Circular Motion in Magnetic Fields
In 1930, Ernest O. Lawrence at the University of California, Berkeley, built the world's first cyclotron — just 11.4 cm across — and accelerated protons to 80 keV using a 1.27 MeV/m magnetic field. His key insight was that the orbital period T = 2πm/qB contains no velocity term: a faster proton spirals in a larger circle but still takes the same time per lap as a slower one. By 1939 his 60-inch cyclotron reached 20 MeV; today's medical cyclotrons produce 18 MeV protons for PET scan radioisotopes, treating 25,000+ cancer patients annually in Australia.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
An electron enters a uniform magnetic field pointing out of the page, moving perpendicular to the field.
- What shape do you think the electron's path will be: straight line, parabola, circle, or spiral?
- If you doubled the magnetic field strength, would the circle become larger or smaller?
- If you doubled the electron's speed, would it complete one full circle faster, slower, or in the same time?
Write your predictions before reading on — you will revisit them at the end.
Warm-up — when a charged particle moves perpendicular to a magnetic field, the magnetic force acts in which direction relative to the velocity?
Know — Circular Motion Equations
- $r = mv/qB$ is derived by equating magnetic force to centripetal force
- Period $T = 2\pi m/qB$ is independent of speed
- Frequency $f = qB/2\pi m$ is the cyclotron frequency
Understand — Why a Circle?
- Magnetic force is always perpendicular to velocity, so it acts as centripetal force
- Speed is constant (no work done), so radius is constant
- The period is independent of speed: faster particles travel larger circles in the same time
Can Do — Calculate & Predict
- Calculate radius and period given $m$, $v$, $q$, $B$
- Predict how changing $v$, $B$, or $m$ affects the path
- Explain why the period is independent of speed
Core Content
Equating magnetic force to centripetal force
Fire a proton sideways into a region of uniform magnetic field — and instead of continuing in a straight line or following a parabola, it sweeps in a perfect circle and returns exactly to where it started. The magnetic force at every instant points inward toward the centre of that circle, always perpendicular to the proton's motion, always the same magnitude. This is uniform circular motion, with the magnetic force playing the role of the centripetal force.
We equate the magnetic force to the required centripetal force:
$F_{\text{magnetic}} = F_{\text{centripetal}}$
$qvB = \dfrac{mv^2}{r}$
$r = \dfrac{mv}{qB}$ — radius increases with momentum ($mv$) and decreases with field strength ($B$)
The period of revolution is the circumference divided by speed:
$T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}$ — period is independent of speed
$f = \dfrac{1}{T} = \dfrac{qB}{2\pi m}$ — cyclotron frequency depends only on $q/m$ and $B$
This independence of period from speed is the principle behind the cyclotron particle accelerator: particles can be accelerated by an electric field and still stay in sync with the alternating voltage because their orbital period does not change as they speed up.
Figure 1 — Top-down view: velocity is tangent to the circle; force points to the centre at every instant
A proton and an electron have the same speed and enter the same uniform magnetic field perpendicular to their velocities. Which one travels in the larger circle? By what factor is the radius larger?
Setting $qvB = mv^2/r$ gives radius $r = mv/qB$ (m) and period $T = 2\pi m/qB$ (s). Period is independent of speed — only $m/q$ and $B$ matter. Cyclotron frequency: $f = qB/2\pi m$.
Pause — copy the highlighted formulas into your book before moving on.
The period of a charged particle orbiting in a magnetic field is given by $T = 2\pi m/qB$. Which statement correctly follows from this formula?
Why faster particles take the same time to complete a loop
We just saw that $r = mv/qB$ and $T = 2\pi m/qB$. That raises a question: how can two particles with very different speeds both complete a loop in the same time? This card answers it → faster particles travel a larger circle, but the period $T$ has no $v$ term — this isochronous property is the basis of the cyclotron.
The fact that $T = 2\pi m/qB$ contains no $v$ is remarkable. A fast particle travels in a large circle at high speed; a slow particle travels in a small circle at low speed — but both take exactly the same time to complete one revolution.
This is the operating principle of the cyclotron. Particles are injected near the centre and accelerated by an electric field every half-revolution. Because the period is constant, the alternating voltage can be tuned to a fixed frequency ($f = qB/2\pi m$) and it stays in sync with the particles even as they gain energy and spiral outward.
When comparing two particles in the same B-field, remember that $r \propto mv/qB$ while $T \propto m/qB$. The period depends only on mass-to-charge ratio, not on speed.
In a cyclotron, protons are accelerated to higher and higher speeds. A student worries that as they speed up, they will fall out of sync with the alternating voltage. Explain why this does not happen in an ideal cyclotron.
In a cyclotron, faster particles spiral into larger circles but the period $T = 2\pi m/qB$ stays constant — independent of speed. The fixed alternating voltage frequency $f = qB/2\pi m$ stays permanently in sync with particles as they accelerate.
Add the highlighted principle to your notes before the check below.
In a cyclotron, the period of a proton's circular orbit is independent of its speed.
Doubling the speed of a particle in a uniform magnetic field doubles its orbital period.
The magnetic force on a moving charged particle does no work on the particle.
Radius, period, and the power of charge-to-mass ratio
We just saw the cyclotron principle — period is independent of speed. That raises a question: how different are the actual orbits of an electron versus a proton in the same field? This card answers it → the proton's radius is 1836× larger and its period 1836× longer due to its greater mass.
An electron and a proton each enter a uniform magnetic field of 0.020 T perpendicular to their velocity. Both have speed 4.0 × 106 m/s.
- (a) Calculate the radius of the electron's path.
- (b) Calculate the radius of the proton's path.
- (c) Calculate the period of each particle's motion.
- (d) Explain why the proton's period is much longer even though both travel at the same speed.
$r_e = \dfrac{m_e v}{eB} = \dfrac{(9.11 \times 10^{-31})(4.0 \times 10^6)}{(1.60 \times 10^{-19})(0.020)}$
$r_e = \mathbf{1.14 \times 10^{-3}}$ m = 1.14 mm
$r_p = \dfrac{m_p v}{eB} = \dfrac{(1.67 \times 10^{-27})(4.0 \times 10^6)}{(1.60 \times 10^{-19})(0.020)}$
$r_p = \mathbf{2.09}$ m
The proton's radius is about 1836 times larger than the electron's because the proton is 1836 times more massive.
Electron: $T_e = \dfrac{2\pi m_e}{eB} = \dfrac{2\pi(9.11 \times 10^{-31})}{(1.60 \times 10^{-19})(0.020)} = \mathbf{1.79 \times 10^{-9}}$ s
Proton: $T_p = \dfrac{2\pi m_p}{eB} = \dfrac{2\pi(1.67 \times 10^{-27})}{(1.60 \times 10^{-19})(0.020)} = \mathbf{3.28 \times 10^{-6}}$ s
The period depends only on $m/qB$, not on speed. The proton has the same charge magnitude as the electron but is 1836 times more massive. Therefore its period is 1836 times longer. Even though both particles travel at the same speed, the proton's much greater inertia means it resists the centripetal change in direction, taking far longer to complete each loop.
Using $r = mv/qB$ and $T = 2\pi m/qB$ in SI units: proton radius $\approx 1836\times$ electron radius at the same speed and field; proton period $\approx 1836\times$ longer regardless of speed.
Pause — write the highlighted ratio into your book before moving on.
An electron travels at 2.0 × 106 m/s perpendicular to a 0.010 T field. The radius of its path is approximately:
A charged particle in a B-field and a satellite in orbit follow the same mathematical logic
We just saw how to calculate radius and period for charged particles in B-fields. That raises a question: how is this the same as — or different from — a satellite orbiting Earth? This card answers it → both use real force = $mv^2/r$, but magnetic radius $\propto v$ while gravitational radius $\propto 1/v^2$.
The syllabus asks you to compare charged particle motion in magnetic fields to other examples of uniform circular motion. The most important comparison is with a satellite in orbit.
| Feature | Charged Particle in B-Field | Satellite in Orbit |
|---|---|---|
| Centripetal force | Magnetic force: $F = qvB$ | Gravitational force: $F = GMm/r^2$ |
| Setting equal to $mv^2/r$ | $qvB = mv^2/r$ | $GMm/r^2 = mv^2/r$ |
| Radius result | $r = mv/qB$ | $r = GM/v^2$ |
| Key difference | Radius increases with speed ($r \propto v$) | Radius decreases with speed ($r \propto 1/v^2$) |
| Key similarity | Both derive radius by equating the real force to the required centripetal force $mv^2/r$ | |
When asked to compare, always state the similarity (both use $F = mv^2/r$) and the difference (how radius depends on speed). Do not just describe each situation separately.
Similarity: both set real force = $mv^2/r$. Difference: magnetic $r \propto v$ (faster → bigger circle); gravitational $r \propto 1/v^2$ (faster → lower orbit). Always start any circular-motion problem with: real force = $mv^2/r$.
Add the highlighted comparison to your notes before the check below.
Three of these are correct comparisons between a charged particle in a B-field and a satellite in orbit. Pick the odd one out.
Predict, then verify using the interactive above
- Set B = 50 mT and v = 3.0 × 106 m/s with an electron. Predict the radius using $r = mv/qB$. Check the simulator readout.
- Double the speed to 6.0 × 106 m/s. What happens to the radius? What happens to the period? Explain why the period stays the same even though the particle is moving faster.
- Switch to a proton at the same speed and field. Predict the radius and period by what factor they change compared to the electron. Verify.
- Switch to an alpha particle (2× charge, ~4× proton mass). Predict the radius and period. Verify your prediction with the simulator.
Fill the gap. For a proton (mass $m_p$, charge $e$) in a field $B$, the cyclotron frequency is $f = eB / (\_\_\_ \cdot m_p)$. The missing factor is _____.
Link charged particle motion to satellite orbital mechanics
Two electrons enter the same magnetic field at different speeds. Electron A has twice the speed of Electron B. Compare: (i) their orbital radii, (ii) their periods, (iii) their cyclotron frequencies, and (iv) the centripetal force required for each.
Which set correctly describes how the radius of a charged particle's circular path changes when each variable is doubled (all others constant)?
A fresh five-question set drawn from this lesson's bank — feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence — that tells the system what to drill next.
ApplyBand 4(3 marks) 1. A proton enters a uniform magnetic field of 0.030 T perpendicular to its velocity of 5.0 × 105 m/s. (a) Calculate the radius of the proton's circular path. (b) Calculate the period of the proton's motion. (c) An electron enters the same field at the same speed. Without calculating, explain how its radius and period compare to the proton's.
1 mark: correct radius with working · 1 mark: correct period with working · 1 mark: correct comparison with explanation
EvaluateBand 6(4 marks) 2. A student claims: "In a cyclotron, faster protons must spin at a higher frequency to complete each smaller circle in less time." Evaluate this claim. In your answer, derive the expression for cyclotron frequency and explain what it depends on.
1 mark: identifies the claim as incorrect · 1 mark: derives $f = qB/2\pi m$ from $qvB = mv^2/r$ and $T = 2\pi r/v$ · 1 mark: states frequency is independent of speed · 1 mark: explains that faster protons travel in larger circles, covering the greater circumference in the same time
Show all answers
Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Short Answer — Model Answers
Q1 (a): $r = mv/qB = (1.67 \times 10^{-27} \times 5.0 \times 10^5) / (1.60 \times 10^{-19} \times 0.030) = 8.35 \times 10^{-22} / 4.80 \times 10^{-21} = 0.174$ m (1 mark).
Q1 (b): $T = 2\pi m/qB = 2\pi \times 1.67 \times 10^{-27} / (1.60 \times 10^{-19} \times 0.030) = 2.19 \times 10^{-6}$ s (1 mark).
Q1 (c): The electron's radius is much smaller than the proton's (by a factor of $m_p/m_e \approx 1836$) because $r \propto m$ and the electron is 1836 times less massive. The electron's period is also much smaller (by the same factor) because $T \propto m$ — the electron zips around its tiny circle in far less time (1 mark).
Q2 (4 marks): The claim is incorrect (1 mark). Derivation: from $qvB = mv^2/r$, the radius $r = mv/qB$. The period $T = 2\pi r/v = 2\pi m/qB$, which contains no $v$ (1 mark). Therefore the frequency $f = 1/T = qB/2\pi m$ is independent of speed (1 mark). Faster protons travel in larger circles: the greater circumference is exactly compensated by the higher speed, so each orbit takes the same time. This is why cyclotrons work with a fixed-frequency alternating voltage (1 mark).
Five timed questions on circular motion in magnetic fields. Beat the boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).
Enter the arenaAt the start you were asked about Lawrence's 1930 cyclotron at UC Berkeley: a proton and an electron entering the same 1.4 T field at the same speed — which curves more tightly, and by what factor?
The answer: the electron curves far more tightly. Since $r = mv/qB$ and both particles have the same charge magnitude but the proton is 1836 times more massive, the proton's radius is 1836 times larger. In Lawrence's cyclotron this is why protons (not electrons) were accelerated — electrons would spiral out almost instantly; protons stay in tight, controllable orbits.