Force on a moving charge
Key idea: H2 Physics lessons on current-produced fields, magnetic forces and crossed-field velocity selection.
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The core idea
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Big question: Why can a magnetic field bend a particle without changing its speed?
A moving charge experiences F = B|q|v sinθ, with direction perpendicular to both velocity and magnetic field. Because the force is perpendicular to displacement, it does no work and changes direction rather than speed. Apply the positive-charge direction rule first, then reverse it for a negative charge.
Find the force direction before its magnitude
A charge Q moving with velocity v in magnetic flux density B experiences force magnitude F = B|Q|v sinθ, where θ is between v and B. The force is zero for parallel motion and greatest when the directions are perpendicular.
Use the positive-charge direction from Fleming's left-hand rule or a vector rule, then reverse it for a negative charge. A stationary charge has v = 0 and therefore no magnetic force.
Check your understanding: A positive and negative particle enter with the same velocity. How do their magnetic forces compare?
For equal charge magnitudes, the forces have equal magnitudes and opposite directions.
Explain why magnetic force changes direction, not speed
Magnetic force is perpendicular to the instantaneous velocity, so it does no work on the particle. It can change the velocity direction, but magnetic force alone cannot change speed or kinetic energy.
Do not draw a force permanently in one page direction: as a path curves, the instantaneous velocity changes and the magnetic-force direction must be reconsidered.
Check your understanding: What happens to kinetic energy in a magnetic field alone?
It remains constant because the perpendicular magnetic force does no work.
Key ideas to keep
- A stationary charge feels no magnetic force.
- A negative charge's force is opposite to the positive-charge rule.
- Magnetic force alone cannot change kinetic energy.
See the reasoning
Worked example
Find magnetic force on an electron
Question: An electron moves east at 6.0 × 10⁶ m s⁻¹ through a 0.25 T field directed vertically upward. Find the force magnitude and direction, then describe what changes. Use e = 1.60 × 10⁻¹⁹ C.
Step 1: Use the perpendicular magnitude
Why: sin90° = 1.
Working: F = B|Q|v = 0.25(1.60×10⁻¹⁹)(6.0×10⁶) = 2.4×10⁻¹³ N.
Step 2: Assign direction
Why: The electron is negative.
Working: A positive charge would be forced south, so the electron force is north.
Step 3: Describe the motion quantity
Why: Force remains perpendicular to instantaneous velocity.
Working: Velocity direction changes; speed and kinetic energy remain constant.
Answer: Force is 2.4 × 10⁻¹³ N north; it changes velocity direction but not speed.
Check: The result has unit C·T·m s⁻¹ = N and no work is done by the magnetic field.
Another worked model
Question
A proton enters a 0.40 T field at 30° to the field with speed 5.0 × 10⁶ m s⁻¹. Find magnetic force.
Check the worked solution
F = BQv sin θ = (0.40)(1.60 × 10⁻¹⁹)(5.0 × 10⁶)sin30° = 1.60 × 10⁻¹³ N. Only the velocity component perpendicular to B contributes.
Use a hint if needed
Practise with support
Try this
A charge moves exactly parallel to a magnetic field. Find its magnetic force.
Hint: Only perpendicular velocity produces magnetic force.
Check your answer
F = BQv sin0° = 0.
Now work without the hint
Practise independently
Your turn
Give a reliable direction method for positive and negative charges and explain why magnetic force does no work.
Check your answer
First determine v × B for a positive charge using a right-hand vector rule, then reverse for negative charge. The force is perpendicular to instantaneous velocity, so F·v = 0 and kinetic energy remains constant.
Avoid these traps
Common mistakes
Common mistake
A negative charge follows the positive-charge force direction.
What is wrong with this reasoning?
Show better thinking
Determine v × B for positive charge, then reverse the direction for negative charge.
Common mistake
The full speed contributes when velocity is oblique to B.
What is wrong with this reasoning?
Show better thinking
Only v sin θ, the component perpendicular to B, contributes to magnetic force.
Write for the examiner
Exam guidance
Use the instantaneous velocity direction, then reverse the result if the charge is negative.
Exam-style practice [7 marks]
A proton moves east at 3.0 × 10⁵ m s⁻¹ through a 0.40 T field directed vertically upward. Find the force magnitude and direction. State the changes to speed, velocity and kinetic energy.
Plan before you answer
- Use sin90° for magnitude.
- Apply the positive-charge direction rule.
- Separate speed from velocity.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
F = 0.40(1.60×10⁻¹⁹)(3.0×10⁵) = 1.92×10⁻¹⁴ N. Using east, north and vertically upward as mutually perpendicular directions, the positive-charge force is south. The force turns the velocity but does no work, so speed and kinetic energy remain constant.
Come back in three days
Check what stayed with you
Recall question 1
State the magnetic-force equation for a moving charge.
Check the answer
F = B|Q|v sinθ in magnitude.
Recall question 2
When is the force zero?
Check the answer
When the charge is stationary or moves parallel or antiparallel to B.
Recall question 3
Why is kinetic energy unchanged?
Check the answer
Magnetic force is perpendicular to displacement and does no work.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 17 states no explicit exclusions. Straight-wire, flat-coil and long-solenoid field equations use their named ideal geometries; a ferrous core changes the air-core model. In F = BIl sin θ, θ is between conventional current and B; in F = BQv sin θ, direction is determined for positive charge then reversed for negative charge. Magnetic force does no work. Circular-path formulae require velocity perpendicular to a uniform B. Velocity selection requires uniform mutually perpendicular E, B and beam velocity with opposing forces.
- GCE A-Level H2 PhysicsTopic 17(j) / Topic 17(k) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027