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Velocity Selector

Key idea: Use crossed electric and magnetic fields to select particles of speed v = E/B, with clear force directions and exam-style calculations (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Apply crossed electric and magnetic fields to velocity selection.

1. Definitions (Must Know)

A. Velocity selector

A velocity selector uses perpendicular electric and magnetic fields to allow only particles with a particular speed to pass through undeflected.

B. Forces used

  • Electric force: F_E = qE
  • Magnetic force (when vector v⊥ vector B): F_B = Bqv

2. Key Ideas (What Earns Marks)

  • For no deflection, forces must balance in magnitude: qE = Bqv ⇒ v = E/B
  • The selected speed does not depend on q or m (they cancel).
  • Direction matters:
    • vector E gives a force along vector E for q > 0 (opposite for q < 0),
    • vector B gives a force perpendicular to both vector v and vector B.
Syllabus scope (9478)

This lesson targets learning outcome 17m: perpendicular electric and magnetic fields used for velocity selection.

3. Detailed Explanations

A. How the selector works (why only one speed passes)

Imagine a beam entering a region where:

  • vector E is vertical,
  • vector B is perpendicular to the page,
  • the beam velocity vector v is horizontal.

Then:

  • electric force has magnitude qE (constant),
  • magnetic force has magnitude Bqv (depends on v),
  • the two forces can be arranged to oppose each other.

Only one speed gives equal magnitudes, so the net force is zero and the beam travels straight.

Velocity selection with crossed electric and magnetic fieldsA positive particle beam travels right through a downward electric field and a magnetic field into the page. Opposing electric and magnetic force arrows balance for speed E divided by B, while slower and faster paths curve in opposite directions.+−B into page (×)Epositive beam+magnetic force Bqvelectric force qEv = E/B: straightfaster: magnetic force winsslower: electric force wins
Scroll diagram horizontally to read all labels.
For a positive charge, the electric and magnetic forces oppose. Only v = E/B makes their magnitudes equal; slower and faster particles bend to opposite sides.

B. What happens if v is too small or too large?

Assuming the forces oppose:

  • if v < E/B, then Bqv < qE → electric force dominates → deflection in the electric-force direction.
  • if v > E/B, then Bqv > qE → magnetic force dominates → deflection in the magnetic-force direction.

Velocity selector: compare E and Bv (example values)

A constant E line and a rising Bv line; their intersection gives the selected speed v = E/B.

Scroll across the graph to read all labels.

A constant E line and a rising Bv line; their intersection gives the selected speed v = E/B.A constant E line and a rising Bv line; their intersection gives the selected speed v = E/B.
Where the lines cross, Bv = E so qE and Bqv balance: the beam is undeflected and v = E/B (here 1.5×10⁵ m s⁻¹).
Open full-size graph
View figure data
Values and uncertainty for Velocity selector: compare E and Bv (example values)
SeriesParticle speed, v (m s⁻¹)Particle speed, v uncertaintyForce per unit charge (N C⁻¹)Force per unit charge uncertainty
Electric field: E = 3.0×10⁴ N C⁻¹030000
Electric field: E = 3.0×10⁴ N C⁻¹30000030000
Magnetic term: Bv (B = 0.20 T)00
Magnetic term: Bv (B = 0.20 T)10000020000
Magnetic term: Bv (B = 0.20 T)15000030000
Magnetic term: Bv (B = 0.20 T)20000040000
Magnetic term: Bv (B = 0.20 T)30000060000

4. Common Mistakes

  • Forgetting that the magnetic force depends on v (so it is the “speed filter”).
  • Using F_B = Bqv when vector v is not perpendicular to vector B (should be Bqv sin θ).
  • Mixing up field directions and force directions (especially for negative charges).
  • Using E in V/cm instead of converting to V/m.

5. Exam Tips

  • Write the balance condition first: qE = Bqv.
  • Cancel q explicitly so you don’t lose signs.
  • If asked for directions, draw a quick sketch of vector v, vector E, and vector B and label the two forces.

6. Worked Examples

Modelled example 1

Find the selected speed

Core

Problem

A velocity selector has E = 3.0 × 10⁴ V m⁻¹ and B = 0.20 T. Find the speed that passes undeflected and state the required force relationship.
Study the worked solution

  1. Set the balance

    Method

    Use |q|E = B|q|v with the forces opposed.

    Reason

    Undeflected motion requires zero resultant transverse force.

    Working

    |q|E = B|q|v

  2. Cancel charge and solve

    Method

    v = E/B.

    Reason

    Both force magnitudes contain the same charge magnitude.

    Working

    v = E/B

  3. Evaluate

    Method

    v = 1.5 × 10⁵ m s⁻¹.

    Reason

    The result depends only on the two field magnitudes under the perpendicular geometry.

    Working

    v = (3.0 × 10⁴)/0.20 = 1.5 × 10⁵ m s⁻¹

Guided practice 2

Find E for a required speed

About 4 min

Problem

Select ions of speed 2.0 × 10⁶ m s⁻¹ using B = 0.50 T. Find the required electric field magnitude.

Try this before viewing the solution

Unit: V m⁻¹

Hints

Hint 1: make electric field the subject
The balanced-force condition gives E = Bv.
View solution step by step

  1. Rearrange the selector condition

    Method

    Use E = vB.

    Reason

    The electric force must match the speed-dependent magnetic force.

    Working

    E = vB

  2. Evaluate

    Method

    E = 1.0 × 10⁶ V m⁻¹.

    Reason

    Multiply the target speed by magnetic flux density.

    Working

    E = (2.0 × 10⁶)(0.50) = 1.0 × 10⁶ V m⁻¹

Common misconception 3

Find B for a required speed

Find and correct the mistake

Learner claim

For E = 5.0 × 10⁴ V m⁻¹ and selected speed 2.0 × 10⁵ m s⁻¹, a learner writes B = Ev. Diagnose the rearrangement and calculate B.

Try this before viewing the solution

Unit: T

View solution step by step

  1. Repair the rearrangement

    Method

    Use B = E/v.

    Reason

    The balance condition is E = Bv.

    Working

    B = E/v

  2. Evaluate

    Method

    B = 0.25 T.

    Reason

    The quotient has the correct field magnitude and units.

    Working

    B = (5.0 × 10⁴)/(2.0 × 10⁵) = 0.25 T

Examiner practice 4

Find v from plate voltage and separation

3 marks

Examination question

Parallel plates have 600 V across 2.0 × 10⁻² m and are crossed with B = 0.15 T. Find the uniform electric field and selected speed, stating the balance condition. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Find electric field

    1 mark

    Method

    E = 3.0 × 10⁴ V m⁻¹.

    Reason

    The approximately uniform plate field is V/d.

    Working

    E = 600/(2.0 × 10⁻²) = 3.0 × 10⁴ V m⁻¹

  2. State force balance

    1 mark

    Method

    |q|E = B|q|v with opposite directions.

    Reason

    Undeflected transmission requires zero net force.

    Working

    v = E/B

  3. Calculate speed

    1 mark

    Method

    v = 2.0 × 10⁵ m s⁻¹.

    Reason

    Divide the electric field by magnetic flux density.

    Working

    v = (3.0 × 10⁴)/0.15 = 2.0 × 10⁵ m s⁻¹

Challenge 5

Net force when the speed is not selected

Minimal support

Independent transfer

A positive ion enters a selector with E = 2.0 × 10⁴ V m⁻¹, B = 0.10 T and v = 1.5 × 10⁵ m s⁻¹, with the two forces arranged to oppose. Determine whether it is deflected and which way relative to vector E.

Try this before viewing the solution

Hints

Hint 1: compare force per charge
Compare E with Bv before using the positive charge sign.
View solution step by step

  1. Compare force magnitudes

    Method

    Bv = 1.5 × 10⁴ V m⁻¹ < E.

    Reason

    Dividing both force magnitudes by |q| leaves E and Bv.

    Working

    Bv = (0.10)(1.5 × 10⁵) = 1.5 × 10⁴ V m⁻¹

  2. Predict deflection

    Method

    The electric force is larger, so the ion deflects in the direction of vector E.

    Reason

    The ion is positive and the opposed magnetic force is too small to balance it.

    Working

    F_E > F_B; net force follows vector E.

7. Mind Stretchers

Mind stretcher 1: Why does the selected speed not depend on charge sign?Extension

Explain why the speed condition v = E/B is the same for positive and negative charges (even though the forces reverse direction).

Show Answer

For a given field arrangement, both F_E = qE and F_B = Bqv reverse direction when q changes sign.

So the forces still oppose each other and balance at the same speed because the magnitudes are still equal when |q|E = B|q|v.

Mind stretcher 2: How velocity selection helps measure m/qExtension

Explain why a velocity selector is often used before a “B-field only” region when an instrument needs to determine m/q for ions.

Show Answer

The selector ensures all ions entering the next region have the same speed v = E/B (known from fields you control).

In a magnetic-only region, the radius is: r = mv/|q|B So measuring r lets you find m/q because v and B are known. Without velocity selection, different speeds would give different radii even for the same m/q.

8. Optional (Enrichment)

See Deflection of Charged Particles for the standard “electric field gives parabolic path; magnetic field gives circular/helical path” comparison.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027