Use magnetic force as centripetal force
You will be able to: Connect perpendicular magnetic force to circular motion and compare radius changes.
What determines the radius of a charged particle’s turn?
A beam enters a uniform field at right angles. The force keeps turning the beam while its speed stays the same. In the ideal magnetic-only model, the path is a circle.
A useful starting point: Magnetic force turns motion without doing work →
Words and symbols before equations
- Radius r
- Distance from the circle’s center to the particle, in m.
- Centripetal acceleration
- Inward acceleration needed for a circular path, v²/r.
- Mass m
- Particle inertia, in kg.
- Uniform field
- Field with the same vector throughout the modeled region.
What this picture assumes
Nonrelativistic proton, m=1.67×10⁻²⁷ kg, q=1.60×10⁻¹⁹ C; perpendicular uniform B and no other forces. Circle radius uses a fixed 50 drawing units per cm. Motion is clockwise viewed from the front; no energy loss.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- r=0.01044 m (1.044 cm). Speed=100000 m/s; K=8.35e-18 J stays constant along this orbit. Stronger B changes radius, not speed.
- 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 perpendicular motion, |q|vB supplies the inward force mv²/r. Equating them gives r=mv/(|q|B). A stronger field bends the same particle more tightly; a faster or more massive particle has a larger radius.
Charge sign controls which way the path curves, while |q| controls radius. The magnetic force remains perpendicular to motion at each location and therefore cannot change the particle’s kinetic energy.
This is an application combining magnetic force with circular-motion ideas. It assumes constant uniform B, nonrelativistic speed and no other forces or energy losses. If motion also has a component parallel to B, that component persists and the path can become helical rather than circular.
A worked example, step by step
A proton with m=1.67×10⁻²⁷ kg and q=1.60×10⁻¹⁹ C moves perpendicular to 0.10 T at 1.0×10⁵ m/s. Find r.
- Use |q|vB=mv²/r.
- Solve r=mv/(|q|B).
- r=(1.67×10⁻²⁷)(1.0×10⁵)/[(1.60×10⁻¹⁹)(0.10)]=0.0104 m.
- The radius is about 1.04 cm; doubling B would halve it without changing speed.
A tighter bend does not imply that the magnetic field has slowed the particle.
Double speed with m, |q| and B fixed. What happens to r?
Compare with an explanation
It doubles because r is proportional to v.
Predict. Change one thing. Explain.
Change speed or field strength. Compare the numerical radius and the circle with its fixed length scale; predict the factor before moving the slider.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
r=0.01044 m (1.044 cm). Speed=100000 m/s; K=8.35e-18 J stays constant along this orbit. Stronger B changes radius, not speed.
Nonrelativistic proton, m=1.67×10⁻²⁷ kg, q=1.60×10⁻¹⁹ C; perpendicular uniform B and no other forces. Circle radius uses a fixed 50 drawing units per cm. Motion is clockwise viewed from the front; no energy loss.
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 questionTwo nonrelativistic particles have equal mass and speed in the same perpendicular B. One has charge magnitude q, the other 2q. (a) Write the radius relation. (b) Compare radii. (c) Compare kinetic energies. (d) Explain the effect of reversing one charge’s sign.
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Compare with the answer and four-point rubric
- 1 point: r=mv/(|q|B).
- 1 point: The 2q particle has half the radius.
- 1 point: Kinetic energies are equal because masses and speeds match.
- 1 point: Its bend direction reverses; its radius magnitude does not.
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 provides the centripetal force?
The magnetic force in the stated ideal model.
RECALL 2Does a magnetic-only circular orbit require constant speed?
Yes; magnetic force does no work.
RECALL 3What happens to a velocity component parallel to B?
It remains unchanged under magnetic force alone.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Use magnetic force as centripetal force
- Perpendicular magnetic-only motion: r=mv/(|q|B).
- Stronger B or larger |q| gives smaller r at fixed m and v.
- Speed and kinetic energy remain constant.
Remember: A tighter bend does not imply that the magnetic field has slowed the particle.
Conditions: Nonrelativistic proton, m=1.67×10⁻²⁷ kg, q=1.60×10⁻¹⁹ C; perpendicular uniform B and no other forces. Circle radius uses a fixed 50 drawing units per cm. Motion is clockwise viewed from the front; no energy loss.
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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