Mutual Inductance

Figure 45: Two inductively coupled circuits.

Consider two arbitrary conducting circuits, labelled 1 and 2. Suppose that $I_1$ is the instantaneous current flowing around circuit 1. This current generates a magnetic field ${\bf B}_1$ which links the second circuit, giving rise to a magnetic flux ${\mit\Phi}_{2}$ through that circuit. If the current $I_1$ doubles, then the magnetic field ${\bf B}_1$ doubles in strength at all points in space, so the magnetic flux ${\mit\Phi}_{2}$ through the second circuit also doubles. This conclusion follows from the linearity of the laws of magnetostatics, plus the definition of magnetic flux. Furthermore, it is obvious that the flux through the second circuit is zero whenever the current flowing around the first circuit is zero. It follows that the flux ${\mit\Phi}_{2}$ through the second circuit is directly proportional to the current $I_1$ flowing around the first circuit. Hence, we can write

$${\mit\Phi}_{2} = M_{21}\,I_1,$$ (231)

where the constant of proportionality $M_{21}$ is called the mutual inductance of circuit 2 with respect to circuit 1. Similarly, the flux ${\mit\Phi}_{1}$ through the first circuit due to the instantaneous current $I_2$ flowing around the second circuit is directly proportional to that current, so we can write

$${\mit\Phi}_{1} = M_{12}\,I_2,$$ (232)

where $M_{12}$ is the mutual inductance of circuit 1 with respect to circuit 2. It is possible to demonstrate mathematically that $M_{12}= M_{21}$. In other words, the flux linking circuit 2 when a certain current flows around circuit 1 is exactly the same as the flux linking circuit 1 when the same current flows around circuit 2. This is true irrespective of the size, number of turns, relative position, and relative orientation of the two circuits. Because of this, we can write

$$M_{12}= M_{21} = M,$$ (233)

where $M$ is termed the mutual inductance of the two circuits. Note that $M$ is a purely geometric quantity, depending only on the size, number of turns, relative position, and relative orientation of the two circuits. The SI units of mutual inductance are called Henries (H). One henry is equivalent to a volt-second per ampere:

$$1\,{\rm H} \equiv 1\,{\rm V\,s\,A}^{-1}.$$ (234)

It turns out that a henry is a rather unwieldy unit. The mutual inductances of the circuits typically encountered in laboratory experiments are measured in milli-henries.

Suppose that the current flowing around circuit 1 changes by an amount $d I_1$ in a time interval $dt$. It follows from Eqs. (231) and (233) that the flux linking circuit 2 changes by an amount $d{\mit\Phi}_{2} = M\,d I_1$ in the same time interval. According to Faraday's law, an emf

$${\cal E}_2 = - \frac{d{\mit\Phi}_{2}}{d t}$$ (235)

is generated around the second circuit due to the changing magnetic flux linking that circuit. Since, $d{\mit\Phi}_{2} = M\,d I_1$, this emf can also be written

$${\cal E}_2 = - M\, \frac{d I_1}{d t}.$$ (236)

Thus, the emf generated around the second circuit due to the current flowing around the first circuit is directly proportional to the rate at which that current changes. Likewise, if the current $I_2$ flowing around the second circuit changes by an amount $d I_2$ in a time interval $dt$ then the emf generated around the first circuit is

$${\cal E}_1 = - M\, \frac{d I_2}{d t}.$$ (237)

Note that there is no direct physical coupling between the two circuits. The coupling is due entirely to the magnetic field generated by the currents flowing around the circuits.

As a simple example, suppose that two insulated wires are wound on the same cylindrical former, so as to form two solenoids sharing a common air-filled core. Let $l$ be the length of the core, $A$ the cross-sectional area of the core, $N_1$ the number of times the first wire is wound around the core, and $N_2$ the number of times the second wire is wound around the core. If a current $I_1$ flows around the first wire then a uniform axial magnetic field of strength $B_1 = \mu_0\,N_1\,I_1/l$ is generated in the core (see Sect. 8.8). The magnetic field in the region outside the core is of negligible magnitude. The flux linking a single turn of the second wire is $B_1\,A$. Thus, the flux linking all $N_2$ turns of the second wire is ${\mit\Phi}_{2} = N_2\,B_1\,A = \mu_0\,N_1\,N_2\,A\,I_1/l$. From Eq. (231), the mutual inductance of the second wire with respect to the first is

$$M_{21}= \frac{{\mit\Phi}_{2}}{I_1} = \frac{\mu_0\,N_1\,N_2\,A}{l}.$$ (238)

Now, the flux linking the second wire when a current $I_2$ flows in the first wire is ${\mit\Phi}_{1} = N_1\,B_2\,A$, where $B_2 = \mu_0\,N_2\,I_2/l$ is the associated magnetic field generated in the core. It follows from Eq. (232) that the mutual inductance of the first wire with respect to the second is

$$M_{12} = \frac{{\mit\Phi}_{1}}{I_2} = \frac{\mu_0\,N_1\,N_2\,A}{l}.$$ (239)

Note that $M_{21} = M_{12}$, in accordance with Eq. (233). Thus, the mutual inductance of the two wires is given by

$$M = \frac{\mu_0\,N_1\,N_2\,A}{l}.$$ (240)

As described previously, $M$ is a geometric quantity depending on the dimensions of the core, and the manner in which the two wires are wound around the core, but not on the actual currents flowing through the wires.