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

This study guide covers the first two parts of section 6.3 (Electromagnetism) of the H556 Physics A specification from OCR (Version 3.0, March 2026): 6.3.1 Magnetic fields and 6.3.2 Motion of charged particles. Flux linkage, Faraday’s law, generators and transformers (6.3.3) are taught separately. For condensed recall use the revision notes, then test yourself with the practice questions. Also see the subject hub, the checklist to print and the free 10-minute diagnostics. For the circular-motion ideas, see circular motion and centripetal force.

Coverage of the outcomes

Outcome	You must be able to
6.3.1(a)	explain that magnetic fields come from moving charges or permanent magnets
6.3.1(b)	use field lines to map magnetic fields
6.3.1(c)	describe the field patterns of a long straight wire, a flat coil and a long solenoid
6.3.1(d)	use Fleming’s left-hand rule
6.3.1(e)(i)	use F = BIL sinθ for the force on a current-carrying conductor
6.3.1(e)(ii)	describe how to find a uniform flux density with a current-carrying wire and a digital balance
6.3.1(f)	define magnetic flux density and the tesla
6.3.2(a)	use F = BQv for a charge moving at right angles to a uniform field
6.3.2(b)	analyse circular orbits of charged particles in a uniform field
6.3.2(c)	analyse charges in electric and magnetic fields together; the velocity selector

6.3.1(a)-(c) Sources and patterns of magnetic fields

A magnetic field is a region where a magnet, a current-carrying conductor or a moving charge feels a force. It is created by permanent magnets and by moving charges. A current is a flow of moving charge, so every current-carrying wire has a magnetic field around it.

A field line is a line drawn so that its tangent shows the direction of the field at that point. The direction is the way a free north pole would be pushed. Lines leave north poles and enter south poles, never cross, and crowd together in the strongest part of the field.

You must know three patterns.

• Long straight wire. The lines are rings centred on the wire, each lying in a plane perpendicular to it. They are further apart as you move away from the wire, so the field weakens with distance. Right-hand grip rule: point the thumb along the conventional current and the curled fingers show the field direction.
• Flat circular coil. Close to each part of the wire the lines are circles. Through the centre of the coil they run straight and along the axis, and the field there is strongest. Outside, the lines loop back round. From far away it looks like a short bar magnet.
• Long solenoid. Inside, away from the ends, the lines are straight, parallel and equally spaced: the field is uniform. Outside, the pattern matches a bar magnet, with lines leaving one end (north) and returning to the other (south). Using the right-hand grip rule on the coils, curl your fingers along the current and your thumb points to the north end. A larger current, or more turns in the same length, makes a stronger field.

Between two flat, parallel pole faces the field is uniform in the middle and bulges outwards near the edges.

6.3.1(d) Fleming’s left-hand rule

The rule gives the direction of the force on a wire carrying current in a field.

Digit	Represents
Thumb	Force (thrust)
First finger	Field, north to south
Second finger	Conventional current, positive to negative

Hold the digits mutually at right angles. For a moving charge, the second finger points along the velocity of a positive charge. For a negative charge such as an electron, point it opposite to the velocity, because the conventional current is the other way.

6.3.1(e)(i) The force on a current-carrying conductor

A straight conductor of length L carrying current I in a uniform field of flux density B, with the wire at angle θ to the field lines, feels a force

F = BIL sinθ

Only the component of the field perpendicular to the wire, B sinθ, produces the force. When θ = 90° the force is a maximum, F = BIL. When the wire is parallel to the field (θ = 0°) there is no force. L is the length of wire that lies inside the field.

Worked example 1. A straight wire carries 4.5 A. A length of 0.080 m lies in a uniform field of 0.35 T at 40° to the field lines. Find the force.

F = BIL sinθ = 0.35 × 4.5 × 0.080 × sin 40° F = 0.126 × 0.6428 = 0.081 N (2 s.f.)

If the same wire were turned to 90°, the force would rise to 0.126 N.

6.3.1(f) Magnetic flux density and the tesla

Rearranging F = BIL sinθ gives

B = F / (IL sinθ)

Magnetic flux density is the force per unit current per unit length on a conductor placed at right angles to the field. The tesla (T) is the flux density for which a 1 m length of conductor, carrying 1 A perpendicular to the field, feels a force of 1 N.

So 1 T = 1 N A⁻¹ m⁻¹. Since 1 N = 1 kg m s⁻², the tesla in base units is kg s⁻² A⁻¹.

6.3.1(e)(ii) Measuring flux density with a current balance

Setup. Place a U-shaped yoke carrying two magnets, with opposite poles facing across a gap, on a digital balance. This gives a roughly uniform field in the gap. Clamp a stiff horizontal wire (or a rigid rectangular loop whose lower side is in the gap) so that it passes through the gap at right angles to the field without touching the yoke. Connect it in series with a power supply, variable resistor and ammeter.

Method.

1. Switch the current off and zero (tare) the balance.
2. Switch on a current I and read the balance. The wire feels a force F from the magnets, so by Newton’s third law the magnets feel an equal and opposite force on the balance.
3. The change in reading Δm (in kg) gives F = Δm g.
4. Repeat for several currents.
5. Measure the length L of wire in the field with a ruler.

Since F = BIL and F = Δm g,

Δm = (BL / g) × I

A graph of Δm against I is a straight line through the origin with gradient BL/g, so B = gradient × g / L.

Worked example 2. The length of wire in the gap is L = 0.060 m. Readings of the increase in mass are:

I / A	1.0	2.0	3.0	4.0
Δm / g	0.74	1.46	2.21	2.95

Gradient = (2.95 − 0.74) / (4.0 − 1.0) = 0.737 g A⁻¹ = 7.37 × 10⁻⁴ kg A⁻¹. B = gradient × g / L = 7.37 × 10⁻⁴ × 9.81 / 0.060 = 0.12 T (2 s.f.)

Sources of error. The field is not uniform near the edges, so the effective length is uncertain. The wire must be exactly perpendicular to the field. A gradient over several currents averages out random error. Keep other magnetic materials away from the balance.

6.3.2(a) The force on a moving charge

A single charge Q moving at speed v at right angles to a uniform field B feels

F = BQv

This follows from F = BIL, since IL = (Q/t)(vt) = Qv.

Being perpendicular to the velocity at every instant, the force does no work on the particle and does not change its speed or kinetic energy. It only changes the direction of motion. A charge moving parallel to the field feels no force.

Worked example 3. An electron moves at 2.4 × 10⁶ m s⁻¹ at right angles to a uniform field of 0.045 T. Find the force and the acceleration. (e = 1.60 × 10⁻¹⁹ C, electron mass 9.11 × 10⁻³¹ kg.)

F = BQv = 0.045 × 1.60 × 10⁻¹⁹ × 2.4 × 10⁶ = 1.7 × 10⁻¹⁴ N a = F/m = 1.728 × 10⁻¹⁴ / 9.11 × 10⁻³¹ = 1.9 × 10¹⁶ m s⁻²

6.3.2(b) Circular orbits in a uniform field

The magnetic force is perpendicular to the velocity and has constant size. A force of constant size that is always at right angles to the motion provides the centripetal force, so the particle moves in a circle in the plane perpendicular to the field. The magnetic force supplies the centripetal force:

BQv = mv² / r

so the radius is

r = mv / (BQ)

The period of one orbit is T = 2πr / v = 2πm / (BQ). Notice that v has cancelled. The period does not depend on the speed or the radius. A faster particle goes round a bigger circle in the same time.

For a particle accelerated from rest through a potential difference V before it enters the field, QV = ½mv², which gives v.

Worked example 4. A proton is accelerated from rest through 1500 V and enters a field of 0.20 T at right angles. Find its radius and period. (e = 1.60 × 10⁻¹⁹ C, proton mass 1.673 × 10⁻²⁷ kg.)

v = √(2eV/m) = √(2 × 1.60 × 10⁻¹⁹ × 1500 / 1.673 × 10⁻²⁷) = 5.4 × 10⁵ m s⁻¹ r = mv/(eB) = 1.673 × 10⁻²⁷ × 5.36 × 10⁵ / (1.60 × 10⁻¹⁹ × 0.20) = 2.8 × 10⁻² m T = 2πm/(eB) = 2π × 1.673 × 10⁻²⁷ / (1.60 × 10⁻¹⁹ × 0.20) = 3.3 × 10⁻⁷ s

An electron through the same 1500 V and field has a radius about 43 times smaller, the ratio √(mp/me), since r is proportional to √m for equal energy gain.

Direction. Fleming’s left-hand rule gives a force towards the centre. Positive and negative particles go round in opposite senses.

6.3.2(c) Electric and magnetic fields together: the velocity selector

A velocity selector gives a beam of particles all with the same speed. It places a uniform electric field and a uniform magnetic field in the same region, at right angles to each other and to the beam.

• The electric force is QE, in the direction of E for a positive charge. For parallel plates a distance d apart with p.d. V, E = V/d.
• The magnetic force is BQv, at right angles to the velocity.
• The fields are arranged so the two forces are in opposite directions.

A particle goes straight through when the forces balance:

QE = BQv, so v = E / B

The charge Q cancels, as does the mass. Any particle with speed E/B passes undeflected, whatever its charge or mass. A particle that is too fast feels the larger magnetic force; one that is too slow feels the larger electric force. A narrow exit slit lets through only the undeflected beam.

Worked example 5. Plates 6.0 mm apart have a p.d. of 900 V across them. A uniform magnetic field of 0.050 T is applied at right angles. Find the speed of undeflected particles.

E = V/d = 900 / 6.0 × 10⁻³ = 1.5 × 10⁵ V m⁻¹ v = E/B = 1.5 × 10⁵ / 0.050 = 3.0 × 10⁶ m s⁻¹

Common errors

• Using the whole wire length instead of the length in the field.
• Taking θ from the wrong line: it is the angle between the wire and the field lines.
• Applying Fleming’s rule to an electron without reversing the direction of conventional current.
• Saying the magnetic force “speeds up” the particle. It changes direction only.
• Forgetting that the balance reading gives mass, so F = Δm g.

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).