University Physics II · Electromagnetic Induction · 10.6
Eddy Currents and Magnetic Braking
Change the flux through a solid conductor and current circulates with no wire to guide it. The same loops that slow a falling magnet cook a transformer core, and the only difference is whether you wanted the heat.
Build the model
Connect the measurement to the mechanism.
Faraday's law does not need a circuit. A changing flux sets up an electric field whose lines close on themselves, and that field exists inside whatever material occupies the region; the local form of Ohm's law, J = σE, turns it into current wherever the metal is continuous. The loops choose themselves — their size is set by the geometry of the conductor and of the field region, not by a wiring diagram.
Two consequences follow. The current dissipates at p = J²/σ, so any bulk conductor in a changing flux heats. And the current sits in the field, so J × B gives a force that opposes whatever caused the change: drag on a moving plate, torque on a spinning disc, lift on a fast one.
Both are paid for by the agent that changes the flux, never by the field itself. Design then splits in two — cut the loops short when the heat is waste, make them large when the force is the point.
- Simple definition
- Eddy currents are the closed loops of current that a changing magnetic flux drives inside a bulk conductor. They dissipate energy as heat and produce a force opposing whatever change created them.
- Example
- Drop a neodymium magnet down a copper pipe and it sinks at a few centimetres per second. Nothing touches it: the current loops it induces in the wall push back, and the lost gravitational energy leaves as heat in the copper.
No wire fixes the path: the conductor's own geometry picks the loops.
E in V m⁻¹, J in A m⁻², σ in S m⁻¹
The field grows outward, so the rim carries the strongest current.
r in m from the axis; B ⟂ to the disc face
I²R per unit volume. The mover's work rate equals the total heating.
p in W m⁻³; F is the drag on the conductor
Copper, R = 40 mm, t = 0.50 mm, B₀ = 50 mT, 50 Hz → 3.7 W.
needs μ₀σωR t / 4 ≪ 1 — the radius matters, not just t ≪ δ
N sheets replacing one slab of the same total thickness: loss/N².
W m⁻³; d is sheet thickness, B lies in the sheet
Linear only while R(m) ≪ 1; past the peak the drag falls as 1/v.
b in N s m⁻¹; C is a geometry factor of order 0.1–1
Faraday's law without a circuit
A wire loop is a convenience, not a requirement. Written as ∮E·dl = −dΦ/dt, Faraday's law says a changing flux produces an electric field whose lines close on themselves, and that field exists in whatever material sits in the region. Put metal there and the local form of Ohm's law, J = σE, converts it into current. Because no wire fixes the path, the current runs on every closed route the conductor offers, weighted towards the low-resistance ones — hence the name. Take a disc of radius R in a uniform field B(t) normal to its face. Symmetry makes the E lines circles about the axis, so a circle of radius r encloses flux πr²B and gives 2πrE = πr²|dB/dt|, or E = (r/2)|dB/dt|. Current density therefore grows linearly outward: the rim carries the strongest loops and the centre almost none. Lenz's law sets the sense — the loops' own field opposes the change that made them.
Heating, and what the loss depends on
Current in a resistive medium dissipates at p = J·E = J²/σ watts per cubic metre, which is I²R written locally. For the disc with B = B₀ sin ωt, E(r) peaks at rωB₀/2, so the time-averaged rate is σr²ω²B₀²/8; integrating over shells 2πr t dr from 0 to R gives P = πσω²B₀²tR⁴/16. Numbers make the scalings concrete. A copper disc (σ = 5.96 × 10⁷ S m⁻¹) of radius 40 mm and thickness 0.50 mm in a 50 mT, 50 Hz field absorbs 3.7 W. Halve the radius and it falls to 0.23 W, a factor of 16, because P ∝ R⁴. Double the frequency and it quadruples. The dependence on σ looks backwards at first: the better the conductor, the more it heats, because the same emf drives proportionally more current. That holds only while the disc's own field barely opposes the applied one, which for a disc means μ₀σωR t / 4 ≪ 1 — about 0.12 here — and not merely that the metal is thinner than the skin depth δ = √(2/(μ₀σω)), 9.2 mm for copper at 50 Hz. Scale the radius up at fixed thickness and the screening grows while δ does not move. Thicker than that and the current crowds into a surface layer, and the loss instead falls roughly as 1/√σ.
Cutting the loops: lamination
When the heat is waste, attack the loop area. A transformer or motor core carries B along the iron, so slicing it into thin sheets laid parallel to B leaves the magnetic path untouched while confining every eddy loop to one sheet. Inside a sheet of thickness d the induced field is E = x dB/dt with x measured from the mid-plane; averaging x² across the thickness gives p = σω²B₀²d²/24 = π²σf²B₀²d²/6 per unit volume. The dependence is d², so N sheets in place of one slab of the same total thickness cut the loss by N², not by N. Grain-oriented silicon steel, σ ≈ 2.0 × 10⁶ S m⁻¹, at 1.5 T and 50 Hz with d = 0.35 mm gives about 2.3 kW m⁻³, near 0.30 W kg⁻¹. The varnish between sheets does not make the steel more resistive — the steel is unchanged — it only stops loops crossing from sheet to sheet. The rule needs d below the skin depth, about 0.80 mm for that steel at 50 Hz with μ(r) ≈ 4000; above a few kilohertz designers move to ferrites and powder cores, where σ itself is tiny.
Drag, damping, and why it is not friction
Move a conductor through a non-uniform field, or sweep the field past the conductor, and the same currents appear — but now they sit in B, so the metal feels F = ∫ J × B dV, directed against the relative motion by Lenz's law. At low speed J ∝ v, so the drag is linear: F = −bv, with b ≈ Cσwa²B² for a sheet of thickness w and a field region of size a, C a geometry factor of order 0.1 to 1. Linear drag behaves nothing like Coulomb friction. Newton's second law gives m dv/dt = −bv, so v = v₀ exp(−t/τ) with τ = m/b: the speed decays exponentially and never reaches zero. The distance is finite, v₀τ, but the time is not, which is why a lorry's eddy-current retarder or a roller-coaster fin brake always sits alongside a friction brake that can hold at rest. Read the other way, the same linearity is a gift. A copper vane in a magnet gap hands an oscillator a clean −bv term with no stiction and no wear, and instrument designers choose b = 2√(km) to bring a balance or a galvanometer needle to rest in the shortest time without overshoot.
Where the energy actually comes from
Magnetic forces do no work on charges: qv × B is always perpendicular to v. The field therefore cannot be the source of the heat that appears in a braked disc; it is a broker. Whatever agent changes the flux — the hand pushing the plate, gravity pulling the magnet, the primary winding driving the core — does work against the induced emf, and that work reappears as J²/σ dissipation in the conductor. The bookkeeping is exact: for a conductor dragged at steady speed, Fv = ∫(J²/σ) dV, with nothing stored once the transient has passed. That equality fixes the design limits. A 1200 kg vehicle at 30 m s⁻¹ carries 540 kJ of kinetic energy; dumped into a 10 kg copper rotor with c = 385 J kg⁻¹ K⁻¹, that is a 140 K temperature rise, so eddy brakes are rated by thermal capacity and cooling rather than by peak force. It also explains their fade: σ drops as the metal warms, and b drops with it.
Fast motion, flux exclusion, and levitation
The linear law assumes the induced currents' own field is negligible beside the applied one. The test is the magnetic Reynolds number R(m) = μ₀σvL. For copper μ₀σ = 74.9 s m⁻², so a magnet drifting at 0.05 m s⁻¹ past a 20 mm feature has R(m) = 0.075 — safely linear. A vehicle at 100 m s⁻¹ over aluminium (μ₀σ = 44.0 s m⁻²) with L = 0.3 m has R(m) ≈ 1.3 × 10³, a different regime: the induced currents largely exclude the field from the conductor, much as a mirror image would. There the drag rises with v, peaks near R(m) ≈ 1, then falls as 1/v, while the repulsive lift climbs to a plateau, so the lift-to-drag ratio grows roughly as R(m) and an electrodynamic suspension gets more efficient the faster it runs. The static case still fails: a magnet cannot hover over ordinary copper, because with no motion there is no dΦ/dt and the loops die away in milliseconds. Levitation needs relative motion, an AC drive whose induced current lags by nearly half a cycle, or a superconductor.
Change one variable at a time
Make the relationship visible.
Start at v = 0.1 m s⁻¹, where the dot sits on the dashed straight line, then drag v upwards: past the R(m) = 1 marker the drag turns over and shrinks as the sheet moves faster. Widen the field region L and the whole peak slides left, so a slow magnet over a broad pole face is already out of the linear regime.
MAGNETIC REYNOLDS R(m)0.75
DRAG ÷ PEAK DRAG0.960 ×
LINEAR LAW WOULD SAY1.498 ×
SPEED AT PEAK DRAG0.67 m s⁻¹
Live interpretationMAGNETIC REYNOLDS R(m): 0.75. DRAG ÷ PEAK DRAG: 0.960 ×. LINEAR LAW WOULD SAY: 1.498 ×. SPEED AT PEAK DRAG: 0.67 m s⁻¹
Catch the common trap
Explain before calculating.
A transformer core of thickness D is replaced by N insulated laminations of thickness d = D/N, stacked parallel to B. Total cross-section, peak flux density and frequency are unchanged, and every sheet stays thinner than the skin depth. What happens to the eddy-current loss?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA transformer core of grain-oriented silicon steel (σ = 2.0 × 10⁶ S m⁻¹) runs at 50 Hz with a peak flux density of 1.2 T. Find the eddy-current loss per cubic metre for 0.50 mm laminations, then for 0.25 mm ones.
- B lies in the plane of each sheet, so the lamination law applies: p = π²σf²B₀²d²/6. It holds while d stays under the skin depth, about 0.80 mm for this steel at 50 Hz, so both thicknesses qualify.
- Gather everything except d: π²σf²B₀²/6 = 9.8696 × 2.0 × 10⁶ × 50² × 1.2² ÷ 6 = 1.184 × 10¹⁰ W m⁻⁵.
- For d = 0.50 mm = 5.0 × 10⁻⁴ m, d² = 2.5 × 10⁻⁷ m², so p = 1.184 × 10¹⁰ × 2.5 × 10⁻⁷ = 2.96 × 10³ W m⁻³.
- Halving d to 2.5 × 10⁻⁴ m quarters d² to 6.25 × 10⁻⁸ m², so p = 1.184 × 10¹⁰ × 6.25 × 10⁻⁸ = 7.40 × 10² W m⁻³.
Answerp = 3.0 × 10³ W m⁻³ at 0.50 mm and 7.4 × 10² W m⁻³ at 0.25 mm. The d² is what makes lamination worth the trouble: ten sheets in place of one slab cut the loss a hundredfold, not tenfold.
MediumA copper disc (σ = 5.96 × 10⁷ S m⁻¹), radius 60 mm and thickness 0.40 mm, lies face-on in a uniform field B = B₀ sin ωt with B₀ = 30 mT at 50 Hz. Find the peak induced electric field at the rim, the mean power dissipated, and what is left if the disc is trimmed to 40 mm radius.
- Faraday round a circle of radius r inside the disc: 2πrE = πr²|dB/dt|, so E(r) = (r/2)|dB/dt| — zero on the axis, largest at the rim.
- |dB/dt| peaks at ωB₀ = 2π(50)(0.030) = 9.42 T s⁻¹, so E(R) = (0.060/2)(9.42) = 0.283 V m⁻¹ and J(R) = σE = 1.69 × 10⁷ A m⁻².
- Check the thin-metal assumption before integrating: δ = √(2/(μ₀σω)) = 9.2 mm for copper at 50 Hz, and t = 0.40 mm is far under it, so J is uniform through the thickness.
- Time-averaging p = σE² gives σr²ω²B₀²/8, and integrating over shells 2πrt dr from 0 to R gives P = πσω²B₀²tR⁴/16.
- Substitute step by step: σω² = 5.96 × 10⁷ × 9.870 × 10⁴ = 5.882 × 10¹²; × B₀² = 9.0 × 10⁻⁴ → 5.294 × 10⁹; × t = 4.0 × 10⁻⁴ → 2.118 × 10⁶; × R⁴ = 1.296 × 10⁻⁵ → 27.44; × π ÷ 16 = 5.39 W.
- Only R changes on trimming, and P ∝ R⁴: P' = 5.39 × (40/60)⁴ = 5.39 × 0.1975 = 1.06 W.
AnswerE = 0.28 V m⁻¹ at the rim, P = 5.4 W, falling to 1.1 W at 40 mm radius. Two thirds of the radius leaves a fifth of the heat, because the rim carries almost all of it.
HardA 0.60 kg carriage runs a copper fin through a magnet gap. At 0.20 m s⁻¹ the fin feels 0.30 N of retarding force; treat the drag as F = −bv. Find b and the time constant, how far the carriage coasts from 0.20 m s⁻¹, how long it takes to fall to 1% of that speed, and where the kinetic energy ends up. Then confirm the linear law is allowed, with a 20 mm pole face and μ₀σ = 74.9 s m⁻² for copper.
- One force at one speed fixes the constant: b = F/v = 0.30/0.20 = 1.5 N s m⁻¹.
- Check the regime before trusting F ∝ v: R(m) = μ₀σvL = 74.9 × 0.20 × 0.020 = 0.30, comfortably under 1, so the induced currents' own field is a small correction and J stays proportional to v.
- m dv/dt = −bv integrates to v = v₀e(−t/τ) with τ = m/b = 0.60/1.5 = 0.40 s.
- The coast is ∫v dt = v₀τ = 0.20 × 0.40 = 0.080 m — finite, even though v never reaches zero; falling to 1% of v₀ takes t = τ ln 100 = 0.40 × 4.605 = 1.8 s.
- Magnetic forces do no work on charges, so every joule the carriage loses reappears as J²/σ heating in the fin: Q = ½mv₀² = ½(0.60)(0.20)² = 0.012 J.
Answerb = 1.5 N s m⁻¹, τ = 0.40 s, a coast of 80 mm, 1.8 s to reach 1% of v₀, and 12 mJ of heat in the copper. Exponential decay means the carriage always creeps, so an eddy brake that must hold a load at rest needs a friction brake beside it.