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AP Physics C: Electricity and Magnetism · Unit 13 · Topic 2 of 6 · 6 min lesson

Faraday’s Law

A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.

The big idea

A changing magnetic flux induces an emf: ε = −N dΦB/dt. It doesn’t matter whether B changes, the area changes, or the loop turns. Behind it is an induced electric field that circles around the changing flux: ∮E · dl = −dΦB/dt.

1. Flux change makes emf

ε = −N dΦB/dtN turns
ΦB = BA cos θchange B, A or θ
ε = NBAω sin ωtcoil spinning at ω
εmax = NBAωgenerator peak
∮E · dl = −dΦB/dtinduced E field
Φ = BA cos ωt ε = NBAω sin ωt ε max when Φ = 0 π 2π 3π ωt
In a generator, emf is the slope of the flux curve: biggest where flux crosses zero, zero where flux peaks.

2. Reading it right

Rate, not amount: a huge steady flux makes no emf. On a Φ–t graph, emf is the slope; steepest slope means largest emf.

Current: if the loop has resistance R, I = |ε|/R.

Induced E is different: its field lines close on themselves, so it isn’t conservative. You can’t define a potential for it, and a charge going around once gains energy qε.

Use symmetry like Ampère’s law: inside a solenoid with changing B, E(2πr) = πr² dB/dt.

3. Example, then try it

Example

A 200-turn coil of area 0.010 m² sits perpendicular to a field that drops steadily from 0.60 T to 0.10 T in 0.25 s. Find the emf. If the coil’s resistance is 5.0 Ω, what current flows?

ε = N ΔΦ/Δt = (200)(0.010)(0.50) ÷ 0.25 = 4.0 V.

I = 4.0 ÷ 5.0 = 0.80 A.

Try it · work it on paper first

A long solenoid of radius 3.0 cm has a field inside it increasing at 0.50 T/s. Find the induced electric field at 2.0 cm and at 6.0 cm from the axis.

1

Path: a circle of radius r centered on the axis. By symmetry E is tangent and constant, so ∮E · dl = E(2πr).

2

Inside (r = 2.0 cm): flux = πr²B, so E(2πr) = πr² dB/dt → E = (r/2)(dB/dt) = (0.010)(0.50) = 5.0 × 10⁻³ V/m.

3

Outside (r = 6.0 cm): flux is only inside the solenoid, πR²B. E = R²(dB/dt)/(2r) = (0.0009)(0.50) ÷ 0.12 ≈ 3.8 × 10⁻³ V/m.

4

Notice: E isn’t zero outside, even though B is.

Answer: 5.0 × 10⁻³ V/m inside at 2.0 cm; about 3.8 × 10⁻³ V/m outside at 6.0 cm.

4. Common mistakes

Expecting emf from a large but steady field.Only a change in flux induces emf.
Forgetting N.Each turn adds its own emf.
Thinking there’s no E field where B = 0.A changing flux nearby still creates a circulating E.
Check yourself

At one instant, the flux through a spinning generator coil is at its maximum. What is the emf at that instant?

The flux is at a peak, so its rate of change is momentarily zero, and so is the emf.

Ready to practice?

5 questions on Faraday’s Law, each with a hint and an explanation for every wrong answer.

Unit 13 topics

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