---
title: "Cyclotron proton frequency and energy"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:cyclotron-proton-radius-frequency-energy"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/cyclotron-proton-radius-frequency-energy
site: BCT Engineering
---

# Cyclotron proton frequency and energy

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2067 Asar, 5 marks
- Video: [Motion of Charge in a Magnetic Field - The Cyclotron | Physics with Professor Matt Anderson | M23-06](https://www.youtube.com/watch?v=HmVGGs6Dxp8) — Physics with Professor Matt Anderson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/cyclotron-proton-radius-frequency-energy
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

In a certain cyclotron a proton moves in a circle of radius 0.5m. The magnitude of the magnetic field is 1.20T. What is the oscillator frequency? What is the kinetic energy of the proton in eV?

## Short answer

Oscillator frequency f = qB/(2πm) ≈ 1.83×10⁷ Hz (18.3 MHz). Kinetic energy K = ½m(qBr/m)² ≈ 2.76×10⁻¹² J = 1.72×10⁷ eV ≈ 17.2 MeV.

## Answer

In a cyclotron, qvB = mv²/r. The proton's circulation frequency is **f = qB/(2πm)**, independent of speed. The speed at radius r is v = qBr/m.

**Given:** r = 0.5 m, B = 1.20 T, m_p = 1.673×10⁻²⁷ kg, q = 1.602×10⁻¹⁹ C.

#### Oscillator frequency
```text
f = (1.602×10⁻¹⁹ × 1.20) / (2π × 1.673×10⁻²⁷)
  = 1.9224×10⁻¹⁹ / 1.0511×10⁻²⁶
  = 1.83×10⁷ Hz
```

#### Kinetic energy
```text
v = qBr/m = (1.602×10⁻¹⁹ × 1.20 × 0.5) / 1.673×10⁻²⁷ = 5.74×10⁷ m/s
K = ½ m v² = ½ × 1.673×10⁻²⁷ × (5.74×10⁷)² = 2.76×10⁻¹² J
K = 2.76×10⁻¹² / 1.602×10⁻¹⁹ = 1.72×10⁷ eV
```

**Answer: f ≈ 18.3 MHz; K ≈ 1.72×10⁷ eV = 17.2 MeV** (non-relativistic, since v ≈ 0.19c).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The proton oscillator frequency is:
   A. f = qBr/(2πm)
   B. f = qB/(2πm)
   C. f = 2πm/(qB)
   D. f = qv/(2πB)
   Answer: B. Independent of speed and radius. (Key point: Frequency)

2. For B = 1.20 T the frequency is about:
   A. 18.3 MHz
   B. 1.83 MHz
   C. 183 MHz
   D. 18.3 kHz
   Answer: A. 1.9224×10⁻¹⁹/1.0511×10⁻²⁶ ≈ 1.83×10⁷ Hz. (Key point: Frequency value)

3. The speed at radius r is:
   A. v = mBr/q
   B. v = qB/(mr)
   C. v = qBr/m ≈ 5.74×10⁷ m/s
   D. v = qr/(mB)
   Answer: C. From r = mv/(qB). (Key point: Speed)

4. The kinetic energy is about:
   A. 1.72 MeV
   B. 172 MeV
   C. 5.7 MeV
   D. 17.2 MeV
   Answer: D. 2.76×10⁻¹² J ÷ 1.602×10⁻¹⁹ ≈ 1.72×10⁷ eV. (Key point: Energy)

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