Eddy-current braking and heating
Key idea: Changing magnetic flux induces circulating eddy currents in metal. They cause resistive heating and oppose the change; laminations reduce them by breaking up their paths.
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Guided lessons · coming soon
Guided lesson
Guided steps
- Follow how changing flux makes currents in a metal.
- Follow the force balance and energy changes at terminal speed.
- Predict how slits change the braking force.
- Compare useful braking and heating with unwanted heating.
- Find how doubling resistance changes power at fixed voltage.
- Decide which plates have changing flux and induced currents.
- Explain the slower fall and the energy changes.
- Find the heat while the magnet is still speeding up.
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ContinueDrop a strong magnet down a copper pipe and it drifts down slowly, as if through syrup, even though copper is not ferromagnetic. The changing flux induces currents that swirl round inside the metal. These eddy currents brake trains without touching the rails, heat pans on induction hobs, and waste energy in transformers unless the cores are designed to stop them.
Eddy currents
Induced currents do not need a wire coil. When the magnetic flux through closed paths inside a piece of metal changes, an e.m.f. is induced round each path and currents circulate within the metal. These are eddy currents.
A magnet falling through a copper tube changes the flux through the rings of tube wall above and below it. Eddy currents flow round the tube. By Lenz’s law their magnetic field opposes the motion that produces them, so they push up on the magnet. As the magnet speeds up, the currents and the braking force grow, until the braking force equals the weight and the magnet falls at a steady, slow terminal speed. In a plastic tube there are no conducting paths, so no currents and no braking.
Scroll across the figure to read all labels.
The eddy currents heat the copper. At terminal speed the magnet’s kinetic energy is constant, so the heat comes from the gravitational potential energy it loses.
Worked example 1
Energy at terminal speed
Problem
A magnet of mass 0.050 kg falls through a long copper tube at its terminal speed and drops 0.20 m. Ignoring air resistance, find the braking force on the magnet and the energy transferred to the tube as heat.
Show full solution
Braking force
Method
Set the braking force equal to the weight.Reason
At terminal speed the velocity is constant, so the resultant force is zero.Working
Answer
F = mg = (0.050)(9.81) = 0.49 N, upwards
Heat
Working
Answer
The kinetic energy is constant, so all the gravitational potential energy lost becomes heat: mgh = (0.050)(9.81)(0.20) = 0.098 J
Quick check
The force on the magnet is upwards while it moves downwards: a force opposing the motion does not have to reverse it.
Guided practice 2
A slotted plate
Problem
A metal plate swinging between the poles of a magnet stops quickly because of eddy-current braking. Slits are cut across the plate, and it is swung again. How does the braking change?
Predict the braking
Hints
Hint 1
current paths
Trace the closed paths the currents can still follow after cutting the slits.
Show full solution
Braking
Working
Answer
Weaker. The slits break the large loops, so the remaining paths enclose less changing flux and carry smaller currents. Smaller currents give a smaller opposing force, so the plate swings for longer.
Using and reducing eddy currents
Braking. In an eddy-current rail brake, magnets on a train pass just above the rail without touching it. As each part of the rail passes through the field, the flux through paths in the rail changes and eddy currents flow; their field opposes the relative motion and slows the train. The train’s kinetic energy becomes internal energy of the rail, without rubbing contact at the brake. The braking force depends on relative motion: it fades as the train slows and is zero at rest, so a friction brake must finish the stop.
Heating. An induction hob has a coil carrying a rapidly alternating current under the glass. The changing flux induces eddy currents in the base of a metal pan, which heat it directly.
Unwanted heating. The iron core of a transformer sits in a changing field, so eddy currents flow in it and waste energy as heat. The core is made from thin sheets, called laminations, insulated from each other. The sheets lie parallel to the main magnetic flux. The insulation stops currents crossing between sheets, so each current path is small: it encloses less changing flux, has a smaller e.m.f. and carries a much smaller current.
Spot the mistake 3
Do laminations increase heating?
Learner claim
A student says: “Laminations increase the resistance of the current paths, and P = I²R, so they must increase the heating.” For a current path with a fixed induced e.m.f., by what factor does the heating power change if the resistance doubles?
Try this before viewing the solution
Show solution step by step
What is fixed
Method
Identify the quantity that induction fixes.Reason
In this example the induced e.m.f. stays fixed; the current then depends on resistance.Working
Answer
I = E/R halves when R doubles.
Power
Working
Answer
P = E²/R halves. Smaller loops also have a smaller e.m.f., so the heating falls further.
Common mistake
P = I²R predicts more heating only if the current stays the same. Here the current falls, because the e.m.f. is what stays fixed.
Try it yourself 4
Three copper plates
Problem
Each plate is perpendicular to the field, so magnetic flux passes through its face. In which of these do eddy currents flow? (a) A copper plate at rest in a steady magnetic field. (b) A copper plate at rest in a magnetic field whose strength is changing. (c) A copper plate moving into a region of magnetic field.
Decide for each plate
Hints
Hint 1
one test
For each plate, consider both the field strength and the area of the plate inside the field.
Show solution step by step
Steady case
Reason
Nothing is changing.Working
Answer
(a) No eddy currents.
Changing cases
Working
Answer
(b) and (c) both have eddy currents. In (b) the flux changes because the field does, and the supply changing the field provides the energy. In (c) the flux changes because the plate moves into the field; the currents brake it, and the energy comes from its kinetic energy or from whatever pushes it.
Exam-style question 1
Copper and plastic tubes
Examination question
Identical magnets are dropped through a vertical copper tube and through a plastic tube of the same size.
(a) Explain why the magnet in the copper tube falls much more slowly. [3 marks]
(b) The magnet in the copper tube reaches a terminal speed. State the energy changes while it falls at this speed. [2 marks]
Write your answer before viewing the mark scheme
Show solution step by step
(a) Slower fall
3 marksMethod
Link the changing flux to currents, then to a force.Reason
Only a conductor provides closed paths for induced currents.Working
Answer
The moving magnet changes the flux through the tube wall, inducing eddy currents in the copper. By Lenz’s law, these currents produce a magnetic field that opposes the motion, giving an upward force on the magnet. Plastic does not conduct, so there are no currents and no braking force.
(b) Energy
2 marksWorking
Answer
The kinetic energy is constant. The gravitational potential energy lost becomes internal energy (heat) in the copper, via the eddy currents.
Self-mark with the mark scheme
Tick each mark point that your answer includes.
Tick each point only if your answer states it clearly.
Check your understanding 1 Before terminal speed
A 0.050 kg magnet falls 0.20 m inside a copper tube. Its speed rises from 0.80 to 1.2 m s⁻¹. Ignore air resistance and take g = 9.81 m s⁻². Find the energy transferred to the tube as heat during this fall.
Show the steps Optional working
Show answer
The gravitational potential energy lost is mgh = 0.050 × 9.81 × 0.20 = 0.0981 J.
The kinetic energy gained is 1/2 m(v²-u²) = (1/2)(0.050)(1.2²-0.80²) = 0.0200 J.
So the tube gains 0.0981-0.0200 = 0.0781 J as heat. Using mgh alone would count the energy that is still in the magnet’s motion as heat. The heat is positive and smaller than the gravitational energy lost.
Common mistakes
- Expecting eddy currents in a metal that is simply sitting in a steady field; the flux must change.
- Thinking a magnet falls slowly in a copper tube because copper attracts it; copper is not ferromagnetic.
- Thinking laminations reduce eddy currents by stopping the core being magnetised; they break up the current paths.
- Expecting an eddy-current brake to hold a stationary vehicle.
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Syllabus and review details
- GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30
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