Motional emf & the Rod on Rails

You’ll be able to

Compute the emf induced in a conductor moving through a magnetic field
Apply Lenz's law to find the direction of the induced current
Explain why a magnetic force opposes the motion that induces the current

Changing flux, not field, induces emf

Faraday's law says the induced emf equals the rate of change of magnetic flux, where flux Φ = BA cos θ. Three things can change it: the field strength, the area of the loop, or the orientation of the loop relative to the field. A loop sitting still in a strong but steady field has large flux and zero induced emf, which is the distinction the exam tests most often.

Faraday and motional emf

ε = −ΔΦ/Δt with Φ = BA cos θ · for a rod of length L moving at speed v perpendicular to B: ε = BLv

The BLv form is the special case where the changing quantity is area, at rate Lv. The minus sign is Lenz's law.

The rod on rails, in full

A conducting rod of length L slides at speed v along two rails connected by a resistor R, all in a field B perpendicular to the plane. The enclosed area grows at rate Lv, so the flux grows and an emf ε = BLv appears. That drives a current I = BLv/R around the loop. The current in the rod then sits in the field and feels a force F = BIL = B²L²v/R — and by Lenz's law that force opposes the motion. Every part of this chain is a separate rubric point on a free response.

Why the opposition is required

If the induced force helped the motion, the rod would speed up, inducing more current, producing more force, and accelerating without limit — energy from nothing. Lenz's law is energy conservation expressed as a direction rule: the induced effect always opposes the change that produced it. So to keep the rod moving at constant speed an external agent must supply work at exactly the rate the resistor dissipates energy, P = ε²/R.

Terminal velocity of a falling loop

Drop a conducting loop through a field region and it falls, inducing a current whose magnetic force opposes the fall. As speed rises the opposing force rises with it, and the loop reaches a terminal velocity where B²L²v/R equals mg. This is the same structure as air-resistance terminal velocity from Physics 1, and it is why a magnet dropped down a copper pipe descends slowly.

Worked example

A rod of length 0.30 m slides at 4.0 m/s along rails connected by a 6.0 Ω resistor, in a 0.50 T field perpendicular to the plane. Find the emf, the current, the retarding force, and the power dissipated.

1.emf: ε = BLv = (0.50)(0.30)(4.0) = 0.60 V.
2.Current: I = ε/R = 0.60/6.0 = 0.10 A.
3.Force on the rod: F = BIL = (0.50)(0.10)(0.30) = 0.015 N, opposing the motion.
4.Power dissipated: P = ε²/R = (0.60)²/6.0 = 0.060 W.
5.Check: mechanical power supplied = Fv = (0.015)(4.0) = 0.060 W. The two agree, as energy conservation requires.

Answer: ε = 0.60 V, I = 0.10 A, a retarding force of 0.015 N, and 0.060 W dissipated. The agreement between Fv and ε²/R is the check that the whole chain is consistent.

Watch out

A large steady flux induces nothing. Only a changing flux produces an emf, so a loop at rest in a strong uniform field has no induced current however strong the field is.

Checkpoint

A conducting loop sits at rest in a strong, constant magnetic field. The induced emf is:

Checkpoint

A rod sliding on rails induces a current. The magnetic force on the rod:

Checkpoint

The speed of a rod on rails is doubled. The power dissipated in the circuit resistor:

On the exam

Answer induction questions in the order emf → current → force → power. Each step feeds the next, they are usually separate rubric points, and the Fv against ε²/R check verifies the whole chain in one line.

Answer the 3 checkpoints as you read.