## What is Induction?

Induction is the magnetic field which is proportional to the rate of change of the magnetic field. This definition of induction holds for a conductor. Induction is also known as inductance. 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.

### Factors Affecting Inductance

Following are the factors that affect the inductance:

- The number of turns of the wire used in the inductor.
- The material used in the core.
- The shape of the core.

Electromagnetic Induction law was given by Faraday which states that by varying the magnetic flux electromotive force is induced in the circuit. From Faraday’s law of electromagnetic induction, the concept of induction is derived. Inductance can be defined as the electromotive force generated to oppose the change in current at a particular time duration.

### According to Faraday’s Law:

Electromotive force = – L \( \frac {\Delta I}{\Delta t}\)

Unit of Inductance = \( \frac {Volt ~Second }{Ampere}\)=Henry

## Types of Inductance

Two types of inductance are there:

- Self Induction
- Mutual Induction

## What is Self Induction?

When there is a change in the current or magnetic flux of the coil, an opposed induced electromotive force is produced. This phenomenon is termed as Self Induction. 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:

\( \phi \)= I

\( \phi \) = L I

Where L is termed as 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 = – \( \frac {d \phi}{dt} \) = – \( \frac {d (LI) }{dt} \)

or e = – L \( \frac {dI}{dt} \)

### Self Inductance Formula

\(L=N\frac{\phi }{I}\) |

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 Induction?

We take two coils, and they are placed close to each other. The two coils are P- coil (Primary coil) and S- coil (Secondary coil). To the P-coil, a battery, and a key is connected wherein the S-coil a galvanometer is connected across it. When there is a change in the current or magnetic flux linked with two coils an opposing electromotive force is produced across each coil, and this phenomenon is termed as Mutual Induction. The relation is given as:

\( \phi \) = I

\( \phi \) = M I

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 = – \( \frac {d \phi}{dt} \) = – \( \frac {d (MI)}{dt}\)

e = – M \( \frac {dI}{dt} \)

### Mutual Inductance Formula

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

Where,

- μ
_{0}is the permeability of free space - μ
_{r}is the relative permeability of the soft iron core - N is the number of turns in coil
- A is the cross-sectional area in m
^{2} - l is the length of the coil in m

### Difference between Self and Mutual Inductance

Self induction |
Mutual induction |

Self inductance is the characteristic of the coil itself. | Mutual inductance is the characteristic of a pair of coils. |

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. |

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. |

### Derivation of Inductance

Consider a DC source with its switch on. 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:

\(\frac{d\varphi }{dt}\)

Apply Faraday’s law of electromagnetic induction,

E = N\(\frac{d\phi }{dt}\)

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\(\frac{d\phi }{dt}\)

The above equation is modified for calculating the value of inductance

E = -N\(\frac{d\phi }{dt}\)

E = -L \(\frac{di }{dt}\)

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

N^{2}BA/Hl = N^{2}μHA/Hl

L = μN^{2}A/l = μN^{2}𝜋r^{2}/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 time 0.4 second.

**Solution:**

Given:

No.of turns, N = 500 turns

Relative permeability, μ_{r} = 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 = μN^{2}Al = μ_{0}μ_{r}N^{2}𝜋r^{2}/l

Substituting the values we get

(4)(3.14)(10^{-7})(800)(500^{2})(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, ε_{2} = 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:

M_{21} = ε_{2}/(di/dt) = 60×10^{-3}×0.4/8 = 3×10^{-3}H

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

ε_{2} = M_{21} di/dt = 3×10^{-3}×12/0.03 = 1.2V

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