OxfordAQA International AS and A-level Physics: Force on conductors and moving charges in magnetic fields (9630)

This study guide covers sections 3.10.1 (Magnetic flux density) and 3.10.2 (Moving charges in a magnetic field) of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, for International AS and A-level exams from May/June 2020 onwards. Section 3.10 is marked International A-level only: it is not part of the International AS content. Flux, flux linkage and induction (3.10.3 onwards) are taught separately. Use the revision notes for condensed recall and the practice questions to test yourself. Also see the subject hub, the printable checklist and the free 10-minute diagnostics.

What the specification asks for

Section	You must be able to
3.10.1	use F = BIL for a wire with the field perpendicular to the current
3.10.1	use Fleming’s left-hand rule to find the direction of the force
3.10.1	explain magnetic flux density B and define the tesla
3.10.2	apply F = BQv to a charge whose velocity is at 90° to the field
3.10.2	give the direction of the force on positive and negative charges
3.10.2	explain the circular path of a charged particle and apply it to devices such as the cyclotron

The specification asks only for the perpendicular case, so every result below assumes the field meets the current or velocity at 90°.

3.10.1 Force on a current-carrying wire

A conductor with a current through it feels a force when it sits in a magnetic field. The size is

F = BIL

where F is the force in newtons, B the magnetic flux density in tesla, I the current in amperes and L the length of conductor inside the field in metres. L is only the part of the wire that sits in the field, not the whole wire.

The force is largest when the wire crosses the field at 90° and is zero when the wire lies along the field lines. The force is perpendicular to both the current and the field.

Fleming’s left-hand rule

Hold the thumb, first finger and second finger of your left hand at right angles to each other.

Digit	Shows
Thumb	direction of the force (motion)
First finger	direction of the field, north pole to south pole
Second finger	direction of the conventional current, positive to negative

Worked example 1. A straight horizontal wire carries a current of 6.4 A. A length of 0.095 m lies in a uniform field of flux density 0.085 T, perpendicular to the wire. Find the force.

F = BIL = 0.085 × 6.4 × 0.095 = 0.0517 N, which is 0.052 N to 2 significant figures.

Worked example 2 (direction). Draw the page as a map. A wire lies along the page and carries current to the right. The field points into the page. Which way is the force?

Point your second finger to the right (current). Point your first finger into the page (field). Your thumb now points up the page. The force is up the page. Reverse either the current or the field and the force reverses. Reverse both and it stays the same.

Magnetic flux density and the tesla

Rearranging F = BIL gives B = F / (IL).

Magnetic flux density measures the strength of the field in terms of the force it produces: it equals the force acting on each metre of conductor for each ampere of current, when the conductor lies across the field.

The tesla (T) is defined from this. A uniform field has a flux density of 1 T when a 1 m straight wire carrying 1 A, set across the field, experiences a 1 N force.

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

Worked example 3. A wire of length 0.045 m, carrying 2.5 A at right angles to a uniform field, feels a force of 3.6 × 10⁻³ N. Find B.

B = F / (IL) = 3.6 × 10⁻³ / (2.5 × 0.045) = 3.6 × 10⁻³ / 0.1125 = 0.032 T.

Measuring B with a current balance

A pair of magnets on a top-pan balance produces a field across a gap. A stiff wire of known length L in the gap, perpendicular to the field, is connected to a power supply. When the current flows, the wire feels a force F. The magnets are pushed with an equal force the other way (Newton’s third law), so the reading shifts by Δm, with F = Δm g.

Combining F = BIL with F = Δm g gives Δm = (BL / g) I. Plotting Δm against I gives a straight line through the origin whose gradient is BL / g, so B = gradient × g / L. Using several currents and a gradient averages out random error. Measure L with a ruler and check that the wire is at 90° to the field.

3.10.2 Force on a moving charge

A current is moving charge, so the force on a wire is the sum of the forces on its moving charge carriers. In a time t a charge Q passing a point in the wire gives I = Q / t. A carrier moving at speed v covers L = vt, so

IL = (Q / t)(vt) = Qv

and F = BIL becomes the force on one charge:

F = BQv

This applies when the field is perpendicular to the velocity. The force is at right angles to both v and B. A charge moving along the field lines feels no force, and a stationary charge feels no magnetic force.

Direction for positive and negative charges

Fleming’s left-hand rule uses conventional current. For a positive charge, point the second finger along the velocity. For a negative charge such as an electron, point the second finger opposite to the velocity, because the equivalent conventional current runs the other way. For the same velocity and field, an electron is therefore pushed in the opposite direction to a proton.

Worked example 4. An electron travels at 3.2 × 10⁶ m s⁻¹ at right angles to a field of 0.062 T. The electron charge has magnitude 1.60 × 10⁻¹⁹ C and its mass is 9.11 × 10⁻³¹ kg. Find the force and the acceleration.

F = BQv = 0.062 × 1.60 × 10⁻¹⁹ × 3.2 × 10⁶ = 3.2 × 10⁻¹⁴ N

a = F / m = 3.1744 × 10⁻¹⁴ / 9.11 × 10⁻³¹ = 3.5 × 10¹⁶ m s⁻²

Worked example 5 (direction). An electron moves to the right, and the field points out of the page. Which way does it deflect?

Treat it as conventional current to the left. Second finger left, first finger out of the page. The thumb points up the page. A proton with the same velocity would be pushed down.

Circular paths

The magnetic force always acts at 90° to the velocity. A force at right angles to the motion does no work, so the speed and kinetic energy stay constant and only the direction changes. The force has constant size, so it acts as a centripetal force so the particle follows a circular path lying at right angles to the field:

BQv = mv² / r

Cancelling one v gives

r = mv / (BQ)

Larger momentum gives a larger circle. A stronger field, or a larger charge, gives a tighter one. The time for one orbit is T = 2πr / v. Substituting r gives

T = 2πm / (BQ), and so f = BQ / (2πm).

The speed has cancelled: the time for a revolution does not depend on speed or radius. A faster particle moves in a bigger circle but takes the same time to go round it. This is the key to the cyclotron.

If a particle is accelerated from rest through a potential difference V before entering the field, QV = ½mv², so v = √(2QV / m).

Worked example 6. An alpha particle (charge 3.20 × 10⁻¹⁹ C, mass 6.64 × 10⁻²⁷ kg) is accelerated from rest through 2400 V and enters a field of 0.15 T at right angles. Find its orbit radius and period.

v = √(2QV / m) = √(2 × 3.20 × 10⁻¹⁹ × 2400 / 6.64 × 10⁻²⁷) = 4.81 × 10⁵ m s⁻¹

r = mv / (BQ) = 6.64 × 10⁻²⁷ × 4.81 × 10⁵ / (0.15 × 3.20 × 10⁻¹⁹) = 0.067 m

T = 2πm / (BQ) = 2π × 6.64 × 10⁻²⁷ / (0.15 × 3.20 × 10⁻¹⁹) = 8.7 × 10⁻⁷ s

Worked example 7. Electrons travel at 4.5 × 10⁶ m s⁻¹ in a field of 0.012 T. Find the radius of the circle.

r = mv / (BQ) = 9.11 × 10⁻³¹ × 4.5 × 10⁶ / (0.012 × 1.60 × 10⁻¹⁹) = 2.1 × 10⁻³ m

The cyclotron

A cyclotron accelerates charged particles to high kinetic energy using a modest voltage many times over.

• Two hollow D-shaped metal boxes, called dees, sit in a vacuum with a small gap between them. A uniform magnetic field acts at right angles to the plane of the dees.
• An alternating potential difference is applied across the gap. The electric field exists only in the gap, and it accelerates the particle each time it crosses.
• Inside a dee there is no electric field, so the particle only feels the magnetic force. It follows a semicircle at constant speed.
• The time for a semicircle is half of 2πm / (BQ), and it does not depend on speed. So the alternating pd can have a fixed frequency f = BQ / (2πm), set so that its polarity reverses just as the particle returns to the gap.
• After each crossing the particle has more kinetic energy, so it moves faster and its radius r = mv / (BQ) grows. The path spirals outwards until it reaches the outer edge, radius R, and leaves.

Maximum speed: v = BQR / m. Maximum kinetic energy: ½mv² = (BQR)² / (2m).

Worked example 8. A cyclotron has a uniform field of 0.80 T and dees of radius 0.25 m. It accelerates protons (charge 1.60 × 10⁻¹⁹ C, mass 1.67 × 10⁻²⁷ kg). Find (a) the frequency of the alternating pd, (b) the maximum speed, (c) the maximum kinetic energy, and (d) roughly how many gap crossings are needed if each gains 30 keV.

(a) f = BQ / (2πm) = 0.80 × 1.60 × 10⁻¹⁹ / (2π × 1.67 × 10⁻²⁷) = 1.2 × 10⁷ Hz

(b) v = BQR / m = 0.80 × 1.60 × 10⁻¹⁹ × 0.25 / 1.67 × 10⁻²⁷ = 1.9 × 10⁷ m s⁻¹

(c) Ek = ½mv² = ½ × 1.67 × 10⁻²⁷ × (1.916 × 10⁷)² = 3.1 × 10⁻¹³ J, which is 1.9 MeV.

(d) 1.916 MeV / 30 keV ≈ 64 crossings, which is about 32 full revolutions.

At speeds close to the speed of light the particle’s mass increases, the revolution time is no longer constant and the simple fixed frequency fails. The calculations above assume speeds well below that.

Common errors

• Using the whole wire length for L instead of the length in the field.
• Pointing the second finger along an electron’s velocity. It must point opposite.
• Writing the tesla as N A m⁻¹ rather than N A⁻¹ m⁻¹.
• Claiming the magnetic force raises the particle’s speed. It only turns the particle. In a cyclotron the electric field in the gap does the accelerating.
• Forgetting that the radius, not the period, depends on speed.
• Not converting grams or centimetres to kilograms and metres.

Test yourself with the practice questions or find your weak spots with a diagnostic. For earlier ideas on forces and circular motion, see the motion and Newton’s laws guide.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), version 4.4, for International AS and A-level exams from May/June 2020 onwards, published by Oxford International AQA Examinations. Sections 3.10.1 and 3.10.2.