A moving rod separates charge
You will be able to: Connect magnetic charge separation in a moving rod with changing loop flux.
How can motion through a field create voltage?
A vertical conducting rod slides right on horizontal rails in a field into the page. Positive charges moving with the rod feel an upward magnetic force. The top becomes positive, producing an emf that can drive a counterclockwise loop current.
A useful starting point: Oppose the change, not always the field →
Words and symbols before equations
- Rod length L
- Separation between rails and active rod length, in m.
- Motional emf
- Potential-driving effect from a conductor moving through a magnetic field.
- Rail circuit
- Conducting return path joining the ends of the moving rod.
- Closed-loop resistance R
- Total resistance used to determine current; includes relevant load and conductor resistance.
What this picture assumes
Uniform B=0.50 T into page, vertical rod length 0.40 m, mutually perpendicular geometry. Closed-loop R=2 Ω. Negligible self-inductance and friction; open-circuit current is zero while motional voltage may remain.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- Motional emf=0.6 V; closed R=2 Ω gives I=0.3 A. Top of rod positive; closed-loop current goes up the rod (counterclockwise front view).
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For mutually perpendicular rod, velocity and field, emf magnitude is BLv. A positive charge feels q(v×B) upward in the stated setup, so the rod’s top becomes positive. Open circuit: charge separation can produce voltage while sustained loop current is zero.
With a closed resistive return path, current magnitude is BLv/R. It flows upward through the rod and counterclockwise around the rectangle as viewed from the front. The moving rod increases area Lx and inward flux; counterclockwise current creates outward field, matching Lenz’s law.
The formula assumes a uniform field over the active rod, rigid translation and the stated perpendicular geometry. For other geometry, identify the force direction and flux change instead of blindly applying BLv.
A worked example, step by step
A 0.40 m rod moves right at 3 m/s in B=0.50 T into the page. The closed circuit has R=2 Ω. Find emf, current and polarity.
- All relevant directions are mutually perpendicular.
- |ε|=BLv=(0.50)(0.40)(3)=0.60 V.
- I=0.60/2=0.30 A in the closed circuit.
- Top of rod is positive; current goes up the rod, counterclockwise around the loop.
An open circuit can have motional voltage even though no sustained current circulates.
Open the return path without changing rod speed. Does the ideal open-circuit voltage vanish?
Compare with an explanation
No. Charge separation still produces motional voltage, while sustained loop current is zero.
Predict. Change one thing. Explain.
Change rod speed and open or close the return path. Compare emf and current, and identify the top/bottom polarity from v×B.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Motional emf=0.6 V; closed R=2 Ω gives I=0.3 A. Top of rod positive; closed-loop current goes up the rod (counterclockwise front view).
Uniform B=0.50 T into page, vertical rod length 0.40 m, mutually perpendicular geometry. Closed-loop R=2 Ω. Negligible self-inductance and friction; open-circuit current is zero while motional voltage may remain.
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 0.20 m rod moves at 5 m/s through perpendicular 0.40 T. (a) Find emf. (b) Find current if R=2 Ω. (c) State open-circuit current. (d) Explain how Faraday’s law gives the same emf magnitude.
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Compare with the answer and four-point rubric
- 1 point: Emf=0.40 V.
- 1 point: Current=0.20 A.
- 1 point: Sustained open-loop current is zero.
- 1 point: Area changes at Lv, so flux changes at BLv and Faraday gives the same magnitude.
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 1What causes the initial separation of rod charges?
Magnetic force on charges moving with the rod.
RECALL 2What changes the rail loop’s flux?
Its enclosed area grows as the rod moves.
RECALL 3Does voltage alone prove a closed current path?
No; the path may be open.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
A moving rod separates charge
- Perpendicular rod geometry: |ε|=BLv.
- Closed resistive loop: I=|ε|/R.
- Polarity follows the magnetic force on positive charges.
Remember: An open circuit can have motional voltage even though no sustained current circulates.
Conditions: Uniform B=0.50 T into page, vertical rod length 0.40 m, mutually perpendicular geometry. Closed-loop R=2 Ω. Negligible self-inductance and friction; open-circuit current is zero while motional voltage may remain.
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.
Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.
Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.
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