# Force on moving charged particle

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u23-force-on-moving-charged-particle/

This sub-topic covers how magnetic fields exert force on moving charged particles, how to calculate force magnitude and direction, and the resulting circular motion of charges in uniform magnetic fields, a core concept for CIE A-Level exams.

**Prerequisites:** [Magnetic flux density](https://www.owlsprep.com/study/cie-9702-u23-magnetic-flux-density/); [Circular motion](https://www.owlsprep.com/study/cie-9702-u12-circular-motion/)

## Learning objectives

- Recall and use the formula for magnetic force on a moving charged particle
- Determine the direction of force for positive and negative charges
- Derive and use the radius formula for circular motion of charges in B fields
- Analyze the operation of common applications like velocity selectors

## Magnitude and Direction of Force

**Magnetic Force on Moving Charge** — Force exerted on a charged particle moving through a magnetic field, proportional to charge, the perpendicular component of velocity, and magnetic flux density.

*Notation:* F

*Example:* A proton moving at $10^5 \text{ m s}^{-1}$ perpendicular to a 0.1 T field experiences ~$1.6 \times 10^{-15} \text{ N}$ force.

The general formula for force magnitude, where $\theta$ is the angle between velocity $\vec{v}$ and magnetic flux density $\vec{B}$, is:

$$F = qvB \sin\theta$$

If $\vec{v}$ is parallel to $\vec{B}$, $\theta = 0^\circ$ so $\sin\theta = 0$, meaning no force acts. If $\vec{v}$ is perpendicular to $\vec{B}$, $\theta = 90^\circ$ so $\sin\theta = 1$, and $F = qvB$.

For positive charges, use Fleming's Left Hand Rule to find direction: for negative charges, the force direction is reversed.

> **FBI Mnemonic**
>
> First finger = Field, Second finger = Current (motion of positive charge), Thumb = Force (motion). This makes it easy to remember the rule for exams.

**Worked example:** An electron with charge $-1.6 \times 10^{-19} \text{ C}$ moves at $2.0 \times 10^6 \text{ m s}^{-1}$ perpendicular to a uniform 0.5 T magnetic field. Calculate the magnitude of the force on the electron.

1. Velocity is perpendicular to B, so $\sin\theta = 1$, and magnitude is $F = |q|vB$:
2. $$F = (1.6 \times 10^{-19}) \times (2.0 \times 10^6) \times 0.5$$
3. Calculate the final result:
4. $$F = 1.6 \times 10^{-13} \text{ N}$$

## Circular Motion of Charges in Uniform B Fields

When a charged particle moves perpendicular to a uniform magnetic field, the force is always perpendicular to velocity. This means the force does no work (so speed remains constant) and acts as a centripetal force, causing uniform circular motion.

**Derivation:** Derive the formula for the radius of the circular path

*Starting from:* Equate magnetic force to centripetal force

1. Magnetic force (perpendicular case):
2. $$F = qvB$$
3. Centripetal force for mass $m$, radius $r$:
4. $$F = \frac{mv^2}{r}$$
5. Equate the two forces:
6. $$qvB = \frac{mv^2}{r}$$
7. Cancel $v$ (non-zero) from both sides:
8. $$qB = \frac{mv}{r}$$

*Conclusion:* Rearranging gives the radius formula: $r = \frac{mv}{qB} = \frac{p}{qB}$, where $p$ is the momentum of the particle.

**Worked example:** A proton of mass $1.67 \times 10^{-27} \text{ kg}$ and charge $1.6 \times 10^{-19} \text{ C}$ moves in a circular path of radius 0.2 m in a 0.15 T uniform magnetic field. Calculate the speed of the proton.

1. Rearrange the radius formula to solve for $v$:
2. $$v = \frac{qBr}{m}$$
3. Substitute given values:
4. $$v = \frac{(1.6 \times 10^{-19}) \times (0.15) \times (0.2)}{1.67 \times 10^{-27}}$$
5. Calculate the final result:
6. $$v \approx 2.9 \times 10^6 \text{ m s}^{-1}$$

## Application: Velocity Selector

**Velocity Selector** — A device with perpendicular (crossed) uniform electric and magnetic fields that only allows charged particles of a specific speed to pass through undeflected.

*Example:* Used in mass spectrometers to filter ions before mass separation.

For a particle to pass through undeflected, the electric force must balance the magnetic force, resulting in zero net force.

**Worked example:** Show that only particles with speed $v = \frac{E}{B}$ pass through a velocity selector undeflected.

1. Electric force on charge $q$: $F_E = qE$ (direction depends on charge sign)
2. Magnetic force (v perpendicular to B): $F_B = qvB$ (opposite direction to $F_E$)
3. For undeflected motion, net force = 0, so force magnitudes are equal:
4. $$qE = qvB$$
5. Cancel non-zero $q$ from both sides to get:
6. $$v = \frac{E}{B}$$

**Check your understanding**

Test your understanding

1. A positive ion moves faster than $v = \frac{E}{B}$ through a velocity selector. What happens to it?

   - Deflected in direction of electric force
   - Deflected in direction of magnetic force
   - Passes through undeflected
   - Stops immediately

   *Answer:* Deflected in direction of magnetic force

   *Why:* For $v > \frac{E}{B}$, $qvB > qE$, so magnetic force is larger than electric force, so the ion deflects in the direction of the magnetic force.

## Common pitfalls

- **Wrong:** Forgetting to reverse force direction for negative charges when using Fleming's Left Hand Rule
  - Why it fails: Fleming's Left Hand Rule is defined for conventional current (positive charge movement), so negative charges have opposite force direction
  - Correct: Always reverse the direction given by Fleming's Left Hand Rule for electrons and other negative charges
- **Wrong:** Using $F = qvB$ when velocity is parallel to the magnetic field
  - Why it fails: The $\sin\theta$ term is zero when velocity is parallel to B, so no force acts
  - Correct: Always use the full formula $F = qvB \sin\theta$ and check the angle between velocity and B
- **Wrong:** Claiming speed increases for a particle in circular motion in a magnetic field
  - Why it fails: Force is always perpendicular to velocity, so it does no work and cannot change kinetic energy or speed
  - Correct: Recognize that only direction changes; speed and kinetic energy remain constant
- **Wrong:** Assuming all undeflected particles in a velocity selector have speed $v = E/B$
  - Why it fails: Neutral particles have zero charge, so experience no force and pass through undeflected regardless of speed
  - Correct: Note that velocity selectors only select speed for charged particles; neutral particles are not affected

## Cheatsheet

| Concept | Formula/Rule | Key Note |
| --- | --- | --- |
| Force magnitude | $F = qvB \sin\theta$ | $\theta$ = angle between $v$ and $B$ |
| Direction (+ve charge) | Fleming's Left Hand Rule | FBI: Field, Current, Force |
| Direction (-ve charge) | Reverse of Fleming's result | Charge sign flips force direction |
| Circular path radius | $r = \frac{mv}{qB} = \frac{p}{qB}$ | $r$ proportional to momentum |
| Velocity selector (undeflected) | $v = \frac{E}{B}$ | Crossed electric and magnetic fields |

## What's next

Understanding force on moving charged particles is the foundation for many key electromagnetism topics in CIE A-Level Physics, including mass spectrometry, cyclotrons, and the Hall effect. This concept is often combined with motion of charges in electric fields in Paper 2 and Paper 4 exam questions, so linking these two topics is critical for exam success. Mastery of this sub-topic also prepares you for electromagnetic induction, the next core unit in the CIE 9702 syllabus.

- [Charged particle motion in B-fields](https://www.owlsprep.com/study/cie-9702-u23-charged-particle-motion-in-b/)
- [Hall effect](https://www.owlsprep.com/study/cie-9702-u23-hall-effect/)
- [Electromagnetic induction](https://www.owlsprep.com/study/cie-9702-u24-overview/)

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