Motional emf

Take a rod of length L = 0.30 m sliding at v = 5.0 m s⁻¹ across a uniform field B = 0.45 T that points out of the page. Every conduction electron inside is carried along at v, so it feels a magnetic force qv × B whose direction, for v to the right and B out of the page, lies along the rod. Charge therefore separates: one end goes positive, the other negative, and the separation grows until the electrostatic field it creates cancels the magnetic push on the carriers still inside. That balance fixes the interior field at E = vB = 2.25 V m⁻¹, and multiplying by the rod's length gives the voltage between its ends, vBL = 0.675 V. Nothing is flowing yet. The equilibrium is reached in nanoseconds, and an isolated rod simply coasts along with 0.675 V across it.

Let the rod's ends ride on rails joined by a resistor, with a loop resistance of R = 1.5 Ω in total. The separated charge now has somewhere to go, and the rod behaves as a 0.675 V source: I = ε/R = 0.675/1.5 = 0.45 A. The direction comes straight from the carrier force, with no flux argument at all. With B out of the page and v to the right, v × B points from the top rail towards the bottom rail, so positive charge is pushed down the rod; conventional current runs down the rod, left along the bottom rail, up through the resistor and back along the top rail. That is the same direction Lenz's law gives, because the loop's area is growing and the induced current's own field opposes the increase. Note what does not depend on R: the emf. Only the current does.

The rod is now a current-carrying wire sitting in the very field that drove the current, so it feels F = B I L = 0.45 × 0.45 × 0.30 = 0.0608 N. Working out I L × B for a current running down the rod in a field out of the page puts that force to the left, against the motion. Substituting I = BLv/R gives the compact form F = B²L²v/R, which is the same 0.0608 N. To hold the speed at 5.0 m s⁻¹ you must pull with exactly that force, delivering P = Fv = 0.0608 × 5.0 = 0.304 W, and the resistor dissipates I²R = 0.45² × 1.5 = 0.304 W. The agreement is exact, not approximate. Notice the asymmetry in the exponents: doubling v doubles the emf and doubles the drag, but doubling B quadruples the drag while only doubling the emf.

Because the drag is proportional to v rather than constant, a rod left to coast obeys m dv/dt = −B²L²v/R, which is the exponential-decay equation with time constant τ = mR/(B²L²). Check that it is a time: write Ω as kg m² s⁻³ A⁻² and T as kg s⁻² A⁻¹, and kg Ω ÷ (T² m²) reduces to s. With m = 0.25 kg, R = 0.60 Ω, B = 0.40 T and L = 0.50 m, B²L² = 0.0400 and τ = 0.25 × 0.60/0.0400 = 3.75 s, so a rod released at 2.0 m s⁻¹ has slowed to 2.0 e^{(−5.0/3.75)} = 0.53 m s⁻¹ after 5.0 s. Add a steady driving force instead, gravity on an inclined pair of rails for instance, and the same linear drag produces a terminal speed: the rod accelerates until B²L²v/R matches the drive, then holds that speed.

Count the same situation with flux instead of forces. In 0.20 s the rod sweeps out ΔA = Lv Δt = 0.30 × 5.0 × 0.20 = 0.30 m², so the flux through the circuit changes by ΔΦ = BΔA = 0.45 × 0.30 = 0.135 Wb and |ε| = ΔΦ/Δt = 0.675 V. The same number, because dΦ/dt = BLv exactly when a rod of length L sweeps at speed v. The two routes are one statement seen from two places: one counts the force on the carriers, the other counts the area swept. One result belongs to the flux route alone. The charge delivered is q = ∫I dt = ΔΦ/R, so it depends on the area swept and not on how fast it was swept: moving this rod 0.20 m along the rails passes 0.45 × 0.30 × 0.20/1.5 = 0.018 C whether it takes a second or an hour.

ε = BLv assumes every part of the rod moves at the same velocity. Pivot the rod at one end instead and spin it at angular speed ω: the element at radius r moves at ωr, so it contributes dε = ωrB dr, and the whole rod gives ε = ∫₀^L ωBr dr = ½ωBL². With ω = 25 rad s⁻¹, B = 0.20 T and L = 0.60 m that is 0.90 V, exactly half of the 1.8 V you would wrongly get by putting the tip speed ωL = 15 m s⁻¹ into BLv. The other assumptions are worth naming too. B must be uniform over the swept region and the rod must stay inside it, or the emf collapses the moment the rod leaves the field; the loop's resistance must swamp its inductance, so the current tracks the emf without lag; and v must be far below c.