---
title: "Cyclotron deuterons, flux density and energy"
subject: "Engineering Physics"
chapter: 8
key: "semester_1/engineering_physics:cyclotron-deuteron-flux-density"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/cyclotron-deuteron-flux-density
site: BCT Engineering
---

# Cyclotron deuterons, flux density and energy

Engineering Physics · Chapter 8: Electromagnetism

- Asked in: 2074 Asoj, 5 marks
- Video: [Deuterons in a cyclotron describes a circle of radius `32.0cm`. Just before emerging from the `D\](https://www.youtube.com/watch?v=3P3bWzTXmNE) — Doubtnut
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/cyclotron-deuteron-flux-density
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8

## Question

Deuterons in a cyclotron describe a circle of radius 0.32 m just before emerging from dees. The frequency of the applied emf's is 10 MHz. Find the flux density of the magnetic field and the energy of deuterons emerging out of the cyclotron. (mass of deuterons = 3.32 × 10⁻²⁷ kg.)

## Short answer

From the cyclotron resonance condition f = qB/(2πm), flux density B = 2πmf/q = 1.3 T. The exit speed v = 2πfR = 2.0×10⁷ m/s gives kinetic energy K = ½mv² = 6.7×10⁻¹³ J ≈ 4.2 MeV.

## Answer

Given: R = 0.32 m, f = 10 MHz = 10⁷ Hz, m = 3.32 × 10⁻²⁷ kg. The deuteron has charge q = +e = 1.6 × 10⁻¹⁹ C.

#### Flux density
The cyclotron resonance condition is f = qB/(2πm).

B = 2πmf/q = (2π × 3.32 × 10⁻²⁷ × 10⁷) / (1.6 × 10⁻¹⁹)
B = (2.086 × 10⁻¹⁹)/(1.6 × 10⁻¹⁹) = 1.30 T

**Answer: B ≈ 1.3 T**

#### Energy of emerging deuterons
1. Speed at the exit: v = 2πfR = 2π × 10⁷ × 0.32 = 2.01 × 10⁷ m/s
2. K = ½ m v² = ½ × 3.32 × 10⁻²⁷ × (2.01 × 10⁷)²
3. K = 6.71 × 10⁻¹³ J
4. In electron volts: K = 6.71 × 10⁻¹³ / 1.6 × 10⁻¹⁹ = 4.19 × 10⁶ eV

Check using K = q²B²R²/(2m) gives the same value.

**Answer: K ≈ 6.7 × 10⁻¹³ J ≈ 4.2 MeV** (v ≈ 2.0 × 10⁷ m/s, about 7% of the speed of light, so relativistic effects are small).

## Check yourself

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

1. B is found from
   A. f = qB/(2πm)
   B. K = qV
   C. v = fR only
   D. F = ma
   Answer: A. The resonance condition gives B = 2πmf/q. (Key point: Method)

2. B equals
   A. 0.13 T
   B. 13 T
   C. 1.3 T
   D. 2.6 T
   Answer: C. B = 2π(3.32e−27)(1e7)/1.6e−19 ≈ 1.30 T. (Key point: Flux density)

3. The exit speed is
   A. 2π f R ≈ 2.0 × 10^7 m/s
   B. f R ≈ 3.2 × 10^6 m/s
   C. 2.0 × 10^5 m/s
   D. f/R
   Answer: A. v = 2π × 10^7 × 0.32. (Key point: Exit speed)

4. The emerging energy is about
   A. 42 MeV
   B. 0.42 MeV
   C. 4.2 MeV
   D. 6.7 MeV in joules ×10^-19
   Answer: C. K = ½mv^2 = 6.7 × 10^−13 J ≈ 4.2 MeV. (Key point: Energy)

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