Cyclotron Frequency

Equivalent forms

\omega_c = \frac{qB}{m}

r = \frac{m v}{q B}

Velocity cancels out of the period — a rare conspiracy where geometry (r ∝ v) and dynamics (F ∝ v) neutralize each other, making magnetic confinement a clock.

Symbol	Name	SI unit	Dimension	Range
$f_c$	Cyclotron frequencyoutput Number of complete orbits per second	Hz	T⁻¹	0 – 100000000
$q$	Charge Magnitude of the particle's electric charge	C	T·I	1.602176634e-19 – 1.602176634e-18
$B$	Magnetic field Uniform field perpendicular to the particle's velocity	T	M·T⁻²·I⁻¹	0.01 – 10
$m$	Mass Mass of the orbiting particle	kg	M	9.1093837015e-31 – 1e-25

Unit systems

Symbol	SI	CGS	Imperial
$f_c$	Hz	Hz	Hz
$q$	C	statC	C
$B$	T	G	T
$m$	kg	g	lb

Where it holds

Valid for non-relativistic speeds (v ≪ c) in a uniform field. At relativistic speeds replace m with

\gamma m

; in non-uniform fields the orbit drifts (basis of plasma confinement theory).

Dimensional analysis

$[q][B]/[m] =$ $(A\cdot s)(kg\cdot s^{-2}\cdot A^{-1})/(kg) = s^{-1} = Hz \checkmark$

Discovery

Ernest O. Lawrence · 1930

Lawrence realized the speed-independence meant a fixed-frequency oscillator could kick particles every half-turn forever. His 1930 cyclotron — 4.5 inches across — exploited this resonance and earned the 1939 Nobel Prize.

Try this

Why can a particle accelerator kick a proton at the exact same rhythm regardless of how fast it's already going?

A proton circles in a 1 T magnetic field. At what frequency does it orbit, and why doesn't the answer depend on its speed?

Research status: stable

Real-world applications

Cyclotrons producing medical isotopes (PET tracers)
Mass spectrometers separating ions by m/q
Electron cyclotron resonance heating in fusion tokamaks
Microwave ovens' magnetrons and radiation belts around Earth

Common misconceptions

The frequency does NOT depend on speed or radius — only on q, B, m (until relativity kicks in)
The magnetic force does no work; the particle's speed is constant, only its direction changes
Cyclotron motion is the particle orbiting field lines, not moving along them — motion along B is unaffected, giving helical paths

Experimental verification

Cyclotron resonance: sweep an RF field's frequency across a magnetized plasma or semiconductor and absorption peaks exactly at

qB/(2\pi m)

. This is precise enough to measure effective electron masses in crystals.

Derivation

Magnetic force supplies centripetal force:

qvB = mv^{2}/r

, so

r = mv/(qB)

.

The period is

T = 2\pi r/v = 2\pi m/(qB)

— v cancels.

Frequency

f_c = 1/T = qB/(2\pi m)

.

Limiting cases

v \to c

⟶ f decreases

as \gamma

growsRelativistic mass increase breaks the speed-independence:

f = qB/(2\pi \gamma m)

. This is why synchrotrons must sweep their RF frequency.

B \to 0

⟶

f_c \to 0

,

r \to \infty

No field means no bending — the particle travels in a straight line (infinite-radius circle).

m \to m/2

(lighter particle)⟶ f_c doublesLighter particles turn faster: an electron in the same field orbits

\sim 1836\times

faster than a proton.

What if…

What if the particle also has velocity along B?

That component is untouched — the particle spirals along the field line in a helix with the same cyclotron frequency.

What if the particle approaches the speed of light?

$\gamma$ grows, the effective mass $\gamma m$ rises, and the orbit frequency drops. Fixed-frequency cyclotrons top out around 25 MeV for protons; synchrocyclotrons and synchrotrons compensate by sweeping frequency.

What if you double B?

Frequency doubles and the orbit radius halves at fixed speed — stronger fields make tighter, faster circles.

1

Proton in a 1 T field

Given ·

q:: 1.602176634e-19
B:: 1
m:: 1.67262192369e-27

Find · f_c

Steps

$f_c = qB/(2\pi m)$
$= (1.602\times 10^{-19} \times 1)/(2\pi \times 1.673\times 10^{-27})$
$= 1.602\times 10^{-19} / 1.051\times 10^{-26} \approx 1.525\times 10^{7} Hz$
$\approx 15.25\,\mathrm{MHz}$ — the RF frequency a proton cyclotron must supply

Result ·

f_c \approx 15.25\,\mathrm{MHz}

2

Electron in the same field

Given ·

q:: 1.602176634e-19
B:: 1
m:: 9.1093837015e-31

Find · f_c

Steps

$f_c = (1.602\times 10^{-19})/(2\pi \times 9.109\times 10^{-31})$
$\approx 2.80\times 10^{10} Hz = 28\,\mathrm{GHz}$
Ratio to proton $= m_p/m_e \approx 1836$ , as expected

Result ·

f_c \approx 28.0\,\mathrm{GHz}

Lorentz Force Law→Magnetic Force on a Current-Carrying Wire→Larmor Formula→