Mutual Induction and Mutual Inductance: Definition, Formula, and Explanation

22.5 MUTUAL INDUCTION AND MUTUAL INDUCTANCE

22.5.1 Mutual Induction

Figure 22.64 Magnetic field lines produced by current $I_1$ in coil 1 pass through coil 2, showing magnetic coupling. A change in current in coil 1 causes a change in magnetic flux in coil 2.

Consider two coils which are placed close to each other. A current I₁ flows through coil 1. Hence, a magnetic field is produced, whose magnetic field lines pass through coil 2, as shown in Figure 22.64. Hence, the two coils are magnetically coupled, even though they are not physically linked to each other. If current I₁ changes with time, then the magnetic flux through coil 2 will change with time simultaneously. Due to the change of flux linkage through coil 2, an emf E₂ is induced in coil 2. This is in accordance to Faraday’s law of electromagnetic induction. In other words, a change of current in one coil leads to the production of an induced emf in a second coil which is magnetically linked to the first coil. This process of producing an induced emf in one circuit or circuit component X due the change of current in another circuit or circuit component Y which is magnetically coupled to X is known as mutual induction.

22.5.2 Emf Induced in Each Magnetically Coupled Coil

Refer to coil 1 and coil 2 mentioned above. According to Faraday’s law of electromagnetic induction, the induced emf E₂ produced in coil 2 due to the change of current I₁ in coil 1 is

Emf induced in coil 2 $(E_2) ∝ −$ rate of change of current in coil 1 $(\frac{dI_1}{dt})$

$E_2 ∝ − \frac{dI_1}{dt}$

or,

$E_2 = −M_{21}\frac{dI_1}{dt}$

where $M_{21}$ is a constant known as the mutual inductance of the two coils.

Next, suppose there is no current in coil 1 but there is a current I₂ flowing in coil 2. If the current changes with time, mutual induction occurs and an emf E₁ will be induced in coil 1. If the rate of change of current I₂ is dI₂/dt, then

$E_1 ∝ − \frac{dI_2}{dt}$

or,

$E_1 = −M_{12}\frac{dI_2}{dt}$

where $M_{12}$ is the mutual inductance of the two coils. If the geometric properties like sizes, shapes and relative positions of the coils are not changed, then we can show that

$M_{21} = M_{12} = M$

Hence, we have

$E_1 = −M \frac{dI_2}{dt}$    and    $E_2 = −M \frac{dI_1}{dt}$

Note: We need to remember that
(a) $\frac{dI_1}{dt}$ in coil 1 induces $E_2$ in coil 2
(b) $\frac{dI_2}{dt}$ in coil 2 induces $E_1$ in coil 1

Figure 22.65 shows how the emf induced in one coil is related to the rate of change of current in the other coil.

Figure 22.65 The diagram shows that a changing current in one coil produces an induced emf in the other coil. Specifically, a rate of change of current in coil 1 induces an emf in coil 2, and similarly, a changing current in coil 2 induces an emf in coil 1, demonstrating the principle of mutual induction.

22.5.3 Mutual Inductance

1. Definition
The mutual inductance between two magnetically coupled circuit components is defined as the property of the circuit component which can induce an emf in the neighbouring magnetically coupled component.

2. Unit
Mutual inductance is also measured in henry.

3. Mutual inductance between two magnetically coupled coaxial coils

Figure 22.66 Two coils are arranged with their axes coinciding, forming a coaxial system. When the current $I_1$ in coil 1 changes with time, it produces a changing magnetic field that passes through coil 2. This changing magnetic flux induces an emf $I_2$ in coil 2 according to Faraday’s law of electromagnetic induction.

Figure 22.66 shows two coils whose axes coincide with each other. A current I₁ which flows through coil 1 changes with time. The emf E₂ induced in coil 2 due to the current changing in coil 1 is given by

$E_2 = −M (\frac{dI_1}{dt})$

According to Faraday’s law of electromagnetic induction, we have

$E_2 = −N_2 (\frac{dΦ_2}{dt})$

where N₂, Φ₂ are the number of turns of coil 2 and the magnetic flux through coil 2 respectively. Hence, we have

$M (\frac{dI_1}{dt}) = N_2 (\frac{dΦ_2}{dt})$

Since M and N₂ are constant, we have

$M I_1 = N_2 Φ_2$

or,

$M = \frac{N_2 Φ_2}{I_1}$

Similarly, we can show that

$M I_2 = N_1 Φ_1$

or,

$M = \frac{N_1 Φ_1}{I_2}$

Notice that the magnetic flux Φ₂ through coil 2 is produced by the current I₁ in coil 1. If the geometrical properties of the coils are not changed, then we have

$Φ_2 ∝ I_1$

$Φ_2 = kI_1$

where k is a constant which is determined by the geometrical properties of the coils. We have

$M = \frac{N_2Φ_2}{I_1}$

$M= \frac{N_2kI_1}{I_1}=N_2k$

This expression indicates the mutual inductance of the magnetically coupled coaxial coils is determined by the geometrical properties of the coils.

EXAMPLE 22.35

Two coils, X and Y, are magnetically coupled. The emf induced in coil Y is 2.5 V when the current flowing through coil X changes at the rate of 5.0 A s⁻¹. Determine

1. the mutual inductance of the coils.
2. the emf induced in coil X if there is a current flowing through coil Y which changes at the rate of 1.5 A s⁻¹.

Answer

(a)

$|E_y| = M (\frac{dI_x}{dt})$

$2.5 = M(5.0)$

$M = 0.50 \ H$

(b)

$|E_x| = M (\frac{dI_y}{dt})$

$= (0.50)(1.5) = 0.75 \ V$

EXAMPLE 22.36

Two coils, X and Y, are magnetically coupled. Coil Y has 50 turns. A current of 2.5 A flowing through coil X produces a magnetic field so that magnetic flux of 35 mWb passes through coil Y.

1. Determine the mutual inductance of the coils.
2. The current of 2.5 A which flows through coil X changes to 4.5 A in 2.0 ms. Estimate the average emf induced in coil Y in the time interval of 2.0 ms.

Answer

(a)

$MI_1 = NΦ_y$

$M(2.5) = (50)(35 \times 10^{-3})$

$M = 0.70 \ H$

(b)

|$E_y| = M (\frac{dI_x}{dt})$

$= M (\frac{ΔI_x}{Δt})$

$= \frac{(0.70) (4.5 − 2.5)}{2.0 \times 10^{-3}} = 700 \ V$

EXAMPLE 22.37

Two coils, X and Y, have mutual inductance of 550 mH. Determine the rate of change of magnetic flux linkage through coil Y at the instant when the current flowing through coil X changes at the rate of 5.5 A s⁻¹.

Answer

At the instant when the current flowing through coil X changes at the rate of 5.5 A s⁻¹, an emf is induced in coil Y, which is given by

$|E_y| = M \frac{dI_x}{dt}$

$= (0.550)(5.5) = 3.03 \ V$

According to Faraday’s law of electromagnetic induction, we have

$|E_y| = N_y \frac{dΦ_y}{dt}$

= rate of change of flux linkage through coil Y

Hence, the rate of change of flux linkage through coil Y is 3.0 Wb s⁻¹.

EXAMPLE 22.38

Figure 22.67 A long ideal air-core solenoid is connected in series with a 12 V battery and a 20 Ω resistor. A secondary coil is tightly wound around the central portion of the solenoid, allowing magnetic coupling so that changes in current in the solenoid can induce an emf in the coil.

Explain the terms mutual induction and mutual inductance of a combination of two magnetically linked coils.

A long ideal air-core solenoid is connected in series to a battery of emf 12 V and a resistor of 20 ohm, as shown in Figure 22.67. A coil is tightly wound over the central portion of the solenoid.

(a) Derive an expression for the mutual inductance of the coil-solenoid combination.

(b) The solenoid has a diameter of 5.0 cm and 60 turns per cm of length. The coil has 100 turns. Determine

(i) the mutual inductance of the combination
(ii) the magnetic flux linkage through the coil
(iii) the average emf induced in the coil when the circuit is switched off and the current drops to zero in 20 ms.

Answer

(a) Suppose current I flows through the solenoid. The current produces a magnetic field. Hence there is magnetic flux Φ_coil passing through the coil of N_coil turns. The mutual inductance M of the coil-solenoid combination is given by

$M = \frac{N_{coil}Φ_{coil}}{I}$

The magnetic flux through the coil Φ_coil is equal to the magnetic flux through the central portion of the solenoid and is given by

$Φ_{coil} = BA \ cos \ 0^0$

where B is the magnetic flux density in the central portion of the solenoid. We have

$B = μ_0 n I$

Hence,

$M = \frac{N_{coil}Φ_{coil}}{I}$

$M = \frac{N_{coil}BA}{I}$

$= \frac{N_{coil}(μ_0 nI)BA}{I}$

$= μ_0 nN_{coil}A$

$M = μ_0 nN_{coil}A$

(b) (i)

where

$n = \frac{60}{1.0 \times 10^{-2}}= 6 \times 10^3 \ m^{-1}$

and

$A = \frac{1}{4} πd^2$

$= \frac{1}{4} π \times (5.0 \times 10^{-2})^2 = 1.96 \times 10^{-3} \ m^2$

Hence,

$M = (4π \times 10^{-7})(6 \times 10^3)(100)(1.96 \times 10^{-3})$

$= 1.48 \times 10^{-3} \ H$

(ii) magnetic flux linkage through coil $= N_{coil}BA$

where

$B = μ_0 nI$

and

$I = \frac{E}{(r + R)}$

$= \frac{12}{(0 + 20)} = 0.60 \ A$

Hence,

$B = (4π \times 10^{-7})(6 \times 10^3)(0.6)$

$= 4.52 \times 10^{-3} \ T$

flux linkage $= N_{coil}BA$

$= (100)(4.52 \times 10^{-3})(1.96 \times 10^{-3})$

$= 8.86 \times 10^{-4} \ Wb$

(iii) Average induced emf

$E_{coil} = −$ (change of flux linkage / time taken)

$ = − \frac{(0 − 8.86 \times 10^{-4})}{(20 \times 10^{-3})} = 0.044 \ V$

EXAMPLE 22.39

A smaller circular coil with $N_S$​ turns and radius $R_S$​ is placed inside a larger circular coil with $N_L$​ turns and radius $R_L$​. The centres of both coils coincide and their planes are parallel, allowing magnetic interaction between the coils for mutual induction.

A circular coil with N_S turns and radius R_S is placed inside a larger circular coil with N_L turns and radius R_L. The centres of the coils coincide and their planes are parallel to each other, as shown in Figure 22.68.

(a) Derive an expression for the mutual inductance of the pair of coils. Explain why the value of the mutual inductance calculated using this expression is only an approximation.

(b) Given the following data:

$R_S = 0.50 \ cm$; $R_L = 10 \ cm$

$N_S = 10$; $N_L = 30$

A current of 3.5 A flows through the large coil. Estimate
(i) the mutual inductance of the coils
(ii) the average emf induced in the small coil if the current in the large coil drops to zero in 10 ms.

Answer

(a) The mutual inductance M of the coils is given by

 $M = \frac{N_SΦ_S}{I}$

where Φ_S is the magnetic flux through the small coil and I is the current in the large coil.

The magnetic flux density B at the centre of the coils produced by the current I is given by

$B = \frac{μ_0N_LI}{2R_L}$

Assume that the magnetic field passing normally through the whole cross section of the small coil is uniform. Then the magnetic flux Φ_S is

$Φ_S = BA_S \ cos 0^0$

$M = \frac{N_SBA_S}{I}$

$= \frac{\left(\frac{N_S}{I}\right) (μ_0N_LI)A_S}{2RL}$

$M= \frac{μ_0N_SN_LA_S}{2R_L}$

In the derivation of the expression above, we make one assumption: The magnetic field is uniform throughout the whole plane of the small coil, and its strength at any point on this plane is equal to $\frac{μ_0N_LI}{2R_L}$. If the radius R_S of the small coil is large so that the area of the cross section is not small, then this assumption is not true. The magnetic field will not be uniform anymore on the plane.

(b) (i)

$A_S = πR_S^2$

$= π(0.50 \times 10^{-2})^2 = 7.85 \times 10^{-5} \ m^2$

$M= \frac{μ_0N_SN_LA_S}{2R_L}$

$= \frac{(4π \times 10^{-7})(10)(30)(7.85 \times 10^{-5})}{2(0.10)}$

$= 1.48 \times 10^{-7}\ H ≅ 0.15 \ μH$

(ii) The emf E_S induced in the small coil is given by

$E_S = −M (\frac{ΔI}{Δt})$

$= −\frac{(1.48 \times 10^{-7})  (0 − 3.5)}{(10 \times 10^{-3}) } = 5.2 \times 10^{-5} \ V$

Application of Mutual Induction and Mutual Inductance

Mutual induction is widely used in modern electrical and electronic systems to transfer energy efficiently without direct contact.

⚡ Transformer

Uses two coils (primary and secondary) to transfer electrical energy. A changing current in the primary coil induces emf in the secondary coil, allowing voltage transformation.

📱 Wireless Charging

Charging pads generate a changing magnetic field that induces current in the device, enabling charging without cables.

🍳 Induction Cooker

Produces heat directly in cookware by inducing currents in the metal, making cooking faster and more efficient.

⚙️ Electric Generators

Changing magnetic fields induce emf in coils, enabling conversion of mechanical energy into electrical energy.

🔒 Electrical Isolation

Allows energy transfer between circuits without direct connection, improving safety and reducing electrical noise.

📡 Sensors & Devices

Used in inductive sensors and communication systems to detect objects or signals without physical contact.

🔍 Conclusion

Mutual induction is a fundamental principle in electromagnetism that enables energy transfer between circuits without direct electrical contact. It plays a crucial role in many modern technologies, including transformers, wireless charging systems, and electronic devices.

By understanding how a changing current in one coil can induce an emf in another, we can design efficient systems for energy transfer, improve electrical safety, and enhance the performance of various electrical and electronic applications.

💡 Key Insight:
A changing current → produces a changing magnetic field → induces emf in another coil.

Frequently Asked Questions (FAQ)

What is mutual induction?

Mutual induction is the process in which a changing current in one coil induces an emf in another nearby coil due to a changing magnetic field.

What is mutual inductance?

Mutual inductance is a property of two coils that determines how effectively a change in current in one coil can induce an emf in the other coil.

What is the formula for mutual inductance?

The formula is M = (NΦ) / I or E = −M (dI/dt), where M is mutual inductance, Φ is magnetic flux, and I is current.

What is the unit of mutual inductance?

The SI unit of mutual inductance is the henry (H).

Why is mutual induction important?

It allows energy transfer between circuits without direct connection and is widely used in transformers, wireless charging, and electronic devices.

Where is mutual induction used?

It is used in transformers, induction cookers, wireless chargers, generators, and inductive sensors.

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