OCR A Level Physics A: Magnetic fields, force on conductors and moving charges (H556) – Revision Notes

These revision notes condense 6.3.1 (Magnetic fields) and 6.3.2 (Motion of charged particles) from the H556 Physics A specification from OCR (Version 3.0, March 2026). For full explanations and longer worked examples use the study guide, then try the practice questions. Also see the subject hub, the checklist to print and the free 10-minute diagnostics.

Key equations

Equation	Meaning	Outcome
F = BIL sinθ	force on length L of wire at angle θ to the field	6.3.1(e)(i)
B = F / (IL sinθ)	definition of flux density	6.3.1(f)
F = BQv	force on charge moving at right angles to the field	6.3.2(a)
r = mv / (BQ)	radius of circular orbit	6.3.2(b)
T = 2πm / (BQ)	period of the orbit	6.3.2(b)
v = E / B	undeflected speed in a velocity selector	6.3.2(c)
E = V / d	uniform electric field between parallel plates	6.3.2(c)

6.3.1(a)-(c) Sources and field patterns

• Magnetic fields are due to moving charges (including currents) or permanent magnets.
• Field lines: tangent gives the field direction (the way a free north pole is pushed). Lines run north to south outside a magnet, never cross, and crowd together where the field is strongest.

Source	Pattern	Strength
Long straight wire	rings around the wire, each in a plane perpendicular to it	weaker with distance, so lines spread out
Flat coil	circles near the wire; straight along the axis through the centre	strongest at the centre
Long solenoid	uniform, parallel lines inside; bar-magnet pattern outside	uniform inside, away from the ends

Right-hand grip for a wire: thumb along the conventional current; the curled fingers give the field direction. Right-hand grip for a coil or solenoid: fingers along the current, thumb points to the north end.

6.3.1(d) Fleming’s left-hand rule

Thumb is Force, first finger is Field, second finger is Current. They are at right angles. Field runs north to south and current is conventional (positive to negative). For an electron, point the second finger opposite to its velocity.

6.3.1(e)(i) and (f) Force on a wire; the tesla

• Use the length of wire that is inside the field.
• θ is the angle between the wire and the field lines. Maximum force at 90°, zero force at 0°.
• Magnetic flux density is force per unit current per unit length, for a conductor at right angles to the field.
• 1 tesla: a 1 m conductor carrying 1 A perpendicular to the field feels 1 N.
• Units: 1 T = 1 N A⁻¹ m⁻¹ = 1 kg s⁻² A⁻¹.

Worked reminder. A wire 0.12 m long carries 6.0 A at 30° to a field of 0.080 T. F = 0.080 × 6.0 × 0.12 × sin 30° = 0.029 N.

6.3.1(e)(ii) Current balance

Steps for the method:

1. Rest magnets on a yoke on a digital balance, with the wire passing through the gap at right angles to the field.
2. Zero the balance with the current off.
3. Switch on a current I and record the change in mass Δm.
4. Force on the wire F = Δm g (the magnets feel an equal and opposite force).
5. Repeat for several I. Plot Δm against I: straight line through the origin, gradient = BL / g.
6. B = gradient × g / L, with L the length of wire in the field.

Check: gradient in kg A⁻¹, not g A⁻¹, before multiplying by g. Reasons for uncertainty: edge effects on L, wire not perpendicular to the field, zero error on the balance.

6.3.2(a) Force on a moving charge

• F = BQv for motion at right angles to B.
• The force is perpendicular to the velocity, so it does no work. Speed and kinetic energy stay the same. Only the direction changes.
• A charge moving parallel to the field feels no force.
• Link to the wire: F = BIL with IL = Qv.

6.3.2(b) Circular orbits

• Magnetic force provides the centripetal force: BQv = mv² / r.
• r = mv / (BQ). Faster, heavier or less charged means a bigger circle. A stronger field means a smaller circle.
• T = 2πm / (BQ), independent of speed and radius.
• After acceleration through V: ½mv² = QV, so v = √(2QV / m).
• The path is in the plane at right angles to the field. Oppositely charged particles circle in opposite senses.

Method in steps for a radius question

1. If accelerated by a p.d., find v from QV = ½mv².
2. Put BQv = mv²/r.
3. Rearrange for r = mv / (BQ) and substitute with units.
4. Check the answer has units of metres and a sensible size.

Worked reminder. A proton at 4.0 × 10⁶ m s⁻¹ in 0.50 T: r = 1.673 × 10⁻²⁷ × 4.0 × 10⁶ / (1.60 × 10⁻¹⁹ × 0.50) = 0.084 m.

6.3.2(c) Velocity selector

• Electric and magnetic fields at right angles, forces opposite.
• Undeflected when QE = BQv, so v = E / B, independent of charge and mass.
• Too fast: magnetic force larger, beam bends towards the magnetic force. Too slow: electric force larger, beam bends towards the electric force.
• E = V / d for parallel plates a distance d apart.

Must-know distinctions

Idea	Do not confuse with
Magnetic force on a moving charge does no work	electric force on a charge, which does work along its direction
Period is independent of speed	radius, which is proportional to speed
Conventional current direction	electron flow direction
Angle θ to the field lines	angle to the plane of the field
Δm from the balance in kg	reading in g

Quick self-test

1. State the two sources of magnetic fields.
2. Describe the field pattern around a long straight current-carrying wire.
3. A wire 0.20 m long carries 3.0 A at right angles to a field of 0.50 T. Find the force. Then find it when the wire is turned to 30° to the field.
4. Define the tesla.
5. A balance reading rises by 0.45 g when a current of 2.5 A flows in a wire with 0.040 m inside the field. Find B.
6. An electron moves at right angles to a field of 0.050 T and feels a force of 1.6 × 10⁻¹⁵ N. Find its speed.
7. Explain why a magnetic field cannot change the kinetic energy of a charged particle.
8. A proton moves at 4.0 × 10⁶ m s⁻¹ in a field of 0.50 T at right angles. Find the radius of its orbit.
9. By what factor does the radius change if the speed doubles? By what factor does the period change?
10. A velocity selector has B = 0.020 T. Find the electric field that passes particles at 5.0 × 10⁵ m s⁻¹, and the p.d. needed across plates 5.0 mm apart.

Answers

1. Moving charges (including currents) and permanent magnets.
2. Rings centred on the wire, each in a plane perpendicular to it, further apart at greater distance; direction by the right-hand grip rule.
3. F = BIL = 0.50 × 3.0 × 0.20 = 0.30 N; at 30°, F = 0.30 × sin 30° = 0.15 N.
4. The flux density for which 1 m of conductor carrying 1 A perpendicular to the field feels 1 N.
5. F = Δm g = 0.45 × 10⁻³ × 9.81 = 4.4 × 10⁻³ N; B = F / (IL) = 4.4 × 10⁻³ / (2.5 × 0.040) = 0.044 T.
6. v = F / (BQ) = 1.6 × 10⁻¹⁵ / (0.050 × 1.60 × 10⁻¹⁹) = 2.0 × 10⁵ m s⁻¹.
7. Being perpendicular to the velocity at every instant, the force does no work.
8. r = mv / (BQ) = 0.084 m.
9. The radius doubles; the period does not change.
10. E = Bv = 0.020 × 5.0 × 10⁵ = 1.0 × 10⁴ V m⁻¹; V = Ed = 1.0 × 10⁴ × 5.0 × 10⁻³ = 50 V.

Where marks are usually lost

• Using g or grams in the balance calculation without converting the mass to kg.
• Using the total length of wire, not the length in the field.
• Giving the angle to the wrong line, so using cosθ instead of sinθ.
• Applying Fleming’s left-hand rule to an electron without reversing the current direction.
• Writing that the magnetic force “speeds up” the particle, or “does work”.
• Stating v = E/B without saying the electric and magnetic forces are equal and opposite.
• Forgetting that a particle accelerated through V gains kinetic energy QV before entering the field.
• Describing a circular orbit as centripetal force “in addition to” the magnetic force, when the magnetic force is the centripetal force.

Official syllabus

Source: the OCR Physics A (H556) specification, Version 3.0 (March 2026), first assessment 2017. Section 6.3 Electromagnetism; outcomes 6.3.1(a)-(f) and 6.3.2(a)-(c).