Mutual inductance

Mutual inductance describes how a varying flow of electric current through one coil induces a voltage in an adjacent coil nearby. This phenomenon arises from the magnetic field produced by the first coil's current interacting with the second coil's windings. This principle is crucial in transformers and inductive coupling, playing a vital role in applications such as power transmission and signal transfer in electronics. The coefficient of mutual inductance quantitatively defines this relationship, influenced by the physical configuration and orientation of the coils relative to each other.

1.0Definition of Mutual Induction

Mutual inductance is an electromagnetic phenomenon where a fluctuating electric current in one coil induces a voltage in a nearby coil. This induction arises from the magnetic field created by the first coil's changing current interacting with the second coil's winding turns. 

Mutual inductance is pivotal for efficient power distribution, reliable signal transmission, and optimizing impedance in electronic circuits.

2.0Coefficient of Mutual Induction

Total flux $\left({\varphi }_2\right)$ linked with the secondary coil due to current in the primary coil (I₁)

$\Rightarrow {\varphi }_2\propto I_1$

$\Rightarrow {\varphi }_2={\mathrm{Ml}}_1$

(∴ M =Coefficient of Mutual Induction)

According to Lenz’s law

$\Rightarrow ϵ=-N\frac{d\varphi }{dt}$ 

(∴ N= Number of turns in the coil)

$\Rightarrow {ϵ}_2=-\frac{d{\varphi }_2}{dt}$

$\Rightarrow {ϵ}_2=-\frac{dM{I}_1}{dt}$

$\Rightarrow M=-\frac{{ϵ}_2}{\frac{d{l}_1}{dt}}$

The S.I. unit of Mutual Induction is Henry.

3.0Factors on which Mutual inductance depends

• Number of turns of both the coils primary(N₁) and secondary ( N₂)
• Area of a cross-section of the coils
• Magnetic Permeability of medium between the coils( r) 
• Nature of material on which two coils are wound
• Distance between two coils
• Orientation between the primary and secondary coil
• Coupling factor (K) between two coils(Primary and Secondary)

4.0Mutual Induction Formula

$\Rightarrow {ϵ}_2=-M\frac{d{I}_1}{dt}$

$\Rightarrow M=-\frac{{ϵ}_2}{\frac{d{I}_1}{dt}}$

5.0Definition of Self- Induction and Coefficient of Self -Induction

Self Induction

When an electric current passes through a coil, the magnetic flux changes, adhering to Faraday's Law of Electromagnetic Induction. This change induces an electromotive force (emf) in the coil or circuit, which opposes the initial change, known as self-induction. The emf induced is referred to as back emf, and the resulting current in the coil is termed induced current.

6.0Coefficient of Self-Induction

The magnetic flux enclosed by the coil changes is directly proportional to the current

$\Rightarrow \varphi \propto I$

$\Rightarrow \varphi =LI$

(∴ L= Coefficient of Self Induction)

According to Lenz's Law

$ϵ=-\frac{d\varphi }{dt}$

$ϵ=-\frac{d(LI)}{dt}$

$ϵ=-L\frac{d(I)}{dt}$

$L=-\frac{\epsilon }{\frac{dl}{dt}}$

The SI unit of Self Induction Is Henry.

7.0Mutual Inductance between two long coaxial coils(coaxial solenoid)

Let S₁ and S₂ be two coaxial coils having length ‘l’, the area of cross-section is A, N_1,N₂ is the number of turns in coils S₁ and S₂ .

The magnetic flux associated with the coil fluctuates directly with the electric current flowing through the primary coil.

${\varphi }_2\propto I_1$

${\varphi }_2=M{I}_1$. …………..(1)

(The magnetic flux linked with the coil is determined by $\varphi$ = NBA)

Magnetic flux linked with the coil S₂ is given by

${\varphi }_2=N_2{B}_1\mathrm{A}$ ……………(2)

The Magnetic field at the center of the solenoid is given by 

$B={\mu }_0\mathrm{n}I$

The Magnetic field in coil S₁ is given by

$B_1={\mu }_0{n}_1{I}_1$ (∴ $\mathrm{n}=\frac{N}{l}$ , number of turns per unit length)

$B_1={\mu }_0\frac{N_1}{l}{I}_1$ …………….(3)

Putting the value of B1 in magnetic flux ${\varphi }_2$

${\varphi }_2={\mu }_0\frac{N_1}{l}{I}_1{\mathrm{N}}_2{\mathrm{I}}_1\mathrm{A}$ ……………(4)

From equation (1) and (4) we get

$\mathrm{M}{I}_1={\mu }_0\frac{N_1}{l}{I}_1{\mathrm{N}}_2{\mathrm{I}}_1\mathrm{A}$

$\mathrm{M}=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$, the mutual inductance of two long coaxial coils.

8.0Solved questions

Q-1. The mutual inductance between two flat, concentric rings of radius r₁ and r₂ with r₁ > r₂ placed in air is.

Sol. The magnetic field, due to the larger coil at its centre is

$\mathrm{B}=\frac{{\mu }_{0}I}{2r_1}$

The flux through the inner coil is

$\varphi =B\times \pi \times r_2^2=\frac{{\mu }_{0}I}{2r_1}\times \pi r_2^2$

$\varphi =MI$

$\mathrm{M}=\frac{{\mu }_0\pi r_2^2}{2r_1}$

Q-2. The mutual inductance coefficient between the primary and secondary windings of a transformer 5H. Calculate the induced emf in the secondary when 3 A current in the primary is cutoff in 2.5 x 10${}^{-4}$ s.

Sol.

${ϵ}_2=-M\frac{d{I}_1}{dt}=-5\times \frac{3}{2.5\times 10^{-4}}=-5\times 3\times 4000=-6\times 10^4\mathrm{V}$

Q-3. List the factors that affect Mutual inductance.

Sol.

1. Number of turns of both the coils primary(N1) and secondary( N2).
2. Area of cross - section of the coils.
3. Magnetic Permeability of medium between the coils $\left({\mu }_r\right)$
4. Nature of material on which two coils are wound.
5. Distance between two coils.
6. Orientation between primary and secondary coil.
7. Coupling factor (K) between two coils(Primary and Secondary).

Q-4.Write the dimensional formula for mutual induction?

Sol.

M = $-\frac{{ϵ}_2}{\frac{d{I}_1}{dt}}$ , the dimensional formula is [M¹L²T^-2A^-2]

Q-5. A conducting wire with 100 turns is wound around the central region of a solenoid. The solenoid itself is 100 cm in length and has a radius of 2 cm, with 600 turns of wire. Calculate the mutual inductance between these two coils.

Sol.

$\mathrm{M}=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$

N₁ = 600, N₂ = 100, L = 100 cm = 1m, r = 2cm = 2 x 10${}^{-2}$m     

$\mathrm{A}=\pi r^2=3.14\times 4\times 10^{-4}{\mathrm{m}}^2$

$M=4\times 3.14\times 10^{-7}\times 600\times 100\times 3.14\times 4\times 10^{-4}$

$M=\frac{4\times 3.14\times 10^{-7}\times 600\times 100\times 3.14\times 4\times 10^{-4}}{1}=9.45\times 10^{-5}\mathrm{H}$