Question

(a) Define self inductance. Write its SI units.

(b) Derive an expression for self inductance of a long solenoid of length l, cross-sectional area A having N number of turns.

Answer 1

(a) Self induction: The phenomenon according to which an opposing emf is induced in a coil as a result of varying current in the coil itself is called self induction. The emf produced in coil is also called back emf.

The effect of self-induction in an electric circuit is similar to the inertia of motion, therefore, it is called electric inertia.

(b) Coefficient of self induction: The magnitude of the magnetic flux through a coil due to its own current depends on the strength of the current and the shape and size of the circuit.

If    I = strength of the current

and Φ = magnetic flux linked with the circuit,

we have

Φ ∝ I

or Φ = L I

This constant, L, is called the coefficient of self induction or simply self inductance of the circuit.

If I = 1, unit of current Φ = L

From this, the coefficient of self-inductance may be defined as follows:

The self inductance or coefficient of self induction of a circuit is equal to the magnetic flux linked with circuit when unit current is flowing in it.

Another definition of L

e = \(-\frac{d\phi}{dt}\) 

or e = \(-\frac{d(LI)}{dt} = -L \frac{dI}{dt}\)

The negative sign indicates that the direction of the induced emf opposes the change in magnetic flux.

If \(-\frac{dI}{dt}\) = 1

Then e = L (numerically) 

The self inductance or the coefficient of self-induction of a

circuit is equal to the emf induced in the circuit when the

current flowing through it is unity.

Unit of Self-inductance. As stated above

e = -L \(\frac{dI}{dt}\)

In SI unit, if e = 1 volt

and \(\frac{dI}{dt}\)  = 1 A s-1, then L = 1 henry (H).

Hence, the self inductance of a coil is (1) henry if a change of current equal to 1 ampere per second causes an induced emf equal to one volt in the circuit.

(c) Expression for Coefficient of self-induction for a Solenoid

Consider a solenoid of length l

Let r = radius of solenoid such that r < l

n = no. of turns per unit length

I = strength of current through the solenoid

The magnetic field produced by the current inside the solenoid is practically constant and B = µ₀ nI

∴ Magnetic flux linked with the loop,

dΦ = µ₀ n IA

Magnetic flux linked with entire solenoid

Φ = nl (µ₀ n IA) = µ₀ n²lAI

Since Φ = LI

∴ LI = µ₀n² l I A

L = µ₀ n² l A ................(i)

If the coil is wound on a material having relative permittivity µr

then L = µ₀ µ_r n² l A

(d) Examples of self induction

(1) Non-inductive coils: In electrical measurements, we have to make use of resistance boxes, post office boxes and other bridges. In these instruments, inductance is a nuisance. To minimize the inductive effect, this is done by double backing the resistance wire on itself by winding it on the same bobbin. The free ends of the wire are connected to binding screws of the box. Fig. shows a non-inductive coil. Note that every part of the coil is traversed by two opposite currents. The magnetic field due to the current in one part is opposed by that in the other. The total field is negligible. Hence, the coil becomes non-inductive.

(2) Electromagnetic damping: When some current is passed through a galvanometer coil, it suffers a few to and fro oscillations about its final equilibrium position before settling down. As the coil gets deflected, it moves in the magnetic field and. therefore, an induced emf is produced in the coil which opposes its motion. It is thus the electromagnetic induction which is mainly responsible for the damping of the galvanometer.

If    N = total number of turns, then

N = nl

or n = \(\frac{N}{l}\) 

So the equations L = M₀n²lA becomes

L = \(\frac{\mu_0N^2 A}{l}\)