AQA · A-Level · Physics
Course version: 7408 · Version from Sept 2016
Circular path of particles; application in devices such as the cyclotron.
Full specification wording
Moving charges in a magnetic field (A-level only): Circular path of particles; application in devices such as the cyclotron.
When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force is always perpendicular to its velocity and has constant magnitude Bqv.
Checked against the AQA 7408 specification (opens in a new tab) on .
- Point 3 of 3
- 4 key terms
Study notes
From the checked teaching notes for this specification point.
The idea
When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force is always perpendicular to its velocity and has constant magnitude Bqv. This force provides the centripetal force, so the particle follows a circular path of radius r given by Bqv = mv²/r, hence r = mv/(Bq). The period T = 2πm/(Bq) is independent of speed and radius, which is the key to the cyclotron: particles cross a gap repeatedly, accelerated by an alternating potential difference whose frequency matches the cyclotron frequency f = Bq/(2πm). Because the period is constant, the alternating voltage stays in step as the radius grows. Example: a proton in a 1.5 T field has a cyclotron frequency of about 23 MHz, so the accelerating voltage must alternate at that frequency.
What earns the marks
- Derive or use Bqv = mv²/r to show that the radius of the circular path is r = mv/(Bq).
- Explain that the magnetic force is perpendicular to velocity and therefore acts as the centripetal force, so the path is circular at constant speed.
- Show that the period T = 2πm/(Bq) and the cyclotron frequency f = Bq/(2πm) are independent of speed and radius.
- Describe how a cyclotron uses a uniform magnetic field and an alternating potential difference across a gap to accelerate particles in expanding semicircles.
- Explain that the alternating voltage frequency must equal the cyclotron frequency so that particles are accelerated each time they cross the gap.
- Relate the maximum radius of the dee to the maximum kinetic energy or speed of the emerging particles.
Common mistakes and how to put them right
Using the magnetic force to change the particle's speed.
Put it right: The force is centripetal, so speed and kinetic energy remain constant while direction changes.
Assuming the cyclotron frequency depends on the particle's speed or radius.
Put it right: f = Bq/(2πm) depends only on charge, mass and magnetic flux density.
Forgetting that the alternating potential difference must reverse each half-cycle.
Put it right: Match its frequency to the cyclotron frequency so acceleration occurs at every gap crossing.
Exam tips
- Start circular-motion questions by equating the magnetic force to the centripetal force, then rearrange for the quantity required.
- When asked about a cyclotron, state the role of the magnetic field (circular motion) and the alternating voltage (acceleration) separately.
- Check units carefully: use kilograms for mass, coulombs for charge and tesla for magnetic flux density before calculating r, T or f.
Try the lesson
The start of the Moving charges in a magnetic field (A-level only): Circular path of particles; application in devices such as the cyclotron. lesson, made from checked teaching notes for this specification point.
Have a go first
A charged particle moves at constant speed perpendicular to a uniform magnetic field. What is the shape of the path followed by the particle?
Listen
When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force is always perpendicular to its velocity and has constant magnitude Bqv. This force provides the centripetal force, so the particle follows a circular path of radius r given by Bqv = mv²/r, hence r = mv/(Bq). The period T = 2πm/(Bq) is independent of speed and radius, which is the key to the cyclotron: particles cross a gap repeatedly, accelerated by an alternating potential difference whose frequency matches the cyclotron frequency f = Bq/(2πm). Because the period is constant, the alternating voltage stays in step as the radius grows. Example: a proton in a 1.5 T field has a cyclotron frequency of about 23 MHz, so the accelerating voltage must alternate at that frequency. A strong answer earns marks for these points. Derive or use Bqv = mv²/r to show that the radius of the circular path is r = mv/(Bq). Explain that the magnetic force is perpendicular to velocity and therefore acts as the centripetal force, so the path is circular at constant speed. Show that the period T = 2πm/(Bq) and the cyclotron frequency f = Bq/(2πm) are independent of speed and radius. Describe how a cyclotron uses a uniform magnetic field and an alternating potential difference across a gap to accelerate particles in expanding semicircles. A common slip: Using the magnetic force to change the particle's speed. Instead, the force is centripetal, so speed and kinetic energy remain constant while direction changes. Exam tip: Start circular-motion questions by equating the magnetic force to the centripetal force, then rearrange for the quantity required.
Key words
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Centripetal force from a magnetic field
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Circular motion of charged particles
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Cyclotron resonance and constant period
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Energy gain in an alternating electric field
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See every definition
- Centripetal force from a magnetic field
- The magnetic force is always perpendicular to velocity and has constant magnitude Bqv, so it provides the centripetal force.
- Circular motion of charged particles
- A particle moving perpendicular to a uniform magnetic field follows a circular path of radius r given by Bqv = mv²/r.
- Cyclotron resonance and constant period
- The period T = 2πm/(Bq) is independent of speed and radius, so the alternating voltage stays in step as the radius grows.
- Energy gain in an alternating electric field
- Particles cross a gap repeatedly, accelerated by an alternating potential difference whose frequency matches the cyclotron frequency f = Bq/(2πm).
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