Electricity & Magnetism
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Cyclotron Frequency

Rotation frequency of a charge in a magnetic field.

Formula

f=qB2πmf = \frac{qB}{2\pi m}

Variables

qCharge
BField
mMass

How it works

A charged particle moving perpendicular to a uniform magnetic field experiences a sideways magnetic force, qvB, which acts as the centripetal force needed to keep it moving in a circle. Setting qvB = mv^2/r gives the circular motion condition, and eliminating r shows the angular speed is qB/m, so the rotation frequency is f = (qB)/(2\pi m). The result is independent of the particle’s speed and radius as long as the motion is non-relativistic and the field is uniform.

When to use it

Use this when analyzing the orbit of ions or electrons in a uniform magnetic field, such as in cyclotrons, mass spectrometers, or plasma physics. It is also useful whenever you need the rotation rate, period, or charge-to-mass ratio from magnetic circular motion.

Worked example, step by step

A proton moves in a uniform magnetic field of magnitude B = 0.85 T. Given: q = 1.60 × 10^-19 C, m = 1.67 × 10^-27 kg Use f = qB / (2\pi m) Substitute: f = (1.60 × 10^-19 × 0.85) / (2\pi × 1.67 × 10^-27) Calculate the numerator: 1.36 × 10^-19 Calculate the denominator: 2\pi × 1.67 × 10^-27 ≈ 1.05 × 10^-26 Divide: f ≈ (1.36 × 10^-19) / (1.05 × 10^-26) Final answer: f ≈ 1.30 × 10^7 Hz

Common mistakes to avoid

  • Using the particle speed in the formula instead of the charge, field strength, and mass
  • Forgetting that the charge should be used as a magnitude, so the frequency is positive
  • Mixing up frequency and angular frequency, or omitting the factor of 2\pi

Did You Know?

A charged particle's cyclotron frequency is independent of its speed, the insight that made particle accelerators possible.

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Reviewed & maintained by the FormulaWon editorial teamLast reviewed September 2026