Mini Physics
Part of Mini EducationSwitch subject
Search

Motion of a Moving Charge in a Uniform Magnetic Field

Key idea: Use F = Bqv sinθ and circular-motion ideas to analyse the motion of a charged particle in a uniform magnetic field, including helical paths (A Level Physics).

  • GCE A-Level H2 Physics 2027
On this page

Learning objectives

  • Analyse forces and paths of moving charges in uniform fields.

1. Definitions (Must Know)

A. Magnetic force on a moving charge

A charge q moving with speed v in a magnetic field of flux density B experiences a magnetic force of magnitude:

F = Bqv sin θ

where θ is the angle between vector v and vector B.

B. Special cases

  • If vector v⊥ vector B, then sin θ = 1 and F = Bqv.
  • If vector v∥ vector B, then sin θ = 0 and F = 0.

2. Key Ideas (What Earns Marks)

  • Direction: the magnetic force is perpendicular to both vector v and vector B (use a right-hand rule / vector v × vector B).
  • A magnetic force does no work on the charge because it is perpendicular to the motion, so the speed stays constant.
  • If vector v⊥ vector B, the motion is uniform circular motion because the magnetic force provides the centripetal force: Bqv = mv²/r ⇒ r = mv/|q|B
  • If vector v has a component parallel to vector B, the motion is helical (circular motion + constant forward motion).
Common shortcut

In calculations, use magnitudes with |q| and state the direction separately (especially for electrons where q < 0).

3. Detailed Explanations

A. Why the force is perpendicular (what it implies)

The magnetic force always acts at right angles to the velocity, so:

  • it changes the direction of the velocity,
  • but not the speed.

So kinetic energy stays constant (no work done by the magnetic force).

B. Circular motion when vector v⊥ vector B

If vector v⊥ vector B, F = Bqv has constant magnitude and is always perpendicular to vector v.

That is exactly the condition for uniform circular motion. Using F_centripetal = mv²/r:

Bqv = mv²/r; r = mv/|q|B

Cyclotron radius vs magnetic flux density (example)

An inverse curve showing that increasing B reduces the radius of curvature for a fixed particle speed and charge.

Scroll across the graph to read all labels.

An inverse curve showing that increasing B reduces the radius of curvature for a fixed particle speed and charge.An inverse curve showing that increasing B reduces the radius of curvature for a fixed particle speed and charge.
Since r ∝ 1/B, doubling B halves the radius. This is a quick proportionality check in radius questions.
Open full-size graph
View figure data
Values for Cyclotron radius vs magnetic flux density (example)
Magnetic flux density, B (T)Proton at v = 3.0×10⁶ m s⁻¹ (r = mv/|q|B)
0.20.156
0.30.104
0.40.078
0.50.063
0.60.052
0.80.039

C. Helical motion when vector v is not perpendicular

Split the velocity into components:

  • v_⊥ (perpendicular to vector B) → circular motion,
  • v_∥ (parallel to vector B) → unchanged.

The combined path is a helix:

Charged-particle paths in a uniform magnetic fieldOne panel shows a positive charge following a circle when velocity is perpendicular to the magnetic field. A second panel shows a helical path when velocity has both perpendicular and parallel components.velocity perpendicular to B+vF toward centrespeed constant; radius r = mv/(|q|B)perpendicular and parallel componentsB; parallel motion →+parallel speed sets pitch; perpendicular speed sets radius
Scroll diagram horizontally to read all labels.
The perpendicular velocity component produces circular motion. A parallel component is unchanged, so the combined path is helical.

4. Common Mistakes

  • Using F = Bqv without checking whether sin θ is needed.
  • Forgetting that q can be negative (direction reverses).
  • Claiming the particle “speeds up” in a magnetic field (it does not, in this model).
  • Mixing up B (tesla) with magnetic flux Φ (weber).

5. Exam Tips

  • Start by stating the angle: “vector v⊥ vector B, so sin θ = 1”.
  • Use magnitudes for calculations, then add one clear direction sentence.
  • If asked for radius in a magnetic field, write: Bqv = mv²/r then rearrange.

6. Worked Examples

Modelled example 1

Force on a charge

Core

Problem

An electron at 2.0 × 10⁷ m s⁻¹ moves perpendicular to a 1.5 × 10⁻³ T field. Find the force magnitude using |q| = 1.60 × 10⁻¹⁹ C, and state how direction would be determined.
Study the worked solution

  1. Apply the perpendicular case

    Method

    Use F = B|q|v.

    Reason

    θ = 90°, so sin θ = 1; charge magnitude gives force magnitude.

    Working

    F = B|q|v

  2. Evaluate

    Method

    F = 4.8 × 10⁻¹⁵ N.

    Reason

    Substitute the field, charge magnitude and speed.

    Working

    F = (1.5 × 10⁻³)(1.60 × 10⁻¹⁹)(2.0 × 10⁷) = 4.8 × 10⁻¹⁵ N

  3. Determine direction

    Method

    Find vector v × vector B for positive charge, then reverse it for the electron.

    Reason

    The electron has negative charge.

    Working

    Direction requires the stated vector orientations.

Guided practice 2

Radius of circular motion

About 5 min

Problem

A proton enters a uniform 0.40 T field perpendicularly at 3.0 × 10⁶ m s⁻¹. Find its path radius using mₚ = 1.67 × 10⁻²⁷ kg and |q| = 1.60 × 10⁻¹⁹ C.

Try this before viewing the solution

Unit: m

Hints

Hint 1: make magnetic force centripetal
Set B|q|v = mv²/r.
View solution step by step

  1. Derive the radius relation

    Method

    Use r = mv/(|q|B).

    Reason

    The perpendicular magnetic force supplies the centripetal force.

    Working

    B|q|v = mv²/r ⇒ r = mv/|q|B

  2. Evaluate

    Method

    r = 7.8 × 10⁻² m.

    Reason

    The speed remains constant while the force turns the velocity.

    Working

    r = ((1.67 × 10⁻²⁷)(3.0 × 10⁶))/((1.60 × 10⁻¹⁹)(0.40)) = 7.8 × 10⁻² m

Common misconception 3

Angled entry

Find and correct the mistake

Learner claim

A charge enters a uniform field at angle θ to vector B. A learner uses the full speed in r = mv/(|q|B) and predicts a planar circle. Diagnose both the radius input and path shape.

Try this before viewing the solution

Circular-motion speed component

View solution step by step

  1. Resolve velocity

    Method

    Use v_⊥ = v sin θ for the circular component.

    Reason

    The parallel component experiences no magnetic force.

    Working

    v_⊥ = v sin θ, v_∥ = v cos θ

  2. Correct the radius

    Method

    r = mv sin θ/(|q|B).

    Reason

    The perpendicular component alone supplies circular motion.

    Working

    r = mv_⊥/|q|B = (mv sin θ)/|q|B

  3. Correct the path

    Method

    The combined motion is helical, not a planar circle.

    Reason

    v_∥ remains constant along the field while v_⊥ circulates.

    Working

    Circular transverse motion + uniform parallel motion ⇒ helix.

Examiner practice 4

Period of circular motion

3 marks

Examination question

A proton moves perpendicular to a uniform 0.20 T field. Derive the period relation and calculate the period using mₚ = 1.67 × 10⁻²⁷ kg and |q| = 1.60 × 10⁻¹⁹ C. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Find the radius

    1 mark

    Method

    r = mv/(|q|B).

    Reason

    Magnetic force supplies centripetal force.

    Working

    B|q|v = mv²/r ⇒ r = mv/(|q|B)

  2. Derive the period

    1 mark

    Method

    T = 2π m/(|q|B).

    Reason

    Use T = 2π r/v; the speed cancels.

    Working

    T = (2π r)/v = (2π m)/|q|B

  3. Evaluate

    1 mark

    Method

    T = 3.28 × 10⁻⁷ s.

    Reason

    The non-relativistic period depends on mass, charge magnitude and field, not speed.

    Working

    T = (2π(1.67 × 10⁻²⁷))/((1.60 × 10⁻¹⁹)(0.20)) = 3.28 × 10⁻⁷ s

Challenge 5

Helical pitch

Minimal support

Independent transfer

An electron enters a 1.0 × 10⁻³ T field with v_⊥ = 2.0 × 10⁷ m s⁻¹ and v_∥ = 1.0 × 10⁷ m s⁻¹. Find the helix pitch using mₑ = 9.11 × 10⁻³¹ kg and |q| = 1.60 × 10⁻¹⁹ C, and identify which velocity component sets it.

Try this before viewing the solution

Hints

Hint 1: advance during one transverse turn
Find T = 2π m/(|q|B), then multiply by the unchanged velocity along vector B.
View solution step by step

  1. Find one-turn time

    Method

    T = 3.58 × 10⁻⁸ s.

    Reason

    The transverse circular period is independent of v_⊥ in this model.

    Working

    T = (2π(9.11 × 10⁻³¹))/((1.60 × 10⁻¹⁹)(1.0 × 10⁻³)) = 3.58 × 10⁻⁸ s

  2. Find axial advance

    Method

    p = 0.358 m ≈ 0.36 m.

    Reason

    Pitch is the distance advanced parallel to the field during one turn.

    Working

    p = v_∥ T = (1.0 × 10⁷)(3.58 × 10⁻⁸) = 0.358 m

  3. Identify the controlling component

    Method

    v_∥ sets the pitch for a given period.

    Reason

    v_⊥ sets the helix radius, not its axial advance per turn.

    Working

    Pitch = v_∥ T.

7. Mind Stretchers

Mind stretcher 1: Why does B not change the speed?Extension

Explain, using work/energy, why a uniform magnetic field does not change a particle’s speed.

Show Answer

Work done is W = vector F · vector s. The magnetic force is always perpendicular to the displacement, so vector F · vector s = 0 and W = 0.

So kinetic energy does not change, and the speed stays constant.

Mind stretcher 2: Why is the period independent of speed?Extension

Show that, for non-relativistic motion with vector v⊥ vector B, the period T of circular motion is independent of the particle’s speed.

Show Answer

For circular motion: B|q|v = mv²/r ⇒ r = mv/|q|B

But v = (2π r)/T ⇒ T = (2π r)/v.

Substitute r = mv/|q|B: T = (2π/v)(mv/|q|B) = (2π m)/|q|B

v cancels, so T is independent of speed (in this model).

8. Optional (Enrichment)

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