# 10.10: Mutual Inductance

- Page ID
- 7894

Consider two coils, not connected to one another, other than being *close together in space*. If the current changes in one of the coils, so will the magnetic field in the other, and consequently an EMF will be induced in the second coil. **Definition**: The ratio of the EMF \(V_2\) induced in the second coil to the rate of change of current \(\dot I_1\) in the first is called the *coefficient of mutual inductance* \(M\) between the two coils:

\[V_2 = M\dot I_1 .\]

The dimensions of mutual inductance can be found from the dimensions of EMF and of current, and are readily found to be \(\text{ML}^2\text{Q}^{ −2}\) .

**Definition**: If an EMF of one volt is induced in one coil when the rate of change of current in the other is 1 amp per second, the coefficient of mutual inductance between the two is 1 *henry*, \(\text{H}\).

*Mental Exercise*: If the current in coil 1 changes at a rate \(\dot I_1\), the EMF induced in coil 2 is \(M\dot I_1\). Now ask yourself this: If the current in coil 2 changes at a rate \(\dot I_2\) is it true that the EMF induced in coil 1 will be \(M\dot I_2\)? (The answer is "yes" – but you are not excused the mental effort required to convince yourself of this.)

*Example*: Suppose that the primary coil is an infinite solenoid having \(n_1\) turns per unit length wound round a core of permeability \(\mu\). Tightly would around this is a plain circular coil of \(N_2\) turns. The solenoid and the coil wrapped tightly round it are of area \(A\). We can calculate the mutual inductance of this arrangement as follows. The magnetic field in the primary is \(\mu n_1I\) so the flux through each coil is \(\mu n_1AI\). If the current changes at a rate \(\dot I\), flux will change at a rate \(\mu n A\dot I\) and the EMF induced in the secondary coil will be \(\mu n_1 N_2 a\dot I\). Therefore the mutual inductance is

\[M=\mu n_1 N_2 A.\label{10.10.2}\]

Several points:

- Verify that this has the correct dimensions.
- If the current in the solenoid changes in such a manner as to cause an increase in the magnetic field towards the right, the EMF induced in the secondary coil is such that, if it were connected to a closed circuit so that a secondary current flows, the direction of this current will produce a magnetic field towards the left – i.e. such as to oppose the rightward increase in \(B\).
- Because of the little mental effort you made a few minutes ago, you are now convinced that, if you were to change the current in the plane coil at a rate \(\dot I\), the EMF induced in the solenoid would be \(M\dot I\), where \(M\) is given by equation \ref{10.10.2}.
- Equation \ref{10.10.2} is the equation for the mutual inductance of the system, provided that the coil and the solenoid are
*tightly coupled*. If the coil is rather loosely draped around the solenoid, or if the solenoid is not infinite in length, the mutual inductance would be rather less than given by equation \ref{10.10.2}. It would be, in fact, \(k\mu n_1N_2 A\), where \(k\), a dimensionless number between 0 and 1, is the*coupling coefficient*. - While we have hitherto expressed permeability in units of tesla metres per amp (\(\text{T m A}^{−1}\) ) or some such combination, equation \ref{10.10.2} shows that permeability can equally well be (and usually is) expressed in henrys per metre, \(\text{H m}^{−1}\) . Thus, we say that the permeability of free space is \(\mu_0= 4\pi \times 10^{-7}\text{ H m}^{-1}\) .

Exercise \(\PageIndex{1}\)

A plane coil of 10 turns is tightly wound around a solenoid of diameter 2 cm having 400 turns per centimeter. The relative permeability of the core is 800. Calculate the mutual inductance. (I make it 0.126 \(\text{H}\).)