Balance electric and magnetic forces
You will be able to: Add independent electric and magnetic forces and identify a no-deflection speed.
Which speed passes through crossed fields without deflection?
A positive beam moves right through an upward electric field and an out-of-page magnetic field. Electric force points up; magnetic force points down. At one speed, the forces balance.
A useful starting point: Use magnetic force as centripetal force →
Words and symbols before equations
- Electric force
- F_E=qE; it can act even when a charge is stationary.
- Crossed fields
- Electric and magnetic field directions are perpendicular in this setup.
- No deflection
- Net transverse force is zero for the specified initial direction.
- Selected speed
- Speed v=E/B at which the two force magnitudes match in this perpendicular geometry.
What this picture assumes
|q|=1 μC; E=2000 N/C upward and B=0.020 T out of page. Readouts are initial forces for the stated rightward velocity, not a computed trajectory after deflection.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- F_E,y=2000 μN; F_B,y=-2000 μN; net=0 μN. No initial deflection. Balanced speed=1.0×10⁵ m/s.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Find each force separately. For positive q moving right with B out of the page, v×B points down. With E upward, electric force points up. The net vertical force is q(E−vB).
Balance requires qE=qvB, giving v=E/B for nonzero charge in this geometry. A negative charge reverses both forces, so the same selected speed still balances them.
A slower or faster particle has a nonzero initial transverse force. The model reports the initial force, not a full trajectory after deflection changes velocity. Do not use E/B when the fields or velocity have a different orientation without checking directions.
| Property | Electric | Magnetic |
|---|---|---|
| Relation | qE | q(v × B) |
| Stationary charge | Can feel force | Zero magnetic force |
| Work | Can change kinetic energy | Magnetic force alone does no work |
A worked example, step by step
E=2000 N/C upward and B=0.020 T out of the page. What rightward speed produces no deflection?
- For positive charge, electric force is up and magnetic force is down.
- Set their magnitudes equal: |q|E=|q|vB.
- v=E/B=2000/0.020=1.0×10⁵ m/s.
- The same speed balances a negative charge, with both force directions reversed.
Equal force magnitudes cancel only if the forces point in opposite directions.
A positive particle is slower than E/B in this setup. Initial force points where?
Compare with an explanation
Up: the electric force is larger than the downward magnetic force.
Predict. Change one thing. Explain.
Move the beam speed below, to and above the selected value. Then reverse charge sign. Explain why the zero-force speed is unchanged.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
F_E,y=2000 μN; F_B,y=-2000 μN; net=0 μN. No initial deflection. Balanced speed=1.0×10⁵ m/s.
|q|=1 μC; E=2000 N/C upward and B=0.020 T out of page. Readouts are initial forces for the stated rightward velocity, not a computed trajectory after deflection.
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 positive charge moves right at 2×10⁴ m/s in E=600 N/C upward and B=0.020 T out of page. (a) Give electric force direction. (b) Give magnetic force direction. (c) Compare E and vB. (d) Find the no-deflection speed.
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Compare with the answer and four-point rubric
- 1 point: Electric force up.
- 1 point: Magnetic force down.
- 1 point: E=600 versus vB=400 N/C; net force initially up.
- 1 point: Selected speed=3×10⁴ m/s.
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 1Can an electric field act on a stationary charge?
Yes.
RECALL 2Can a magnetic field alone act on a stationary point charge?
No.
RECALL 3Why is the selected speed independent of |q|?
Both force magnitudes contain the same factor |q|, which cancels.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Balance electric and magnetic forces
- F_net=q(E+v×B).
- This perpendicular opposing-force setup: v_selected=E/B.
- Find electric and magnetic directions separately.
Remember: Equal force magnitudes cancel only if the forces point in opposite directions.
Conditions: |q|=1 μC; E=2000 N/C upward and B=0.020 T out of page. Readouts are initial forces for the stated rightward velocity, not a computed trajectory after deflection.
Refresh Kid · AP Physics 2 Unit 4 (official Unit 12) · Objectives 12.2.B · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 12.2, objectives 12.2.B. 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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