Mutual Inductance (M)

The set-up

• Consider two coils placed side by side. Suppose a fixed current $I_1$ flows through coil 1; because of it a magnetic field $\vec{B_1}(x,y,z)$ is produced in the region surrounding coil 1, and part of it threads coil 2.
• Diagram in words: two coaxial coils facing each other a short distance apart; coil 1 carries $I_1$ from a source, and the field lines it produces spread out, some of them passing through the area $S_2$ enclosed by coil 2.
• The circuit in which the current is provided by an external source is called the primary; the other is the secondary.

Definition and derivation

• The magnetic flux $\phi_{21}$ linked with the surface area $S_2$ of coil 2 due to $\vec{B_1}$ is

$$\phi_{21} = \int_{S_2}\vec{B_1}\cdot\vec{da} \qquad \ldots (1)$$

• If the positions of the coils are fixed in space, then $\phi_{21}\propto I_1$, so

$$\phi_{21} = M_{21}I_1 \qquad \ldots (2)$$

• where $M_{21}$ is a constant of proportionality termed the mutual inductance (coefficient of mutual induction) of coil 2 with respect to coil 1.
• Suppose $I_1$ changes slowly with time. Then the field $B_1$ in the vicinity of coil 2 is related to $I_1$ in the same way as it would be for a steady current, and $\phi_{21}$ changes in proportion as $I_1$ changes. The induced emf in coil 2 is

$$e_{21} = -\dfrac{d\phi_{21}}{dt} = -M_{21}\dfrac{dI_1}{dt}$$

• Now let a current $I_2$ flow through coil 2 instead. The flux $\phi_{12}$ linked with coil 1 is proportional to $I_2$:

$$\phi_{12}\propto I_2,\qquad \phi_{12} = M_{12}I_2,\qquad M_{12} = \dfrac{\phi_{12}}{I_2} \qquad \ldots (3)$$

• and the induced emf in coil 1 is

$$e_{12} = -M_{12}\dfrac{dI_2}{dt}$$

• By symmetry, $M_{12} = M_{21} = M$.

Alternative definitions of mutual inductance

• From $\phi_{21} = MI_1$ and $\phi_{12} = MI_2$,

$$M = \dfrac{\phi_{21}}{I_1} = \dfrac{\phi_{12}}{I_2}$$

• Hence the mutual inductance of two circuits is equal to the magnetic flux linked with one circuit per unit current in the other. In primary-secondary language,

$$M = \dfrac{\phi_s}{I_p}\qquad\text{or}\qquad \phi_s = MI_p$$

• The mutual inductance $M$ of two circuits is the magnetic flux linked with the secondary circuit per unit current in the primary circuit.
• From Faraday's law applied to the secondary,

$$e_s = -\dfrac{d\phi_s}{dt} = -\dfrac{d}{dt}(MI_p) = -M\dfrac{dI_p}{dt} \qquad\Rightarrow\qquad M = \dfrac{\lvert e_s\rvert}{\lvert dI_p/dt\rvert}$$

• Hence mutual inductance is also defined as the value of the induced emf produced in the secondary circuit per unit rate of change of current in the primary circuit.
• Mutual potential energy: it can be shown that the mutual potential energy of two circuits is $W = MI_1I_2$, so $M$ may also be defined as the mutual potential energy of two circuits corresponding to unit current flowing in each: $M = \dfrac{W}{I_1I_2}$, and $M = W$ when $I_1 = I_2 = 1\ \text{A}$.

Unit of mutual inductance

• The unit is the henry (H).

$$\text{henry} = \dfrac{\text{volt}}{\text{ampere/second}} = \text{ohm}\cdot\text{second},\qquad 1\ \text{henry} = 1\ \text{ohm}\cdot\text{second}$$

• If, corresponding to a rate of change of current of $1\ \text{A s}^{-1}$ in the primary circuit, the induced emf produced in the secondary circuit is 1 volt, then the mutual inductance of the two circuits is 1 henry.

Self induction and mutual induction compared

Coefficient of coupling

• The coefficient of coupling $K$ is a measure of the portion of the flux of coil 1 that actually reaches coil 2. The greater the coefficient of coupling, the greater the mutual inductance.
• Since the inductance of any circuit is proportional to the induced voltage it can develop, $M\propto e_{21}$, where $e_{21}$ is the emf developed in coil 2 due to the portion $K\phi_1$ of coil 1's flux reaching it.
• The induced emf is also proportional to the number of turns, so $e_{21}\propto N_2K\phi_1$; and $\phi_1\propto N_1$, so $e_{21}\propto KN_1N_2$.
• Also $L\propto N^2$, so $N\propto\sqrt{L}$, giving $N_1N_2\propto\sqrt{L_1L_2}$. Replacing $e_{21}$ by $M$,

$$M = K\sqrt{L_1L_2}\qquad\Rightarrow\qquad K = \dfrac{M}{\sqrt{L_1L_2}}$$

• $K$ is usually less than unity. If $K = 1$ the two coils are perfectly coupled and $M = \sqrt{L_1L_2}$.
• $K>0.5$: tightly coupled. $K<0.5$: loosely coupled. If $L_1 = L_2 = L$, a coil coupled to itself has $M = \sqrt{L\cdot L} = L$.
• Expressed as a percentage, $K$ is the percentage of the flux from one coil that reaches the other.
• Theoretically the coupling is never perfect. Two coils wound on a common iron core may be taken to have $K\approx1$. For air-core coils or coils on separate iron cores, $K$ depends on the distance between the coils and the angle between their axes: $K$ is maximum when the coils are parallel and in line (coaxial), and minimum when their axes are at right angles and in line. To prevent interaction, coils should be at right angles to each other and as far apart as possible. For radio-frequency and intermediate-frequency transformer coils, $K$ lies between 0.001 and 0.05.
• A large $M$ is desirable in a transformer but not in a home appliance. In an electric clothes dryer a dangerous emf could be induced on the metallic case if the mutual inductance between the heating coils and the case were large, so the heating coils are counter-wound: their magnetic fields cancel one another and $M$ with the case is reduced.

Standard geometries

• Solenoid with a small coil wound round its centre: with $N_1$ turns on the solenoid of length $l$ and area $A$, and $N_2$ turns on the secondary, $B = \mu_0\dfrac{N_1}{l}I_1$, the flux linkage of the secondary is $N_2BA$, and

$$M = \dfrac{\mu_0 N_1N_2A}{l}$$

• Wireless battery charger: the base unit is a solenoid (coil B) of length $l$ with $N_B$ turns carrying current $i_B$ and cross-section $A$; the handle coil (coil H) of $N_H$ turns surrounds it completely. With $B_{\text{solenoid}} = \mu_0\left(\dfrac{N_B}{l}\right)i_B$ and $\phi_H = B_{\text{solenoid}}A$, the flux linkage is $N_H\phi_H$ and

$$M = \dfrac{N_H\phi_H}{i_B} = \dfrac{\mu_0N_BN_HA}{l}$$

Why two coils in parallel have less inductance than either

• For a parallel combination, the current through each parallel inductor is only a fraction of the total current, while the voltage across each parallel inductor is the same.
• As a result, a given change in the total current develops less voltage across the parallel array than it would across any one of the individual inductors.
• There is therefore less voltage drop across the parallel inductors for a given rate of change of current than across any of the individual inductors.
• Less voltage for the same rate of change of current means less inductance.
• Hence the total inductance of two coils in parallel is less than the inductance of either coil, consistent with $\dfrac{1}{L_{\text{total}}} = \dfrac{1}{L_1}+\dfrac{1}{L_2}$.

Standard results

• $\phi_{21} = MI_1$, $e_{21} = -M\dfrac{dI_1}{dt}$, $M_{12} = M_{21} = M$.
• $M = \dfrac{\phi_s}{I_p} = \dfrac{\lvert e_s\rvert}{\lvert dI_p/dt\rvert} = \dfrac{W}{I_1I_2}$; unit henry $=$ ohm second.
• $M = K\sqrt{L_1L_2}$ with $K\le1$; $M = \dfrac{\mu_0N_1N_2A}{l}$ for a solenoid and a coil wound round it.