Motional EMF

Electromotive Force closed with a catalog of pumps — chemical, thermal, photonic — and a promissory note: the grandest pumps are made of nothing but motion and magnetic field. This article cashes the first half of that note. Drag a conducting rod through a static magnetic field and every mobile charge inside it feels the magnetic force $q\mathbf{v}\times\mathbf{B}$ — a force per unit charge, $\mathbf{v}\times\mathbf{B}$ , that is exactly the $\mathbf{f}_s$ of the previous article's definition. Its round-trip circulation $\varepsilon = \oint(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}$ is the motional EMF: the pump that turns pulling into current, the physics of every generator that has ever fed a grid. The flagship is the rod-on-rails machine, run end-to-end below: EMF $\varepsilon = BLv$ , current, retarding force, and the power balance $Fv = \varepsilon I = I^2R$ — exact, not approximate — which is the generator theorem: mechanical work in, electrical heat out, the field taking no cut. That exactness forces the article's central paradox. Magnetic forces do no work — the audit Lorentz Force Law performed stands unrepealed — so who pays for the heat? The resolution, a two-component decomposition worth doing with care, shows the magnetic force acting as a frictionless crank arm: it redirects the puller's work into the circuit, contributing nothing and stealing nothing. Two theorems finish the story. The swept-ribbon argument proves that for any loop moving or deforming through a static field, $\varepsilon = -d\Phi/dt$ — the flux rule, derived for the motional case rather than postulated. And Faraday's spinning disk — a machine the naive flux rule stumbles on, solved cleanly by the line integral — stands as the warning that the rule is a theorem with premises, whose autopsy belongs to The Flux Rule and Its Exceptions. What this article deliberately does not touch: EMF from a changing field threading a stationary circuit. That is new physics, not motion — the next article's revolution.

Key takeaways

A conductor moving through a static magnetic field carries a built-in pump: force per unit charge $\mathbf{f}_s = \mathbf{v}\times\mathbf{B}$ , giving the motional EMF $\varepsilon = \oint(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}$ — the magnetic species of Electromotive Force's catalog.
The rail generator end-to-end: $\varepsilon = BLv$ , current $I = BLv/R$ , retarding force $F = B^2L^2v/R$ , and the exact balance $Fv = \varepsilon I = I^2R$ : every joule of heat is a joule the puller supplied. Generators convert; they do not create.
The work paradox resolved: magnetic forces do no work (Lorentz Force Law's audit, spent here). Decompose each carrier's velocity into rod motion $\mathbf{v}$ and drift $\mathbf{u}$ : the $\mathbf{v}$ -component of the force drives the current, the $\mathbf{u}$ -component resists the pull — and their works cancel exactly. The magnet is a crank arm; the puller pays.
The flux rule proved for motion: for any loop moving or deforming in a static field, the swept-ribbon argument gives $\varepsilon = -d\Phi/dt$ — a theorem here, with $\nabla\cdot\mathbf{B} = 0$ as its engine.
Faraday's disk ( $\varepsilon = \tfrac12\omega BR^2$ ) is solved by the line integral in three lines but embarrasses the naive flux rule — the autopsy is owned by The Flux Rule and Its Exceptions.
Sign reading of the induced current (opposing the change) is Lenz's territory, next-but-one; the changing-field half of induction is Faraday's Law of Induction, next — zero overlap: here the field is static and the circuit moves.
KEY IDEA: motional EMF is the magnetic force moonlighting as a pump — $\varepsilon = \oint(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell} = -d\Phi/dt$ for moving circuits — converting mechanical work into electrical energy exactly, with the field as an unpaid intermediary.

The core insight

Motional EMF is the circulation of $\mathbf{v}\times\mathbf{B}$ around a moving circuit: it equals $-d\Phi/dt$ for any loop moving or deforming through a static magnetic field, it converts mechanical to electrical power exactly ( $Fv = \varepsilon I$ ), and the magnetic force that produces it does no net work — the mover pays for every joule.

Where this lives today

Motional EMF is the grid, full stop — synchronous generators converting turbine torque at 98%+ efficiency are the rail derivation wound into rotating geometry — but its 2026 growth areas are at the extremes. Small: energy-harvesting for sensor nodes (vibrating magnets in coils powering bridge-strain monitors), regenerative-braking arithmetic in every EV (the drag force $B^2L^2v/R$ is the brake, with $R$ replaced by battery charging impedance — recovering the crank-arm-redirected watts instead of heating discs). Large: superconducting homopolar motors for ship propulsion revive Faraday's disk at megawatt scale, exploiting exactly the low-voltage/huge-current character worked example 2 exposed. And at planetary scale, the physics runs itself: conducting seawater through Earth's field generates measurable EMFs across ocean channels (the Dover–Calais cable monitored tidal flow this way for decades), and Jupiter's moon Io, sweeping through the Jovian field, drives a five-million-amp circuit along flux tubes to the planet's poles — a rail generator with a moon for the rod.

Origins · 1 minJoseph Henry and how this was discovered

Joseph Henry

1797–1878 · Albany, Princeton, Washington

Henry built the machines that made motional EMF discoverable, and met induction from the other side of the Atlantic divide. An Albany schoolteacher who electromagnetized iron better than anyone alive — his insulated multi-turn windings lifted a record 750 pounds by 1831 — he observed self-induction's spark and mutual induction's kick in 1831–32, independently of Faraday and essentially simultaneously; publication delays cost him formal priority for induction but won him the SI unit of inductance. His electromagnets and relays begat the practical telegraph (Morse leaned on his advice and his relay concept), and his powerful field magnets were exactly the hardware early generator builders spun their disks and coils inside. As founding Secretary of the Smithsonian Institution he built American science's first national infrastructure — weather telegraphy that seeded the Weather Service, research grants, publication series. In this article's ledger he stands for the engineering half of induction's discovery: Faraday asked what the field does; Henry built apparatus strong enough to make it answer.

History

1821

— Faraday's rotation experiments make a current-carrying wire circle a magnet: the motor, before the generator

1831

— Faraday spins a copper disk between magnet poles and draws continuous current: the homopolar generator

1831

— Henry's giant electromagnets and induction observations supply the machine shop of early electromagnetism

1832

— Pixii's rotating-magnet machine delivers alternating current from a coil, the alternator's ancestor

1833

— Lenz states the opposition rule for the induced current's direction

1867

— Siemens's self-excited dynamo bootstraps its own field: generators leave the laboratory

1873

— Gramme's machine runs backwards as a motor at the Vienna exhibition: conversion works both ways

The physical picture

Take a copper rod, lay it across two parallel conducting rails connected through a resistor, thread a uniform magnetic field up out of the table, and pull the rod along the rails. Every conduction electron in the rod is being carried sideways through the field at the rod's velocity $\mathbf{v}$ , so each feels $q\mathbf{v}\times\mathbf{B}$ — directed along the rod. The magnetic force has become, for the charges trapped in the moving conductor, exactly what the chemical force was for charges in a battery: a non-electrostatic push per unit charge, localized in one branch of the circuit, driving circulation. The rod is a battery whose electrolyte is motion. Current flows; the resistor warms. Now the field presents its second face: the rod carries current $I$ across the field $B$ , so it feels $F = BIL$ — and the geometry conspires so that this force always opposes the pull. (Always: reverse $\mathbf{v}$ , and $\varepsilon$ , $I$ , and the force all reverse together. The direction-reading — induced effects oppose their cause — is Lenz's law, previewed here, owned by Lenz's Law two doors down; here energy conservation is diagnosis enough, since a force helping the pull would make a runaway free-energy machine.) To keep the rod moving you must pull with $Fv$ watts of honest mechanical power, and the arithmetic below shows the resistor receives exactly that. Nothing is free; the machine is a converter. This little rig, scaled and rotated, is the electrical grid. A turbine blade is the pulled rod; the stator's field is the tabletop's $\mathbf{B}$ ; the retarding force is the load torque the steam must overcome. When demand rises on the grid, more current flows in the stators, the retarding torque on every generator rises within milliseconds, and governors admit more steam — the physics of this article, at gigawatt scale, sixty times a second. The engineering superstructure (windings, commutation, synchronous operation) belongs to the power track; the conversion theorem is here.

Figure

The rail generator - a rod pulled through a static field is a battery whose electrolyte is motion

The pump, formalized

Continuity with the last article is the whole point, so state it as a definition-by-substitution. For charges riding a conductor whose element at position $\mathbf{r}$ moves with velocity $\mathbf{v}(\mathbf{r})$ , the magnetic force per unit charge is $\mathbf{v}\times\mathbf{B}$ ; feeding it into Electromotive Force's definition $\varepsilon = \oint\mathbf{f}_s\cdot d\boldsymbol{\ell}$ :

\varepsilon = \oint\left(\mathbf{v}\times\mathbf{B}\right)\cdot d\boldsymbol{\ell} \qquad (1)

where: $\mathbf{v}$ — the velocity of the circuit element (not of the drifting carriers; that distinction is the paradox section's business); the integral runs around the circuit; only segments that both move and sit in the field contribute. Equation (1) is not new dynamics — it is the Lorentz force (Lorentz Force Law, recalled) read through the EMF definition. Its species name, motional EMF, marks the premise: static $\mathbf{B}$ , moving conductor. When the field itself changes in time, nothing in this equation responds — that gap in the account is precisely what Faraday's Law of Induction will fill with genuinely new physics.

Derivation AThe rail generator end-to-end - EMF, current, drag, and the exact power balance

The goal is the complete energy audit of the flagship machine: EMF $BLv$ , the induced current, the retarding force, and the theorem $Fv = \varepsilon I = I^2R$ exactly — mechanical power in equals electrical power dissipated. The strategy is equation (1) on the simplest possible geometry, then the magnetic force on the current-carrying rod, then multiplication. Watch the exactness claim: no approximation enters anywhere, which is why the result deserves the name generator theorem.

Step 1. The EMF. Rod of length $L$ along $\hat{\mathbf{y}}$ , moving at $\mathbf{v} = v\hat{\mathbf{x}}$ , field $\mathbf{B} = B\hat{\mathbf{z}}$ . In the rod, $\mathbf{v}\times\mathbf{B} = vB\,(\hat{\mathbf{x}}\times\hat{\mathbf{z}}) = -vB\,\hat{\mathbf{y}}$ — uniform, along the rod. The rails and resistor are stationary ( $\mathbf{v} = 0$ there), so the circulation collapses to the rod segment:

\varepsilon = \oint(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell} = B L v

Step 2. The current, by the loop theorem of Electromotive Force ( $\varepsilon = IR$ , total loop resistance $R$ ):

I = \frac{BLv}{R}

Step 3. The drag. The rod now carries $I$ across $\mathbf{B}$ ; the force on a current-carrying segment (Magnetic Force on Current-Carrying Wires, recalled) is $\mathbf{F} = I\mathbf{L}\times\mathbf{B}$ , magnitude $BIL$ , directed against $\mathbf{v}$ :

F = BIL = \frac{B^2L^2}{R}\,v

Why the minus-v direction is guaranteed: run the sign through and the drag opposes the motion for either polarity of B or v — as energy conservation demands, since an assisting force would amplify its own cause without limit. This opposition-reading generalizes to Lenz's law, owned two articles on.

Step 4. The balance. Pulling at constant $v$ against the drag requires mechanical power $P_{\rm in} = Fv = B^2L^2v^2/R$ . The circuit dissipates $P_{\rm out} = I^2R = (BLv)^2/R$ . They are identical:

Fv = \varepsilon I = I^2R \qquad (2)

where: every joule the resistor radiates was supplied by the hand on the rod, routed through the field without loss or gain. The velocity-proportional drag also means an unpowered rod decays exponentially, $v(t) = v_0e^{-t/\tau}$ with $\tau = mR/B^2L^2$ — magnetic braking, whose bulk-conductor version is gestured here and owned by Eddy Currents at the chapter's end. Verified: numbers of worked example 1 (0.5 T, 0.2 m, 3 m/s, 0.5 Ω) give 0.30 V, 0.60 A, 0.060 N, 0.18 W by BOTH routes of equation (2); direction audit consistent with Lenz reading; the exactness matches Griffiths §7.1.3's discussion — the magnetic field is an ideal, non-dissipative transmission.

Who does the work? The paradox, resolved

Equation (2) says mechanical power becomes electrical power. But Lorentz Force Law proved — and this track has spent repeatedly — that magnetic forces do no work: $\mathbf{F} = q\mathbf{v}\times\mathbf{B}$ is always perpendicular to $\mathbf{v}$ . Yet here is a magnetic force $BIL$ , apparently eating the puller's watts, and a magnetic pump $\mathbf{v}\times\mathbf{B}$ , apparently feeding them to the resistor. Contradiction?

Go deeper · Derivation B · 2 minThe crank-arm theorem - the magnetic force redirects the work and does none itself

Derivation BThe crank-arm theorem - the magnetic force redirects the work and does none itself

The goal is the exact resolution of the work paradox: show that the magnetic force on each carrier does zero net work, while decomposing into two components whose separate works are equal and opposite — one driving the current, one resisting the pull — so that the puller's mechanical work is redirected into the circuit with the field as intermediary. The strategy is to use the carrier's true velocity, which is neither the rod's $\mathbf{v}$ nor the drift $\mathbf{u}$ alone but their sum. Watch the geometry: everything follows from the cross product's perpendicularity, applied honestly to the total velocity.

Step 1. A carrier in the rod moves with total velocity $\mathbf{w} = \mathbf{v} + \mathbf{u}$ : carried sideways at the rod speed $v$ ( $\hat{\mathbf{x}}$ ) and drifting along the rod at $u$ ( $-\hat{\mathbf{y}}$ , the current direction for positive carriers). The magnetic force is $q\mathbf{w}\times\mathbf{B}$ , and its work rate is $q(\mathbf{w}\times\mathbf{B})\cdot\mathbf{w} = 0$ identically — the audit stands.

Step 2. Decompose the force by source:

q\mathbf{w}\times\mathbf{B} = \underbrace{q\mathbf{v}\times\mathbf{B}}_{\text{from being carried}} + \underbrace{q\mathbf{u}\times\mathbf{B}}_{\text{from drifting}}

The first term has magnitude $qvB$ , directed along the rod: this is the pump, driving the drift. The second has magnitude $quB$ , directed against the rod's motion ( $-\hat{\mathbf{x}}$ ): summed over all carriers it is exactly the drag $BIL$ of the rail derivation.

Step 3. Audit each component's work rate. The pump component works on the drift motion: $q(vB)(u)$ per carrier, positive — it feeds the circuit. The drag component works on the carried motion: $q(uB)(-v)$ per carrier, negative — it fights the rod. Sum:

quvB - quvB = 0

Exactly zero, carrier by carrier, instant by instant. The two books balance identically because both entries are the same cross product read against different velocity components.

Step 4. Find the payer. The drag component's negative work on the carriers would decelerate the rod — unless the puller supplies $+BILv$ through the lattice (the carriers transmit the magnetic drag to the rod body via collisions and the same internal fields that hold them in the wire). So the energy path is: puller → rod lattice → carriers (via the constraint forces) → pushed along the wire by the pump component → dissipated in $R$ . The magnetic field's role: it takes the puller's sideways watts and hands them out lengthwise, taking nothing. A crank arm — rigid, frictionless, unpaid. Verified: the decomposition reproduces Griffiths §7.1.3's Fig. 7.12 analysis and the work-paradox resolution of Lorentz Force Law recalled by name; the zero is exact (both terms are quvB); consistency check — summing the drag component over the carrier density recovers F = BIL, and the pump component's total work rate equals εI, matching equation (2) both ways.

The moral generalizes far beyond rails: whenever a magnetic force appears to do work — lifting a bar magnet, spinning a motor — a careful decomposition finds the true payer (whoever maintains the currents or the motion) with the field as an exchange medium. This track's standing phrase: magnetic forces are brokers, not benefactors.

The flux rule, proved for motion

The rail machine's EMF, $BLv$ , happens to equal the rate at which the circuit's enclosed flux shrinks: the loop loses area $Lv$ per second, hence flux $BLv$ per second. Coincidence of the simple geometry? No — theorem.

Go deeper · Derivation C · 1 minThe swept-ribbon argument - EMF equals minus dPhi/dt for any moving loop in a static field

Derivation CThe swept-ribbon argument - EMF equals minus dPhi/dt for any moving loop in a static field

The goal is the motional flux rule: for a circuit of any shape, moving and deforming arbitrarily through a static but arbitrarily nonuniform field $\mathbf{B}(\mathbf{r})$ , the motional EMF of equation (1) equals $-d\Phi/dt$ . The strategy is geometric bookkeeping: compare the flux through the loop at two nearby instants using the closed surface formed by the two loop positions plus the ribbon swept between them, and let $\nabla\cdot\mathbf{B} = 0$ do the accounting. Watch where solenoidality enters — the theorem is a gift of the no-monopole law (Divergence of B and Absence of Magnetic Monopoles, spent here).

Step 1. Let the loop occupy curve $C(t)$ at time $t$ and $C(t+dt)$ at $t+dt$ ; each element $d\boldsymbol{\ell}$ moves with velocity $\mathbf{v}(\mathbf{r})$ (which may vary around the loop — deformation included). The two loop positions, capped by any surfaces $S(t)$ and $S(t+dt)$ , together with the narrow ribbon traced by the moving curve, form a closed surface.

Step 2. Apply $\oint_{\rm closed}\mathbf{B}\cdot d\mathbf{a} = 0$ (no monopoles): flux in through $S(t)$ plus flux in through the ribbon equals flux out through $S(t+dt)$ . Hence the change is exactly the ribbon's flux:

d\Phi = \Phi(t+dt) - \Phi(t) = \Phi_{\rm ribbon}

Step 3. Parametrize the ribbon. The element $d\boldsymbol{\ell}$ sweeps the parallelogram with sides $\mathbf{v}\,dt$ and $d\boldsymbol{\ell}$ : area element $d\mathbf{a} = (\mathbf{v}\,dt)\times d\boldsymbol{\ell}$ (orientation consistent with the outward closure). So

d\Phi = \oint_C \mathbf{B}\cdot\left[(\mathbf{v}\,dt)\times d\boldsymbol{\ell}\right] = dt\oint_C \mathbf{B}\cdot(\mathbf{v}\times d\boldsymbol{\ell})

Step 4. Cycle the scalar triple product, $\mathbf{B}\cdot(\mathbf{v}\times d\boldsymbol{\ell}) = -(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}$ :

\frac{d\Phi}{dt} = -\oint_C(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell} = -\,\varepsilon \qquad\Longrightarrow\qquad \varepsilon = -\frac{d\Phi}{dt} \qquad (3)

where: $\Phi = \int\mathbf{B}\cdot d\mathbf{a}$ through any surface spanning the loop (any: that freedom is again $\nabla\cdot\mathbf{B} = 0$ ). Equation (3) is, for the motional case, a derived theorem: geometry plus the Lorentz force plus no-monopoles. Nothing about induction was assumed. Verified: reproduces Griffiths §7.1.3's general proof; the rail machine checks instantly (area shrinks at Lv, ε = BLv ✓); premises audited — the derivation used a well-defined material loop whose elements' velocities are unambiguous, and machines that break that premise (sliding contacts, exchanged circuit paths) are exactly the flux-rule pathologies owned by The Flux Rule and Its Exceptions.

The theorem's power is computational: for moving machinery, differentiate the flux instead of chasing $\mathbf{v}\times\mathbf{B}$ around the circuit — one scalar function of time instead of a vector line integral. Its danger is scope creep, and the classic trap is next.

The disk that stumps the rule

Faraday's first generator (1831) was not a rod on rails but a copper disk spinning in an axial field, current drawn between the axle and a brush on the rim. Solve it with the honest line integral: a charge at radius $r$ rides at $\mathbf{v} = \omega r\,\hat{\boldsymbol{\varphi}}$ , feels $(\mathbf{v}\times\mathbf{B})\cdot\hat{\mathbf{r}} = \omega rB$ , and the radial path from axle to rim collects

\varepsilon = \int_0^R \omega B\,r\,dr = \tfrac12\,\omega B R^2 \qquad (4)

where: $\omega$ — angular speed; $R$ — disk radius. For 0.1 T, $R = 0.1$ m, $\omega = 300$ rad/s: $\varepsilon = 0.15$ V — real, brush-measurable, and the ancestor of every dynamo. Now try the flux rule naively: pick the obvious circuit (axle → radius → rim → brush → external wire → axle)... and the enclosed flux is constant — the disk looks the same at every instant, $d\Phi/dt = 0$ , yet the EMF is patently nonzero. The rule has not failed; the premise has: "the circuit" through a solid rotating conductor with sliding contacts is not a well-defined material loop of the swept-ribbon derivation. Which loop, whose velocity, what counts as sweeping — the full autopsy, with this homopolar generator as exhibit A, belongs to The Flux Rule and Its Exceptions. This article's verdict is only the safe rule: the line integral $\oint(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}$ never lies; the flux rule is its shadow, faithful when the loop is honest.

Figure

Faraday's disk - solved by the line integral, embarrassing to the naive flux rule

Limiting cases

Case	Result	Meaning
stationary circuit, static field	$\varepsilon = 0$	no motion, no pump — and no induction either; the quiet baseline
rod ⊥ field, speed $v$	$\varepsilon = BLv$	the flagship; linear in every factor
$R \to \infty$ (open circuit)	charge separation, no current	$\mathbf{v}\times\mathbf{B}$ piles charge at the rod ends until the electrostatic field balances it: a battery on open circuit, the standstill of Electromotive Force
$R \to 0$ (short)	$I \to BLv/R$ grows, drag $\propto 1/R$	the rod becomes immovable — magnetic braking's ideal limit
unpowered rod	$v = v_0e^{-t/\tau}$ , $\tau = mR/B^2L^2$	exponential magnetic braking; bulk version → Eddy Currents
rotating disk	$\varepsilon = \tfrac12\omega BR^2$	line integral succeeds where the naive flux rule stalls → The Flux Rule and Its Exceptions
field changes, circuit still	equation (1) gives zero	motional physics is silent; the missing EMF is Faraday's Law of Induction — new physics, next

Common trapThe flux rule is a theorem with premises - and the moving-charge force still does no work

Two executions, both scheduled above but worth a clean statement. First: $\varepsilon = -d\Phi/dt$ was derived here for a well-defined material loop moving through a static field. Machines with sliding contacts, solid rotating conductors, or switch-swapped paths (homopolar disk, Faraday's paradoxes) can defeat the premise while $\oint(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}$ remains flawless — when the two disagree, trust the line integral and send the corpse to The Flux Rule and Its Exceptions. Second: no part of this article repealed the no-work theorem. The phrase "the magnetic force drives the current" is shorthand for the crank-arm mechanism — the $\mathbf{v}$ -component force pushes carriers along the wire with power drawn, via the $\mathbf{u}$ -component drag, from whoever moves the conductor. Write "the field did work on the charges" in an exam and the decomposition above is the red ink you will receive.

Worked examples

1 · The rail generator, audited to the milliwatt (confidence builder)

Rod of length $L = 0.2$ m on rails, $B = 0.5$ T out of the table, pulled at $v = 3$ m/s through total circuit resistance $R = 0.5$ Ω. Find everything.

Step 1. EMF: $\varepsilon = BLv = (0.5)(0.2)(3) =$ 0.30 V.

Step 2. Current: $I = \varepsilon/R = 0.30/0.5 =$ 0.60 A.

Step 3. Drag on the rod: $F = BIL = (0.5)(0.60)(0.2) =$ 0.060 N, opposing the pull.

Step 4. Power, both routes: mechanical in, $Fv = 0.060\times3 = 0.18$ W; electrical out, $I^2R = (0.60)^2(0.5) = 0.18$ W; pump output, $\varepsilon I = 0.30\times0.60 = 0.18$ W. All three agree: 0.18 W — equation (2) to the digit, the generator theorem in miniature.

2 · Faraday's disk at the brush (exam level)

A copper disk of radius $R = 0.10$ m spins at $\omega = 300$ rad/s in a uniform axial field $B = 0.10$ T. Find the axle-to-rim EMF, the current through a 10 mΩ load (brush and disk resistance negligible), and the torque the motor must supply.

Step 1. EMF, equation (4): $\varepsilon = \tfrac12\omega BR^2 = \tfrac12(300)(0.10)(0.10)^2 =$ 0.15 V. Note the machine's character: tiny voltage, but the source is a solid copper disk — internal resistance in micro-ohms, so the current ceiling is enormous. Homopolar generators are the natural low-voltage, kilo-ampere machines.

Step 2. Current: $I = \varepsilon/R_{\rm load} = 0.15/0.010 =$ 15 A.

Step 3. Torque, by the power balance (the disk version of equation (2)): $\tau_{\rm mech}\,\omega = \varepsilon I$ , so $\tau = \varepsilon I/\omega = (0.15)(15)/300 =$ $7.5\times10^{-3}$ N·m. (Microscopically it is the $I\,d\boldsymbol{\ell}\times\mathbf{B}$ drag on the radial current path, integrated — the crank-arm theorem guarantees the two bookkeepings match without checking, but checking is good hygiene: $\int_0^R I\,B\,r\,dr = \tfrac12 IBR^2 = 7.5\times10^{-3}$ N·m ✓.)

Step 4. Scaling read: $\varepsilon \propto \omega BR^2$ — the disk pays quadratically for radius but only linearly for speed and field. Real homopolar machines (naval and pulsed-power units) choose big radii and superconducting fields, and live with brush engineering at kilo-amps.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §7.1.3 (the motional-EMF section this article tracks, including the work-paradox figure) · Purcell & Morin, Electricity and Magnetism, 3rd ed., ch. 7 · Jackson, Classical Electrodynamics, 3rd ed., §5.15 · Zangwill, Modern Electrodynamics, §14.2 · Feynman Lectures II ch. 17 (the flux-rule exceptions discussion behind the disk's deferral).

Provenance. Equations standard from Griffiths Introduction to Electrodynamics 4th ed. §7.1.2-7.1.3 (motional EMF, flux rule, energy argument), Purcell & Morin 3rd ed. Ch.7, Jackson 3rd ed. §5.15. A verifier should recompute: (1) $\mathcal E=BvL$ from $\oint(\mathbf v\times\mathbf B)\cdot\mathrm d\boldsymbol\ell$ and from $-\mathrm d\Phi/\mathrm dt$ ; (2) the homopolar result $\tfrac12 B\omega R^2$ by integration; (3) the drag force $B^2L^2v/R$ and power balance $P=I^2R$ ; (4) the numerical check $\mathcal E=0.4$ V. All numerical examples use round illustrative inputs, not physical constants. Addendum for this edition: flagship numbers recomputed on the lane's rail set (0.5 T, 0.2 m, 3 m/s, 0.5 Ω → 0.30 V, 0.60 A, 0.060 N, 0.18 W verified by all three routes of the generator theorem) and the disk (0.1 T, 0.1 m, 300 rad/s → 0.15 V; torque $7.5\times10^{-3}$ N·m cross-checked by direct $\int IBr\,dr$ ). The changing-field half of induction is deliberately absent (zero overlap with Faraday's Law of Induction); the disk's flux-rule embarrassment is flagged, not autopsied (The Flux Rule and Its Exceptions owns it).

Builds on: Electromotive Force · Ohm's Law and Conductivity · Maxwell's Equations in Vacuum

Leads to: Faraday's Law of Induction · The Flux Rule and Its Exceptions · Lenz's Law · Self-Inductance · Induced Electric Fields

An airplane with a 60 m wingspan flies at 250 m/s through Earth's vertical field component of about 5e-5 T. Is there really a voltage across the wingtips - and can the plane harvest it?

The EMF is real: $\varepsilon = BLv = (5\times10^{-5})(60)(250) = 0.75$ V tip-to-tip — charge separates until the electrostatic field balances $\mathbf{v}\times\mathbf{B}$ , exactly the open-circuit rod of the limiting-cases table. Harvesting it is impossible in principle, and the reason is instructive: to draw current you must close the circuit, and any return wire is aboard the same airplane, moving at the same $\mathbf{v}$ through the same $\mathbf{B}$ — its $\oint(\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}$ around the closed loop is zero. You would need a stationary return path (or a field gradient): the Dover–Calais seawater measurements and the space-tether experiments work precisely because the return circuit (the sea floor, the ionosphere) does not ride along. A pump needs a frame difference; the airplane carries its whole circuit in one frame.

In the rod's own rest frame there is no velocity - what drives the current there?

An electric field. Transform to the rod frame and the charges are at rest, so $q\mathbf{v}\times\mathbf{B}$ vanishes — but the current is a frame-independent fact, so something pushes. In that frame there genuinely is an electric field $\mathbf{E}' \approx \mathbf{v}\times\mathbf{B}$ (to first order in $v/c$ ): what one frame calls a magnetic force on moving charges, another calls an electric field acting on static ones. This is not a paradox but a preview — the observation that launched Einstein's 1905 paper opens the next article, and the full transformation machinery is the relativistic chapter (E10). Motional EMF is the subject's first honest hint that $\mathbf{E}$ and $\mathbf{B}$ are one object viewed from different chairs.

Why does the drag force always oppose the motion - could clever geometry make it assist?

No, and the prohibition is thermodynamic before it is geometric. Suppose some configuration produced an assisting force: the rod would accelerate, increasing the EMF, increasing the current, increasing the assistance — a runaway converting nothing into kinetic energy plus resistor heat. Energy conservation forbids the sign, and the mathematics enforces it: the drag is $-\,(B^2L^2/R)\,\mathbf{v}$ -shaped in every honest geometry, quadratic in the field so that reversing $B$ leaves it opposing. This "induced effects oppose their cause" reading is Lenz's law — stated properly, with its microscopic bookkeeping, in Lenz's Law; here it operates as a corollary of the exact power balance.

Where does the charge pile up on an open-circuited moving rod, and how big is the effect?

The pump component $qvB$ pushes positive carriers toward one end of the rod (which end: right-hand rule on $\mathbf{v}\times\mathbf{B}$ ). With no circuit, they accumulate until the internal electrostatic field cancels the pump — the same standstill anatomy as an open-circuit battery in Electromotive Force, with $E_{\rm int} = vB$ inside and terminal "voltage" $BLv$ across the ends. The amounts are minuscule: for the flagship rod, $E_{\rm int} = 1.5$ V/m needs surface charge densities around $\varepsilon_0E \sim 10^{-11}$ C/m² — a few million electrons displaced along the whole rod. As always in circuit physics, the fields are the headline and the rearranged charge is a footnote too small to see, yet nothing works without it.

Is the EMF different if the magnet moves and the rod stays still? The relative motion is the same.

The measured current is identical — but the mechanism, in the lab frame's description, is completely different, and the difference is the hinge of the whole chapter. Rod moves, field static: magnetic force $q\mathbf{v}\times\mathbf{B}$ , this article, no new physics. Magnet moves, rod static: the charges have $\mathbf{v} = 0$ , the magnetic force is zero, and equation (1) predicts nothing — yet current flows. The field at the rod is changing in time, and a changing $\mathbf{B}$ makes an electric field: Faraday's law, genuinely new physics, the next article. That the two descriptions give the same answer for the same relative motion is exactly the "asymmetry" Einstein found intolerable, and relativity is where it resolves. Keep the bookkeeping clean: this article = things move; next article = fields change.

Real generators use coils rotating in the field. How does the rail-and-rod analysis carry over?

Directly, with geometry as the only novelty. A rectangular coil rotating at $\omega$ in uniform $B$ has flux $\Phi = BA\cos\omega t$ ; the (honest, material-loop) flux rule gives $\varepsilon = NBA\omega\sin\omega t$ — alternating, which is why grids are AC. Segment by segment it is still $\mathbf{v}\times\mathbf{B}$ : the two sides moving across the field are pulled rods, the ends contribute nothing, and the retarding torque $\tau = NIAB\sin\omega t$ is the drag force's rotational avatar, satisfying $\tau\omega = \varepsilon I$ — the generator theorem again. Slip rings take the AC out (or a commutator rectifies it, making a DC dynamo). Everything beyond — three phases, synchronous operation, field excitation — is arrangement, owned by the power-engineering track; the physics never leaves this article.

Does the induced current's magnetic field matter - when can I ignore the circuit acting back on itself?

You have been ignoring it legitimately because the loads here are resistive and the currents small: the induced current's own flux through the circuit is $LI$ with $L$ the self-inductance, and for the flagship numbers it is nanowebers against the applied field's milliwebers. But the back-flux grows with current and with winding count, and in any fast transient it fights the change of current, storing energy $\tfrac12LI^2$ in its own field. That bookkeeping — the circuit's induction on itself and on its neighbors — is Self-Inductance and Mutual Inductance, three articles on; the energy it parks was already audited from the field side by Magnetic Energy in Materials. For steady pulls and modest loops, the approximation "applied field only" is honest; every generator engineer eventually meets the regime where it is not (armature reaction), and that is where those articles pick up.