Describe charged-particle motion in a magnetic field
Explain the circular path of a charged particle moving perpendicular to a uniform magnetic field.
Section 20 · Magnetic fields
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 20. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use current, charge, force, velocity, vectors, area and SI units; distinguish conventional current from electron flow.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For perpendicular entry into a uniform magnetic field, what supplies centripetal force?
What you will learn
- Describe the path of a charge moving perpendicular to uniform B
- Relate magnetic force to centripetal force
- Explain why the speed stays constant while direction changes
Perpendicular magnetic force supplies centripetal force
If a charged particle enters a uniform magnetic field with velocity perpendicular to B, the magnetic force is always perpendicular to its velocity. Its magnitude stays constant when speed and B are constant, while its direction continually turns with the particle.
The force therefore acts as the centripetal force and the path is circular. Equating |q|vB = mv²/r gives r = mv/(|q|B). The orbit radius increases with momentum and decreases with charge magnitude or field strength.
The field bends motion without changing speed
Because magnetic force is perpendicular to the instantaneous displacement, it does no work in the ideal model. The particle's speed and kinetic energy remain constant while its velocity direction changes.
A positive and negative particle with the same velocity bend in opposite directions. If a velocity component is parallel to B as well as a perpendicular component, the parallel component is unchanged and the full path is helical; the syllabus case here is perpendicular entry and a circular path.
Pause and explain
What happens to the speed of an ideal charged particle moving perpendicular to a uniform magnetic field?
Worked example
A proton with mass 1.67 × 10⁻²⁷ kg and speed 2.0 × 10⁶ m s⁻¹ enters perpendicular to a 0.20 T field. Find the orbit radius. Use q = 1.60 × 10⁻¹⁹ C.
Show the worked solution
- Use r = mv/(|q|B).
- Substitute the proton mass, speed, charge magnitude and field.
- State metres.
Answer r = (1.67 × 10⁻²⁷ × 2.0 × 10⁶)/(1.60 × 10⁻¹⁹ × 0.20) ≈ 0.104 m.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
An electron moves at 3.0 × 10⁶ m s⁻¹ perpendicular to B = 2.0 × 10⁻³ T. Find its orbit radius using m = 9.11 × 10⁻³¹ kg and |q| = 1.60 × 10⁻¹⁹ C.
Give me a hint
Use r = mv/(|q|B), then substitute the electron values and keep the charge magnitude positive.
Compare my reasoning
- r = 9.11 × 10⁻³¹ × 3.0 × 10⁶/(1.60 × 10⁻¹⁹ × 2.0 × 10⁻³).
- Numerator = 2.733 × 10⁻²⁴ kg m s⁻¹.
- Denominator = 3.20 × 10⁻²²; r ≈ 8.54 × 10⁻³ m.
r ≈ 8.5 mm.
Look for these in your work
- Correct orbit relation used
- Momentum calculated
- Radius and unit stated
Two particles with equal momentum magnitude enter the same uniform field perpendicularly. One has twice the charge magnitude of the other. Compare their radii.
Give me a hint
Use r = p/(|q|B), then compare the charge magnitudes while momentum and field stay fixed.
Compare my reasoning
- B and p are common.
- Radius is inversely proportional to |q|.
- The particle with twice the charge magnitude has half the radius.
The more highly charged particle follows a circle with half the radius.
Look for these in your work
- Common quantities identified
- Inverse charge dependence used
- Radius ratio given
A particle curves in a uniform magnetic field while its speed is measured to remain constant. Explain the observations using force and energy.
Give me a hint
Relate force direction to velocity and work.
Compare my reasoning
- Magnetic force is perpendicular to the instantaneous velocity.
- A perpendicular force changes direction but does no work.
- The particle follows a curved path at constant speed and kinetic energy.
The field continually turns the velocity without changing its magnitude, so the path curves while speed and kinetic energy stay constant.
Look for these in your work
- Perpendicular force stated
- Work consequence stated
- Path and speed reconciled
Common mistakes
- Drawing the force parallel to the particle velocity
- Claiming the magnetic field increases speed
- Using signed q to calculate a negative radius
- Confusing the circular perpendicular-entry path with a helix that has a parallel velocity component
What can you explain now?
For perpendicular entry into a uniform magnetic field, what supplies centripetal force?
Compare with the explanation
Magnetic force
The force is perpendicular to velocity and bends the path.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
How does the magnetic force produce a circular path without changing the particle's kinetic energy?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.