Changing flux produces an emf
You will be able to: Relate average induced emf to flux change rate and coil turns.
What does a flux–time graph tell you?
A coil’s flux per turn rises from 0.010 Wb to 0.030 Wb in 0.20 s. More rapid change produces a larger induced emf. Holding the final flux steady produces no continuing emf from this mechanism.
A useful starting point: Flux depends on field through an area →
Words and symbols before equations
- Induced emf ε
- Energy-per-charge driving effect associated with changing magnetic flux, in volts.
- Turns N
- Number of identical coil turns linking the stated flux.
- Flux per turn
- Flux through one turn; multiply its change by N only once.
- Average rate
- ΔΦ/Δt over an interval; on a graph, the secant slope.
What this picture assumes
Initial flux per turn is +0.010 Wb; flux changes linearly to the selected endpoint. Emf is signed relative to the circulation associated with the chosen positive normal. Self-inductance neglected; graph is synthetic, and emf is the interval value.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- ΔΦ=0.02 Wb per turn in 0.2 s. Rate=0.1 Wb/s; ε_avg=−N(rate)=-2 V. The sign refers to the chosen normal and positive loop circulation.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Faraday’s law gives ε_avg=−N ΔΦ_B/Δt when all N turns link the same flux. The magnitude measures how quickly flux changes, not how large the flux happens to be.
A straight segment on a flux–time graph has constant slope and therefore constant emf during that segment. A flat segment gives zero emf. A steeper segment gives larger magnitude; reversing slope reverses the signed emf for a fixed normal and loop direction.
Emf can exist without a closed conducting path. Current requires a path and depends on resistance and other circuit properties. The simple I=|ε|/R relation applies to our resistive closed-loop model with self-inductance neglected.
A worked example, step by step
A 20-turn coil has flux per turn increasing from 0.010 Wb to 0.030 Wb in 0.20 s. Find average emf magnitude.
- ΔΦ=0.030−0.010=0.020 Wb per turn.
- Rate ΔΦ/Δt=0.020/0.20=0.10 Wb/s.
- |ε_avg|=20×0.10=2.0 V.
- If flux then stays at 0.030 Wb, its rate is zero and induced emf becomes zero.
Do not multiply by N twice when a problem already supplies total flux linkage NΦ.
Double the change duration with the same endpoints and N. What happens to average emf magnitude?
Compare with an explanation
It halves because the flux-change rate halves.
Predict. Change one thing. Explain.
Change the duration of the same flux change, then change N. Predict slope and emf; compare the sign when final flux is below initial flux.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
ΔΦ=0.02 Wb per turn in 0.2 s. Rate=0.1 Wb/s; ε_avg=−N(rate)=-2 V. The sign refers to the chosen normal and positive loop circulation.
Initial flux per turn is +0.010 Wb; flux changes linearly to the selected endpoint. Emf is signed relative to the circulation associated with the chosen positive normal. Self-inductance neglected; graph is synthetic, and emf is the interval value.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant field, force, flux or energy relationship to justify your prediction.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA 5-turn coil’s flux per turn decreases from 0.040 Wb to 0.010 Wb in 0.10 s. (a) Find ΔΦ. (b) Find its average rate. (c) Find signed emf relative to the chosen positive convention. (d) Explain the effect of twice the time.
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Compare with the answer and four-point rubric
- 1 point: ΔΦ=−0.030 Wb.
- 1 point: Rate=−0.30 Wb/s.
- 1 point: ε_avg=−5(−0.30)=+1.5 V.
- 1 point: Same change in twice the time gives half the emf magnitude, 0.75 V.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Does a large constant flux imply emf?
No; its change rate is zero.
RECALL 2What does flux–time slope determine?
The signed induced emf after multiplying by −N.
RECALL 3Can an open loop have induced emf without current?
Yes; emf and conduction current are different quantities.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Changing flux produces an emf
- ε_avg=−N ΔΦ/Δt for equal linked flux per turn.
- 1 Wb/s=1 V.
- Induced current additionally requires a conducting path.
Remember: Do not multiply by N twice when a problem already supplies total flux linkage NΦ.
Conditions: Initial flux per turn is +0.010 Wb; flux changes linearly to the selected endpoint. Emf is signed relative to the circulation associated with the chosen positive normal. Self-inductance neglected; graph is synthetic, and emf is the interval value.
Refresh Kid · AP Physics 2 Unit 4 (official Unit 12) · Objectives 12.4.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 12.4, objectives 12.4.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Refresh Kid calls this the fourth AP Physics 2 unit; College Board numbers it Unit 12; the first unit in this course is official Unit 9. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.
Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.
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