {d\phi }{dt} \\ \end{align*}$$

Here E is the magnitude of induced electric field.

Energy Stored in an Inductor

Consider an inductor of inductance L having initially zero current. It is assumed that an inductor has zero resistance so that there is no dissipation of energy with in inductor. Let I be the current at any instant of time so that di/dt is the rate of change of current. Here current is increasing. The voltage between the terminals a and b of the inductor at this instant is \(V_{ab} = L\frac {di}{dt} \) and the rate P at which energy is being delivered to the inductor is given by

$$ P = v_{ab}I = Li\frac {di}{dt} $$

The energy supplied to the inductor in small amount of time is small which is written as

$$\begin{align*} dU &= P\:dt \\ &= Li \frac {di}{dt}dt \\ dU &= Lidt \dot (i) \end{align*}$$

To obtain total amount of energy stored in the inductor, we have to integrate it from 0 to 1.

$$\begin{align*} \text {i.e}\: U &= \int_0^I L.i.di \\ \text {or,}\: U &= L\left [\frac {i^2}{2}\right ]_0^I = \frac 12 LI^2 \\\therefore \text {Total energy stored in an inductor,} \:U = \frac 12 LI^2 \end{align*}$$

When current has reached final steady value I, \(\frac {di}{dt} =0 \) so no more energy is input to the inductor. Where there is no charge then energy is not stored, the energy is \(\frac LI^2\).

Energy stored in Toroid

The self inductance of the toroidal solenoid with vacuum within its coils is

$$\begin{align*} L &= \frac {\mu_0 N^2A}{2\pi r} \\ \text {Thus energy stored in the toroidal solenoid} \\ U &= \frac 12LI^2 = \frac 12\frac {\mu_0N^2A}{2\pi r}I^2 \\ \therefore U &= \frac 12 \frac {\mu_0N^2A}{2\pi r}I^2 \\\end{align*}$$