8.13.3.2. Cyclotron Frequency

Table of Contents

Circular motion in a magnetic field

A cyclotron works because a charged particle moving perpendicular to a magnetic field follows a circular path. The magnetic force provides the centripetal force. If a particle has charge $q$, mass $m$, speed $v$, and moves in a uniform magnetic field $B$, then

$$|q|vB = \frac{mv^2}{r}$$

where $r$ is the radius of the circular path. Solving for the speed and radius relation gives

$$r = \frac{mv}{|q|B}$$

This means that faster particles move in larger circles, but the key result is that the time for one revolution does not depend on the radius.

Derivation of the cyclotron frequency

The circumference of the circular path is $2\pi r$, so the period of one full revolution is

$$T = \frac{2\pi r}{v}$$

Substituting $r = \frac{mv}{|q|B}$ gives

$$T = \frac{2\pi}{v}\frac{mv}{|q|B} = \frac{2\pi m}{|q|B}$$

So the cyclotron frequency, which is the number of revolutions per second, is

$$f = \frac{1}{T} = \frac{|q|B}{2\pi m}$$

The angular frequency is

$$\omega = 2\pi f = \frac{|q|B}{m}$$

For a nonrelativistic charged particle in a uniform magnetic field, the cyclotron frequency is
$$f = \frac{|q|B}{2\pi m}$$
and the angular cyclotron frequency is
$$\omega = \frac{|q|B}{m}$$
These formulas are fundamental for the operation of a cyclotron.

Why this frequency matters in a cyclotron

In a cyclotron, the accelerating electric field must reverse at just the right rate so that the particle is accelerated each time it crosses the gap between the dees. That required rate is the cyclotron frequency. Since $f$ does not depend on the particle speed in the nonrelativistic case, the same alternating voltage can keep accelerating the particle over many turns.

This is the special feature that makes the cyclotron possible. As the particle gains speed, the orbit radius increases, but the revolution frequency remains the same.

Dependence on particle properties

The cyclotron frequency depends on the magnetic field strength, the particle charge, and the particle mass.

$$f \propto B$$

$$f \propto |q|$$

$$f \propto \frac{1}{m}$$

A stronger magnetic field gives a higher frequency. A particle with larger charge also has a higher frequency. A more massive particle has a lower frequency.

The sign of the charge changes the direction of rotation, but not the value of the frequency.

Comparison of different particles

For the same magnetic field, lighter particles rotate faster than heavier ones.

Particle	Charge magnitude	Mass	Cyclotron frequency
Electron	$e$	small	high
Proton	$e$	much larger	lower
Alpha particle	$2e$	about $4m_p$	lower than proton

For an alpha particle,

$$f = \frac{2eB}{2\pi(4m_p)} = \frac{eB}{4\pi m_p}$$

which is half the proton cyclotron frequency in the same magnetic field.

Numerical example

Suppose a proton moves in a magnetic field of magnitude $B = 1.5\ \text{T}$. Using

$$f = \frac{|q|B}{2\pi m}$$

with $q = 1.60 \times 10^{-19}\ \text{C}$ and $m = 1.67 \times 10^{-27}\ \text{kg}$,

$$f \approx \frac{(1.60 \times 10^{-19})(1.5)}{2\pi(1.67 \times 10^{-27})}$$

$$f \approx 2.29 \times 10^7\ \text{Hz}$$

So the proton revolves at about $22.9\ \text{MHz}$.

Visual picture

The particle spirals outward because its speed increases while the frequency stays nearly constant. Each larger circle takes the same time as the smaller one, as long as the motion remains nonrelativistic.

Rendering scientific sketch…

Cyclotron motion and constant revolution frequency

Important limitation

The formula for cyclotron frequency is exact only for nonrelativistic motion. At very high speeds, the particle mass in relativistic dynamics effectively increases, and the frequency decreases. Then the particle can fall out of step with the alternating electric field.

The simple cyclotron frequency formula
$$f = \frac{|q|B}{2\pi m}$$
is valid only when relativistic effects are negligible.

Summary relation

Cyclotron frequency is the natural orbital frequency of a charged particle in a magnetic field. It sets the operating frequency of the accelerating voltage in a cyclotron and is given by

$$f = \frac{|q|B}{2\pi m}, \qquad \omega = \frac{|q|B}{m}$$

This is why a cyclotron can repeatedly accelerate particles using a fixed oscillating electric field, at least for speeds where relativistic effects are small.

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