Eddy currents

A plate that will not swing freely

Hold a thick copper plate by its handle and push it quickly between the poles of a strong magnet. The plate does not slip through. It moves as if it were in thick honey. Pull it out and you feel the same drag.

Copper is not a magnetic metal, and no wire touches the plate. So where does the drag come from? The answer is a current that nobody connected.

Currents without wires

In a coil, a changing flux makes an emf, and the emf drives a current round the wire. A solid piece of metal has no wire. But it is full of free electrons. Draw any loop you like inside the metal. If the flux through that loop changes, the electrons are pushed round it.

So the currents swirl through the metal, like eddies in a stream. We call them eddy currents.

Two things make the flux change:

• The metal moves into, out of, or through a field that is not the same everywhere.
• The metal stays still, and the field changes with time.

You know the two laws that apply. Faraday's law gives the size of the emf. Lenz's law gives the direction: the eddy currents oppose the change that makes them.

Why a moving plate is braked

Look at the first picture. A plate moves to the left into a field that points out of the page.

1. More and more of the plate is in the field. The flux out of the page through the plate grows.
2. By Lenz's law the eddy current opposes this growth. It must make a field into the page, so it goes clockwise.
3. The left side of the loop lies inside the field, and its current goes up the page. The force on a current in a field is $I\vec{l} \times \vec{B}$. For this current it points to the right.
4. The plate moves to the left. So the force on the plate is against its motion.

Pull the plate out again. The flux now falls, the current turns the other way, and the force again opposes the motion. Eddy currents always fight the motion that makes them.

What happens to the energy? You do work against that force. The work does not vanish. The eddy currents heat the plate, at the rate $I^2R$. At a steady speed, the rate at which you do work, $Fv$, equals the rate of heating. This is Lenz's law keeping energy safe. If the force helped the motion, the plate would speed up and heat itself, with nobody paying for it.

The emf is in proportion to the speed. So at a moderate speed the drag is nearly in proportion to the speed too, $F = kv$. At rest the drag is zero. A magnet does not grip a plate that stands still.

A magnet falling in a tube

Drop a strong magnet down a copper tube. It does not fall freely. It sinks slowly, as if the tube were full of oil. The faster it moves, the bigger the drag. So it soon reaches a terminal speed $v_t$, where the drag equals the weight:

A worked example. A magnet of mass $0.045$ kg falls through a copper tube $0.60$ m long, at a steady speed of $0.12$ m/s.

1. The drag constant. We want $k$, so $k = \dfrac{mg}{v_t} = \dfrac{0.045\ \text{kg} \times 9.8\ \text{m/s}^2}{0.12\ \text{m/s}} = 3.7\ \text{kg/s}$.
2. The heating. The weight does work at the rate $mgv_t = 0.441\ \text{N} \times 0.12\ \text{m/s} = 0.053\ \text{W}$. All of it goes into heat.
3. The time in the tube. $\dfrac{0.60\ \text{m}}{0.12\ \text{m/s}} = 5.0\ \text{s}$. A free fall over the same distance takes $\sqrt{2 \times 0.60\ \text{m} / 9.8\ \text{m/s}^2} = 0.35\ \text{s}$.

Check. Without the eddy currents the magnet would fall in a third of a second. With them it takes about $14$ times longer. The potential energy it loses each second goes into the tube as heat.

Practise with new numbers

How much heat?

How large are the eddy currents? Take a flat disc of radius $a$ and thickness $t$. The field is at right angles to the disc, and it changes at the rate $\dfrac{dB}{dt}$.

Think of a thin ring of radius $r$ and width $dr$ in the disc.

1. The flux through the ring is $B\pi r^2$. So the emf round it is $\varepsilon = \pi r^2 \dfrac{dB}{dt}$.
2. The ring is $2\pi r$ long, and its cross section is $t\,dr$. So its resistance is $dR = \dfrac{\rho \cdot 2\pi r}{t\,dr}$.
3. The heat each second in the ring is $dP = \dfrac{\varepsilon^2}{dR} = \dfrac{\pi t r^3}{2\rho}\left(\dfrac{dB}{dt}\right)^2 dr$.

Add up all the rings from $r = 0$ to $r = a$. The sum is

Look at the radius. It appears to the fourth power. A disc of twice the radius loses $16$ times as much. Big loops carry big emf and a lot of current, and the heat grows very fast.

A worked example. A copper disc has a radius of $4.0$ cm and a thickness of $2.0$ mm. The resistivity of copper is $1.7 \times 10^{-8}\ \Omega\,\text{m}$. The field changes at $3.0$ T/s.

1. The emf at the rim. $\varepsilon = \pi a^2 \dfrac{dB}{dt} = \pi \times (0.040\ \text{m})^2 \times 3.0\ \text{T/s} = 0.015\ \text{V}$.
2. The heating. $P = \dfrac{\pi \times 2.0 \times 10^{-3}\ \text{m} \times (0.040\ \text{m})^4 \times (3.0\ \text{T/s})^2}{8 \times 1.7 \times 10^{-8}\ \Omega\,\text{m}} = 1.06\ \text{W}$.

What if the disc is cut in pieces? The loops cannot be bigger than a piece. Slots stop the large loops. That is the second picture above: each tooth of the plate holds only a small loop. The emf is smaller, the path is narrower, and the force is much smaller.

Practise with new numbers

Thin sheets in a transformer

A transformer or a motor has an iron core. The field in it changes fifty or sixty times each second. If the core were one solid block, large eddy currents would heat it and waste power.

So the core is built from thin sheets, coated with varnish and stacked. Each sheet is a thin strip, and the eddy loop must stay inside it.

Each loop encloses very little flux, so its emf is small. The loss in a thin sheet of thickness $d$, in a field of peak value $B_m$ and frequency $f$, is for each cubic metre

We do not work it out here. The idea is the same as for the disc. It holds for sheets that are thin. Read it as a rule of proportion:

• The loss goes as $d^2$. Halve the thickness, and the loss falls to a quarter.
• The loss goes as $f^2$ and as $B_m^2$.
• A larger resistivity $\rho$ lowers the loss. Steel with a little silicon in it has a high resistivity, about $4.7 \times 10^{-7}\ \Omega\,\text{m}$.

A worked example. The sheets are $0.50$ mm thick. The peak field is $1.4$ T, at $50$ Hz.

1. The loss in each cubic metre. $\dfrac{P}{V} = \dfrac{\pi^2 \times (50\ \text{Hz})^2 \times (1.4\ \text{T})^2 \times (0.50 \times 10^{-3}\ \text{m})^2}{6 \times 4.7 \times 10^{-7}\ \Omega\,\text{m}} = 4.3\ \text{kW/m}^3$.
2. A core of $0.015\ \text{m}^3$. The loss is $4287\ \text{W/m}^3 \times 0.015\ \text{m}^3 = 64\ \text{W}$.

Set the thickness to $1.0$ mm and then to $0.5$ mm. Does the loss really fall to a quarter?

Practise with new numbers

Putting eddy currents to work

Sometimes we want eddy currents.

• Magnetic brakes. A train or a roller coaster carries metal fins that pass between magnets. The drag needs no contact, so nothing wears out. But the drag falls as the speed falls, so it cannot hold a train at rest. Ordinary brakes finish the job.
• Damping. The coil of a moving-coil meter is wound on a metal frame. When the pointer swings, eddy currents in the frame slow it. The pointer settles quickly and does not keep oscillating.
• The induction furnace. A coil carries a large current that changes very fast. The metal inside it is heated by its own eddy currents, until it melts. No flame is needed.
• The induction cooker. A coil under the glass top carries a current that changes tens of thousands of times each second. Eddy currents heat the iron base of the pan. A pan of glass cannot be heated this way. Glass has no free electrons to carry eddy currents.
• The metal detector. A coil makes a changing field. A coin in the ground carries eddy currents, and they change the signal in the coil.

Try it in Python

The program finds the loss in a core of $0.015\ \text{m}^3$, for sheets of five different thicknesses.

Common mistakes

• Thinking only iron has eddy currents. Copper and aluminium show them strongly. Any conductor does.
• Thinking a magnet at rest heats a plate. It does not. The flux must change.
• Thinking the force can help the motion. It cannot. It always opposes the change, and this is what keeps energy safe.
• Taking the loss as in proportion to $d$. It goes as $d^2$.
• Taking the drag as a fixed force. It grows with the speed, and it is zero at rest.

Remember these

• A changing flux drives eddy currents in a solid conductor. They oppose the change that makes them (Lenz's law).
• A moving plate is braked. The work done against the drag becomes heat.
• A disc of radius $a$ and thickness $t$ loses $P = \dfrac{\pi t a^4}{8\rho}\left(\dfrac{dB}{dt}\right)^2$. The loss goes as $a^4$.
• Slots and thin insulated sheets break up the loops. In a sheet, $\dfrac{P}{V} = \dfrac{\pi^2 f^2 B_m^2 d^2}{6\rho}$.
• Uses: brakes, damping, induction furnace and cooker, metal detector.

Check yourself

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

1. A copper plate is let go between the poles of a strong magnet. It falls slowly. Why?

2. A strong magnet lies still on an aluminium plate. Is the plate heated?

3. The sheets of a core are replaced by sheets of half the thickness. The frequency and the field are the same. How does the loss in each cubic metre change?

4. A disc loses $0.30$ W in a changing field. A disc of the same metal and thickness has three times the radius. The field changes at the same rate. How much does it lose?

5. Why can an induction cooker not heat a glass pan?