A proton enters a uniform magnetic field with speed v. A second proton enters the same field with speed 2v. Which statement correctly describes their orbital periods?
AThe second proton has twice the orbital period, since it travels a larger circle
BThe second proton has half the orbital period, since it moves faster
CBoth protons have the same orbital period — period depends only on charge-to-mass ratio and field strength, not speed
DThe second proton has a period √2 times larger, since radius grows with speed
The cyclotron period is T = 2πm/(qB), which contains no velocity term. The second proton has twice the gyroradius (r = mv/(qB) doubles when v doubles), but also travels twice as fast around that larger circle — these effects cancel exactly. T = 2πr/v = 2π(mv/qB)/v = 2πm/(qB). The common misconception is to notice the larger radius and conclude the period must be longer, forgetting that speed also increased proportionally.
Question 2 Multiple Choice
Why can a cyclotron accelerator use a fixed-frequency alternating electric field to continuously accelerate particles, even as those particles gain energy and spiral outward?
AThe magnetic field increases as the particles spiral outward, compensating for the speed increase
BThe cyclotron frequency depends only on charge-to-mass ratio and field strength — as particles gain speed, their larger radius and higher speed cancel, keeping the orbital period constant
CThe electric field frequency is automatically adjusted by feedback from the particle beam
DOnly very slow particles can be accelerated by a cyclotron; fast particles require a different device
The key insight is that the cyclotron frequency f_c = qB/(2πm) is independent of velocity. As a particle gains energy and speed, its gyroradius grows — but the larger circle is traversed at proportionally higher speed, keeping the period exactly constant. A fixed-frequency electric field therefore stays in perfect resonance with the orbiting particle at every turn, continuously accelerating it. This elegant self-synchronization is the operating principle of the cyclotron. It breaks down only at relativistic speeds, where effective mass increases.
Question 3 True / False
A proton moving at speed 2v in a magnetic field traces a larger circle than a proton at speed v, so the faster proton takes longer to complete one full orbit.
TTrue
FFalse
Answer: False
This is the central misconception about cyclotron motion. The faster proton does trace a larger circle (r = mv/(qB) is proportional to v), but it also travels around that circle proportionally faster. The period T = 2πr/v = 2πm/(qB) — when you substitute r = mv/(qB), the v in the numerator and denominator cancel completely. The period is identical for both protons. This velocity-independence is the defining and counterintuitive feature of cyclotron motion.
Question 4 True / False
Cyclotron motion eventually breaks down at relativistic particle speeds because the particle's effective mass increases with velocity, shifting the orbital frequency away from its non-relativistic value.
TTrue
FFalse
Answer: True
The cyclotron frequency formula f_c = qB/(2πm) treats mass m as constant. At relativistic speeds, the effective mass increases as m_rel = γm₀ (where γ = 1/√(1 − v²/c²) increases with speed). As particles accelerate, the increasing effective mass causes the orbital period T = 2πm_rel/(qB) to grow — the particles fall out of sync with a fixed-frequency electric field. This is precisely why synchrotrons were developed: they vary the magnetic field or RF frequency to maintain resonance as the relativistic mass increases.
Question 5 Short Answer
Derive why the cyclotron frequency is independent of a charged particle's velocity, starting from the condition that the magnetic Lorentz force provides the centripetal force.
Think about your answer, then reveal below.
Model answer: Setting the Lorentz force equal to the centripetal force: qvB = mv²/r. Solving for gyroradius: r = mv/(qB). The period is T = circumference/speed = 2πr/v = 2π(mv/qB)/v = 2πm/(qB). The velocity v cancels in the last step. Therefore the cyclotron frequency f_c = 1/T = qB/(2πm) contains no v — it depends only on the charge-to-mass ratio q/m and the magnetic field strength B.
The cancellation is exact and non-trivial: a faster particle has a larger radius (more v in the numerator of r) but also moves faster (more v in the denominator of T = 2πr/v), and these effects precisely cancel. This is not a coincidence but a consequence of the Lorentz force being linear in v — the centripetal acceleration requirement (v²/r) also contains a v that cancels with the v in the force. The result is one of the most elegant and practically important facts in classical electromagnetism.