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

These revision notes condense 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. Section 3.10 is International A-level only. For full explanations and derivations, read the study guide, then try the practice questions. Related pages: the subject hub, the printable checklist and the free 10-minute diagnostics.

Both sections assume the field is perpendicular to the current or to the velocity.

Key definitions

• Magnetic flux density, B: the force on each metre of conductor per ampere of current, with the conductor lying across the field.
• Tesla (T): the flux density of a uniform field that exerts 1 N on 1 m of straight wire carrying 1 A and lying across the field.
• Base units of the tesla: 1 T = 1 N A⁻¹ m⁻¹ = 1 kg s⁻² A⁻¹.
• Cyclotron: a device that accelerates charged particles in a spiral path using a magnetic field to bend the path and an alternating electric field in a gap to speed the particle up.

Formulae

Situation	Equation	Notes
Wire in a field	F = BIL	L is the length inside the field
Rearranged	B = F / (IL)	gives the tesla
Moving charge	F = BQv	Q is the magnitude of the charge
Circular orbit radius	r = mv / (BQ)	from BQv = mv² / r
Orbit period	T = 2πm / (BQ)	independent of speed
Orbit frequency	f = BQ / (2πm)	the cyclotron frequency
After accelerating through V	v = √(2QV / m)	starting from rest
Cyclotron maximum speed	v = BQR / m	R is the dee radius
Cyclotron maximum kinetic energy	(BQR)² / (2m)	equals ½mv²

Direction rules

Fleming’s left-hand rule: thumb gives the force, first finger the field (north to south), second finger the conventional current.

Particle	Where the second finger points
Positive charge	along the velocity
Negative charge (electron)	opposite to the velocity
Current in a wire	along the conventional current

• The force is perpendicular to both the field and the velocity.
• The force on an electron is opposite to the force on a proton with the same velocity in the same field.
• A particle moving along the field direction feels no force.
• The force does no work, so kinetic energy and speed are unchanged.

Method in steps: force on a wire

1. Identify L, the length of wire inside the field, in metres.
2. Check that the current and field are perpendicular.
3. Calculate F = BIL.
4. Use Fleming’s left-hand rule for the direction. Reversing the current or the field reverses the force.

Reminder. B = 0.18 T, I = 3.5 A, L = 0.20 m gives F = 0.18 × 3.5 × 0.20 = 0.126 N, or 0.13 N to 2 significant figures.

Reminder. F = 0.0420 N on a wire with I = 6.0 A and L = 0.35 m gives B = 0.0420 / (6.0 × 0.35) = 0.020 T.

Method in steps: current balance

1. Zero the balance with the current off.
2. Switch on the current I and record the change in mass Δm.
3. The wire’s force is F = Δm g; the magnets receive the same force the other way (Newton’s third law).
4. Plot Δm against I. The line is straight through the origin with gradient BL / g.
5. B = gradient × g / L.

Method in steps: a charged particle in a circle

1. Write the centripetal force condition: BQv = mv² / r.
2. Rearrange to r = mv / (BQ).
3. If the particle was accelerated through a potential difference V, find v from QV = ½mv² first.
4. Find the period with T = 2πm / (BQ) if asked.
5. Direction of circulation: use the left-hand rule for the force at one point and sketch the centre on that side.

Reminder. An electron is accelerated from rest through 120 V. Then v = √(2 × 1.60 × 10⁻¹⁹ × 120 / 9.11 × 10⁻³¹) = 6.5 × 10⁶ m s⁻¹. In a field of 4.0 × 10⁻³ T, r = 9.11 × 10⁻³¹ × 6.5 × 10⁶ / (4.0 × 10⁻³ × 1.60 × 10⁻¹⁹) = 9.2 × 10⁻³ m.

Cyclotron in six lines

• Two D-shaped dees in a vacuum, field at right angles to the dees.
• Electric field only in the gap; no electric field inside the dees.
• Inside a dee the particle moves in a semicircle at constant speed.
• The time per semicircle is πm / (BQ), independent of speed, so the pd frequency is fixed at f = BQ / (2πm).
• Each gap crossing adds kinetic energy QV, so the radius grows and the path spirals outwards.
• The particle leaves at radius R with speed v = BQR / m.

At speeds approaching the speed of light these simple results stop working, because the mass of the particle is no longer constant. The quantities above assume much lower speeds.

Must-know distinctions

Compare	Point to remember
Force on a wire and force on a charge	F = BIL and F = BQv are the same law: IL = Qv for the charge carriers
Magnetic and electric force	the electric field in a gap speeds the particle up; the magnetic field only turns it
Radius and period	radius depends on speed; period does not
Proton and electron	same field and velocity, opposite force direction
Mass and charge	a heavier particle has a larger orbit; a larger charge gives a smaller orbit

Quick self-test

1. Define the tesla.
2. A wire carries 3.5 A and 0.20 m of it lies perpendicular to a field of 0.18 T. What is the force?
3. A force of 0.0420 N acts on 0.35 m of wire carrying 6.0 A at right angles to a field. What is B?
4. Take east, north and up as three perpendicular directions. A current flows north in a field pointing upward. Which way is the force?
5. An electron moves at 1.8 × 10⁶ m s⁻¹ perpendicular to a field of 0.30 T. Find the force (e = 1.60 × 10⁻¹⁹ C).
6. An electron moves to the right across the page in a field directed into the page. Which way is it pushed?
7. A proton moves at 2.4 × 10⁶ m s⁻¹ perpendicular to a field of 0.50 T. Find the radius of its path. What is the radius if the field doubles? (proton mass 1.67 × 10⁻²⁷ kg)
8. A proton and an electron have equal kinetic energy and enter the same field at right angles. What is the ratio of their orbit radii?
9. Find the orbit frequency of a proton in a field of 0.65 T.
10. In a cyclotron, why can the alternating pd have a fixed frequency even though the particle speeds up?

Answers

1. The flux density of a uniform field that exerts 1 N on 1 m of wire carrying 1 A and lying across the field.
2. F = 0.18 × 3.5 × 0.20 = 0.126 N, so 0.13 N.
3. B = 0.0420 / (6.0 × 0.35) = 0.020 T.
4. Current north (second finger), field up (first finger), so the thumb points east.
5. F = 0.30 × 1.60 × 10⁻¹⁹ × 1.8 × 10⁶ = 8.6 × 10⁻¹⁴ N.
6. Treat as conventional current to the left, field into the page. The thumb points down the page.
7. r = 1.67 × 10⁻²⁷ × 2.4 × 10⁶ / (0.50 × 1.60 × 10⁻¹⁹) = 0.050 m. Doubling B halves r to 0.025 m.
8. With equal kinetic energy, mv = √(2mEk), so r is proportional to √m. The ratio r(proton) / r(electron) = √(1.67 × 10⁻²⁷ / 9.11 × 10⁻³¹) = 43 (the proton’s orbit is larger).
9. f = BQ / (2πm) = 0.65 × 1.60 × 10⁻¹⁹ / (2π × 1.67 × 10⁻²⁷) = 9.9 × 10⁶ Hz.
10. The time for each semicircle is πm / (BQ), which does not depend on speed. A faster particle goes round a bigger semicircle in the same time, so it always reaches the gap when the pd reverses.

Where marks are usually lost

• Writing the unit of B as N A m⁻¹ instead of N A⁻¹ m⁻¹, or giving the base units wrongly.
• Using the length of the whole wire rather than only the length in the field.
• Getting the direction of the force on an electron the wrong way round.
• Saying the magnetic field accelerates a particle in the cyclotron, rather than the electric field in the gap.
• Stating that the force changes the speed. It changes only the direction, because it does no work.
• Failing to show that the period is independent of speed when the question asks you to show it.
• Mixing up charge and mass, for example using the proton mass for an alpha particle.
• Giving too many or too few significant figures, or leaving out the unit.
• Describing a definition with no mention of right angles, current, length or force.

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.