# 14.1: Mutual Inductance

**Inductance** is the property of a device that tells us how effectively it induces an emf in another device. In other words, it is a physical quantity that expresses the effectiveness of a given device.

When two circuits carrying time-varying currents are close to one another, the magnetic flux through each circuit varies because of the changing current *I* in the other circuit. Consequently, an emf is induced in each circuit by the changing current in the other. This type of emf is therefore called a *mutually induced emf*, and the phenomenon that occurs is known as **mutual inductance ( M)**. As an example, let’s consider two tightly wound coils (Figure). Coils 1 and 2 have \(N_1\) and \(N_2\) turns and carry currents \(I_1\) and \(I_2\) respectively. The flux through a single turn of coil 2 produced by the magnetic field of the current in coil 1 is \(\Phi_{12}\), whereas the flux through a single turn of coil 1 due to the magnetic field of \(I_2\) is \(\Phi_{12}\).

**Figure 14.1.1**. Some of the magnetic field lines produced by the current in coil 1 pass through coil 2.

The mutual inductance \(M_{21}\) of coil 2 with respect to coil 1 is the ratio of the flux through the \(N_2\) turns of coil 2 produced by the magnetic field of the current in coil 1, divided by that current, that is,

\[M_{21} = \frac{N_2\Phi_{21}}{I_1}.\]

Similarly, the mutual inductance of coil 1 with respect to coil 2 is

\[M_{12} = \frac{N_1\Phi_{12}}{I_2}.\]

Like capacitance, mutual inductance is a geometric quantity. It depends on the shapes and relative positions of the two coils, and it is independent of the currents in the coils. The SI unit for mutual inductance *M* is called the **henry (H)** in honor of Joseph **Henry** (1799–1878), an American scientist who discovered induced emf independently of Faraday. Thus, we have \(1 \space H = 1 \space V \cdot s/A\). From Equation and Equation, we can show that \(M_{21} = M_{12}\), so we usually drop the subscripts associated with mutual inductance and write

Note

\[M = \frac{N_2\Phi_{21}}{I_1} = \frac{N_1 \Phi_{12}}{I_2}.\]

The emf developed in either coil is found by combining Faraday’s law and the definition of mutual inductance. Since \(N_2\Phi_{21}\) is the total flux through coil 2 due to \(I_1\), we obtain

\[\epsilon_2 = - \frac{d}{dt} (N_2 \Phi_{21}) = - \frac{d}{dt} (MI_1) = - M\frac{dI_1}{dt}\]

where we have used the fact that *M* is a time-independent constant because the geometry is time-independent. Similarly, we have

Note

\[\epsilon_1 = - M\frac{dI_2}{dt}.\]

In Equation, we can see the significance of the earlier description of mutual inductance (*M*) as a geometric quantity. The value of *M* neatly encapsulates the physical properties of circuit elements and allows us to separate the physical layout of the circuit from the dynamic quantities, such as the emf and the current. Equation defines the mutual inductance in terms of properties in the circuit, whereas the previous definition of mutual inductance in Equation is defined in terms of the magnetic flux experienced, regardless of circuit elements. You should be careful when using Equation and Equation because \(\epsilon_1\) and \(\epsilon_2\) do not necessarily represent the total emfs in the respective coils. Each coil can also have an emf induced in it because of its *self-inductance* (self-inductance will be discussed in more detail in a later section).

A large mutual inductance *M* may or may not be desirable. We want a transformer to have a large mutual inductance. But an appliance, such as an electric clothes dryer, can induce a dangerous emf on its metal case if the mutual inductance between its coils and the case is large. One way to reduce mutual inductance is to counter-wind coils to cancel the magnetic field produced (Figure).

**Figure 14.1.2.** The heating coils of an electric clothes dryer can be counter-wound so that their magnetic fields cancel one another, greatly reducing the mutual inductance with the case of the dryer.

**Digital signal processing** is another example in which mutual inductance is reduced by counter-winding coils. The rapid on/off emf representing 1s and 0s in a digital circuit creates a complex time-dependent magnetic field. An emf can be generated in neighboring conductors. If that conductor is also carrying a digital signal, the induced emf may be large enough to switch 1s and 0s, with consequences ranging from inconvenient to disastrous.

Example

**Mutual Inductance**

Figure shows a coil of \(N_2\) turns and radius \(R_2\) surrounding a long solenoid of length \(l_1\), radius \(R_1\), and \(N_1\) turns. (a) What is the mutual inductance of the two coils? (b) If \(N_1 = 500 \space turns, \space N_2 = 10 \space turns, \space R_1 = 3.10 \space cm, \space l_1 = 75.0 \space cm\), and the current in the solenoid is changing at a rate of 200 A/s, what is the emf induced in the surrounding coil?

**Figure 14.1.3**. A solenoid surrounded by a coil.

**Strategy**

There is no magnetic field outside the solenoid, and the field inside has magnitude \(B_1 = \mu_0(N_1/l_1)I_1\) and is directed parallel to the solenoid’s axis. We can use this magnetic field to find the magnetic flux through the surrounding coil and then use this flux to calculate the mutual inductance for part (a), using Equation. We solve part (b) by calculating the mutual inductance from the given quantities and using Equation to calculate the induced emf.

**Solution**

- The magnetic flux \(\Phi_{21}\) through the surrounding coil is \[\Phi_{21} = B_1 \pi R_1^2 = \frac{\mu_0 N_1I_1}{l_1}\pi R_1^2.\] Now from Equation, the mutual inductance is \[M = \frac{N_2\Phi_{21}}{I_1} = \left(\frac{N_2}{I_1}\right)\left(\frac{\mu_0N_1I_1}{l_1}\right) \pi R_1^2 = \frac{\mu_0N_1N_2 \pi R_1^2}{l_1}.\]
- Using the previous expression and the given values, the mutual inductance is \[M = \frac{(4\pi \times 10^{-7} \space T \cdot m/A)(500)(10)\pi (0.0310 \space m)^2}{0.750 \space m}\]\[=2.53 \times 10^{-5} \space H.\] Thus, from Equation, the emf induced in the surrounding coil is \[\epsilon_2 = - M\frac{dI_1}{dt} = - (2.53 \times 10^{-5} H)(200 \space A/s)\]\[= - 5.06 \times 10^{-3}V.\]

**Significance**

Notice that *M* in part (a) is independent of the radius \(R_2\) of the surrounding coil because the solenoid’s magnetic field is confined to its interior. In principle, we can also calculate *M* by finding the magnetic flux through the solenoid produced by the current in the surrounding coil. This approach is much more difficult because \(\Phi_{12}\) is so complicated. However, since \(M_{12} = M_{21}\), we do know the result of this calculation.

Note

**Check Your Understanding**

A current \(I(t) = (5.0 \space A) \space sin \space ((120\pi \space rad/s)t)\) flows through the solenoid of part (b) of Example. What is the maximum emf induced in the surrounding coil?

[Hide Solution]

\(4.77 \times 10^{-2} \space V\)

# Summary

- Inductance is the property of a device that expresses how effectively it induces an emf in another device.
- Mutual inductance is the effect of two devices inducing emfs in each other.
- A change in current \(dI_1/dt\) in one circuit induces an emf \((\epsilon_2)\) in the second: \[\epsilon_2 = - M\frac{dI_1}{dt},\] where
*M*is defined to be the mutual inductance between the two circuits and the minus sign is due to Lenz’s law. - Symmetrically, a change in current \(dI_2/dt\) through the second circuit induces an emf \((\epsilon_1\) in the first: \[\epsilon_1 = - M\frac{dI_2}{dt},\] where
*M*is the same mutual inductance as in the reverse process.

# Conceptual Questions

Exercise

Show that \(N\Phi_m/I\) and \(\epsilon/(dI/dt)\), which are both expressions for self-inductance, have the same units.

[Hide Solution]

\(\frac{W_b}{A} = \frac{T \cdot m^2}{A} = \frac{V \cdot s}{A} = \frac{V}{A/s}\)

Exercise

A 10-H inductor carries a current of 20 A. Describe how a 50-V emf can be induced across it.

Exercise

The ignition circuit of an automobile is powered by a 12-V battery. How are we able to generate large voltages with this power source?

The induced current from the 12-V battery goes through an inductor, generating a large voltage.

Exercise

When the current through a large inductor is interrupted with a switch, an arc appears across the open terminals of the switch. Explain.

# Problems

Exercise

When the current in one coil changes at a rate of 5.6 A/s, an emf of \(6.3 \times 10^{-3} V\) is induced in a second, nearby coil. What is the mutual inductance of the two coils?

Exercise

An emf of \(9.7 \times 10^{-3} V\) is induced in a coil while the current in a nearby coil is decreasing at a rate of 2.7 A/s. What is the mutual inductance of the two coils?

[Hide Solution]

\(M = 3.6 \times 10^{-3} H\)

Exercise

Two coils close to each other have a mutual inductance of 32 mH. If the current in one coil decays according to \(I = I_0 e^{-\alpha t}\), where \(I_0 = 5.0 \space A\) and \(\alpha = 2.0 \times 10^3 s^{-1}\), what is the emf induced in the second coil immediately after the current starts to decay? At \(t = 1.0 \times 10^{-3} \space s\)?

Exercise

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area \(7.5 \times 10^{-3} \space m^2\). The solenoid is 0.50 m long and has 500 turns. (a) What is the mutual inductance of this system? (b) The outer coil is replaced by a coil of 40 turns whose radius is three times that of the solenoid. What is the mutual inductance of this configuration?

[Hide Solution]

a. \(3.8 \times 10^{-4} H\); b. \(3.8 \times 10^{-4} H\)

Exercise

A 600-turn solenoid is 0.55 m long and 4.2 cm in diameter. Inside the solenoid, a small \((1.1 \space cm \times 1.4 \space cm)\), single-turn rectangular coil is fixed in place with its face perpendicular to the long axis of the solenoid. What is the mutual inductance of this system?

Exercise

A toroidal coil has a mean radius of 16 cm and a cross-sectional area of \(0.25 \space cm^2\); it is wound uniformly with 1000 turns. A second toroidal coil of 750 turns is wound uniformly over the first coil. Ignoring the variation of the magnetic field within a toroid, determine the mutual inductance of the two coils.

[Hide Solution]

\(M_{21} = 2.3 \times 10^{-5} H\)

Exercise

A solenoid of \(N_1\) turns has length \(l_1\) and radius \(R_1\), and a second smaller solenoid of \(N_2\) turns has length \(l_2\) and radius \(R_2\). The smaller solenoid is placed completely inside the larger solenoid so that their long axes coincide. What is the mutual inductance of the two solenoids?

## Glossary

- henry (H)
- unit of inductance, \(1 \space H = 1 \space \Omega \cdot s\); it is also expressed as a volt second per ampere

- inductance
- property of a device that tells how effectively it induces an emf in another device

- mutual inductance
- geometric quantity that expresses how effective two devices are at inducing emfs in one another