Motional EMF
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
A conductor moving through a magnetic field has an emf across it: ε = BLv. The field pushes the rod’s charges along it, or equivalently, the circuit’s area and flux grow. If a current flows, the field then pushes back on the rod with F = ILB, opposing its motion.
1. A rod on rails
2. Two ways to see it
Force on charges: each charge in the rod feels qv × B along the rod. Charges pile up at the ends until qE = qvB, so ΔV = EL = BLv.
Flux: the area grows at Lv per second, so dΦ/dt = BLv. Both views agree.
Energy: to keep the rod at constant speed, a force must balance FB. Its power, B²L²v²/R, equals the I²R heating in the resistor.
Left alone, the rod slows down, since the drag is proportional to v. A falling rod speeds up until the drag equals its weight.
3. Example, then try it
A 0.50 m rod slides at 4.0 m/s on rails in a 0.60 T field, with a 3.0 Ω resistor. Find the emf, current, magnetic force, and power.
ε = (0.60)(0.50)(4.0) = 1.2 V; I = 1.2 ÷ 3.0 = 0.40 A.
F = ILB = (0.40)(0.50)(0.60) = 0.12 N; P = Fv = 0.48 W = I²R.
Vertical rails 0.25 m apart are joined by a 0.50 Ω resistor at the top, with a horizontal 0.40 T field perpendicular to them. A 20 g rod slides down them without friction. Find its terminal speed and the current then.
Forces: weight mg down; magnetic drag B²L²v/R up, growing with v.
Terminal speed: mg = B²L²v/R → v = mgR/(B²L²) = (0.020)(10)(0.50) ÷ [(0.16)(0.0625)] = 10 m/s.
Current: I = BLv/R = (0.40)(0.25)(10) ÷ 0.50 = 2.0 A.
Check: gravity’s power mgv = 2.0 W equals I²R = (4.0)(0.50) = 2.0 W.
Answer: 10 m/s; 2.0 A.
4. Common mistakes
A rod slides at constant speed v on rails in a field B, with resistor R. If v doubles, what happens to the power dissipated in R?
ε = BLv doubles, and P = ε²/R, so P becomes four times as large.
Ready to practice?
4 questions on Motional EMF, each with a hint and an explanation for every wrong answer.