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AP Physics 2 · Unit 12 · Topic 4 of 4 · 6 min lesson

Electromagnetic Induction

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 8 practice questions.

The big idea

A changing magnetic flux through a loop induces an emf, and so a current: ε = −N ΔΦ/Δt (Faraday’s law). The induced current always flows so that its own field opposes the change (Lenz’s law). No change in flux, no emf.

1. Flux and Faraday’s law

Φ = BA cos θmagnetic flux (Wb = T·m²)
|ε| = N ΔΦ/ΔtFaraday’s law
ε = BLvrod sliding on rails
Change B, A or θany of these changes Φ
× × × × × × × × × × × × × × × × × × × × × × × × R v FB I I B into page; area grows → I counterclockwise and FB opposes the motion
The induced current pushes back: the force on the rod opposes its motion, so you have to keep pushing to keep it moving.

2. Using Lenz’s law

1. Which way does the flux through the loop point, and is it growing or shrinking?

2. The induced field points opposite to a growing flux, or the same way as a shrinking one.

3. Use the right-hand rule to find the current that makes that field.

A loop moving at constant speed completely inside a uniform field has constant flux, so no current. Current flows only while it’s entering or leaving.

A magnet dropped through a copper pipe falls slowly: currents induced in the pipe push back on it.

3. Example, then try it

Example

A 200-turn coil of area 0.010 m² sits perpendicular to a field that rises from 0 to 0.50 T in 0.10 s. Find the average emf.

ΔΦ per turn = (0.50)(0.010) = 0.0050 Wb.

|ε| = NΔΦ/Δt = (200)(0.0050) ÷ 0.10 = 10 V.

Try it · work it on paper first

A rod 0.40 m long slides at 5.0 m/s on rails in a 0.50 T field pointing into the page, as shown below. The resistor is 2.0 Ω. Find the emf, the current and its direction, and the force needed to keep the rod moving steadily.

× × × × × × × × × × × × × × × × × × × × × × × × R v B into page
1

emf: ε = BLv = (0.50)(0.40)(5.0) = 1.0 V.

2

Current: I = ε/R = 1.0 ÷ 2.0 = 0.50 A. The flux into the page is growing, so the induced field points out of the page: counterclockwise (up through the rod).

3

Magnetic force on the rod: F = ILB = (0.50)(0.40)(0.50) = 0.10 N, pointing left, against the motion.

4

Applied force: 0.10 N to the right. Check: its power Fv = 0.50 W equals I²R = (0.50)²(2.0) ✓.

Answer: 1.0 V; 0.50 A counterclockwise; push with 0.10 N.

4. Common mistakes

“A big field means a big emf.”Only a changing flux induces an emf.
Opposing the field instead of the change.If the flux is shrinking, the induced field points the same way to keep it up.
Forgetting N.A coil multiplies the emf by its number of turns.
Check yourself

The north pole of a bar magnet is pushed toward a wire loop. Seen from the magnet’s side, which way does the induced current flow?

The flux through the loop (pointing away from you, out of the N pole) is growing, so the induced field points back toward you, toward the magnet. A field toward you comes from a counterclockwise current. The loop acts like a north pole repelling the magnet.

Ready to practice?

8 questions on Electromagnetic Induction, each with a hint and an explanation for every wrong answer.

Unit 12 topics

Every topic has a short lesson or quick review and its own quiz.