# Charged particle motion in B-fields

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u23-charged-particle-motion-in-b/

This module explains how magnetic fields exert force on moving charged particles, describes the resulting circular and helical motion, and applies core theory to common exam devices including velocity selectors and cyclotrons.

**Prerequisites:** [Magnetic force on current-carrying wires](https://www.owlsprep.com/study/cie-9702-u23-magnetic-force-on-current-wires/); [Uniform circular motion](https://www.owlsprep.com/study/cie-9702-u13-uniform-circular-motion/)

## Learning objectives

- Calculate the magnetic force on a moving charged particle
- Derive the radius, period and frequency of circular motion in uniform B-fields
- Explain the operation of velocity selectors and cyclotrons
- Apply core concepts to standard exam problems

## Force on a Moving Charged Particle

**Magnetic force on a moving charge** — The force on a particle of charge $q$ moving with velocity $v$ in a magnetic field of flux density $B$, where $\theta$ is the angle between the velocity vector and the magnetic field. Direction is given by Fleming's Left Hand Rule (FLHR) for positive charges.

*Notation:* F = Bqv\sin\theta

*Example:* An electron moving parallel to B experiences zero net magnetic force.

The magnetic force is always perpendicular to both velocity and the magnetic field vector. Because the force is always perpendicular to displacement, it does no work on the particle. This means the particle's speed and kinetic energy remain constant, only the direction of motion changes.

> **Fleming's Left Hand Rule**
>
> First finger = Field direction, seCond finger = Current direction (positive charge motion), thuMb = Motion (force direction).

**Worked example:** A proton with speed $2.0 \times 10^6 \text{ m s}^{-1}$ enters a uniform magnetic field of $0.10 \text{ T}$, at an angle of $90^\circ$ to the field lines. Calculate the magnitude of the force on the proton ($e = 1.6 \times 10^{-19} \text{ C}$).

1. List known values: $q = e = 1.6 \times 10^{-19} \text{ C}$, $v = 2.0 \times 10^6 \text{ m s}^{-1}$, $B = 0.10 \text{ T}$, $\sin 90^\circ = 1$
2. Substitute into the force formula:
3. $$F = Bqv\sin\theta = 0.10 \times 1.6 \times 10^{-19} \times 2.0 \times 10^6 \times 1$$
4. Calculate the final force:
5. $$F = 3.2 \times 10^{-14} \text{ N}$$

## Circular Motion in Uniform B-Fields

When a charged particle enters a uniform B-field with velocity perpendicular to the field lines, the constant perpendicular magnetic force provides the centripetal force required for uniform circular motion.

**Derivation:** Derive the radius of a charged particle's circular orbit in a uniform perpendicular B-field

*Starting from:* Equate magnetic force to centripetal force

1. Magnetic force (for perpendicular $\theta = 90^\circ$, $\sin\theta = 1$): $F_B = Bqv$
2. Centripetal force for mass $m$, radius $r$: $F_c = \frac{mv^2}{r}$
3. Equate the two forces: $Bqv = \frac{mv^2}{r}$
4. Rearrange for $r$, cancelling $v$ from both sides:

*Conclusion:* $r = \frac{mv}{Bq} = \frac{p}{Bq}$, where $p$ is particle momentum. Radius is proportional to momentum, and inversely proportional to $B$ and $q$.

We can also derive period $T$ (time for one full orbit) and frequency $f$. Substituting $v = \frac{2\pi r}{T}$ into $r = \frac{mv}{Bq}$ gives $T = \frac{2\pi m}{Bq}$, so frequency $f = \frac{Bq}{2\pi m}$. Critically, frequency is independent of velocity and orbit radius.

**Worked example:** An electron with kinetic energy $1.2 \times 10^{-17} \text{ J}$ moves perpendicular to a uniform B-field of $0.50 \text{ T}$. Calculate the orbit radius ($m_e = 9.11 \times 10^{-31} \text{ kg}$, $e = 1.6 \times 10^{-19} \text{ C}$).

1. First find electron speed from kinetic energy:
2. $$E_k = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2E_k}{m}}$$
3. Calculate $v$:
4. $$v = \sqrt{\frac{2 \times 1.2 \times 10^{-17}}{9.11 \times 10^{-31}}} \approx 5.1 \times 10^6 \text{ m s}^{-1}$$
5. Use the orbit radius formula:
6. $$r = \frac{m_e v}{B e} = \frac{9.11 \times 10^{-31} \times 5.1 \times 10^6}{0.50 \times 1.6 \times 10^{-19}} \approx 5.8 \times 10^{-5} \text{ m}$$

## Velocity Selectors: Crossed E and B Fields

A velocity selector uses perpendicular (crossed) uniform electric and magnetic fields to filter charged particles. Only particles with a specific specific velocity travel through undeflected; all others are deflected and filtered out.

**Undeviated condition for velocity selectors** — For a particle to travel straight through, the electric force must exactly balance the magnetic force, giving zero net force.

*Notation:* v = \frac{E}{B}

**Worked example:** A velocity selector has an electric field of $3.0 \times 10^4 \text{ V m}^{-1}$ and magnetic field of $0.20 \text{ T}$. What speed of particles passes through undeflected?

1. Set electric force equal to magnetic force:
2. $$qE = Bqv$$
3. Cancel charge $q$ from both sides and rearrange:
4. $$v = \frac{E}{B} = \frac{3.0 \times 10^4}{0.20} = 1.5 \times 10^5 \text{ m s}^{-1}$$

> **tip**
>
> Charge cancels out, so the selected velocity does not depend on particle mass or charge. This is a very common exam question point.

## Cyclotrons

A cyclotron is a particle accelerator that leverages the fact that cyclotron frequency is independent of velocity and radius. It consists of two hollow D-shaped electrodes in a uniform B-field, with an alternating potential difference between the electrodes.

Particles are injected at the centre, and are accelerated by the electric field every time they cross the gap between the D electrodes. The alternating voltage matches the cyclotron frequency, so it is always in phase to accelerate particles. Because frequency is independent of radius, the frequency does not need to be adjusted as particles speed up.

**Worked example:** A cyclotron accelerates protons, with a magnetic field of $1.2 \text{ T}$. Calculate the required frequency of the alternating voltage ($m_p = 1.67 \times 10^{-27} \text{ kg}$, $e = 1.6 \times 10^{-19} \text{ C}$).

1. The alternating voltage frequency must match the cyclotron frequency:
2. $$f_c = \frac{Bq}{2\pi m_p}$$
3. Substitute values:
4. $$f_c = \frac{1.2 \times 1.6 \times 10^{-19}}{2\pi \times 1.67 \times 10^{-27}} \approx 18 \times 10^6 \text{ Hz} = 18 \text{ MHz}$$

> **exam_tip**
>
> The maximum kinetic energy of output particles depends on the maximum radius of the D electrodes: $E_{k(max)} = \frac{(B q r_{max})^2}{2m}$.

## Common pitfalls

- **Wrong:** Using Fleming's Left Hand Rule for electrons without reversing the current direction
  - Why it fails: Electrons are negatively charged, so conventional current is opposite to their direction of motion
  - Correct: Reverse the velocity direction when applying FLHR to negative charges
- **Wrong:** Claiming magnetic force does work on the particle to change its kinetic energy
  - Why it fails: Force is always perpendicular to displacement, so work done is zero
  - Correct: Kinetic energy and speed remain constant; only direction of motion changes
- **Wrong:** Stating cyclotron frequency depends on particle velocity
  - Why it fails: The derived formula $f_c = \frac{Bq}{2\pi m}$ has no velocity term
  - Correct: Remember cyclotron frequency is independent of particle speed and orbit radius
- **Wrong:** Using the wrong angle in $F = Bqv\sin\theta$
  - Why it fails: $\theta$ is the angle between $v$ and $B$, not between $v$ and $F$
  - Correct: Always measure $\theta$ between velocity and magnetic field direction
- **Wrong:** Claiming orbit radius is inversely proportional to momentum
  - Why it fails: From $r = \frac{mv}{Bq}$, radius is directly proportional to momentum $p = mv$
  - Correct: Higher momentum particles move in larger orbits

## Cheatsheet

| Quantity | Formula | Key Note |
| --- | --- | --- |
| Magnetic force | $F = Bqv\sin\theta$ | $\theta$ = angle between $v$ and $B$ |
| Orbit radius (perpendicular B) | $r = \frac{mv}{Bq} = \frac{p}{Bq}$ | Proportional to momentum |
| Orbit period | $T = \frac{2\pi m}{Bq}$ | Independent of velocity |
| Cyclotron frequency | $f_c = \frac{Bq}{2\pi m}$ | Matches alternating voltage |
| Velocity selector speed | $v = \frac{E}{B}$ | Independent of $m$, $q$ |
| Max cyclotron KE | $E_{k(max)} = \frac{(B q r_{max})^2}{2m}$ | Depends on maximum radius |

## What's next

Understanding charged particle motion in magnetic fields is a core foundation for many applied physics topics examined in CIE A-Level, including mass spectrometry and particle accelerator physics. It connects your prior knowledge of uniform circular motion, forces, and electromagnetism, and is frequently combined with electric field concepts in multi-part exam questions. Many exam questions test your ability to derive key formulas and apply them to new situations, so mastering the derivations here is critical for high marks. The concepts you learn here also underpin understanding of electron beams, which are common exam contexts.

- [Hall effect](https://www.owlsprep.com/study/cie-9702-u23-hall-effect/)
- [Electromagnetic induction](https://www.owlsprep.com/study/cie-9702-u24-overview/)
- [Magnetic flux](https://www.owlsprep.com/study/cie-9702-u24-magnetic-flux/)

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