What is Inductance?

Inductance is the tendency of an electrical conductor to oppose a change in the electric current flowing through it. L is used to represent the inductance, and Henry is the SI unit of inductance. 1 Henry is defined as the amount of inductance required to produce an emf of 1 volt in a conductor when the current change in the conductor is at the rate of 1 Ampere per second.

An electric current flowing through a conductor creates a magnetic field around it. The strength of the field depends upon the magnitude of the current. The generated magnetic field follows any changes in the current, and from Faraday’s law of induction, we know that changing the magnetic field induces an electromotive force in the conductor. Considering this principle, inductance is defined as the ratio of the induced voltage to the rate of change of current causing it. The electronic component designed to add inductance to a circuit is an inductor.

Factors Affecting Inductance

Following factors affect the inductance in a circuit:

  • Number of Wire Turns in the Coil
    Inductance is greater when the number of turns of wire in the coil is greater. More coils of wires indicate a greater amount of magnetic field force for a given amount of coil current.
  • Coil Area
    Inductance is proportional to the coil area. Greater the coil area, the greater the inductance. Greater coil area presents less opposition to the formation of magnetic field flux for a given amount of field force
  • Core Material
    The greater the magnetic permeability of the core to which the coil is wrapped around, the greater the inductance.
  • Coil Length
    The longer the coil’s length, the lesser the inductance. The shorter the coil’s length, the greater the inductance.

Types of Inductance

Inductance is classified into two types as:

  • Self Inductance
  • Mutual Inductance

What is Self Inductance?

When there is a change in the current or magnetic flux of the coil, an electromotive force is induced. This phenomenon is termed Self Inductance. When the current starts flowing through the coil at any instant, it is found that, that the magnetic flux becomes directly proportional to the current passing through the circuit. The relation is given as:

\(\begin{array}{l} \phi \end{array} \)
= L I

Where L is termed self-inductance of the coil or the coefficient of self-inductance, the self-inductance depends on the cross-sectional area, the permeability of the material, or the number of turns in the coil.

The rate of change of magnetic flux in the coil is given as,

e = –

\(\begin{array}{l} \frac {d \phi}{dt} \end{array} \)
= –
\(\begin{array}{l} \frac {d (LI) }{dt} \end{array} \)

or e = – L

\(\begin{array}{l} \frac {dI}{dt} \end{array} \)

Self Inductance Formula

\(\begin{array}{l}L=N\frac{\phi }{I}\end{array} \)

Where,

  • L is the self inductance in Henries
  • N is the number of turns
  • Φ is the magnetic flux
  • I is the current in amperes
Related Articles:

What is Mutual Inductance?

Consider two coils: P – coil (Primary coil) and S – coil (Secondary coil). A battery and a key are connected to the P-coil, whereas a galvanometer is connected across the S-coil. When there is a change in the current or magnetic flux linked with the two coils, an opposing electromotive force is produced across each coil, and this phenomenon is termed as Mutual Inductance.

This phenomenon is given by the relation:

\(\begin{array}{l} \phi = M I \end{array} \)

Where M is termed as the mutual inductance of the two coils or the coefficient of the mutual inductance of the two coils.

The rate of change of magnetic flux in the coil is given as,

e = –

\(\begin{array}{l} \frac {d \phi}{dt} \end{array} \)
= –
\(\begin{array}{l} \frac {d (MI)}{dt}\end{array} \)

e = – M

\(\begin{array}{l} \frac {dI}{dt} \end{array} \)

Mutual Inductance Formula

\(\begin{array}{l}M=\frac{\mu _{0}\mu_{r} N_{1}N_{2}A}{l}\end{array} \)

Where,

  • μ0 is the permeability of free space
  • μr is the relative permeability of the soft iron core
  • N is the number of turns in coil
  • A is the cross-sectional area in m2
  • l is the length of the coil in m

Difference between Self and Mutual Inductance

Self inductance Mutual inductance
In self inductance, the change in the strength of current in the coil is opposed by the coil itself by inducing an e.m.f. In mutual inductance out of the two coils one coil opposes change in the strength of the current flowing in the other coil.
The induced current opposes the growth of current in the coil when the main current in the coil increases. The induced current developed in the neighboring coil opposes the growth of current in the coil when the main current in the coil increases.
The induced current opposes the decay of current in the coil when the main current in the coil decreases. The induced current developed in the neighboring coil opposes the decay of the current in the coil when the main current in the coil decreases.

Derivation of Inductance

Consider a DC source. When the switch is turned on, the current flows from zero to a certain value such that there is a change in the rate of current flowing. Let φ be the change in flux due to current flow. The change in flux is with respect to time which is given as:

\(\begin{array}{l}\frac{d\varphi }{dt}\end{array} \)

Apply Faraday’s law of electromagnetic induction,

E = N

\(\begin{array}{l}\frac{d\phi }{dt}\end{array} \)

Where, 

  • N is the number of turns in the coil
  • E is the induced EMF across the coil

From Lenz’s law, we can write the above equation as

E = -N

\(\begin{array}{l}\frac{d\phi }{dt}\end{array} \)

The above equation is modified for calculating the value of inductance

E = -N

\(\begin{array}{l}\frac{d\phi }{dt}\end{array} \)

E = -L

\(\begin{array}{l}\frac{di }{dt}\end{array} \)

N = dΦ = L di

NΦ = Li

Therefore,

Li = NΦ = NBA

Where,

  • B is the flux density
  • A is the area of the coil

Hl = Ni

Where,

  • H is the magnetizing force due to magnetic flux

B = μH

Li = NBA

L = NBA/i = N2BA/Ni

N2BA/Hl = N2μHA/Hl

L = μN2A/l = μN2?r2/l

Where,

  • r is the radius of the coil

Examples of Self Inductance and Mutual Inductance

Example 1. 

Consider a solenoid with 500 turns which are wound on an iron core whose relative permeability is 800. 40 cm is the length of the solenoid, while 3 cm is the radius. The change in current is from 0 to 3 A. Calculate the average emf induced for this change in the current for a time of 0.4 seconds.

Solution:

Given:

No.of turns, N = 500 turns

Relative permeability, μr = 800

Length, l = 40 cm = 0.4 m

Radius, r = 3 cm = 0.03 m

Change in current, di = 3 – 0 = 3 A

Change in time, dt = 0.4 sec

Self-inductance is given as

L = μN2Al = μ0μrN2?r2/l

Substituting the values we get

(4)(3.14)(10-7)(800)(5002)(3.14)(3×10-2)2/0.4

L = 1.77 H

Magnitude of induced emf, ε = L di/dt = 1.77×3/0.4

ε = 13.275 V

Example 2.

There are two coils such that the current flowing through the first coil experiences a change in current flow from 2 A to 10 A in 0.4 sec. Calculate mutual inductance between the two coils when 60 mV emf is induced in the second coil. Determine the induced emf in the second coil if the current changes from 4 A to 16 A in 0.03 sec in the first coil. 

Solution:

Given:

Case 1:

Change in current, di = 10 -2 = 8 A

Change in time, dt = 0.4 sec

Magnitude of induced emf, ε2 = 60×10-3 V

Case 2:

Change in current, di = 16 – 4 = 12 A

Change in time, dt = 0.03 sec

Mutual inductance of the second coil with respect to the first coil is given as:

M21 = ε2/(di/dt) = 60×10-3×0.4/8 = 3×10-3H

Induced emf in the second coil due to change in the rate of current in the first coil is given as:

ε2 = M21 di/dt = 3×10-3×12/0.03 = 1.2V

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  1. Very helping material. Go the way chosen for the success achiever!