Audited 31 Jul 2026·Last updated 28 Oct 2026·3 citations·Tier 3·0 uses

Cyclotron Frequency Calculator

Find how many times a second a charged particle circles in a magnetic field, f = |q|B/(2πm), from its charge in coulombs, field in tesla and mass in kg.

Cyclotron Frequency Calculator

C
T
kg
Cyclotron frequency
27,992,489,834.2
Orbits completed per second, in hertz, by a nonrelativistic particle moving perpendicular to the uniform field.
Model scope
Nonrelativistic magnitude for uniform B perpendicular to velocity; relativistic mass-energy, field gradients, collisions, and sign-dependent rotation direction are outside it.

Background.

A charged particle moving across a magnetic field is pushed sideways, at right angles to both its motion and the field, and that push bends its path into a circle. The surprising part, which Ernest Lawrence built the first cyclotrons around in the early 1930s, is how long each lap takes. A faster particle swings round a proportionally bigger circle, so the time per orbit does not depend on speed at all. It depends only on the charge, the field strength and the mass: f = |q|B/(2πm).

That independence from speed is what makes the frequency so useful. In a cyclotron, an alternating voltage timed to it gives particles a kick every half-turn as they spiral outwards. In a Penning trap or a Fourier-transform ion cyclotron resonance mass spectrometer, measuring how fast a trapped ion circles is a way of weighing it with great precision. In a plasma, microwaves tuned to the electron cyclotron frequency are absorbed strongly wherever the field has just the right strength, which is how electron cyclotron resonance heating delivers energy to a chosen spot inside a fusion device.

The numbers are worth knowing. An electron in a 1-tesla field circles about 28 billion times a second, squarely in the microwave band; a proton in the same field manages about 15.2 million, a radio frequency, because it is roughly 1,836 times heavier.

The formula used here is the nonrelativistic one. As a particle approaches the speed of light its inertia grows by the Lorentz factor γ and the true frequency falls by the same factor, already about 2% for a 10 keV electron, which is why fixed-frequency cyclotrons cannot take electrons far. It also assumes a uniform field at right angles to the motion; gradients, collisions and the direction of rotation, which depends on the sign of the charge, are outside the calculation.

What is cyclotron frequency calculator?

The cyclotron frequency is the number of complete circles per second a charged particle makes when it moves perpendicular to a uniform magnetic field. It is fixed by the particle's charge-to-mass ratio and the field strength, f_c = |q|B/(2πm), and does not depend on the particle's speed or the size of its orbit, provided the speed is well below that of light. Multiplying by 2π gives the angular frequency, ω_c = |q|B/m, in radians per second.

How to use this calculator.

  1. Enter the charge magnitude in coulombs. An electron or a proton carries 1.602176634 × 10⁻¹⁹ C; for an ion in charge state z, multiply that by z. The sign is not needed, since it only decides which way the particle turns.
  2. Enter the magnetic flux density in tesla. Fields quoted in gauss convert at 10,000 G per tesla, so 875 G is 0.0875 T.
  3. Enter the rest mass in kilograms: 9.1093837139 × 10⁻³¹ kg for the electron, about 1.6726 × 10⁻²⁷ kg for the proton, or an ion's mass in atomic mass units multiplied by 1.6605 × 10⁻²⁷ kg.
  4. Read the frequency in hertz. For a particle moving at a sizeable fraction of light speed, divide by its Lorentz factor γ to get the real orbital frequency.

The formula.

f_c = |q|B / (2πm)

f_c = |q|B / (2πm). A charge q moving at speed v perpendicular to a field B feels a magnetic force |q|vB, and that force supplies the centripetal force mv²/r needed for circular motion. Setting the two equal gives the orbit radius, r = mv/(|q|B). One lap is 2πr long and takes 2πr/v = 2πm/(|q|B) seconds — the speed has cancelled — and the frequency is the reciprocal of that period. The result is proportional to charge and field and inversely proportional to mass, so doubling B doubles f_c, and a particle twice as heavy with the same charge circles half as fast. The units work out cleanly: coulombs times tesla divided by kilograms reduce to reciprocal seconds, or hertz.

A worked example.

Example

Take an electron, with charge 1.602176634 × 10⁻¹⁹ C and mass 9.1093837139 × 10⁻³¹ kg, in a 1 T field. The numerator is |q|B = 1.602176634 × 10⁻¹⁹ × 1 = 1.602176634 × 10⁻¹⁹. The denominator is 2π × 9.1093837139 × 10⁻³¹ = 5.72359 × 10⁻³⁰ kg. Dividing one by the other gives f_c = 2.79925 × 10¹⁰ Hz, or 27,992,489,834.2 Hz: about 27.99 GHz, a microwave frequency with a free-space wavelength of about 1.071 cm. Swap in a proton, mass about 1.6726 × 10⁻²⁷ kg, and the same field gives roughly 1.5245 × 10⁷ Hz, about 15.25 MHz. The charge is the same size, so the 1,836-fold drop comes entirely from the heavier mass.

magnetic Field Tesla1
particle Mass Kg0
charge Magnitude C0

Frequently asked questions.

Why does the frequency not depend on how fast the particle is moving?
Because speed affects both the distance and the pace of each lap equally. A particle twice as fast travels a circle with twice the radius, so the circumference doubles too and the lap takes exactly as long. That cancellation holds until relativistic effects set in, which is what made the original cyclotron possible.
Some textbooks give the answer in radians per second. How do the two relate?
The angular cyclotron frequency is ω_c = |q|B/m, which is 2π times the frequency in hertz shown here. For an electron in a 1 T field, ω_c is about 1.7588 × 10¹¹ rad/s, and dividing by 2π gives the 2.79925 × 10¹⁰ Hz figure.
How much does relativity change the result?
The true frequency is the value shown divided by the Lorentz factor γ. For an electron with 10 keV of kinetic energy, γ is about 1.0196, so the real frequency is about 1.92% lower. Electrons reach that point at modest energies because their rest energy is only about 511 keV; heavy ions tolerate far higher energies before the shift matters.
What field puts electrons in resonance with a 2.45 GHz microwave source?
Rearranging the formula, B = 2πmf/|q|. For an electron and 2.45 GHz that comes to about 0.0875 T, or 875 gauss. Microwave plasma sources running at 2.45 GHz therefore shape their magnets so that a surface of roughly this strength sits inside the plasma, where the microwaves are absorbed.
How do I enter an ion such as doubly charged helium?
Give the charge as the charge state times the elementary charge, 2 × 1.602176634 × 10⁻¹⁹ = 3.204353268 × 10⁻¹⁹ C, and the mass of the ion in kilograms. For a bare helium-4 nucleus that is about 6.6447 × 10⁻²⁷ kg. Ions with the same charge-to-mass ratio circle at the same frequency.

References& sources.

  1. [1]U.S. Naval Research Laboratory, 2019 NRL Plasma Formulary.
  2. [2]OpenStax, University Physics Volume 2, section 11.4, Motion of a Charged Particle in a Magnetic Field.
  3. [3]BIPM, The International System of Units (SI Brochure), 9th ed., version 3.01, coherent derived units and quantity equations.

How this page was produced

Published by
Quanta Calculator
Primary sources
3 cited below
Method
f_c = |q|B / (2πm)
Published
Last verified

Built with AI assistance and verified by automated tests against the cited sources — every worked example on this page is computed by the same code that runs the calculator. How we build and check calculators.

In this category

Embed

Quanta Pro

Paid features are coming later.

  • All 1560 calculators remain free
  • No billing is enabled
Coming soon