Mutual Inductance

According to Faraday’s law, a current-carrying wire produces a magnetic field. If the current changes with time, the magnetic field will also change. Suppose the wire is wound into a coil and another coil is brought closer to it. Then, the changing magnetic field in the first coil will induce an emf (electromotive force) in the second. This phenomenon is known as mutual induction. The property that links the two coils by a common magnetic field is known as mutual inductance.

The induced emf will cause a current to flow in the second coil. Lenz’s law determines the direction of the induced current. This direction is such that it will oppose the change of current in the first coil. Now, suppose both the coils carry current. Then, emf will be induced in each coil due to changing current in the other. Hence, mutual inductance can also be defined as the property that describes the effect on one coil’s magnetic field with another coil’s magnetic field.

Mutual Inductance Equation

Suppose two coils are placed next to each other. The first coil has N₁ turns and carries current I₁ giving rise to magnetic field B₁. The second coil has N₂ turns. Since the two coils are close, some of the magnetic field lines through coil 1 will also pass through coil 2. Let φ₂₁ denote the magnetic flux through one turn of coil 2 due to I₁. According to Faraday’s law, changing the current I₁ changes the magnetic flux φ₂₁. It results in an induced emf ε₂₁ given by the following formula.

\[ \epsilon = -N_2 \frac{\Delta \phi_{21}}{\Delta t} \]

The rate of magnetic flux change in coil 2 is proportional to the rate of current change in coil 1. Therefore,

\[ N_2 \frac{\Delta \phi_{21}}{\Delta t} \propto \frac{\Delta I_1}{\Delta t} \\ \Rightarrow N_2 \frac{\Delta \phi_{21}}{\Delta t} = M_{21} \frac{\Delta I_1}{\Delta t} \]

The constant of proportionality M₂₁ is the mutual inductance. It has the SI unit of Henry or H, named after American scientist Joseph Henry who discovered electromagnetic induction. 1 H = 1 Wb/A or 1 T·m²/A or 1 V·s/A.

Rewriting the induced emf equation

\[ \epsilon_{21} = -M_{21}\frac{\Delta I_1}{\Delta t} \\ \Rightarrow M_{21} = \frac{\epsilon_{21}}{\left| \frac{\Delta I_1}{\Delta t} \right| } \]

Thus, the mutual inductance can be defined as the ratio between the emf generated in coil 2 and the rate of current change in coil 1. It is also known as the coefficient of mutual induction.

Now, consider the following equation again.

\[ N_2 \frac{\Delta \phi_{21}}{\Delta t} = M_{21} \frac{\Delta I_1}{\Delta t} \]

The above equation is discrete in time. So, summing over all time increments we get

\[ N_2 \Sigma \Big ( \frac{\Delta \phi_{21} }{\Delta t}\Big ) \Delta t = M_{21} \Sigma \Big ( \frac{\Delta I_1}{\Delta t} \Big ) \Delta t \\ \Rightarrow N_2 \phi_{21} = M_{21} I_1 \\ \Rightarrow M_{21} = \frac{N_2 \phi_{21}}{I_1} \]

Thisequation is used to calculate the mutual inductance between any two coils, as discussed below.

1. Mutual Inductance of Two Coils

Suppose two coils 1 & 2 with N₁ and N₂ turn have currents I₁ and I₂ flowing through them, as shown in the figure. The flux due to I₁ and I₂ are φ₂₁ and φ₁₂. The mutual inductance of coil 2 due to coil 1 is given by

\[ M_{21} = \frac{N_2 \phi_{21}}{I_1} \]

Similarly, the mutual inductance of coil 1 due to coil 2 is given by

\[ M_{12} = \frac{N_1 \phi_{12}}{I_2} \]

From the reciprocity theorem, the two mutual inductances are the same and can be represented by M. Therefore,

\[ M = M_{21} = M_{12} \\ \Rightarrow M = \frac{N_2 \phi_{21}}{I_1} = \frac{N_1 \phi_{12}}{I_2} \]

The induced emfs in the two coils are given by

\[ \epsilon_{21} = -M\frac{\Delta I_1}{\Delta t} \hspace{5mm} \text{and} \hspace{5mm} \epsilon_{12} = -M\frac{\Delta I_2}{\Delta t} \]

Note that ε₂₁ and ε₁₂ do not represent the total emf in the circuit. Each coil can have its emf due to self-inductance.

The mutual inductance depends upon how close the two coils are placed. If the coils are close enough, all the flux from coil 1 passes through coil 2. Then the mutual inductance is high. The mutual inductance is low if the coils are far. Also, the mutual inductance can be increased by increasing the number of turns of the coils or by enclosing the coils in a soft iron core. The advantage of having a soft iron core is that the magnetic flux leakage is reduced.

Mutual Inductance Formula Between Two Coils

2. Mutual Inductance of a Coil Wrapped Around a Solenoid

Suppose a long solenoid with length d and cross-sectional area A consists of N₁ turns of wire carries current I₁, as shown in the figure. The magnetic field B inside a solenoid is given by

\[ B = \frac{ \mu_o N_1 I_1}{d} \]

Where μ_o (= 1.26 x 10^-6 H/m) is called the permeability of air.

A coil of N₂ turns is wrapped around the solenoid such that the two are coaxial. The magnetic flux φ₂₁ passing through each turn of the coil due to the solenoid is

\[ \phi_{21} = BA = \frac{\mu_o N_1 I_1 A}{d} \]

Thus, the mutual inductance is

\[ M = \frac{N_2 \phi_{21}}{I_1} \\ \Rightarrow M = \frac{\mu_o N_1N_2 A}{d} \]

The self-inductance of a solenoid is

\[ L_1 = \frac{\mu_o N_1^2 A}{d} \]

Similarly, the self-inductance of the coil can be calculated and shown to be

\[ L_2 = \frac{\mu_o N_2^2 A}{d} \]

Multiplying L₁ with L₂, we get

\[ L_1 L_2 = \Big ( \frac{\mu_o N_1^2 A}{d} \Big ) \Big ( \frac{\mu_o N_2^2 A}{d} \Big ) = \Big ( \frac{\mu_o N_1N_2 A}{d} \Big )^2 = M^2 \\[10 pt] \Rightarrow M = \sqrt{L_1L_2} \]

However, this equation assumes no flux leakage and perfect magnetic coupling between the two coils, L₁ and L₂. In reality, there is going to be some leakage due to alignment and the coupling can never be 100%. For this reason, we introduce a term calling coupling constant, k, such that the mutual inductance becomes

\[ M = k \sqrt{L_1L_2} \]

k takes a value between 0 and 1. When k = 1, it means that all of the magnetic flux produced by the solenoid passes through the outer coil and vice versa.

Mutual Inductance of a Coil Wrapped Around a Solenoid

Applications of Mutual Inductance

Mutual inductance is widely used in daily life, especially in the following devices.

Transformers
Generators
Motors
Metal detectors
Pacemakers
Radio receivers

Problems and Solutions

Problem 1: Two coils have a mutual inductance of 4.75 × 10^-4 H. The current in the first coil increases at a uniform rate of 940 A/s. What is the magnitude of induced emf in the second coil?

Solution:

Given

M = 4.75 × 10^-4 H

ΔI/Δt = 940 A/s

The induced emf is given by

|ε| = M ΔI/Δt = 4.75 × 10^-4 H x 940 A/s = 0.45 V

Problem 2: Calculate the mutual inductance between a solenoid of length 30 cm and cross-sectional area of 6.25 cm² with 500 turns and a circular coil of 250 turns wound at the solenoid’s center (μ_o = 1.26 x 10^-6 H/m).

Solution:

Given

d = 30 cm = 0.3 m

A = 6.25 cm² = 6.25 x 10^-4 m²

N₁ = 500

N₂ = 250

μ_o = 1.26 x 10^-6 H/m

The mutual inductance is given by

M = μ_oN₁N₂A/d

Or, M = 1.26 x 10^-6 H/m x 500 x 250 x 6.25 x 10^-4 m²/0.3 m

Or, M = 0.000328 H or 3.28 x 10^-4 H