In short

Move a straight conducting rod of length \ell through a magnetic field \vec{B} at velocity \vec{v}, keeping the rod, velocity, and field mutually perpendicular. The free electrons inside the rod experience a magnetic Lorentz force, drift to one end, and an EMF appears across the rod:

\boxed{\;\varepsilon \;=\; B\ell v\;}

More generally, for any infinitesimal segment of wire moving with velocity \vec{v} in a field \vec{B}, the motional EMF per unit length is (\vec{v}\times\vec{B})\cdot d\vec{\ell}, and the total EMF around a circuit is

\varepsilon \;=\; \oint (\vec{v}\times\vec{B})\cdot d\vec{\ell}

The sliding-rod-on-rails circuit. If the rod of resistance zero (plus a resistor R across the rails) slides at v across rails separated by \ell in a field B, the EMF drives a current I = B\ell v/R, which in turn exerts a retarding force F = BI\ell = B^2\ell^2 v/R on the rod. The mechanical power you supply equals the electrical power dissipated: Fv = I^2R. Energy is conserved exactly; this is Lenz's law at work.

The rotating-coil generator. A flat coil of N turns and area A spinning at angular frequency \omega in a uniform field B produces a sinusoidal EMF

\boxed{\;\varepsilon(t) \;=\; NBA\omega\sin(\omega t)\;}

peak \varepsilon_0 = NBA\omega, RMS \varepsilon_\text{rms} = \varepsilon_0/\sqrt{2}. Slip rings + brushes deliver AC; split rings (commutators) flip the connection every half-cycle to produce pulsating DC. Every turbine at NTPC Singrauli, every hydro generator at Tehri, every wind turbine in Tamil Nadu is a machine of exactly this kind.

Why it matters. Motional EMF is the physics of every electricity generator on Earth. It is also what Indian Railways exploits in regenerative braking (the traction motor becomes a generator, feeding energy back to the overhead line). A conductor moving across Earth's 50 μT field even generates a small EMF — which is why ISRO's electrodynamic-tether experiments can produce power in orbit without batteries.

Hold a short aluminium rod, about as long as a 30 cm ruler, in your hand. The room is full of Earth's magnetic field — 50 microteslas, pointing down at an angle of about 45° to the horizontal in Kolkata, 60° in Leh. Now swing the rod briskly sideways. A few microvolts appear across its ends for the instant you are swinging it. Swing it faster and the voltage grows; swing it the other direction and the sign flips; hold it still and the voltage goes to zero. No magnet is close by. No battery is attached. Yet there is a measurable EMF. Where did it come from?

The answer is the same Lorentz force you met in force on a moving charge. The free electrons inside the aluminium, being carried along with the rod, feel a magnetic push perpendicular to their motion. They drift, accumulate at one end, and set up an electric field inside the rod that eventually balances the magnetic push. The resulting charge separation across the ends is the EMF, and it is motional — it vanishes the moment the rod stops moving.

Scale this up by a million. A 40-tonne rotor of a generator at NTPC Dadri spins at 3000 rpm inside a magnetic field of about 1 tesla, its windings sweeping past the poles thousands of times per second. The motional EMF in those windings is 15,000 volts, feeding (via step-up transformers) the 400 kV high-tension lines that march across North India. Every fluorescent lamp in the country, every Mumbai suburban train, every AC in your flat in Ahmedabad — all draw power that started as motional EMF. This article is about that mechanism, from the single electron in the rod up to the AC generator.

Deriving \varepsilon = B\ell v — the Lorentz-force argument

Put a straight conducting rod of length \ell perpendicular to a uniform magnetic field \vec{B}, and move it with constant velocity \vec{v}, such that \vec{v}, \vec{\ell}, and \vec{B} are mutually perpendicular. Pick coordinates: let \vec{B} = B\hat{z} point out of the page, let the rod lie along \hat{y} (so \vec{\ell} = \ell\hat{y}), and let the rod move in the \hat{x} direction at \vec{v} = v\hat{x}.

A rod moving through a magnetic field — the Lorentz force on a free electron A vertical conducting rod of length l moves to the right at velocity v. A uniform magnetic field B points out of the page (shown as dots). A free electron inside the rod is labelled with velocity v (rightward). The magnetic force qv×B on the electron points downward along the rod, driving electrons toward the bottom end. After a brief transient, an electric field inside the rod points up to balance the magnetic force. B (out of page) + end − end ℓ v e⁻ F on +q = qv×B F on e⁻ (up)
A rod moving at $\vec{v}$ in a field $\vec{B}$. Each free electron is carried along with the rod and feels a Lorentz force $\vec{F} = -e\,\vec{v}\times\vec{B}$ pointing along the rod (upward here). Electrons pile up at the top end; conventional "positive" charge accumulates at the opposite end. The top of the rod is the **+ terminal** of this motional EMF source.

Step 1. Identify the velocity of a free electron inside the rod. The bulk velocity is \vec{v} = v\hat{x} — the whole rod moves rigidly — plus a tiny thermal motion that averages to zero. Use \vec{v} = v\hat{x}.

Why: each electron inside the metal is, on average, moving with the rod. Any random thermal motion washes out when you average over many electrons.

Step 2. Compute the magnetic force on one electron, charge -e.

\vec{F}_\text{mag} \;=\; (-e)\,\vec{v}\times\vec{B} \;=\; (-e)(v\hat{x}\times B\hat{z}) \;=\; (-e)(vB)(\hat{x}\times\hat{z}) \;=\; (-e)(vB)(-\hat{y}) \;=\; evB\,\hat{y}

Why: \hat{x}\times\hat{z} = -\hat{y} (cyclic order xyz reversed). The two minus signs combine to a plus — the force on an electron is in the +\hat{y} direction, i.e. upward along the rod.

Step 3. Electrons accumulate at the top end. As soon as a few electrons pile up there, an electric field \vec{E} builds up inside the rod pointing from the + (bottom) end to the − (top) end — call it E\hat{y} with E > 0 pointing up. (Remember: the top is negative because electrons are there, so the field inside the rod points from the bottom positive end up to the top negative end? Careful — the electric field always points from + to −, so inside the rod it points from the bottom (+) up toward the top (−), i.e. +\hat{y}. But this field pushes electrons against +\hat{y}, so downward. That is what we want — the electric field opposes the pileup.)

Step 4. Equilibrium. In steady state the magnetic and electric forces on each electron exactly cancel:

-e\vec{E} + evB\hat{y} \;=\; 0 \;\Longrightarrow\; \vec{E} \;=\; vB\,\hat{y}

Why: setting the net force on electrons to zero defines equilibrium. The electric field that does this has magnitude vB pointing in +\hat{y}. No more electrons migrate; the charge separation is stable.

Step 5. The EMF across the rod. The electric potential difference between the ends of the rod is

V_\text{top} - V_\text{bottom} \;=\; -\int_\text{bot}^\text{top} \vec{E}\cdot d\vec{\ell} \;=\; -E\ell \;=\; -vB\ell

So V_\text{bottom} > V_\text{top} — the bottom is the + terminal. The magnitude is what we call the motional EMF:

\boxed{\;\varepsilon \;=\; B\ell v\;}

Why: the EMF is defined as the work per unit charge that the non-electrostatic (magnetic) force does on a test charge pushed from one terminal to the other inside the source. Here, pushing a positive test charge from top to bottom against the field \vec{E} = vB\hat{y}, the work per unit charge done by the magnetic-induced force is vB\cdot\ell.

Step 6. Consistency with Faraday's law. Imagine closing the circuit with two horizontal rails at the top and bottom of the rod, and a fixed resistor at the far end. As the rod moves at v, the enclosed area grows at rate dA/dt = \ell v. Flux growth rate is d\Phi_B/dt = B\ell v, and Faraday's law gives |\varepsilon| = d\Phi_B/dt = B\ell v. Same answer — the Lorentz-force and flux-form viewpoints agree, as they must.

The sliding-rod-on-rails circuit

The most important motional-EMF setup in textbook physics is the rod sliding on frictionless rails, closing a circuit through an external resistor R.

Setup. Two parallel rails, distance \ell apart, lie in the xy-plane. A uniform magnetic field \vec{B} points along +\hat{z} (out of the page). A conducting rod of negligible resistance rests across the rails and slides along the +\hat{x} direction at velocity v. A resistor R connects the far ends of the rails.

Animated: sliding-rod current and retarding force over time A horizontal axis shows time from 0 to 5 seconds. Two curves: the induced current I(t) in the circuit, proportional to the rod's velocity, and the retarding magnetic force on the rod, proportional to the square of velocity (quadratic in velocity for the constant-v idealisation: the rod slows because of the retarding force unless pushed). time (seconds) current I and velocity v (normalised) 1.4 0 1 2 3 4 5
A rod given initial velocity $v_0$ and then released (no external push) slows exponentially: $v(t) = v_0 e^{-t/\tau}$ with $\tau = mR/(B^2\ell^2)$. The current $I = B\ell v/R$ (red) follows the same shape. The retarding force $F = B^2\ell^2 v/R$ (dark) also decays exponentially, because it is proportional to $v$. All the rod's initial kinetic energy ends up as $I^2R$ heat in the resistor.

The four quantities — EMF, current, force, power

EMF. We just derived it: \varepsilon = B\ell v.

Current. Ohm's law around the loop, with the rod's resistance set to zero and external resistance R:

I \;=\; \frac{\varepsilon}{R} \;=\; \frac{B\ell v}{R}

Direction of current. Lenz's law — the induced current opposes the change that produces it. Since the flux (with \vec{B} out of the page, normal out of the page) is increasing, the induced current circulates clockwise as viewed from the front (to produce a field into the page inside the loop, opposing the increase). See Lenz's Law for the sign work.

Force on the rod. The current-carrying rod of length \ell sits in the field \vec{B}, so it feels a force \vec{F} = I\vec{\ell}\times\vec{B}. With the current direction we just determined, the force on the rod is

F \;=\; BI\ell \;=\; B\cdot\frac{B\ell v}{R}\cdot\ell \;=\; \frac{B^2\ell^2 v}{R}

pointing in -\hat{x} — opposite to the velocity. This is the retarding force. Lenz's law again: the induced effect opposes the motion that created it.

Mechanical power you supply. To keep the rod moving at constant v you must apply an external force F_\text{ext} = +B^2\ell^2 v/R to cancel the retarding force. The mechanical power you deliver is

P_\text{mech} \;=\; F_\text{ext}\cdot v \;=\; \frac{B^2\ell^2 v^2}{R}

Electrical power dissipated. The current I flowing through R dissipates heat at rate

P_\text{elec} \;=\; I^2 R \;=\; \left(\frac{B\ell v}{R}\right)^2 R \;=\; \frac{B^2\ell^2 v^2}{R}

Exact match. P_\text{mech} = P_\text{elec}. Every joule of mechanical work you do on the rod becomes a joule of heat in the resistor. Energy is conserved. This is the content of Lenz's law.

What if the rod is not pushed?

Give the rod initial speed v_0 and let go. The retarding force now decelerates it:

m\frac{dv}{dt} \;=\; -\frac{B^2\ell^2}{R}v

This is a first-order linear ODE whose solution is

v(t) \;=\; v_0\,e^{-t/\tau}\;,\qquad \tau \;=\; \frac{mR}{B^2\ell^2}

The rod slows exponentially, its kinetic energy being entirely converted to heat in the resistor. The time constant \tau gives the characteristic time for the rod to lose most of its speed.

The AC generator — rotating coil in a uniform field

A flat rectangular coil of N turns and area A rotates at angular frequency \omega about an axis perpendicular to a uniform field \vec{B}. This is the heart of every AC generator.

Geometry. Let the rotation axis be the \hat{z}-axis, let \vec{B} = B\hat{x} point along +\hat{x}, and let the normal to the coil \hat{n} make angle \theta(t) = \omega t + \phi_0 with \vec{B}. Pick \phi_0 = 0 so that at t = 0 the coil's plane is perpendicular to \vec{B} (\hat{n} parallel to \vec{B}, maximum flux).

Flux through the coil. For one turn:

\Phi_1(t) \;=\; BA\cos(\omega t)

For N turns in series, the total flux linkage is N\Phi_1(t) = NBA\cos(\omega t).

EMF. Faraday's law:

\varepsilon(t) \;=\; -\frac{d}{dt}[NBA\cos(\omega t)] \;=\; NBA\omega\sin(\omega t)

Why: derivative of \cos is -\sin; the minus from Faraday's law and the minus from the derivative combine to a plus.

Amplitude and RMS.

\boxed{\;\varepsilon_0 \;=\; NBA\omega\;,\qquad \varepsilon_\text{rms} \;=\; \frac{\varepsilon_0}{\sqrt{2}}\;=\; \frac{NBA\omega}{\sqrt{2}}\;}

The output is pure sinusoidal alternating voltage — AC.

Motional-EMF view of the same result

A nice consistency check: compute the EMF around the coil directly from (\vec{v}\times\vec{B})\cdot d\vec{\ell} on the two moving sides of the coil.

Let the coil have sides a (along the rotation axis, stationary direction) and b (the sides that sweep through \vec{B}). At time t, the two sides parallel to \hat{z} are at radius b/2 from the axis, moving tangentially with speed v = (b/2)\omega. Their velocity is perpendicular to \hat{z} and lies in the rotation plane.

Decompose: when the normal to the coil is at angle \theta = \omega t to \vec{B}, the component of \vec{v} perpendicular to \vec{B} is v\sin\theta. The motional EMF per side is (speed \perp to \vec{B}) \times B \times (side length a):

\varepsilon_\text{per side} \;=\; B\,a\,\cdot\,(b/2)\omega\sin\theta

Both long sides contribute in the same sense around the loop:

\varepsilon_\text{per turn} \;=\; 2\cdot B a (b/2)\omega\sin(\omega t) \;=\; BA\omega\sin(\omega t)

where A = ab. With N turns: \varepsilon = NBA\omega\sin(\omega t). Same answer. The motional picture and the flux picture agree.

AC vs DC — slip rings vs split rings

The coil inside rotates; the two ends of the coil have to be brought out to the external circuit. Two ways to do this:

Indian domestic supply is single-phase 230 V, 50 Hz AC, generated by huge three-phase machines at power plants and delivered through transformers (see Faraday's Law). The 50 Hz comes from generators spinning at exactly 3000 rpm (= 2\pi\times 50 rad/s for a 2-pole machine). Indian Railways overhead catenaries run at 25 kV, 50 Hz single-phase — and the train's onboard transformer steps this down before rectifying to DC for the traction motors.

Worked examples

Example 1: A rod sliding on rails — find EMF, current, and power

Two parallel rails are 40 cm apart and lie horizontally in a uniform vertical magnetic field of 0.80 T. A conducting rod of resistance 0.20 Ω (the only resistance in the circuit) is pulled along the rails at a constant 5.0 m/s. Find (a) the EMF, (b) the current, (c) the force you must apply to keep the rod at constant speed, (d) the mechanical power you deliver, and (e) the rate at which heat is dissipated in the rod.

Sliding-rod setup with numerical values Two horizontal rails 40 cm apart, a conducting rod pulled at 5 m/s. The magnetic field is 0.8 T pointing into the page, shown as a pattern of crosses. B = 0.8 T (into page) rod (R = 0.2 Ω) v = 5 m/s ℓ = 0.4 m
A rod on rails, $\ell = 0.4$ m, pulled at $v = 5$ m/s in a field $B = 0.8$ T. The rod's 0.2 Ω is the only resistance in the loop.

Step 1. EMF.

\varepsilon \;=\; B\ell v \;=\; (0.8)(0.4)(5) \;=\; 1.6\text{ V}

Step 2. Current.

I \;=\; \frac{\varepsilon}{R} \;=\; \frac{1.6}{0.2} \;=\; 8.0\text{ A}

Why: Ohm's law applied to the full circuit; here the rod's resistance is the only resistance.

Step 3. Force on the current-carrying rod (which is the retarding force).

F_\text{retard} \;=\; BI\ell \;=\; (0.8)(8.0)(0.4) \;=\; 2.56\text{ N}

To keep the rod at constant velocity, the external force you apply must match this: F_\text{ext} = 2.56 N in the direction of motion.

Step 4. Mechanical power you supply.

P_\text{mech} \;=\; F_\text{ext}\cdot v \;=\; (2.56)(5) \;=\; 12.8\text{ W}

Step 5. Electrical power dissipated.

P_\text{elec} \;=\; I^2 R \;=\; (8)^2 (0.2) \;=\; 12.8\text{ W}

Result. \varepsilon = 1.6 V, I = 8.0 A, F = 2.56 N, P_\text{mech} = P_\text{elec} = 12.8 W.

What this shows. The mechanical power goes exactly into electrical power dissipated in the rod. Every joule of work you do pulling the rod ends up as heat. This is the generator principle: mechanical energy in, electrical energy out, conversion efficiency limited only by the resistance of the rest of the circuit (here, zero aside from the rod itself, so efficiency is 100% from rod-kinetic to rod-heat).

Example 2: An aeroplane wingtip in Earth's field

An Indian Air Force aeroplane flies due east at 300 m/s over Jabalpur. The vertical component of Earth's magnetic field there is B_V = 42 μT. The wingspan (tip to tip) is 20 m. What motional EMF appears between the two wingtips?

Aeroplane flying east in Earth's vertical field An aeroplane viewed from behind, wings horizontal, flying away from the viewer (east). A downward arrow labelled B_V = 42 μT represents Earth's vertical field component. The two wingtips are marked + (left, south) and − (right, north). + (south tip) − (north tip) wingspan = ℓ = 20 m B_V = 42 μT (down) v (east, into page)
Rear view of an aeroplane flying east. The vertical component of Earth's field, $B_V$, is perpendicular to both the wing (horizontal) and the velocity (horizontal east). The motional EMF $\varepsilon = B_V\ell v$ appears between the two wingtips.

Step 1. The relevant field component is the one perpendicular to both \vec{v} (east) and the wing direction (north-south). For a horizontal velocity and horizontal wings, only the vertical component B_V contributes.

Why: motional EMF is (\vec{v}\times\vec{B})\cdot\vec{\ell}. With \vec{v} eastward and \vec{\ell} along the wing (say southward), \vec{v}\times\vec{B} has a component along the wing only if \vec{B} has a vertical component. The horizontal component of \vec{B} is parallel to \vec{v} (both eastward, roughly) and contributes nothing.

Step 2. Apply \varepsilon = B_V\ell v directly.

\varepsilon \;=\; (42\times 10^{-6}\text{ T})(20\text{ m})(300\text{ m/s})
= (42\times 10^{-6})(6000) \;=\; 0.252\text{ V} \;\approx\; 0.25\text{ V}

Step 3. Which wingtip is the + terminal? Use the right-hand rule for \vec{v}\times\vec{B} with \vec{v} eastward (into the page in the figure) and \vec{B}_V downward. Fingers point into the page, curl toward down — thumb points south. So \vec{v}\times\vec{B} points south. A positive test charge drifts south under this force, so the south wingtip is the + terminal.

Result. Motional EMF \approx 0.25 V between the wingtips; the south wingtip is at higher potential (in the northern hemisphere, where B_V points down).

What this shows. Even Earth's tiny field produces a measurable EMF on a moving aircraft. The effect cannot drive a sustained current (there is no return path in the air), but it can be measured with a voltmeter across the wingtips. ISRO-built electrodynamic tethers deployed from satellites use this same principle — a 20 km tether moving at 7 km/s in Earth's 30 μT field generates several kV of EMF, which can drive current through the ionosphere and produce usable thrust or electrical power.

Example 3: The Indian-grid generator — rotor RMS voltage

A three-phase AC generator at NTPC Dadri has a single phase winding of N = 1200 turns, the rotor area is A = 1.2 m², and the field in the air gap is B = 1.4 T. The rotor spins at 3000 rpm (synchronous with the 50 Hz grid). Find the peak EMF, the RMS EMF, and the power delivered to a load of resistance R = 15 Ω across one phase (ignore stator resistance and reactance).

Step 1. Angular frequency.

\omega \;=\; 2\pi\cdot\frac{3000}{60} \;=\; 2\pi\times 50 \;=\; 314.16\text{ rad/s}

Why: \omega in rad/s = 2\pi\times (frequency in Hz). 3000 rpm / 60 = 50 rev/s = 50 Hz — matching the grid frequency, as required for synchronous operation.

Step 2. Peak EMF.

\varepsilon_0 \;=\; NBA\omega \;=\; (1200)(1.4)(1.2)(314.16)
= (1200)(1.4)(1.2)(314.16) \;=\; 6.33\times 10^5\text{ V} \;\approx\; 633\text{ kV}

(In a real generator, the coil is distributed across multiple slots rather than concentrated, and the winding factor reduces the effective N by about 10–15%; real per-phase peak is typically 15–25 kV, brought up to 400 kV by step-up transformers at the plant. We keep the ideal calculation here for clarity.)

Step 3. RMS EMF.

\varepsilon_\text{rms} \;=\; \frac{\varepsilon_0}{\sqrt{2}} \;=\; \frac{6.33\times 10^5}{1.414} \;\approx\; 4.48\times 10^5\text{ V} \;\approx\; 448\text{ kV}

Step 4. Power delivered to the load. For a pure resistive load,

P \;=\; \frac{\varepsilon_\text{rms}^2}{R} \;=\; \frac{(4.48\times 10^5)^2}{15} \;=\; \frac{2.006\times 10^{11}}{15} \;\approx\; 1.34\times 10^{10}\text{ W} \;\approx\; 13.4\text{ GW}

(The numbers are unrealistically high because the idealised calculation ignores the winding factor; a real phase of a real generator delivers roughly 200–300 MW.)

Result. \varepsilon_0 \approx 633 kV (ideal), \varepsilon_\text{rms} \approx 448 kV (ideal), P \approx 13 GW (ideal). Real plants deliver per-phase power in the 200–300 MW range after winding-factor corrections.

What this shows. Scaling up the "spin a coil in a field" idea to power-plant dimensions directly gives hundreds of megawatts of real electrical power. Every coal burner at NTPC Dadri, every steam turbine, every transmission line from Dadri to Delhi, exists to implement motional EMF on an industrial scale. The two textbook-looking formulas \varepsilon = B\ell v and \varepsilon = NBA\omega\sin(\omega t) are the entire power sector.

Common confusions

You have the standard formulas and the generator physics. What follows is JEE-advanced-level extensions: the rod's terminal velocity on a tilted track, the generator's back-EMF and motor physics, and the Lorentz-force subtlety of who does the work.

Rod on a tilted rail — terminal velocity

Tilt the rail-system downhill at angle \alpha. A rod of mass m slides freely under gravity; rails have negligible resistance; external resistor R is at the top. The field \vec{B} is perpendicular to the inclined plane. As the rod slides, motional EMF drives a current, which produces a retarding force up the slope. At some speed the retarding force equals mg\sin\alpha and the rod stops accelerating — the terminal velocity:

mg\sin\alpha \;=\; \frac{B^2\ell^2 v_t}{R} \;\Longrightarrow\; v_t \;=\; \frac{mgR\sin\alpha}{B^2\ell^2}

At terminal velocity, the mechanical power of gravity (mgv_t\sin\alpha) exactly matches the electrical power dissipated (I^2R = (B\ell v_t)^2/R). A clean energy-conservation problem.

Back-EMF in motors — the generator hiding inside the motor

Run a DC motor in reverse: apply a mechanical torque to the rotor while the armature is still connected to the supply. The rotating armature generates a motional EMF — the back-EMF — that opposes the supply voltage. Kirchhoff's law for the motor circuit:

V_\text{supply} \;=\; IR_\text{arm} \;+\; \varepsilon_\text{back}

where \varepsilon_\text{back} = k\omega depends linearly on rotor speed. At startup \omega = 0, \varepsilon_\text{back} = 0, and the full supply voltage appears across the armature resistance — huge starting current. As the motor accelerates, \varepsilon_\text{back} grows, the current drops, and the motor finds an equilibrium speed where the mechanical load balances the magnetic torque. This is why Indian Railways locomotives have a resistor bank in series with the armature at startup: to limit the inrush current before the back-EMF builds. And it is why a DC fan draws a huge current for the first fraction of a second when you switch it on.

Regenerative braking — the motor becomes a generator

If you disconnect the supply and connect the armature to a resistor (or a battery to be charged), the spinning rotor still generates back-EMF, which now drives current through the external load. The mechanical torque on the rotor is now a retarding torque (Lenz's law) — the motor decelerates, its kinetic energy going into the resistor as heat, or into the battery as stored chemical energy. This is regenerative braking — used on the Delhi Metro, on every Indian Railways EMU (electrical multiple unit), and on every electric two-wheeler from Bengaluru. 20–30% of the traction energy is recovered during braking.

The three-phase generator

A single coil gives single-phase AC. Three coils offset by 120° around the rotor axis give three-phase AC: three sinusoidal voltages \varepsilon_a, \varepsilon_b, \varepsilon_c displaced by 120° in time. Three-phase has two key advantages: (i) the instantaneous total power P_a + P_b + P_c is constant in time (no 100 Hz ripple as in single-phase), and (ii) three-phase motors are self-starting. Every power plant in India generates 3-phase AC; every industrial motor runs on 3-phase; only domestic circuits run single-phase, taken from one of the three generator phases.

Who does the work in motional EMF?

The magnetic force on a moving charge does no work — \vec{F}\perp\vec{v} always. So when a current flows in a rod pulled at velocity v, where does the electrical energy come from? Answer: the external agent pulling the rod. Here is the bookkeeping.

Inside the rod, each electron has two components of velocity: the bulk velocity \vec{v} (rightward, with the rod) and the drift velocity \vec{u} (upward, along the rod). The total velocity is \vec{v} + \vec{u}. The magnetic force is \vec{F}_\text{mag} = -e(\vec{v}+\vec{u})\times\vec{B}. Break into two pieces:

\vec{F}_\text{mag} \;=\; -e\vec{v}\times\vec{B} \;-\; e\vec{u}\times\vec{B}

The first piece is along \vec{u} (the direction of drift along the rod) — it drives the current. The second piece is along -\vec{v} (opposing the rod's motion) — it is the retarding force on the rod.

The first piece does work on the electron in its drift motion; the second piece does work against the electron (its rod-direction component of velocity is along \vec{u}, and the second force is perpendicular to \vec{u}, so no work there) — but the second piece is transmitted through the lattice to the external agent, who must push harder to maintain \vec{v}. So the external agent's mechanical work goes into the electrons via the rod-lattice friction (the agent pushes the lattice, the lattice pushes the electrons), and the electrons' kinetic energy then dissipates as heat in the resistor. The magnetic force just reroutes the energy — it never supplies or absorbs any. The Lorentz force is a perfect redirection mechanism, and this is exactly why Lenz's law works: no free lunch is possible because the magnetic force cannot do work.

Motional EMF in non-uniform fields — the dynamo effect

In a uniform field, all points on a moving conductor see the same \vec{B}. In a non-uniform field (for example, near one pole of a bar magnet), different parts of the same rod see different fields, and the motional EMF is \int_\text{rod} (\vec{v}\times\vec{B})\cdot d\vec{\ell}, which can be complicated. The homopolar generator (Faraday disk, 1831) is the simplest non-uniform example: a conducting disk spinning in an axial field develops an EMF between its centre and its rim. The EMF is \varepsilon = \frac{1}{2}B\omega R^2, derivable by integrating vB\,dr from r=0 to r=R with v = \omega r. This geometry gives DC without any commutator — but the current is huge and the voltage small, which is why Faraday disks never made commercial generators.

The "railgun" — motional EMF run in reverse

Drive a current through a rod-on-rails in a magnetic field, and the force \vec{F} = I\vec{\ell}\times\vec{B} accelerates the rod. This is the railgun — used in some military applications. The back-EMF B\ell v grows with speed, limiting the current and therefore the acceleration. The maximum projectile speed in a constant-voltage supply is v_\text{max} = V_\text{supply}/(B\ell), reached as the back-EMF matches the supply. Realistically the setup is limited by the rails melting (because currents are in the mega-amp range), so railguns exist only as physics demonstrations and military prototypes, not as household equipment.

Where this leads next