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

# Cyclotron frequency and maximum energy

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2071 Chaitra, 5 marks
- Video: [4.Cyclotron- Time Period, Frequency, Energy](https://www.youtube.com/watch?v=ZiAsBhxIYmc) — Mind Maps
- 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-dees-field-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

A particular cyclotron is designed with Dees of radius R = 75cm and with magnets that can provide a field of 1.5T. (a) To what frequency should the oscillator be set if deuterons are to be accelerated? (b) What is the maximum energy of deuterons that can be obtained? (Mass of deuterons is 3.32×10⁻²⁷ kg)

## Short answer

Oscillator frequency f = qB/(2πm) ≈ 11.5 MHz. Maximum kinetic energy K_max = q²B²R²/(2m) ≈ 4.9×10⁻¹² J ≈ 30.5 MeV.

## Answer

In a cyclotron, qvB = mv²/r gives r = mv/(qB). The orbital frequency is independent of speed and radius:

- **f = qB/(2πm)**
- Maximum kinetic energy at r = R: **K_max = q²B²R²/(2m)**

**Given:** R = 0.75 m, B = 1.5 T, deuteron m = 3.32×10⁻²⁷ kg, q = e = 1.602×10⁻¹⁹ C.

#### (a) Oscillator frequency
```text
f = (1.602×10⁻¹⁹ × 1.5) / (2π × 3.32×10⁻²⁷)
  = 2.403×10⁻¹⁹ / 2.086×10⁻²⁶
  = 1.15×10⁷ Hz
```

#### (b) Maximum energy
```text
v_max = qBR/m = (1.602×10⁻¹⁹ × 1.5 × 0.75) / 3.32×10⁻²⁷ = 5.43×10⁷ m/s
K_max = ½ m v² = ½ × 3.32×10⁻²⁷ × (5.43×10⁷)²
      = 4.89×10⁻¹² J
      = 4.89×10⁻¹² / 1.602×10⁻¹³ MeV
      = 30.5 MeV
```

**Answer: (a) f ≈ 11.5 MHz; (b) K_max ≈ 4.9×10⁻¹² J ≈ 30.5 MeV**

## Check yourself

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

1. Equating qvB with mv²/r gives the orbit radius:
   A. r = qB/(mv)
   B. r = mvB/q
   C. r = mv/(qB)
   D. r = qv/(mB)
   Answer: C. Magnetic force supplies the centripetal force. (Key point: Radius in a cyclotron)

2. The oscillator frequency of a cyclotron is:
   A. f = qB/(2πm), independent of speed and radius
   B. f = qBR/(2πm), rising with radius
   C. f = 2πm/(qB)
   D. f = mv/(2πqB)
   Answer: A. Period is 2πm/(qB), the same for every orbit, so the oscillator is fixed. (Key point: Frequency)

3. The maximum kinetic energy is:
   A. K_max = qBR/(2m)
   B. K_max = q²B²R²/(2m)
   C. K_max = q²BR/(2m)
   D. K_max = qB²R/m
   Answer: B. v_max = qBR/m, so ½mv² = q²B²R²/(2m). (Key point: Max energy formula)

4. For deuterons (m = 3.32×10⁻²⁷ kg), B = 1.5 T, R = 0.75 m, f and K_max are about:
   A. 11.5 MHz and 4.9 MeV
   B. 1.15 MHz and 30.5 MeV
   C. 72 MHz and 30.5 MeV
   D. 11.5 MHz and 30.5 MeV
   Answer: D. f = 1.15×10⁷ Hz and K = 4.89×10⁻¹² J ≈ 30.5 MeV. (Key point: Numerical results)

## Related in Engineering Physics

- [Cyclotron deuterons, flux density and energy](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acyclotron-deuteron-flux-density) — Engineering Physics, chapter 8
- [Cyclotron proton frequency and energy](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acyclotron-proton-radius-frequency-energy) — Engineering Physics, chapter 8
- [Cyclotron protons, charge-to-mass ratio](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acyclotron-protons-charge-mass-ratio) — Engineering Physics, chapter 8