Mutual inductance and stored energy

Power without a wire

An electric toothbrush sits on a charging stand. No wire joins them. Yet the brush charges. A transformer passes power from one coil to another, and the two coils are not joined by any wire either.

These devices work because a changing current in one coil makes a changing flux through the other. This is mutual induction.

Mutual inductance

Coil 1 carries a current $I_1$. Some of its flux passes through coil 2, which has $N_2$ turns. The flux linkage of coil 2 is in proportion to $I_1$:

The constant $M$ is the mutual inductance of the two coils. Its unit is the henry, like $L$. Now let $I_1$ change. By Faraday's law, the emf in coil 2 is

The same $M$ works the other way round. A changing current in coil 2 makes an emf in coil 1, with the same $M$. So we speak of one mutual inductance of the pair.

A worked example. A long solenoid has $n = 800$ turns in each metre. A coil of $N_2 = 150$ turns and radius $1.5$ cm sits inside it, on the same axis.

1. The flux. The field in the solenoid is $B_1 = \mu_0 nI_1$. It goes through every turn of the coil, each of area $\pi r_2^2$. So $N_2\phi_2 = \mu_0 nN_2\pi r_2^2\,I_1$.
2. The mutual inductance. $M = \mu_0 nN_2\pi r_2^2 = 4\pi \times 10^{-7}\ \text{T m/A} \times 800\ \text{m}^{-1} \times 150 \times \pi \times (0.015\ \text{m})^2 = 1.07 \times 10^{-4}\ \text{H}$. That is $0.107$ mH.
3. The emf. Say $I_1$ changes at $50$ A/s. Then $|\varepsilon_2| = M\left|\dfrac{dI_1}{dt}\right| = 1.07 \times 10^{-4}\ \text{H} \times 50\ \text{A/s} = 5.3\ \text{mV}$.

What if the coil is wider than the solenoid? The field of the solenoid exists only inside it. A coil wider than the solenoid still has flux only through the area of the solenoid. So $M$ stops growing when the radius of the coil reaches the radius of the solenoid. Try it in the simulation below.

Practise with new numbers

How tightly are two coils linked?

The self-inductances of the coils are $L_1$ and $L_2$. The flux that coil 1 sends through coil 2 can be at most all of its flux. This gives a limit:

We write $M = k\sqrt{L_1L_2}$. The coupling coefficient $k$ lies between $0$ and $1$. If the coils are far apart, $k$ is close to $0$. If one coil is wound over the other on an iron core, $k$ is close to $1$.

A worked example. $L_1 = 0.18$ H, $L_2 = 0.050$ H, and $k = 0.85$.

Two coils in series. Join them in series, so that the same current goes through both. Each coil also has the flux of the other. The inductance of the pair is

The plus sign holds when the two fields point the same way (aiding). The minus sign holds when they point against each other (opposing).

A worked example. $L_1 = 40$ mH, $L_2 = 90$ mH and $M = 30$ mH. The largest possible $M$ is $\sqrt{40 \times 90}\ \text{mH} = 60$ mH, so this is allowed.

• Aiding: $40\ \text{mH} + 90\ \text{mH} + 2 \times 30\ \text{mH} = 190$ mH.
• Opposing: $40\ \text{mH} + 90\ \text{mH} - 2 \times 30\ \text{mH} = 70$ mH.

Energy stored in an inductor

To set up a current in a coil, the source must push against the back emf. When the current is $i$ and it grows at $\dfrac{di}{dt}$, the back emf is $L\dfrac{di}{dt}$. The source gives power $L\dfrac{di}{dt} \times i$. In a short time $dt$ the work done is $dW = Li\,di$.

Add up the work as the current grows from $0$ to $I$:

The energy is not lost. It stays in the magnetic field, and it comes back when the current falls. This is the energy of the spark. Compare it with $\tfrac{1}{2}mv^2$ for a moving body, and with $\tfrac{1}{2}CV^2$ for a capacitor.

A worked example. A coil of $2.4$ H carries $3.0$ A.

That is enough to lift a mass of $1.1$ kg through $1$ m.

The energy is in the field

For a long solenoid, $L = \mu_0 n^2 Al$, with $Al$ the volume inside it. The field is $B = \mu_0 nI$. Put both into $U = \tfrac{1}{2}LI^2$:

So each cubic metre of the field holds the energy

This holds for any magnetic field, not only in a solenoid.

A worked example. A field of $1.5$ T, as in a strong electromagnet.

Compare an electric field of $3 \times 10^6$ V/m, the greatest that air can hold. Its energy density is $\tfrac{1}{2}\varepsilon_0E^2 = 40\ \text{J/m}^3$. A strong magnetic field stores about $20\,000$ times more energy in the same volume.

Practise with new numbers

Try it in Python

The program joins two coils in series, for different values of the coupling. It also finds the energy stored when a current of $2.0$ A flows.

Common mistakes

• Using $\sqrt{L_1L_2}$ as if it were always $M$. It is the largest possible value. Usually $M$ is smaller.
• Forgetting the sign of $2M$. Aiding coils add, and opposing coils subtract.
• Leaving out the half in $\tfrac{1}{2}LI^2$.
• Forgetting that the flux stops at the solenoid. A wider coil does not collect more flux.
• Thinking the energy is in the wire. It is stored in the magnetic field.

Remember these

• Mutual inductance: $N_2\phi_2 = MI_1$ and $\varepsilon_2 = -M\dfrac{dI_1}{dt}$. The same $M$ works both ways.
• Coaxial solenoid and coil: $M = \mu_0 nN_2\pi r_2^2$, for a coil inside the solenoid.
• Coupling: $M = k\sqrt{L_1L_2}$, with $0 \leq k \leq 1$. In series, $L = L_1 + L_2 \pm 2M$.
• Energy stored: $U = \dfrac{1}{2}LI^2$. Energy in each cubic metre of field: $u = \dfrac{B^2}{2\mu_0}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Two coils have a mutual inductance of $0.020$ H. The current in coil 1 changes at $15$ A/s. What is the emf in coil 2?

2. Why can the coupling coefficient never be larger than $1$?

3. The current in an inductor is doubled. How does the stored energy change?

4. A coil of $0.50$ H carries $4.0$ A. How much energy does it store?

5. Where does the energy of the spark at an opening switch come from?