AQA A-level Physics 7408 · Paper 2 · 3.7.5.1–3.7.5.2
Magnetic fields and charged particles
Master force on wires and moving charges, direction rules, circular paths and cyclotron physics. Diagnose the signs first, use the three-mode simulator, then complete a fully worked 33-mark assessment.
Content diagnostic
Can you separate field, charge and force?
Exam route
A reliable order for every magnetic-force question
Check the geometry
The AQA equations F = BIl and F = BQv apply when current or velocity is perpendicular to the field. Parallel motion gives zero magnetic force.
Find the positive direction
Use conventional current in Fleming’s left-hand rule. A positive moving charge follows the same direction rule; reverse the answer for a negative charge.
Identify what the force does
A perpendicular magnetic force changes direction but does no work. For a free particle it is the centripetal resultant, not an extra force.
3.7.5.1
Force on a current-carrying wire
Perpendicular wire
F is force in newtons, B is magnetic flux density in tesla, I is conventional current in amperes and l is the active wire length in metres inside the field.
Rearranging gives B = F/(Il). Magnetic flux density is therefore the force per unit current per unit length on a straight conductor perpendicular to the field.
The tesla
A uniform field has flux density 1 T if it produces a force of 1 N on a 1 m wire carrying 1 A perpendicular to the field.
Direction rules
Fleming’s left-hand rule without sign mistakes
First finger: field
Point from magnetic north to south. On a page, crosses mean into the page and dots mean out of the page.
Second finger: current
Use conventional current, the direction positive charge would move. Electron drift is opposite to conventional current.
Thumb: force
The thumb gives force or motion. Reverse field or current alone and force reverses; reverse both and the original force returns.
Moving charges
Treat a positive charge’s velocity as conventional current. For a negative charge, find the positive-charge result and reverse it.
3.7.5.2
Force on a moving charge
Magnitude
This AQA relationship is for motion perpendicular to the field. Use the magnitude |Q| for force size, then handle the sign separately for direction.
What stays constant?
The magnetic force is always perpendicular to instantaneous velocity. Work done is Fs cos 90° = 0, so speed and kinetic energy remain constant even while velocity direction changes.
Useful simulation
Magnetic force, particle paths and cyclotron explorer
Start with the wire model to practise direction reversals. Then compare proton, electron, alpha and deuteron paths. In cyclotron mode, connect the fixed alternating frequency with energy gained at every gap crossing.
Circular particle motion
Derive radius, period and frequency
- For perpendicular entry, magnetic force provides the centripetal resultant: BQv = mv²/r.
- Cancel one factor of v and rearrange: r = mv/(BQ).
- Use T = 2πr/v and substitute for r: T = 2πm/(BQ).
- Invert the period: f = BQ/(2πm).
| Change | Radius r | Period T | Curve direction |
|---|---|---|---|
| Increase speed v | Increases in direct proportion | Unchanged | Unchanged |
| Increase mass m | Increases | Increases | Unchanged |
| Increase |Q| | Decreases | Decreases | Unchanged if sign stays the same |
| Reverse charge sign or field | Same magnitude | Same magnitude | Reverses |
Cyclotron application
Why the cyclotron stays in step
Inside a dee
The metal dee shields the particle from the alternating electric field. The uniform magnetic field bends the particle through a semicircle but does no work, so its speed is constant during that semicircle.
Across the gap
The alternating p.d. reverses every half-turn. The electric field across the gap accelerates the particle, increasing its kinetic energy by QV on each crossing.
Growing path
After each acceleration the speed is greater. Since r = mv/(BQ), the next semicircle has a larger radius, producing the outward spiral.
Fixed frequency
In the non-relativistic model, frequency is independent of speed and radius. That is why one alternating frequency can repeatedly accelerate the chosen ion.
Misconception repair
Six errors that lose Paper 2 marks
“Use the right-hand rule”
For motor-force direction, AQA expects Fleming’s left-hand rule. State what each digit represents.
“Electron flow is the current”
Fleming’s rule uses conventional current. Reverse the electron-drift direction before using it as current.
“F = BIl always”
The displayed AQA equation is for a perpendicular wire. A parallel wire experiences no magnetic force.
“Centripetal force is extra”
Magnetic force is the centripetal resultant. Do not add mv²/r as another physical force.
“Curving means speeding up”
A magnetic force is perpendicular to velocity, so it changes direction without changing kinetic energy.
“The cyclotron magnet accelerates”
The electric field across the gap supplies energy. The magnetic field bends the path and does zero work.
Exam technique
A compact answer checklist
- Write “perpendicular” before using F = BIl or F = BQv.
- Give force magnitude and direction as separate steps.
- Use |Q| in radius and period magnitudes; use charge sign for curvature direction.
- When deriving r, explicitly equate BQv to mv²/r and cancel v.
- For a cyclotron, separate magnetic bending inside a dee from electric acceleration across the gap.
- Include units: T, N, m, s, Hz and J.
Original exam-style assessment
Magnetic fields and charged particles: 33 marks
Attempt every question before opening its mark scheme. Each question has been worked independently, exact-tag checked and designed for typed or calculated responses; none depends on drawing.
1. A straight wire of active length 0.145 m carries a current of 6.40 A at right angles to a uniform magnetic field of flux density 0.380 T. Calculate the force on the wire. Then state the definition of magnetic flux density and hence explain what is meant by one tesla.[4 marks]
Mark scheme
- (M1) Uses F = BIl because the wire is perpendicular to the field.
- (A1) F = 0.380 × 6.40 × 0.145 = 0.353 N.
- (B1) Magnetic flux density is force per unit current per unit length on a straight conductor perpendicular to the field.
- (B1) One tesla gives a force of 1 N on each metre of wire carrying a current of 1 A perpendicular to the field.
Exact AQA content: 3.7.5.1(a), 3.7.5.1(c)
2. A horizontal wire carries conventional current from west to east through a magnetic field directed vertically into the page. Use Fleming’s left-hand rule to state the force direction. State the direction after the current alone is reversed, and state the force if the wire is rotated until its current is parallel to the field.[4 marks]
Mark scheme
- (B1) First finger represents magnetic field, second finger conventional current and thumb force or motion.
- (B1) With current east and field into the page, the force is towards the north or top of the page.
- (B1) Reversing the current alone reverses the force towards the south or bottom of the page.
- (B1) The force is zero when the current is parallel to the magnetic field.
Exact AQA content: 3.7.5.1(a), 3.7.5.1(b)
3. An alpha particle of charge +3.20 × 10⁻¹⁹ C moves from left to right at 2.50 × 10⁶ m s⁻¹ through a 0.220 T magnetic field directed into the page. Calculate the magnetic force and state its direction. An electron moves with the same velocity in the same field: calculate its force magnitude and state how its direction compares. Use electron charge magnitude e = 1.60 × 10⁻¹⁹ C.[4 marks]
Mark scheme
- (M1) Uses F = BQv for perpendicular motion.
- (A1) Alpha-particle force = 1.76 × 10⁻¹³ N, directed towards the top of the page.
- (A1) Electron force magnitude = 8.80 × 10⁻¹⁴ N.
- (B1) The electron force is towards the bottom of the page because its charge is negative.
Exact AQA content: 3.7.5.2(a), 3.7.5.2(b)
4. A proton enters perpendicular to a uniform magnetic field of flux density 0.750 T at speed 4.20 × 10⁶ m s⁻¹. Derive an expression for the radius of its path and calculate the radius. Explain why its speed remains constant. Use proton mass 1.67 × 10⁻²⁷ kg and proton charge 1.60 × 10⁻¹⁹ C.[4 marks]
Mark scheme
- (M1) Equates magnetic force and centripetal resultant: BQv = mv²/r.
- (A1) Cancels v to obtain r = mv/(BQ).
- (A1) r = 5.85 × 10⁻² m, accept 0.0585 m.
- (B1) Magnetic force is perpendicular to velocity, so it does no work and cannot change the speed or kinetic energy.
Exact AQA content: 3.7.5.2(a), 3.7.5.2(c)
5. Singly charged positive neon-20 and neon-22 ions enter the same uniform magnetic field with equal speeds, perpendicular to the field. Determine the ratio r₂₂/r₂₀. State the effect on radius if the neon-20 ion instead has charge +2e, and state the effect on curvature if its charge is negative.[4 marks]
Mark scheme
- (B1) Uses r = mv/(BQ), so at fixed v, B and Q the radius is proportional to mass.
- (A1) r₂₂/r₂₀ = 22/20 = 1.10.
- (B1) Doubling the charge magnitude halves the radius for unchanged mass and speed.
- (B1) A negative ion curves in the opposite sense; the radius magnitude is unchanged if |Q| is unchanged.
Exact AQA content: 3.7.5.2(b), 3.7.5.2(c)
6. Starting with r = mv/(BQ), show that the non-relativistic cyclotron frequency is f = BQ/(2πm). Calculate the required alternating-frequency for protons in a 0.820 T field, and explain why one fixed frequency continues to accelerate the protons as their path radius grows. Use proton mass 1.67 × 10⁻²⁷ kg and proton charge 1.60 × 10⁻¹⁹ C.[4 marks]
Mark scheme
- (M1) Uses T = 2πr/v and substitutes r = mv/(BQ).
- (A1) Obtains T = 2πm/(BQ) and therefore f = BQ/(2πm).
- (A1) f = 1.25 × 10⁷ Hz, accept 12.5 MHz.
- (B1) T and f are independent of speed and radius, so the particle returns to the gap in step with the reversing p.d.
Exact AQA content: 3.7.5.2(c), 3.7.5.2(d)
7. A proton starts with negligible kinetic energy at the centre of a cyclotron. It crosses a 2.20 kV gap 160 times. Use energy gained per crossing = QV to calculate its final kinetic energy and speed. Explain which field supplies this energy and why the orbit radius increases. Use proton mass 1.67 × 10⁻²⁷ kg and proton charge 1.60 × 10⁻¹⁹ C.[5 marks]
Mark scheme
- (M1) Uses total energy gained = 160QV with V = 2200 V.
- (A1) Final kinetic energy = 5.63 × 10⁻¹⁴ J.
- (M1) Uses ½mv² = 5.63 × 10⁻¹⁴ J.
- (A1) v = 8.21 × 10⁶ m s⁻¹.
- (B1) The electric field in the gap does work; the magnetic field does no work, and r = mv/(BQ) increases as speed increases.
Exact AQA content: 3.7.5.2(c), 3.7.5.2(d)
8. Protons leave a cyclotron at speed 7.50 × 10⁶ m s⁻¹ in a magnetic field of flux density 0.680 T. Calculate the extraction radius and the time for the final semicircle. State how both values would change for deuterons of twice the proton mass but the same charge and speed. Use proton mass 1.67 × 10⁻²⁷ kg and proton charge 1.60 × 10⁻¹⁹ C.[4 marks]
Mark scheme
- (A1) r = mv/(BQ) = 0.115 m.
- (M1) Uses half-turn time t = πr/v, equivalently t = πm/(BQ).
- (A1) t = 4.82 × 10⁻⁸ s.
- (B1) Doubling mass doubles both radius and half-turn time for unchanged B, Q and v.
Exact AQA content: 3.7.5.2(c), 3.7.5.2(d)
Specification coverage
Exact AQA 7408 coverage: 3.7.5.1–3.7.5.2
This page deliberately stops before magnetic flux, electromagnetic induction, alternating currents and transformers, which are covered by 3.7.5.3–3.7.5.6.
- 3.7.5.1(a) Use F = BIl for force on a current-carrying wire perpendicular to a magnetic field.
- 3.7.5.1(b) Use Fleming's left-hand rule.
- 3.7.5.1(c) Define magnetic flux density B and the tesla.
- 3.7.5.2(a) Use F = BQv for a charged particle moving perpendicular to a magnetic field.
- 3.7.5.2(b) Determine force direction for positive and negative charged particles in a magnetic field.
- 3.7.5.2(c) Describe circular motion of charged particles in magnetic fields.
- 3.7.5.2(d) Apply charged-particle circular motion to devices such as the cyclotron.
Continue through AQA Fields and Paper 2
Return to the AQA 3.7 fields module hub. Compare charge motion in an electric field with electric fields, potential and particle motion, revisit the centripetal model in AQA circular motion, and review proton, electron and alpha-particle properties in particles and radiation. Continue through the AQA Paper 2 revision hub, problem-solving practice and MCQ practice.
Written against all seven AQA Physics 7408 points in sections 3.7.5.1 and 3.7.5.2. Questions, values and contexts are original. The charged-particle stimulus is an original generated PNG; the interactive model is the original PhysicsUK magnetic-field explorer.