Mutual Inductance

Mutual inductance is a fundamental concept in electromagnetism, crucial to the operation of transformers, inductors, and many types of electrical circuits. It occurs when the magnetic field created by the current flowing through one coil induces a voltage in a nearby coil. This phenomenon not only forms the basis of many electrical devices but also has practical applications in everyday life. For example, the wireless charging of smartphones and electric toothbrushes relies on mutual inductance to transfer energy from a charging pad to the device without direct electrical contact. In this article, we will understand mutual inductance is essential for both designing efficient electrical systems and appreciating the invisible forces at work in our technologically driven world.

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This Story also Contains

1. What is Mutual Induction?
2. Solved Examples Based on Mutual Inductance
3. Summary

What is Mutual Induction?

Mutual induction is the process by which a change in the electric current in one coil induces an electromotive force (EMF) in a nearby coil through electromagnetic induction. This occurs due to the magnetic field created by the current in the first coil (the primary coil) affecting the second coil (the secondary coil). Whenever the current passing through a coil or circuit changes, the magnetic flux linked with a neighbouring coil or circuit will also change. Hence an emf will be induced in the neighboring coil or circuit. This phenomenon is called ‘mutual induction’. or The phenomenon of producing an induced emf in a coil due to the change in current in the other coil is known as mutual induction.

Coefficient of Mutual Induction

The coefficient of mutual induction (also known as mutual inductance) is a measure of how effectively a change in current in one coil induces a voltage in another coil. It is denoted by the symbol MMM and is defined as the ratio of the induced electromotive force (EMF) in the secondary coil to the rate of change of current in the primary coil. If two coils (P-primary coil or coil 1, S-secondary coil or coil 2) are arranged as shown in the figure below.

If we change the current through coil P then flux passing through Coil S will change.

i.e $N_2{\varphi }_2\alpha i_1\Rightarrow N_2{\varphi }_2=M_{21}{i}_1=M{i}_1$

where

$M_{21}=$ mutual induction of Coil 2 w.r. t Coil 1
$N_1=$ Number of turns in the primary coil
$N_2=$ Number of turns in the secondary coil
$i_1=$ current through the primary coil or coil 1

Similarly, if we exchange the position of Coil 1 and Coil 2

then

If we change the current through coil S then flux passing through Coil P will change.

i.e $N_1{\varphi }_1\alpha i_2\Rightarrow N_1{\varphi }_1=M_{12}{i}_2=M{i}_2$

where

$\begin{array}{rl}& M_{12}=\text{mutual induction of Coil}1\text{w.r. t Coil}2\\ & N_1=\text{Number of turns in the primary coil}\\ & N_2=\text{Number of turns in the secondary coil}\\ & i_2=\text{current through the Coil}2\text{or Coil}\mathrm{S}\end{array}$

As $N_2{\varphi }_2=M{i}_1$

If $i_1=1\mathrm{amp},N_2=1$ then, $M={\varphi }_2$

I.e coefficient of mutual induction of two coils is numerically equal to the magnetic flux linked with one coil when unit current flows through the neighbouring coil.

Using Faraday's Second Law of Induction emf we get

${\epsilon }_2=-N_2\frac{d{\varphi }_2}{dt}=-M\frac{d{i}_1}{dt}$

If $\frac{d{i}_1}{dt}=1\frac{amp}{sec}$ and $N_2=1$ then $\left|{\epsilon }_2\right|=M$

I.e The coefficient of mutual induction of two coils is numerically equal to the emf induced in one coil when the rate of change of current through the other coil is unity.

Units and dimensional formula of ‘M’

S.I. Unit - Henry (H)

And $1H=\frac{1V\cdot sec}{Amp}$

Its dimensional formula is $M{L}^2{T}^{-2}{A}^{-2}$

Dependence of Mutual Inductance

• Number of turns (N₁, N₂) of both coils
• The coefficient of self inductances (L₁, L₂) of both the coilsand the relation between them is given as$M=K\sqrt{L_1{L}_2}$

where K = coefficient of coupling.

If L=0 then M = 0

If K = 0 i.e case of No coupling then M = 0.

• Distance(d) between two coils (i.e As d increases then M decreases)
• The magnetic permeability of the medium between the coils $\left({\mu }_r\right)$

Consider two long co-axial solenoids of the same length $l$. Let A₁ and A₂ be the area of the cross-section of the solenoids with A₁ being greater than A₂ as shown in the figure below.

The turn density of these solenoids are n₁ and n₂ respectively are given as $n_1=\frac{N_1}{l}$ and $n_2=\frac{N_2}{l}$

Let i₁ be the current flowing through solenoid 1, then the magnetic field produced inside it is given as

$B_1={\mu }_o{n}_1{i}_1$

As the field lines of $\overrightarrow{B_1}$ are passing through the area A₂

So the magnetic flux linked with each turn of solenoid 2 due to solenoid 1 and is given by

${\mathrm{\Phi }}_{21}={\int }_{A_2}{\overline{B}}_1\cdot d\overrightarrow{A}=B_1{A}_2=\left({\mu }_0{n}_1{i}_1\right)A_2$

The total flux linkage of solenoid 2 with total turns N₂ is

$\begin{array}{rl}& {\left({\varphi }_{21}\right)}_{\text{total}}=N_2{\mathrm{\Phi }}_{21}=\left(n_{2}l\right)\left({\mu }_0{n}_1{i}_1\right)A_2\\ & \Rightarrow {\left({\varphi }_{21}\right)}_{\text{total}}=N_2{\mathrm{\Phi }}_{21}=\left({\mu }_0{n}_1{n}_2{A}_{2}l\right)i_1\end{array}$

And Using ${\left({\varphi }_{21}\right)}_{\text{total}}=N_2{\mathrm{\Phi }}_{21}=M_{21}{i}_1$ we get

$M_{21}={\mu }_0{n}_1{n}_2{A}_{2}l$

Where M₂₁ is the mutual inductance of the solenoid 2 with respect to solenoid 1.

Similarly, M₁₂ =mutual inductance of solenoid 1 with respect to solenoid 2 is given as

$M_{12}={\mu }_0{n}_1{n}_2{A}_{2}l$

Hence $M_{21}=M_{12}=M$

So, In general, the mutual inductance between two long co-axial solenoids is given by

$M={\mu }_0{n}_1{n}_2{A}_{2}l$

If a dielectric medium of permeability is present inside the solenoids, then

$\mathrm{M}=\mu {\mathrm{n}}_1{\mathrm{n}}_2{\mathrm{A}}_2\mathrm{l}$ or $\mathrm{M}={\mu }_0{\mu }_{\mathrm{r}}{\mathrm{n}}_1{\mathrm{n}}_2{\mathrm{A}}_2\mathrm{l}$

Consider two circular coils one of radius 'r₁' and the other of radius' r₂'placed coaxially with their centres coinciding as shown in the below figure.

Since $r_1\gg \gg >r_2$ so we can assume coil 2 is at the center of coil 1.

If Suppose a current $i_1$ flows through the outer circular coil. Then Magnetic field at the center of coil 1 is given as

$B_1=\frac{{\mu }_0{N}_1{i}_1}{2r_1}$

So the total flux passing through coil 2 will be given as

${\left({\varphi }_2\right)}_{\text{total}}=N_2{B}_1{A}_2=\frac{{\mu }_0{N}_1{N}_2{i}_1{A}_2}{2r_1}$

And using ${\left({\varphi }_2\right)}_{\text{total}}=M{i}_1$

we get $M=\frac{{\mu }_0{N}_1{N}_2{A}_2}{2r_1}=\frac{{\mu }_0{N}_1{N}_2\left(\pi r_2^2\right)}{2r_1}$

Where M=mutual inductance between two concentric coils.

Recommended Topic Video

Solved Examples Based on Mutual Inductance

Example 1: Two coils 'P' and 'Q' are separated by some distance. When a current of 3 A flows through coil 'P', a magnetic flux of ${10}^{-3}\mathrm{Wb}$ passes through 'Q'. No current is passed through 'Q'.When no current passes through 'P' and a current of 2A passes through 'Q', the flux through 'P' is :

1) $6.67\times {10}^{-4}\mathrm{Wb}$
2) $3.67\times {10}^{-3}\mathrm{Wb}$
3) $6.67\times {10}^{-3}\mathrm{Wb}$
4) $3.67\times {10}^{-4}\mathrm{Wb}$

Solution:

As

$\varphi =Mi$

As given in the question

When a current of 3 A flows through coil 'P', a magnetic flux of ${10}^{-3}$ $Wb$ passes through 'Q'

So

${10}^{-3}=M(3)-----(1)$

Now let the flux through 'P' is $\varphi$ when a current of 2A passes through 'Q

So

$\varphi =M(2)-----(2)$

from equation (1) and (2)

$\begin{array}{rl}& \Rightarrow \frac{{10}^{-3}}{\varphi }=\frac{3}{2}\\ & \Rightarrow \varphi =\frac{2}{3}\times {10}^{-3}=6.67\times {10}^{-4}\mathrm{Wb}\end{array}$

Hence, the answer is the option (1).

Example 2:

A time-varying current $I(t)=2\cos wt$ is flowing in the primary coil of 200 turns with a frequency of 40 Hz. The coefficient of mutual induction is 10 m H. Find the emf induced (max) (in mV) is a secondary coil of 400 turns

1) 50.24

2) 12.56

3) 50.44

4) 12.56

Solution:

$$\begin{array}{rl}& {\epsilon }_2=-N_2\frac{d{\varphi }_2}{dt}=-M\frac{d{I}_1}{dt}\\ & \text{Induced emf.}\\ & \epsilon =\frac{-\mu d{I}_1}{dt}\\ & \epsilon =\frac{-\mu d}{dt}(2\cos wt)\\ & \epsilon =\mu 2w\sin wt\end{array}$$

For Max emf $|\sin wt|=1$
$$\begin{array}{rl}& {\epsilon }_{max}=2Mw\\ & {\epsilon }_{max}=2M\times 2\pi f\\ & {\epsilon }_{max}=2\times 10\times {10}^{-3}\times 2\times 3.14\times 40\\ & {\epsilon }_{max}=50.24\mathrm{mV}\end{array}$$

Hence, the answer is the option (1).

Example 3: Find the mutual inductance in the arrangement, when a small circular loop of wire of radius 'R ′ is placed inside a large square loop of wire of side (L >> R). The loops are coplanar and their centres coincide :

1) $M=\frac{\sqrt{2}{\mu }_o{R}^2}{L}$
2) $M=\frac{2\sqrt{2}{\mu }_{o}R}{L^2}$
3) $M=\frac{\sqrt{2}{\mu }_{o}R}{L^2}$
4) $M=\frac{2\sqrt{2}{\mu }_o{R}^2}{L}$

Solution:

$$\begin{array}{rl}& \varphi =\mathrm{MI}\\ & {\varphi }_2={\mathrm{MI}}_1\\ & {\mathrm{B}}_1{\mathrm{A}}_2={\mathrm{MI}}_1\\ & M=\frac{{\mathrm{B}}_1{\mathrm{A}}_2}{{\mathrm{I}}_1}\end{array}$$
${\mathrm{B}}_1\to$ magnetic field due to square frame ${\mathrm{A}}_2\to$ Area of circle ${\mathrm{I}}_1\to$ current in square frame.
$$\begin{array}{rl}& {\mathrm{B}}_1\to \\ & {\mathrm{B}}_1=4\cdot {\mathrm{B}}_{\mathrm{AB}}\\ & =4\left[\frac{{\mu }_0{\mathrm{I}}_1}{2\pi \frac{1}{2}}[\sin {45}^{\circ }+\sin 45]\right]\\ & {\mathrm{B}}_1=2\frac{{\mu }_0{\mathrm{I}}_1}{\pi \mathrm{L}}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})=2\sqrt{2}\frac{{\mu }_0{I}_1}{\pi \mathrm{L}}\\ & \mathrm{M}=\frac{{\mathrm{B}}_1\cdot {\mathrm{A}}_2}{{\mathrm{I}}_{\mathrm{L}}}\\ & \mathrm{M}=\left(\frac{2\sqrt{2}{\mu }_0{\mathrm{I}}_1}{\pi \mathrm{L}}\right)\times \frac{\pi {\mathrm{R}}^2}{{\mathrm{I}}_1}=\frac{2\sqrt{2}{\mu }_0{\mathrm{R}}^2}{\mathrm{L}}\end{array}$$

Hence, the answer is the option (4).

Example 4: Two concentric circular coils with radii 1 cm and 1000 cm, and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be _______ × 10^-8 H. (Take , = 10)

1) 4

2) 5

3) 6

4) 7

Solution:

Given
$$\begin{array}{rl}& a=1000\mathrm{cm}\\ & \mathrm{b}=1\mathrm{cm}\\ & \text{or}b\ll a\end{array}$$
we will take a larger coil as the primary
$$\begin{array}{rl}& B=\frac{{\mu }_0{i}_{p}N}{2a}\\ & \text{flux}{\varphi }_s=BA=\frac{{\mu }_0{i}_{p}N}{2a}\times \pi b^2\times n\\ & \text{Mutual inductance}\mathrm{M}\\ & =\frac{{\varphi }_{\mathrm{s}}}{i_{\mathrm{p}}}\end{array}$$

Mutual inductance $M$
$$\begin{array}{rl}& \mathrm{M}=\frac{{\mu }_0\mathrm{Nn}\pi {\mathrm{b}}^2}{2\times \mathrm{a}}\\ & \text{or}\mathrm{M}=\frac{4\pi \times {10}^{-7}\times 200\times 10\times \pi \times 1\times {10}^{-4}}{2\times 1000\times {10}^{-2}}\\ & =4{\pi }^2\times {10}^{-9}\\ & \text{or}\mathrm{M}=4\times {10}^{-8}\\ & \text{(usin}{\pi }^2=10\text{)}\end{array}$$

Hence, the answer is the option (1).

Example 5: The mutual inductance of a pair of coils is 2 H. If the current in one of the coils changes from 10 A to zero in 0.1 s, the emf induced in the other coil is

1) 2 V

2) 20 V

3) 0.2 V

4) 200 V

Solution:

The induced emf in the other coil (coil 2 ) is
$$\begin{array}{rl}{e}_2& =-M\frac{d{i}_1}{dt}=-M\frac{\mathrm{\Delta }{i}_1}{\mathrm{\Delta }t}=-M\frac{(i_2-i_1)}{\mathrm{\Delta }t}=-\frac{2(0-10)}{0.1}\\ & =200\mathrm{V}\end{array}$$

Hence, the answer is the option (4).

Summary

Mutual inductance is the process where a changing current in one coil induces an electromotive force (EMF) in a nearby coil, foundational to devices like transformers and wireless chargers. The coefficient of mutual induction, denoted as M, quantifies this effect, influenced by factors such as coil turns, distance, and magnetic permeability. Various practical examples and problems illustrate the principles and calculations of mutual inductance in different configurations.