Self Inductance

Self Inductance

Self-inductance is the property of a coil by virtue of which it opposes the growth or decay of the current flowing through it.

Consider a coil connected to a battery through a key(K) as shown in the figure. As the key is on, current in the coil starts increasing. Due to it, the magnetic field and hence flux linkage around the coil also increases. The direction of induced e.m.f. is such that it opposes the growth of current in the coil. This delays the current to acquire the maximum value.

When the key (K) is switched off, the current in the coil starts decreasing. So the magnetic flux linked with the coil decreases as a result. Due to change in magnetic flux e.m.f is also produced. According to lenz’s law, the direction of induced e.m.f is such that it opposes the decay of current in the coil. This delays the current to acquire minimum or zero value.

Such property of the coil which opposes the growth or decay of the current in the circuit.

Demonstration of the Self Inductance Effect

1. Connect two lamps in parallel to each other, one through an ohmic resistor (R) and next one through a coil (N). A battery B with one-way key (K) is connected to them as shown in the figure. When the key is closed, it is found that the lamp B₁, glows immediately with brilliance but lamp B₂glows slowly.
   When current flows through the coil, an e.m.f. is induced in it, which opposes the growth of current in the circuit. Hence the glow of B₂ is slow. On the other hand, no induced e.m.f is produced in the resistor. Hence the lamp B₁receives maximum current as so as the key is closed and it glows at once.
   When the key is opened, lamp B₁ stops growing at once but the lamp B₂ takes some time to stop glowing. In this case, current begins to decrease through a coil, so again e.m.f. is induced in it. This induced e.m.f. opposes the decay of current in the circuit of lamp B₂.
2. Spark is produced in the electric switch when the light is switched off:
   When the current is switched off, the current in the circuit starts to decrease rapidly. So, large induced e.m.f is set up across the switch contacts, which tries to maintain the current. This e.m.f. is sufficient to break down the insulation of the air between the switch contacts and hence spark is produced.
3. Non-inductive Coil
   Each coil of resistance box is doubled back on itself while being coiled up. It is done to decrease the self-induction. The current in every part of the coil is equal and opposite. So the magnetic field around one part is canceled by the magnetic field around the opposite part. Thus self-induction becomes minimum.

Coefficient of Self Induction

Let I be the current flowing through a coil then the magnetic flux \(\phi \) linked with the coil is found to be proportional to the strength of current (I).

$$\begin{align*} \: \text {i.e} \: \phi &\propto I \\ \text {or,}\: \phi &= LI \\ \end{align*}$$

where L is coefficient of self-inductance. From Faraday’s law of electromagnetic induction, the induced e.m.f. in the coil is given by

$$\begin{align*} E &= -\frac {d\phi }{dt} \\ &= -\frac {d}{dt}(LI) = - L \frac {dI}{dt} \\ \therefore E &= -L \frac {dI}{dt} \dots (i) \\ \end{align*}$$

If \(\frac {dI}{dt} = I\), i.e. rate of decrease of current is unity then equation (i) becomes,

$$ E = L$$

Thus coefficient of self induction of a coil is the e.m.f. induced in the coil through which the rate of decrease of current is unity.

Dimensional Formula for Self Inductance (L)

$$\begin{align*} E &= L \frac {dI}{dt} \\ \text {or,} \: L &= \frac {E}{dI/dt} = \frac {W/dq}{dI/dt}\:\:\:\:\: [\because \text {e.m.f.} = \frac {\text {work done}}{\text {charge}} ]\\ &= \frac {W}{\frac {dq}{dt} dI} = \frac {W}{IdI} \\ &= \frac {[ML^2T^{-2}]}{[A^2]} \\ &= \therefore [L] = [M^1L^2T{-2}A^{-2}] \end{align*}$$

An ideal inductance has high value of self inductance and zero ohmic resistance.

Self Inductance of a Solenoid

Let a long solenoid of length l, the area of cross-section A, the current flowing through solenoid I and a number of turns per unit length n.

The magnetic field inside the solenoid is uniform and given by

$$B = \mu _0nI$$

Total number of turns in the solenoid, N = nl. Now the magnetic flux linked with each turn of the solenoid\(= BA = \mu_0\: nIA\).

total magnetic flux with the whole solenoid, \(\phi \) = magnetic flux linked with each turn × number of turns in the solenoid.

$$\begin{align*} \therefore \: \phi = \mu_0 n IA\times nl = \mu_o n^2 IAl \\ \text {or,} \: LI &= \mu_0 n^2 IAl \:\:\: [\because \phi = Li ] \\ \therefore L = \mu_0\: n^2 Al\dots (i) \\ \text {Since,}\: n = \frac Nl. \text {So equation} \:(i) \: \text {becomes,} \\ L &= \mu _0 \frac {N^2}{l^2}Al \\ L &= \mu_0\frac {N^2}{l}A \\ \end{align*}$$

Thus, self inductance of an air cored solenoid (L) depends on

1. Total number of turns (N) of the solenoid
2. Length of the solenoid
3. Area f cross-section of the solenoid (A).