Explain velocity selection with electric and magnetic fields
Balance electric and magnetic forces so only particles at one speed pass through undeflected.
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
What condition lets a charged particle pass through crossed fields undeflected?
What you will learn
- Explain the opposing electric and magnetic forces in a velocity selector
- Use v = E/B for undeflected motion
- Explain how slower or faster particles are deflected
Crossed fields can balance for one speed
In a velocity selector, an electric field and a magnetic field are arranged so their forces on a moving charged particle point in opposite directions. Their magnitudes are |q|E and |q|vB when the motion is perpendicular to B.
For an undeflected particle, |q|E = |q|vB, so v = E/B. The charge magnitude cancels: the selected speed depends on field strengths, not on the particle's mass or charge magnitude.
Other speeds leave a residual force
If v is below E/B, the electric force is larger than the magnetic force; if v is above E/B, the magnetic force is larger. The particle is deflected in the direction of the larger force. Which plate or side it reaches depends on the field orientations and the sign of the charge.
The force directions both reverse for a negative charge, so the balance speed is unchanged. Set the geometry before naming the direction of any residual deflection.
Pause and explain
For E = 2.0 × 10⁴ V m⁻¹ and B = 0.10 T, what speed is selected?
Worked example
A selector uses E = 3.0 × 10⁴ V m⁻¹ and B = 0.20 T. Find the selected speed.
Show the worked solution
- Use v = E/B.
- Substitute the field strengths.
- Check units: (V m⁻¹)/T = m s⁻¹.
Answer v = 3.0 × 10⁴/0.20 = 1.5 × 10⁵ m s⁻¹.
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.
A selector has E = 4.5 × 10⁴ V m⁻¹ and B = 0.15 T. Calculate the speed that passes undeflected.
Give me a hint
Set |q|E = |q|vB and cancel |q|.
Compare my reasoning
- v = E/B.
- v = 4.5 × 10⁴/0.15.
- v = 3.0 × 10⁵ m s⁻¹.
v = 3.0 × 10⁵ m s⁻¹.
Look for these in your work
- Force balance written
- Charge cancelled
- Selected speed stated
A particle travels more slowly than the selected speed. Which force is larger in magnitude?
Give me a hint
Compare |q|E with |q|vB when v is below E/B.
Compare my reasoning
- At the selected speed the forces balance.
- Reducing v reduces magnetic-force magnitude |q|vB.
- The electric force is therefore larger and the particle deflects toward its electric-force direction.
The electric force is larger; the particle deflects in the electric-force direction.
Look for these in your work
- Speed comparison made
- Magnetic force dependence used
- Residual force direction described
Would a velocity selector pass particles with different charge magnitudes at the same selected speed? Explain.
Give me a hint
Inspect the force-balance equation before cancelling charge.
Compare my reasoning
- Balance requires |q|E = |q|vB.
- For a charged particle, |q| appears on both sides and cancels.
- Thus the selected speed is E/B, independent of charge magnitude.
Yes, in the ideal selector the same speed E/B is selected independent of charge magnitude.
Look for these in your work
- Force balance used
- Charge cancellation shown
- Conclusion stated
Common mistakes
- Adding the opposing force magnitudes for the undeflected case
- Including charge in v = E/B after it cancels
- Assuming particles at every speed pass straight through
- Naming a plate direction without specifying field orientation and charge sign
What can you explain now?
What condition lets a charged particle pass through crossed fields undeflected?
Compare with the explanation
Electric and magnetic force magnitudes are equal and opposite
The resultant transverse force must be zero.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Why does a velocity selector choose speed E/B independently of a particle's charge magnitude?
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.