Charged-particle beams in uniform fields
Key idea: H2 Physics lessons on current-produced fields, magnetic forces and crossed-field velocity selection.
Continue where you stopped
The core idea
Build the idea
Learn the idea
Big question: How do uniform fields deflect a beam of charged particles?
A uniform electric field gives constant force qE and can change a particle's speed and direction. A uniform magnetic field gives perpendicular force qvB and circular curvature without changing speed. Beam paths therefore reveal charge sign, momentum and field direction when the entry velocity is known.
Treat electric deflection as constant acceleration
In a uniform electric field, a particle has force F = QE and constant acceleration a = QE/m. Force follows E for a positive charge and opposes E for a negative charge.
A beam entering perpendicular to E keeps its along-plate velocity component while accelerating across the plates, producing a parabolic path in the ideal uniform region. The electric field can do work and change speed.
Check your understanding: Why is the ideal electric-field path parabolic?
It combines constant velocity in one direction with constant acceleration in the perpendicular direction.
Treat magnetic deflection as constant-speed curvature
For v perpendicular to B, magnetic force B|Q|v stays perpendicular to motion and supplies centripetal force. Equating B|Q|v = mv²/r gives r = mv/(|Q|B).
A magnetic field changes direction without changing speed. Larger momentum produces a larger radius; reversing charge reverses the curvature.
Check your understanding: What happens to magnetic curvature if momentum doubles at fixed Q and B?
The radius doubles, so the path curves less sharply.
Key ideas to keep
- Electric and magnetic deflections need not point the same way.
- Magnetic radius r = mv/(|q|B) grows with momentum.
- A beam contains many particles, but each follows the same ideal path if its velocity is the same.
See the reasoning
Worked example
Calculate electric deflection between plates
Question: An electron enters horizontally at 3.0 × 10⁷ m s⁻¹ between plates 0.080 m long. A downward uniform field of 2.0 × 10⁴ N C⁻¹ acts between them. Find its vertical deflection. Use e = 1.60 × 10⁻¹⁹ C and mₑ = 9.11 × 10⁻³¹ kg.
Step 1: Find time in the field
Why: Horizontal velocity remains constant.
Working: t = 0.080/(3.0×10⁷) = 2.67×10⁻⁹ s.
Step 2: Find vertical acceleration
Why: The electron force is opposite the downward field.
Working: a = eE/m = 3.51×10¹⁵ m s⁻² upward.
Step 3: Use perpendicular kinematics
Why: Initial vertical velocity is zero.
Working: y = ½at² = 0.0125 m upward.
Answer: The electron is deflected about 1.25 cm upward.
Check: Its direction is opposite E because the electron is negative.
Find a magnetic-beam radius
Question: A proton of momentum 4.0 × 10⁻²⁰ kg m s⁻¹ enters perpendicular to a 0.50 T field. Find the path radius.
Step 1: Identify the radial force
Why: The perpendicular magnetic force continually turns the proton.
Working: Bqv = mv²/r.
Step 2: Replace mv by momentum
Why: The question supplies p directly.
Working: r = p/(qB).
Step 3: Substitute
Why: A proton has charge magnitude 1.60 × 10⁻¹⁹ C.
Working: r = 4.0×10⁻²⁰/[1.60×10⁻¹⁹(0.50)] = 0.50 m.
Answer: The magnetic path radius is 0.50 m.
Check: Greater momentum or weaker field would give a larger radius.
Another worked model
Question
Derive the radius of a charged particle moving perpendicular to a uniform magnetic field.
Check the worked solution
The magnetic force supplies centripetal force: B|Q|v = mv²/r. Hence r = mv/(B|Q|). The force changes direction, not speed, so the path is circular while the field is uniform.
Use a hint if needed
Practise with support
Try this
A proton's speed doubles in the same perpendicular magnetic field. State the factors by which magnetic force and path radius change.
Hint: Apply the speed dependence in each stated equation.
Check your answer
F = BQv doubles and r = mv/(BQ) also doubles, so the path bends less sharply.
Now work without the hint
Practise independently
Your turn
Analyse a charged beam entering a uniform electric field with transverse velocity, then a uniform magnetic field.
Check your answer
In E, constant qE gives constant transverse acceleration and a parabola while longitudinal velocity stays constant. In perpendicular B, qvB is centripetal, giving a circular arc of radius mv/(B|q|). Electric deflection can change speed; magnetic deflection cannot.
Avoid these traps
Common mistakes
Common mistake
Magnetic force changes a particle's kinetic energy.
What is wrong with this reasoning?
Show better thinking
It is perpendicular to velocity and does no work; it changes direction, not speed.
Common mistake
Electric and magnetic fields always produce the same path shape.
What is wrong with this reasoning?
Show better thinking
A uniform transverse E gives constant acceleration and a parabola; a uniform perpendicular B gives a circular arc.
Write for the examiner
Exam guidance
Sketch the initial velocity and force at entry before choosing a path shape.
Exam-style practice [8 marks]
Compare the deflection of a positive particle entering perpendicular to a uniform electric field and to a uniform magnetic field. Include force, path shape, speed and energy, and derive the magnetic radius expression.
Plan before you answer
- Write each force equation.
- State whether each force can do work.
- Use centripetal force for the magnetic radius.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
The electric field gives constant force QE and acceleration QE/m. With perpendicular entry, uniform motion along the plates combines with accelerated motion across them to give a parabola; electric work can change speed and kinetic energy. A perpendicular magnetic field gives force BQv at right angles to velocity, so the path is a circular arc and speed is unchanged. From BQv = mv²/r, r = mv/(QB) in magnitude.
Come back in three days
Check what stayed with you
Recall question 1
What path results from perpendicular entry into a uniform electric field?
Check the answer
A parabola in the ideal model.
Recall question 2
What path results from perpendicular entry into a uniform magnetic field?
Check the answer
A circular arc at constant speed.
Recall question 3
State the magnetic radius relation.
Check the answer
r = mv/(|Q|B).
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(l) · 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