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Deflection of Charged Particles (Electric & Magnetic Fields)

Key idea: Analyse charged-particle beam deflection in uniform electric and magnetic fields using qE, Bqv and circular/projectile motion methods (A Level Physics).

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Analyse forces and paths of moving charges in uniform fields.
  • Apply crossed electric and magnetic fields to velocity selection.

1. Definitions (Must Know)

  • Electric force on a charge in a uniform electric field: F_E = qE
  • Magnetic force on a moving charge in a uniform magnetic field: F_B = Bqv sin θ
    • θ is the angle between vector v and vector B.
    • If vector v⊥ vector B, then F_B = Bqv.
  • Centripetal force condition for uniform circular motion:
    • F_centripetal = mv²/r

2. Key Ideas (What Earns Marks)

A. Electric vs magnetic deflection (compare)

FeatureUniform electric fieldUniform magnetic field
Force directionalong vector E (for q > 0)perpendicular to both vector v and vector B
Speed changes?yes (field does work)no (force is perpendicular to motion)
Typical pathparabolic (projectile)circular (or helical if there is a component v_∥)
Key equationsF = qE, a = qE/mF = Bqv, r = mv/qB

B. Workflow for exam questions

  1. Determine the force formula (qE or Bqv sin θ).
  2. Use the sign of q to decide direction (or work with magnitudes, then state direction separately).
  3. Choose the motion model:
    • electric field → constant acceleration (SUVAT / projectile)
    • magnetic field with v⊥ B → circular motion
  4. Calculate the required deflection, radius, time, or exit angle.

3. Detailed Explanations

A. Deflection in a uniform electric field (between parallel plates)

If the particle enters with horizontal speed u and the field is vertical:

  • acceleration magnitude: a = qE/m
  • time inside the field region of length L: t = L/u
  • vertical deflection on leaving the field (initial vertical speed 0): y = (1/2)at² = (1/2)(qE/m)(L/u)²
  • vertical exit speed: v_y = at

B. Deflection in a uniform magnetic field (beam entering perpendicular to B)

If vector v⊥ vector B, the magnetic force has constant magnitude F = Bqv and is always perpendicular to vector v. This produces circular motion.

Set magnetic force equal to centripetal force:

Bqv = mv²/r; r = mv/qB

Direction is obtained from a right-hand rule (or Fleming’s left-hand rule if you treat it like “current”).

C. When the beam is not perpendicular to B (helical motion)

If the velocity has components:

  • v_⊥ (perpendicular to vector B) → circular motion
  • v_∥ (parallel to vector B) → constant speed along the field

The combined motion is a helix.

4. Common Mistakes

  • Forgetting that an electron has q = -1.60 × 10⁻¹⁹ C (direction reverses).
  • Using F = Bqv when vector v is not perpendicular to vector B (missing sin θ).
  • Treating magnetic force as doing work (it does not change speed).
  • In electric-field deflection: using the wrong time (it is based on horizontal motion: t = L/u).

5. Exam Tips

  • Write the force equation first, then state the motion model (“constant acceleration” or “circular motion”).
  • Quote a sign convention or use magnitudes and state the direction separately.
  • For electric fields, a short sketch of the velocity components helps avoid sign errors.

6. Worked Examples

Modelled example 1

Electric field deflection (exit displacement)

Core

Problem

A proton enters a 0.050 m long uniform electric-field region horizontally at 2.0 × 10⁶ m s⁻¹. With E = 1.0 × 10⁴ V m⁻¹, find its exit deflection. Use mₚ = 1.67 × 10⁻²⁷ kg and q = 1.60 × 10⁻¹⁹ C.
Study the worked solution

  1. Find transverse acceleration

    Method

    a = qE/m = 9.58 × 10¹¹ m s⁻².

    Reason

    The uniform electric force is constant.

    Working

    a = ((1.60 × 10⁻¹⁹)(1.0 × 10⁴))/(1.67 × 10⁻²⁷) = 9.58 × 10¹¹ m s⁻²

  2. Find time in the field

    Method

    t = L/u = 2.5 × 10⁻⁸ s.

    Reason

    The electric force is vertical, so horizontal speed remains constant.

    Working

    t = 0.050/(2.0 × 10⁶) = 2.5 × 10⁻⁸ s

  3. Find exit deflection

    Method

    y = 3.0 × 10⁻⁴ m.

    Reason

    Initial vertical speed is zero, so y = (1/2)at².

    Working

    y = (1/2)(9.58 × 10¹¹)(2.5 × 10⁻⁸)² = 3.0 × 10⁻⁴ m

Guided practice 2

Magnetic field deflection (radius of curvature)

About 5 min

Problem

An electron enters a 2.0 × 10⁻³ T field perpendicularly at 3.0 × 10⁷ m s⁻¹. Find the radius using mₑ = 9.11 × 10⁻³¹ kg and |q| = 1.60 × 10⁻¹⁹ C.

Try this before viewing the solution

Unit: m

Hints

Hint 1: use circular motion
Equate B|q|v with mv²/r.
View solution step by step

  1. Use the radius relation

    Method

    r = mv/(|q|B).

    Reason

    Perpendicular magnetic force is centripetal and does not change speed.

    Working

    r = mv/|q|B

  2. Evaluate

    Method

    r = 8.5 × 10⁻² m.

    Reason

    Charge sign affects curvature direction, not radius magnitude.

    Working

    r = ((9.11 × 10⁻³¹)(3.0 × 10⁷))/((1.60 × 10⁻¹⁹)(2.0 × 10⁻³)) = 8.5 × 10⁻² m

Common misconception 3

Direction check

Find and correct the mistake

Learner claim

A positive ion moves right while vector B is into the page. A learner points the magnetic force down by reversing the cross-product order. Diagnose the direction.

Try this before viewing the solution

Magnetic-force direction

View solution step by step

  1. Use the positive-charge rule

    Method

    Apply vector F = q(vector v × vector B).

    Reason

    The ion is positive, so no direction reversal is needed.

    Working

    vector v: right; vector B: into page.

  2. Evaluate the cross product

    Method

    The force is upward.

    Reason

    Right crossed into the page points upward.

    Working

    vector v × vector B: upward.

Examiner practice 4

Exit angle from an electric field region

3 marks

Examination question

For Example 1, t = 2.5 × 10⁻⁸ s, a = 9.58 × 10¹¹ m s⁻² and horizontal speed is 2.0 × 10⁶ m s⁻¹. Find the exit angle. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Find vertical speed

    1 mark

    Method

    v_y = 2.40 × 10⁴ m s⁻¹.

    Reason

    Initial vertical speed is zero.

    Working

    v_y = at = (9.58 × 10¹¹)(2.5 × 10⁻⁸) = 2.40 × 10⁴ m s⁻¹

  2. Form the direction ratio

    1 mark

    Method

    tan θ = 1.20 × 10⁻².

    Reason

    The horizontal component remains u.

    Working

    tan θ = v_y/u = 2.40 × 10⁴/(2.0 × 10⁶)

  3. Find the angle

    1 mark

    Method

    θ ≈ 0.69°.

    Reason

    Take the inverse tangent of the component ratio.

    Working

    θ = tan⁻¹ (1.20 × 10⁻²) ≈ 0.69°

Challenge 5

Compare deflection for different particles

Minimal support

Independent transfer

Proton and alpha beams enter the same uniform electric-field region with equal horizontal speed. Protons have (+e,mₚ) and alpha particles (+2e,4mₚ). Compare their exit deflections and directions.

Try this before viewing the solution

Hints

Hint 1: compare charge-to-mass ratios
The time in the field is common, so y ∝ q/m.
View solution step by step

  1. Establish the scaling

    Method

    y ∝ q/m.

    Reason

    a = qE/m and the common L/u gives the same transit time.

    Working

    y = (1/2)(qE/m)(L/u)²

  2. Compare ratios

    Method

    The alpha q/m is half the proton value.

    Reason

    (2e)/(4mₚ) = e/(2mₚ).

    Working

    ((q/m)_α)/((q/m)ₚ) = 1/2

  3. State deflection

    Method

    The alpha beam deflects half as much, in the same direction.

    Reason

    Both charges are positive, so their electric forces point the same way.

    Working

    y_α = (1/2)yₚ.

7. Mind Stretchers

Mind stretcher 1: Example: Speed needed for no deflection in crossed fieldsExtension

A beam passes through perpendicular electric and magnetic fields and is undeflected. Show that its speed must be v = E/B.

Show Answer

Undeflected means the forces balance in magnitude: qE = Bqv ⇒ v = E/B

Mind stretcher 2: Example: Why does the magnetic field path depend on mass?Extension

In a uniform magnetic field with vector v⊥ vector B, explain why heavier particles curve less (larger radius) if they have the same charge and speed.

Show Answer

For vector v⊥ vector B, the radius is: r = mv/|q|B

With v, |q| and B fixed, r ∝ m. So a larger mass gives a larger radius (less curvature).

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