Velocity selection in crossed fields
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: How can crossed fields select particles with one speed?
In perpendicular electric and magnetic fields, a charge travels undeflected when electric and magnetic forces oppose and qE = qvB. The selected speed is v = E/B, independent of charge magnitude and mass. Other speeds curve because one force dominates.
Arrange crossed fields so their forces oppose
A velocity selector uses mutually perpendicular electric field, magnetic field and particle velocity. For one charge sign, choose the field directions so electric force QE and magnetic force BQv act oppositely.
An undeflected particle still experiences both forces; their vector sum is zero. Reversing the particle charge reverses both forces, so the balance condition is unchanged.
Check your understanding: Does undeflected motion mean neither field exerts force?
No. Electric and magnetic forces are equal and opposite.
Derive the selected speed and predict other paths
For undeflected motion, |Q|E = B|Q|v, giving v = E/B. Charge magnitude and mass cancel, so the selector passes any particle at that speed if its velocity and the fields have the required directions.
For a slower particle, electric force is unchanged but magnetic force is smaller; for a faster one, magnetic force is larger. This predicts opposite deflections on either side of the selected speed.
Check your understanding: What happens to a particle moving faster than E/B?
The magnetic force exceeds the electric force, so it deflects in the magnetic-force direction.
Key ideas to keep
- The fields, velocity and both force directions must be mutually consistent.
- Undeflected does not mean no forces act; the forces balance.
- Reversing charge reverses both forces, so the selected speed is unchanged.
See the reasoning
Worked example
Select one speed and classify nearby particles
Question: Crossed fields have E = 4.5 × 10⁴ N C⁻¹ and B = 0.18 T. Find the selected speed. State which force dominates for particles moving at 0.80 and 1.20 times this speed.
Step 1: Balance force magnitudes
Why: Undeflected particles have zero resultant transverse force.
Working: |Q|E = B|Q|v.
Step 2: Cancel charge and calculate
Why: Both forces act on the same charge.
Working: v = E/B = 4.5×10⁴/0.18 = 2.50×10⁵ m s⁻¹.
Step 3: Compare nearby speeds
Why: Electric force is speed-independent but magnetic force is proportional to v.
Working: At 0.80v electric force dominates; at 1.20v magnetic force dominates.
Answer: Selected speed is 2.50 × 10⁵ m s⁻¹; slower particles deflect with the electric force and faster particles with the magnetic force.
Check: The unit (N C⁻¹)/T reduces to m s⁻¹.
Another worked model
Question
Explain why a crossed-field selector transmits a narrow speed independent of charge magnitude and mass.
Check the worked solution
With forces opposed, |Q|E = B|Q|v, so v = E/B and |Q| cancels. Mass never enters the balance. Charge sign reverses both forces, leaving their opposition unchanged for the same geometry.
Use a hint if needed
Practise with support
Try this
E doubles and B halves. State the selected-speed factor.
Hint: Treat numerator and denominator changes separately.
Check your answer
v = E/B, so the factor is 2/(1/2) = 4.
Now work without the hint
Practise independently
Your turn
State all geometric and idealising conditions behind v = E/B.
Check your answer
Uniform E and B must be mutually perpendicular and both perpendicular to the selected beam velocity, with electric and magnetic forces opposite. Other forces and collisions are neglected, and particles must traverse the common field region.
Avoid these traps
Common mistakes
Common mistake
Every particle passes undeflected through crossed fields.
What is wrong with this reasoning?
Show better thinking
Only particles with the geometry and speed v = E/B have opposing forces of equal magnitude.
Write for the examiner
Exam guidance
Prove the forces oppose with a direction diagram before equating their magnitudes.
Exam-style practice [7 marks]
A beam travels undeflected through crossed fields E = 6.0 × 10⁴ N C⁻¹ and B = 0.24 T. Find its speed and explain why the result is independent of charge sign, charge magnitude and mass. Predict the effect of increasing speed by 10%.
Plan before you answer
- Draw opposing force directions.
- Equate magnitudes and cancel common factors.
- Compare the speed dependence of the forces.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
For no deflection, |Q|E = B|Q|v, so v = E/B = 6.0×10⁴/0.24 = 2.5×10⁵ m s⁻¹. Charge magnitude cancels, reversing charge reverses both force directions, and mass is absent from the force balance. At 10% greater speed the magnetic force is 10% larger while electric force is unchanged, so the beam bends in the magnetic-force direction.
Come back in three days
Check what stayed with you
Recall question 1
State the selector speed.
Check the answer
v = E/B.
Recall question 2
Why does charge sign not change the selected speed?
Check the answer
Both electric and magnetic forces reverse when charge sign reverses.
Recall question 3
Which force dominates above the selected speed?
Check the answer
The magnetic force.
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(m) · 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