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Drop a strong magnet down a copper pipe and it drifts down slowly, as if through honey, even though copper is not magnetic. The cause is eddy currents: induced currents that swirl inside solid metal. This post explains their cause, why they brake and heat, how engineers reduce them, and which exams include them, with two worked problems.
Which exams include eddy currents?
The JEE (Main) 2026 syllabus lists, under Electromagnetic Induction, "Faraday's law, induced emf and current, Lenz's law, eddy currents, self and mutual inductance". The JEE (Advanced) 2026 brochure lists "Faraday's law, Lenz's law" without naming eddy currents, though problems built on those laws can still involve them.
For CBSE students, the Physics curriculum for 2026-27 lists Faraday's laws, Lenz's law and self and mutual induction in Chapter 6, and does not name eddy currents. The current NCERT Electromagnetic Induction chapter has no separate section on them. They appear in the NCERT Alternating Current chapter as one of the energy losses in a transformer. Check your own exam's syllabus for the current year.
What causes an eddy current?
Faraday's law says a changing magnetic flux through a loop induces an emf, e = −dΦ/dt. A solid sheet of metal is like a huge number of loops packed together. When the flux through the sheet changes, because a magnet moves near it, the sheet moves through a field, or the field itself changes, an emf appears around every small loop in the metal. Since the metal conducts, currents flow in closed swirls. Those are eddy currents.
Lenz's law gives their direction: they flow so as to oppose the change in flux that made them. Our post on Lenz's law induction problems gives a step-by-step method for finding that direction.
Why do they brake and damp motion?
The NCERT chapter explains that the induced current must oppose the change; otherwise a magnet pushed towards a coil would speed up by itself and create energy from nothing. The work you do against that opposition is "dissipated by Joule heating produced by the induced current".
Physics teachers have measured this carefully. Hahn and colleagues (1998) drove a magnet on a spring through pipes and found the damping matched a model combining the damped-oscillator equation with Faraday's law, giving "strong evidence" for the eddy-current explanation of the magnet-down-a-pipe demonstration. Muiznieks and Dudareva (2012) suspended a magnet above aluminium foil and found the energy lost per oscillation rose linearly with the number of foil layers: more metal, more eddy current, more damping.
How do engineers reduce eddy currents?
Break up the paths the currents can flow along.
Laminated cores. NCERT's Alternating Current chapter lists eddy currents among transformer losses: "The alternating magnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by using a laminated core." Thin insulated sheets force the current into narrow strips, cutting the loops short.
Slots and slits. Zengel and colleagues (2025) note that experiments show the drag on a magnet falling through a copper pipe is smaller when the pipe has vertical slits, because the slits remove metal and block the horizontal current loops.
Worked example 1: terminal speed of a falling loop
A square loop of side l = 0.10 m, mass m = 0.020 kg and resistance R = 0.10 Ω falls vertically out of a horizontal magnetic field B = 0.50 T, with its top edge still inside the field and the plane of the loop perpendicular to B. Find its terminal speed. Take g = 9.8 m/s².
Motional emf in the top edge: e = Blv.
Current: I = Blv/R.
Magnetic force on the top edge, opposing the fall: F = BIl = B²l²v/R.
Terminal speed when F = mg: v = mgR/(B²l²).
v = (0.020 × 9.8 × 0.10) / (0.25 × 0.010) = 0.0196 / 0.0025 = 7.84 m/s.
Doubling B cuts the terminal speed to a quarter, because B appears squared.
Worked example 2: heating a thin disc
A thin copper disc of radius a = 5.0 cm and thickness d = 1.0 mm lies flat in a uniform field perpendicular to it, which rises at dB/dt = 0.10 T/s. Taking the conductivity given in the problem as σ = 6.0 × 10⁷ S/m, find the power lost as heat.
Split the disc into thin rings of radius r and width dr.
Emf round a ring: e = πr² (dB/dt).
Resistance of the ring: length 2πr, cross-section d·dr, so R = 2πr / (σ d dr).
Power in the ring: e²/R = (π/2) σ d (dB/dt)² r³ dr.
Add the rings from 0 to a: P = (π/8) σ d (dB/dt)² a⁴.
Numbers: P = (π/8) × 6.0 × 10⁷ × 1.0 × 10⁻³ × 0.010 × (0.050)⁴. Since (0.050)⁴ = 6.25 × 10⁻⁶, P ≈ 1.5 × 10⁻³ W, about 1.5 mW.
The a⁴ term is the lesson. A disc twice as wide loses 16 times the power, which is why cutting metal into narrow laminations works so well.
Which mistakes cost marks?
Saying copper is attracted to magnets. The pipe slows the magnet through induced currents, not magnetism.
Forgetting Lenz's direction. Eddy currents always oppose the change, which is why they brake.
Thinking laminations stop flux. They block current paths, not the field.
A practice routine
Day 1: Explain the magnet-in-pipe demonstration in five lines using Faraday's and Lenz's laws.
Day 2: Redo worked example 1 with B = 1.0 T and check that v falls to a quarter.
Day 3: Redo worked example 2 for a disc of radius 10 cm and confirm the factor of 16.
Day 4: Write why transformer cores are laminated, then check it against NCERT's text.
Day 5: Mix five induction problems and solve them without notes. Self-testing and spaced practice were the two techniques rated most useful in a review of study methods by Dunlosky and colleagues (2013).
For the history behind the law, read about Faraday and field lines.
If you cannot picture the swirling currents, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Find a magnet and an aluminium or copper tube this week, time a drop through it, then compare with a drop beside it.
