Force on moving charged particle

CIE A-Level Physics· Unit 23: Magnetic Fields, 23.3 Force on moving charges· 15 min read

1. Magnitude and Direction of Force★★☆☆☆⏱ 5 min

📘 Definition

Magnetic Force on Moving Charge

F

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

Example:

A proton moving at $1 0^{5} m s^{- 1}$ perpendicular to a 0.1 T field experiences ~ $1.6 \times 1 0^{- 15} N$ force.

The general formula for force magnitude, where $θ$ is the angle between velocity $v$ and magnetic flux density $B$ , is:

F = qvB \sin\theta

If $v$ is parallel to $B$ , $θ = 0^{\circ}$ so $sin θ = 0$ , meaning no force acts. If $v$ is perpendicular to $B$ , $θ = 9 0^{\circ}$ so $sin θ = 1$ , and $F = q v B$ .

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

📐 Worked Example

An electron with charge $- 1.6 \times 1 0^{- 19} C$ moves at $2.0 \times 1 0^{6} m s^{- 1}$ 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 $sin θ = 1$ , and magnitude is $F = ∣ q ∣ v B$ :
2
$F = (1.6 \times 10^{-19}) \times (2.0 \times 10^6) \times 0.5$
3
Calculate the final result:
4
$F = 1.6 \times 10^{-13} \text{ N}$

2. Circular Motion of Charges in Uniform B Fields★★★☆☆⏱ 5 min

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

Goal:

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
$qB = \frac{mv}{r}$

Result:

Rearranging gives the radius formula: $r = \frac{m v}{q B} = \frac{p}{q B}$ , where $p$ is the momentum of the particle.

📐 Worked Example

A proton of mass $1.67 \times 1 0^{- 27} kg$ and charge $1.6 \times 1 0^{- 19} C$ 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
$v \approx 2.9 \times 10^6 \text{ m s}^{-1}$

3. Application: Velocity Selector★★★☆☆⏱ 5 min

📘 Definition

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 $q$ : $F_{E} = q E$ (direction depends on charge sign)
2
Magnetic force (v perpendicular to B): $F_{B} = q v B$ (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
$v = \frac{E}{B}$

✓ Quick check

Test your understanding

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
Reveal answer
1 —
For $v > \frac{E}{B}$ , $q v B > q E$ , so magnetic force is larger than electric force, so the ion deflects in the direction of the magnetic force.

4. Common Pitfalls

Wrong move:

Forgetting to reverse force direction for negative charges when using Fleming's Left Hand Rule

Why:

Fleming's Left Hand Rule is defined for conventional current (positive charge movement), so negative charges have opposite force direction

Correct move:

Always reverse the direction given by Fleming's Left Hand Rule for electrons and other negative charges

Wrong move:

Using $F = q v B$ when velocity is parallel to the magnetic field

Why:

The $sin θ$ term is zero when velocity is parallel to B, so no force acts

Correct move:

Always use the full formula $F = q v B sin θ$ and check the angle between velocity and B

Wrong move:

Claiming speed increases for a particle in circular motion in a magnetic field

Why:

Force is always perpendicular to velocity, so it does no work and cannot change kinetic energy or speed

Correct move:

Recognize that only direction changes; speed and kinetic energy remain constant

Wrong move:

Assuming all undeflected particles in a velocity selector have speed $v = E / B$

Why:

Neutral particles have zero charge, so experience no force and pass through undeflected regardless of speed

Correct move:

Note that velocity selectors only select speed for charged particles; neutral particles are not affected

5. Quick Reference Cheatsheet

Concept	Formula/Rule	Key Note
Force magnitude	$F = q v B sin θ$	$θ$ = 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	$r = \frac{m v}{q B} = \frac{p}{q B}$	$r$ proportional to momentum
Velocity selector (undeflected)	$v = \frac{E}{B}$	Crossed electric and magnetic fields

6. Frequently Asked

Why is the force always perpendicular to velocity?

Magnetic force is given by the cross product of $v$ and $B$ , so it has no component parallel to velocity. This means it only changes the direction of motion, not speed.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · Paper 4
Circular motion of alpha particles
2022 · Paper 2
Velocity selector calculation
2021 · Paper 1
Force direction for electrons

Going deeper

syllabusOfficial CIE 9702 Physics SyllabusFor full syllabus details

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.