In short

A cyclotron is a device that uses a uniform magnetic field to bend a charged particle into circles, and a rapidly reversing electric field in a narrow gap to push the particle faster each half-turn.

Geometry. Two hollow D-shaped electrodes (the dees) sit in a vacuum chamber between the poles of a strong electromagnet. The gap between the dees is tiny — millimetres.

How it works.

  1. Inside each dee the particle feels only \vec{B} (the electric field is screened out). It executes a half-circle of radius r = mv/(qB) at the cyclotron frequency f_c = qB/(2\pi m).
  2. Every time the particle crosses the gap, the voltage on the dees is reversed — so the electric field across the gap always points in the direction that speeds the particle up.
  3. Faster particle ⇒ bigger radius, so each successive half-circle is wider. The trajectory is an outward spiral of half-circles connected by straight gap-crossings.

The resonance condition. The voltage reversal frequency must equal f_c = qB/(2\pi m). This is the deepest result in cyclotron physics: the time to complete a half-circle, T/2 = \pi m/(qB), is independent of the speed or radius. Fast particles travel bigger arcs, slow particles travel smaller arcs, but they all take the same time.

Maximum kinetic energy. If the exit radius is R, the speed at extraction satisfies qvB = mv^2/R, giving v = qBR/m. The kinetic energy is

\boxed{\;K_\text{max} \;=\; \tfrac{1}{2}mv^2 \;=\; \frac{q^2B^2R^2}{2m}\;}.

The relativistic limit. At high energies the particle's relativistic mass \gamma m grows and the cyclotron frequency drops. The particle falls behind the fixed RF — and the resonance breaks. For protons this happens near 20 MeV, which is why plain cyclotrons top out there; to go further you need a synchrocyclotron (sweep the RF frequency) or an isochronous cyclotron (shape the field B(r) so it increases with radius and compensates \gamma).

Inside a shielded vault at VECC Kolkata, a hydrogen ion the size of a proton climbs into a spiral staircase made of magnetic field. Every half-turn it jumps across a slit a few millimetres wide and a few kilovolts deep; every jump adds about ten kilo-electron-volts to its energy. Sixteen hundred half-turns later it emerges on the other side of a thin aluminium window, a beam of protons at 20 MeV, travelling at about one-fifth the speed of light. Doctors at Variable Energy Cyclotron Centre point that beam at a patient's tumour. The patient goes home.

A handful of machines in India — VECC Kolkata, BARC Mumbai's Pelletron, the cyclotrons being built for proton therapy at Tata Memorial — run on the same piece of physics. One uniform magnetic field, one oscillating electric field, and the remarkable fact that a charged particle in a uniform magnetic field completes a full circle in the same time no matter how fast it is moving. That "no matter how fast" is the loophole that makes the whole machine possible.

The one magical fact — cyclotron frequency is independent of speed

Start by proving the single result on which the cyclotron depends.

A charge q moves with speed v perpendicular to a uniform magnetic field of magnitude B. Neglect gravity and any electric field. The magnetic force is the only force:

\vec{F} \;=\; q\vec{v}\times\vec{B}, \qquad |\vec{F}| = qvB.

Why: \vec{v}\perp\vec{B} so \sin\theta = 1 in F = qvB\sin\theta. The force is perpendicular to \vec{v}, so the charge moves in a circle — a force that stays at 90° to the velocity provides exactly the centripetal pull needed for uniform circular motion.

Step 1. Set the magnetic force equal to the centripetal force mv^2/r:

qvB \;=\; \frac{mv^2}{r}.

Why: circular motion at constant speed requires an inward force of magnitude mv^2/r. Here that force is supplied entirely by the magnetic term.

Step 2. Solve for the radius.

\boxed{\;r \;=\; \frac{mv}{qB}\;}

Why: the v on both sides does not cancel completely — one survives. The faster the particle, the bigger the orbit; the heavier, the bigger; the stronger the field, the tighter.

Step 3. Now write the period T — the time to complete one full circle.

The circumference of the orbit is 2\pi r. At constant speed v, the period is T = 2\pi r / v. Substitute r = mv/(qB):

T \;=\; \frac{2\pi}{v}\cdot\frac{mv}{qB} \;=\; \frac{2\pi m}{qB}.

Why: the v in the numerator cancels the v in the denominator. The period depends only on m, q, and B — not on how fast the particle is moving.

Step 4. The frequency — revolutions per second — is f = 1/T:

\boxed{\;f_c \;=\; \frac{qB}{2\pi m}\;}

This is the cyclotron frequency. A 1 T magnetic field makes every proton in it orbit at f_c = (1.6\times 10^{-19})(1)/(2\pi\times 1.67\times 10^{-27}) \approx 1.5\times 10^7 Hz = 15 MHz. Every proton — whether moving at 1 m/s or 10^7 m/s, whether just injected or nearly relativistic — circles the field at 15 MHz. (Until relativistic effects creep in; more on that later.)

Why this is magic: in ordinary circular motion — a stone on a string, a car on a banked road — faster means bigger circle and shorter period. A charged particle in a magnetic field is the one exception: faster means bigger circle and exactly the same period. That is the cancellation in step 3.

The angular frequency \omega_c = 2\pi f_c = qB/m is called the cyclotron angular frequency; you will meet it as often as the ordinary f_c.

The anatomy of a cyclotron

Now build the machine around this fact. Two hollow, flat, D-shaped metal electrodes — the dees — sit face-to-face with a small gap between them (a few millimetres wide, like the slit between two books on a shelf). The entire pair is sandwiched between the pole faces of a strong electromagnet that produces a uniform vertical field \vec{B}, perpendicular to the plane of the dees.

Inside each dee the interior is a hollow conductor, so any static electric field from the dee's charged surface cannot reach a charge inside — the electric field is screened. The only force on the particle inside a dee is the magnetic force, and the particle moves on a circular arc.

Across the gap between the dees, a radio-frequency (RF) oscillator drives an alternating voltage V(t) = V_0\cos(2\pi f_\text{RF} t). This sets up an electric field \vec{E} across the gap that reverses direction every half-cycle. The electric field is confined almost entirely to the narrow gap — negligible penetration into the dees — so the particle feels \vec{E} only while crossing the gap and \vec{B} only inside a dee.

Top-down schematic of a cyclotron: two dees and the gap Two D-shaped electrodes separated by a narrow gap. Inside each dee, the magnetic field bends the particle into a half-circle. Across the gap, the electric field kicks the particle to a higher speed. The spiral trajectory grows outward with each kick. top view; B points into the page (×) ×× ×× D₁ D₂ gap ion source (centre) RF oscillator
Top view of a cyclotron. The two dees $D_1$ and $D_2$ are hollow D-shaped electrodes separated by a narrow gap. A particle injected at the centre (ion source) executes half-circles inside each dee and picks up energy each time it crosses the gap. The magnetic field points into the page (uniform, vertical). The RF oscillator reverses the dee voltage at the cyclotron frequency.

The ion source sits at the centre of the gap — a small chamber that strips electrons off hydrogen atoms (or other gas) to produce ions at low energy. These ions start at essentially zero speed at the very centre and begin to drift under the electric field in the gap.

The RF timing. The oscillator frequency is set exactly to the cyclotron frequency: f_\text{RF} = f_c = qB/(2\pi m). This is the resonance condition. Every time the particle arrives at the gap — after each half-circle, which takes T/2 = \pi m/(qB) — the voltage has reversed polarity, so \vec{E} now points in the direction that accelerates the particle further. The particle always sees "green light" at the gap, no matter how many times it has crossed before.

Walking through one revolution

Trace the particle's path from birth to first half-circle.

Step 1. The particle starts at the source in the gap. Let the voltage at t = 0 be such that D_1 is negative and D_2 is positive. A positive ion is pushed from D_2 toward D_1 (electric field from + to −, ion moves opposite to... wait — positive ion moves along \vec{E}, from high potential to low, i.e. from D_2 to D_1). The particle enters D_1 with some small speed v_1.

Step 2. Inside D_1 the electric field is zero (the hollow-conductor screening). Only \vec{B} acts. The particle curves in a half-circle of radius r_1 = mv_1/(qB), taking time T/2 = \pi m/(qB). It returns to the gap on the other end of the dee.

Step 3. In the time T/2 between entries to the gap, the RF voltage has completed exactly half a cycle — the polarity has reversed. Now D_1 is positive and D_2 is negative. The electric field in the gap now pushes the positive ion from D_1 toward D_2 — the direction it was already heading. So it gets another kick; its speed rises from v_1 to v_2.

Step 4. Inside D_2, the particle makes a half-circle of larger radius r_2 = mv_2/(qB), again taking time T/2. Back to the gap.

Step 5. Another polarity flip, another kick. Speed rises to v_3. Into D_1 at radius r_3 = mv_3/(qB). And so on.

Each half-circle takes the same time T/2 (the cyclotron-frequency result) regardless of the radius. After N full revolutions, the particle has crossed the gap 2N times and gained 2N kicks of energy.

The kinetic energy after N revolutions. If each gap crossing increases the energy by qV_0 (where V_0 is the peak voltage across the gap — assuming the particle crosses in phase with the maximum voltage), then after 2N crossings:

K_N \;=\; 2N\,qV_0.

For VECC Kolkata's 20 MeV proton cyclotron, typical numbers are V_0 \approx 50 kV, so each crossing adds about q\times 50\ \text{kV} = 50 keV. To reach 20 MeV takes 20\,000/50 = 400 crossings — 200 full revolutions.

The maximum kinetic energy — derive K_\text{max} = q^2B^2R^2/(2m)

A cyclotron cannot grow arbitrarily large; its magnet has a finite pole-face radius R. Once the particle's orbital radius reaches R, it hits the extraction point and leaves the machine. What is its energy at that moment?

Step 1. At the maximum radius R, the magnetic force provides the centripetal force:

qv_\text{max}B \;=\; \frac{mv_\text{max}^2}{R}.

Why: same circular-motion balance as step 1 of the cyclotron-frequency derivation, applied at the extraction radius.

Step 2. Solve for the maximum speed.

v_\text{max} \;=\; \frac{qBR}{m}.

Why: invert the previous equation. The maximum attainable speed depends on three things: the field strength, the machine size, and the charge-to-mass ratio of the particle.

Step 3. The kinetic energy at that speed.

K_\text{max} \;=\; \tfrac{1}{2}m v_\text{max}^2 \;=\; \tfrac{1}{2}m\left(\frac{qBR}{m}\right)^2 \;=\; \frac{q^2B^2R^2}{2m}.

Why: substitute v_\text{max} into K = \tfrac{1}{2}mv^2. The mass in the denominator comes from the 1/m^2 in v^2 meeting the m in \tfrac{1}{2}m v^2.

\boxed{\;K_\text{max} \;=\; \frac{q^2B^2R^2}{2m}\;}

Notice what this formula says:

Watch it spiral — an animated cyclotron trajectory

Animated: a charged particle spiralling outward through a cyclotron A positive charge starts at the centre, moves in a spiral of expanding half-circles, gaining energy at each gap crossing. The trajectory is traced in red; the radius grows with the square root of the number of kicks. x (arb. units) y
A positive charge, injected at rest near the centre, spirals outward through the cyclotron. Each turn is slightly wider than the previous, because each gap crossing adds energy. The time per half-turn is constant — the hallmark of cyclotron motion. The trail shows the path; the radius grows as the square root of the energy, which grows linearly with the number of gap crossings.

The animation uses a smooth Archimedean approximation rather than literal half-circles plus straight gap crossings (the gap is millimetres wide while the orbit is metres across — far too small to see). What you watch is the envelope of the real trajectory: widening turns, constant period, outward drift.

Worked examples

Example 1: Proton cyclotron at VECC Kolkata — peak energy

VECC Kolkata's main cyclotron produces protons with peak magnetic field B = 1.67 T across a pole-face radius of R = 0.95 m. The proton mass is m_p = 1.67\times 10^{-27} kg and charge q = 1.6\times 10^{-19} C. Find the maximum kinetic energy of the extracted proton beam, both in joules and in MeV.

Proton at extraction radius in a 1.67 T, 0.95 m cyclotron Schematic of a proton at the extraction radius R = 0.95 m with centripetal force balancing the magnetic force qvB. p R = 0.95 m v (tangent) F = qvB B into page (1.67 T)
At extraction, the proton moves tangentially at the edge of the pole face; the magnetic force supplies the centripetal pull.

Step 1. Identify the inputs.

q = 1.6\times 10^{-19} C, B = 1.67 T, R = 0.95 m, m = 1.67\times 10^{-27} kg.

Step 2. Plug directly into K_\text{max} = q^2B^2R^2/(2m).

q^2 \;=\; (1.6\times 10^{-19})^2 \;=\; 2.56\times 10^{-38}\ \text{C}^2.
B^2 \;=\; (1.67)^2 \;=\; 2.79\ \text{T}^2.
R^2 \;=\; (0.95)^2 \;=\; 0.9025\ \text{m}^2.

Why: compute each squared factor separately to keep the arithmetic tidy. Exponents add when multiplying.

Step 3. Combine.

K_\text{max} \;=\; \frac{(2.56\times 10^{-38})(2.79)(0.9025)}{2\,(1.67\times 10^{-27})}\ \text{J}.

Numerator: 2.56\times 2.79\times 0.9025 = 6.445. So numerator is 6.445\times 10^{-38}.

Denominator: 2\times 1.67\times 10^{-27} = 3.34\times 10^{-27}.

K_\text{max} \;=\; \frac{6.445\times 10^{-38}}{3.34\times 10^{-27}} \;=\; 1.93\times 10^{-11}\ \text{J}.

Why: divide numerator by denominator; the exponents subtract, -38 - (-27) = -11.

Step 4. Convert to MeV using 1\ \text{eV} = 1.6\times 10^{-19} J, so 1\ \text{MeV} = 1.6\times 10^{-13}\ \text{J}.

K_\text{max} \;=\; \frac{1.93\times 10^{-11}}{1.6\times 10^{-13}}\ \text{MeV} \;=\; 121\ \text{MeV}.

Why: divide joules by 1.6\times 10^{-13} to get MeV.

Result. The peak kinetic energy is K_\text{max} \approx 121 MeV — but in practice VECC extracts its proton beam at a lower energy (typically below 25 MeV for nuclear physics, up to 80 MeV in the K-130 configuration) because relativistic effects break the resonance at 20 MeV and above. That is the subject of the next section. The formula q^2B^2R^2/(2m) gives the non-relativistic upper bound; the relativistic machine has to work harder.

Example 2: Tuning the RF frequency for a proton-therapy cyclotron at Tata Memorial

A proton-therapy facility at Tata Memorial Hospital uses a cyclotron with magnetic field B = 2.5 T. At what RF frequency must the dee voltage oscillate to stay in resonance with the protons? What is the period of one full revolution?

Step 1. Use the cyclotron frequency formula.

f_c \;=\; \frac{qB}{2\pi m} \;=\; \frac{(1.6\times 10^{-19})(2.5)}{2\pi (1.67\times 10^{-27})}.

Why: this is the resonance condition. f_\text{RF} = f_c is required for every gap crossing to coincide with the peak voltage.

Step 2. Evaluate numerator and denominator.

Numerator: 1.6\times 2.5\times 10^{-19} = 4.0\times 10^{-19}.

Denominator: 2\pi\times 1.67\times 10^{-27} = 10.49\times 10^{-27} = 1.049\times 10^{-26}.

f_c \;=\; \frac{4.0\times 10^{-19}}{1.049\times 10^{-26}} \;=\; 3.81\times 10^{7}\ \text{Hz} \;\approx\; 38.1\ \text{MHz}.

Why: divide; exponents subtract, -19 - (-26) = +7. So f_c \sim 38 MHz — in the standard FM radio band of frequencies.

Step 3. Period is the reciprocal.

T \;=\; \frac{1}{f_c} \;=\; \frac{1}{3.81\times 10^{7}\ \text{Hz}} \;=\; 2.62\times 10^{-8}\ \text{s} \;\approx\; 26\ \text{ns}.

Why: 1/f gives the period. Twenty-six nanoseconds per revolution means the electronics controlling the RF voltage have to reverse polarity every 13 ns — which is why cyclotrons need precision RF amplifiers and low-inductance transmission lines.

Step 4. Sanity-check the radius at clinical energy.

Proton therapy needs \sim 230 MeV for deep tumours. At that energy the proton is mildly relativistic (\gamma \approx 1.24), so the simple formula is only approximate. Using it anyway: v \approx 1.85\times 10^8 m/s, giving radius

r \;=\; \frac{mv}{qB} \;=\; \frac{(1.67\times 10^{-27})(1.85\times 10^8)}{(1.6\times 10^{-19})(2.5)} \;\approx\; 0.77\ \text{m}.

Why: the pole-face diameter of a 230 MeV therapy cyclotron ends up around 1.5 m, which matches the physical size of machines like the IBA C235 used in clinical proton-therapy installations worldwide. A true relativistic calculation would give a slightly larger radius because \gamma m replaces m.

Result. The RF frequency is 38.1 MHz, the period is 26 ns, and the extraction radius at clinical energy is roughly 0.77 m. These are the numbers a proton-therapy cyclotron designer juggles — a compact, precisely-timed machine that can deliver a monoenergetic proton beam to a patient's tumour.

Common confusions

If you came here to learn how a cyclotron works and to solve the standard formulas, you have what you need. What follows is for readers who want the relativistic correction, the isochronous fix, and the deeper symmetry that links the cyclotron to the general theory of accelerators.

The relativistic breakdown — derive when the resonance fails

In special relativity, a particle's momentum is \vec{p} = \gamma m\vec{v} where \gamma = 1/\sqrt{1 - v^2/c^2}. The Lorentz force still reads \vec{F} = q\vec{v}\times\vec{B}, and Newton's second law reads d\vec{p}/dt = \vec{F}. For perpendicular motion, |\vec{v}| is constant (magnetic force does no work), so \gamma is constant along one orbit, and the centripetal-force balance becomes

qvB \;=\; \frac{\gamma m v^2}{r}, \qquad r \;=\; \frac{\gamma m v}{qB} \;=\; \frac{p}{qB}.

Why: replace m with \gamma m (relativistic mass) everywhere. The radius is now p/(qB), which is the more fundamental form — it works at any speed.

The cyclotron frequency becomes

f_c^\text{rel} \;=\; \frac{qB}{2\pi\gamma m}.

As the particle gains energy, \gamma grows, so f_c^\text{rel} shrinks. The fixed-frequency RF of a conventional cyclotron (f_\text{RF} = qB/(2\pi m), set for \gamma = 1) becomes too fast for the particle — the voltage has already reversed back before the particle reaches the gap. The resonance breaks.

When does this become a problem? Define the fractional lag per revolution: \Delta f/f_c = 1 - 1/\gamma \approx (\gamma - 1) for small \gamma - 1. For protons, \gamma = 1 + K/(mc^2) with mc^2 = 938 MeV. So at K = 20 MeV, \gamma - 1 = 0.021 — a 2% mismatch per revolution. After 50 revolutions the accumulated phase error is a full half-cycle, and kicks start landing at the wrong phase. Practically, conventional cyclotrons for protons cap out around 20 MeV.

Isochronous cyclotrons — fix the frequency by shaping B(r)

If you want f_c^\text{rel} = qB/(2\pi\gamma m) to stay constant as \gamma grows, make B grow in lockstep with \gamma:

B(r) \;=\; B_0\,\gamma(r).

Since \gamma depends on the particle's energy, and the energy depends on the radius (bigger radius = more accumulated kicks = higher energy), this prescription gives an azimuthally averaged field that increases with r. Machines that implement this are called isochronous cyclotrons — "same time" cyclotrons, keeping every orbit on the same period. VECC's K-130 cyclotron and the K-500 superconducting cyclotron under commissioning at VECC are both isochronous.

A subtlety: an axially symmetric field B_z(r) that increases with r is vertically defocusing. To compensate, isochronous cyclotrons use an azimuthally varying field (AVF) — alternating "hills" and "valleys" around the orbit — that provides vertical focusing through the Thomas mechanism. This is why modern isochronous cyclotron magnets have that distinctive spiral-sector pole-face pattern.

Synchrocyclotrons — the easier (but pulsed) fix

Alternative fix: leave B(r) uniform but sweep the RF frequency downward as the beam accelerates. Start at f_c = qB/(2\pi m) (particles non-relativistic), decrease to f_c/\gamma_\text{max} (particles at peak energy). Each RF sweep corresponds to one "bunch" of particles being accelerated together — so the beam is pulsed, not continuous. Synchrocyclotrons are the simplest way to exceed 20 MeV without isochronous field shaping; they were how energies up to several hundred MeV were first achieved, and they are still used in some compact proton-therapy machines (the Mevion S250 is a synchrocyclotron).

The energy limit of any cyclotron — the synchrotron frontier

As \gamma grows without bound, the orbit radius r = \gamma m v/(qB) grows too. Eventually the required magnet becomes impossibly large. A conventional or isochronous cyclotron topping out at a few hundred MeV for protons uses a magnet 5 m across; pushing to GeV energies requires the synchrotron — a ring of separate bending magnets surrounding a fixed-radius orbit, with the field B(t) ramped in time as the particle accelerates. The synchrotron is the next generation beyond the cyclotron, and it is what drives RRCAT Indore's Indus-2 light source (2.5 GeV electrons) and, in principle, the proposed Indian synchrotron-radiation source upgrade.

The Bhabhatron — not a cyclotron, but related

A common confusion worth addressing: the Bhabhatron, India's indigenous telecobalt unit for cancer radiotherapy (developed at BARC, deployed across dozens of Indian hospitals), is not a cyclotron. It uses cobalt-60 gamma rays, not an accelerator. However, the cobalt-60 source used to be produced by neutron bombardment in a reactor; similar isotopes for PET scanning (fluorine-18, carbon-11) are produced by cyclotrons — typically compact 10–20 MeV proton or deuteron cyclotrons installed in or near hospitals. Indian examples: the cyclotrons at Apollo Hospitals, AIIMS Delhi, and VECC Kolkata (for basic research and isotope production).

The link to the Lorentz force

Every step of the cyclotron's operation traces back to the two-term Lorentz force \vec{F} = q\vec{E} + q\vec{v}\times\vec{B}. The magnetic term q\vec{v}\times\vec{B} supplies the bending inside the dees (does no work, just steers). The electric term q\vec{E} supplies the energy in the gap (does all the work). The resonance condition f_\text{RF} = qB/(2\pi m) ties them together in time. The cyclotron is, in one sentence, the simplest possible machine that separates the steering from the accelerating, using the remarkable speed-independence of the cyclotron frequency to make that separation clean.

Where this leads next