11.1 Mutual Inductance

Samuel J. Ling; Jeff Sanny; William Moebs; Daryl Janzen

11.1 Mutual Inductance

LEARNING OBJECTIVES

By the end of this section, you will be able to:

Correlate two nearby circuits that carry time-varying currents with the emf induced in each circuit
Describe examples in which mutual inductance may or may not be desirable

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 11.1.1). 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 $\Phi_{21}$ , whereas the flux through a single turn of coil 1 due to the magnetic field of $I_2$ is $\Phi_{12}$ .

(Figure 11.1.1) $\begin{gather*}.\end{gather*}$

Figure shows the cross sections of two coils. In each one, the cross sections of the wire of the coil are shown as two circles, one at the top and the other at the bottom. Dots in the upper circles and crosses in the lower ones indicate the direction of flow of current. Coil 1 has field lines labeled B1 passing from between the two circles, going right. Some of these pass through coil 2, which is smaller than coil 1. — **Figure 11.1.1** 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,

(11.1.1) $\begin{equation*}M_{21}=\frac{N_2\Phi_{21}}{I_1}.\end{equation*}$

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

(11.1.2) $\begin{equation*}M_{12}=\frac{N_1\Phi_{12}}{I_2}.\end{equation*}$

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 honour of Joseph Henry (1799–1878), an American scientist who discovered induced emf independently of Faraday. Thus, we have $1~\mathrm{H}=1~\mathrm{V}\cdot\mathrm{s/A}$ . From Equation 11.1.1 and Equation 11.1.2, we can show that $M_{21}=M_{12}$ , so we usually drop the subscripts associated with mutual inductance and write

(11.1.3) $\begin{equation*}M=\frac{N_2\Phi_{21}}{I_1}=\frac{N_1\Phi_{12}}{I_2}.\end{equation*}$

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

(11.1.4) $\begin{equation*}\mathcal{E}_2=-\frac{d}{dt}(N_2\Phi_{21})=-\frac{d}{dt}(MI_1)=-M\frac{dI_1}{dt}\end{equation*}$

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

(11.1.5) $\begin{equation*}\mathcal{E}_1=-M\frac{dI_2}{dt}.\end{equation*}$

In Equation 11.1.5, 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 11.1.5 defines the mutual inductance in terms of properties in the circuit, whereas the previous definition of mutual inductance in Equation 11.1.1 is defined in terms of the magnetic flux experienced, regardless of circuit elements. You should be careful when using Equation 11.1.4 and Equation 11.1.5 because $\mathcal{E}_1$ and $\mathcal{E}_2$ 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 (Equation 11.1.3).

(Figure 11.1.2) $\begin{gather*}.\end{gather*}$

Figure a shows a heating coil within a metal case of a clothes dryer. Figure b shows the same coil, enlarged. The coil is wound on a cylinder in such a way that one wire is wound all the way to the other side, twisted around and wound all the way back. Thus, two adjacent windings have current flowing in opposite directions. — **Figure 11.1.2** 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 $1~\mathrm{s}$ and $0~\mathrm{s}$ in a digital circuit creates a complex time-dependent magnetic field. An emf can be generated in neighbouring conductors. If that conductor is also carrying a digital signal, the induced emf may be large enough to switch $1~\mathrm{s}$ and $0~\mathrm{s}$ , with consequences ranging from inconvenient to disastrous.

EXAMPLE 11.1.1

Mutual Inductance

Figure 11.1.3 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 $N_1=500~\mathrm{turns}$ , $N_2=10~\mathrm{turns}$ , $R_1=3.10~\mathrm{cm}$ , $l_1=75.0~\mathrm{cm}$ and the current in the solenoid is changing at a rate of $200~\mathrm{A/s}$ , what is the emf induced in the surrounding coil?

(Figure 11.1.3) $\begin{gather*}.\end{gather*}$

Figure shows a solenoid, in the form of a long coil with a small diameter, that is concentrically arranged with another, bigger coil. The radius of the solenoid is R1 and that of the coil is R2. The length of the solenoid is l1. — **Figure 11.1.3** 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 11.1.3. We solve part (b) by calculating the mutual inductance from the given quantities and using Equation 11.1.4 to calculate the induced emf.

Solution

a. The magnetic flux $\Phi_{21}$ through the surrounding coil is

$\Phi_{21}=B_1\piR_1^2=\frac{\mu_0N_1I_1}{l_1}\piR_1^2.$

Now from Equation 11.1.3, the mutual inductance is

$M=\frac{N_2\Phi_{21}}{I_1}=\left(\frac{N_2}{I_1}\right)\left(\frac{\mu_0N_1I_1}{l_1}\piR_1^2=\frac{\mu_0N_1N_2\pi R_1^2}{l_1}.\right)$

b. Using the previous expression and the given values, the mutual inductance is

$\begin{eqnarray*}M&=&\frac{(4\pi\times10^{-7}~\mathrm{T}\cdot\mathrm{m/A})(500)(10)\pi(0.0310~\mathrm{m})^2}{0.750~\mathrm{m}}\\&=&2.53\times10^{-3}~\mathrm{H}.\end{eqnarray*}$

Thus, from Equation 11.1.4, the emf induced in the surrounding coil is

$\begin{eqnarray*}\mathcal{E}_2&=&-M\frac{dI_1}{dt}=-(2.53\times10^{-5}~\mathrm{H})(200~\mathrm{A/s})\\&=&-5.06\times10^{-3}~\mathrm{V}.\end{eqnarray*}$

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.

CHECK YOUR UNDERSTANDING 11.1

A current $I(t)=(5.0~\mathrm{A})\sin((120\pi~\mathrm{rad/s})t)$ flows through the solenoid of part (b) of Example 11.1.1. What is the maximum emf induced in the surrounding coil?

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