Charged particle motion in B-fields

CIE A-Level Physics· Unit 23: Magnetic Fields· 45 min read

1. Force on a Moving Charged Particle★★☆☆☆⏱ 10 min

📘 Definition

Magnetic force on a moving charge

F = Bqv\sin\theta

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

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.

📐 Worked Example

A proton with speed $2.0 \times 1 0^{6} m s^{- 1}$ enters a uniform magnetic field of $0.10 T$ , at an angle of $9 0^{\circ}$ to the field lines. Calculate the magnitude of the force on the proton ( $e = 1.6 \times 1 0^{- 19} C$ ).

1
List known values: $q = e = 1.6 \times 1 0^{- 19} C$ , $v = 2.0 \times 1 0^{6} m s^{- 1}$ , $B = 0.10 T$ , $sin 9 0^{\circ} = 1$
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
$F = 3.2 \times 10^{-14} \text{ N}$

2. Circular Motion in Uniform B-Fields★★★☆☆⏱ 15 min

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

Goal:

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 $θ = 9 0^{\circ}$ , $sin θ = 1$ ): $F_{B} = B q v$
2
Centripetal force for mass $m$ , radius $r$ : $F_{c} = \frac{m v ^{2}}{r}$
3
Equate the two forces: $B q v = \frac{m v ^{2}}{r}$
4
Rearrange for $r$ , cancelling $v$ from both sides:

Result:

$r = \frac{m v}{B q} = \frac{p}{B q}$ , 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 $v = \frac{2 π r}{T}$ into $r = \frac{m v}{B q}$ gives $T = \frac{2 π m}{B q}$ , so frequency $f = \frac{B q}{2 π m}$ . Critically, frequency is independent of velocity and orbit radius.

📐 Worked Example

An electron with kinetic energy $1.2 \times 1 0^{- 17} J$ moves perpendicular to a uniform B-field of $0.50 T$ . Calculate the orbit radius ( $m_{e} = 9.11 \times 1 0^{- 31} kg$ , $e = 1.6 \times 1 0^{- 19} C$ ).

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
$r = \frac{m_e v}{B e} = \frac{9.11 \times 10^{-31} \times 5.1 \times 10^6}{0.50 \times 1.6 \times 10^{-19}} \approx 5.8 \times 10^{-5} \text{ m}$

3. Velocity Selectors: Crossed E and B Fields★★★☆☆⏱ 10 min

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.

📘 Definition

Undeviated condition for velocity selectors

v = \frac{E}{B}

For a particle to travel straight through, the electric force must exactly balance the magnetic force, giving zero net force.

📐 Worked Example

A velocity selector has an electric field of $3.0 \times 1 0^{4} V m^{- 1}$ and magnetic field of $0.20 T$ . 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
$v = \frac{E}{B} = \frac{3.0 \times 10^4}{0.20} = 1.5 \times 10^5 \text{ m s}^{-1}$

4. Cyclotrons★★★★☆⏱ 10 min

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 $1.2 T$ . Calculate the required frequency of the alternating voltage ( $m_{p} = 1.67 \times 1 0^{- 27} kg$ , $e = 1.6 \times 1 0^{- 19} C$ ).

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

5. Common Pitfalls

Wrong move:

Using Fleming's Left Hand Rule for electrons without reversing the current direction

Why:

Electrons are negatively charged, so conventional current is opposite to their direction of motion

Correct move:

Reverse the velocity direction when applying FLHR to negative charges

Wrong move:

Claiming magnetic force does work on the particle to change its kinetic energy

Why:

Force is always perpendicular to displacement, so work done is zero

Correct move:

Kinetic energy and speed remain constant; only direction of motion changes

Wrong move:

Stating cyclotron frequency depends on particle velocity

Why:

The derived formula $f_{c} = \frac{B q}{2 π m}$ has no velocity term

Correct move:

Remember cyclotron frequency is independent of particle speed and orbit radius

Wrong move:

Using the wrong angle in $F = B q v sin θ$

Why:

$θ$ is the angle between $v$ and $B$ , not between $v$ and $F$

Correct move:

Always measure $θ$ between velocity and magnetic field direction

Wrong move:

Claiming orbit radius is inversely proportional to momentum

Why:

From $r = \frac{m v}{B q}$ , radius is directly proportional to momentum $p = m v$

Correct move:

Higher momentum particles move in larger orbits

6. Quick Reference Cheatsheet

Quantity	Formula	Key Note
Magnetic force	$F = B q v sin θ$	$θ$ = angle between $v$ and $B$
Orbit radius (perpendicular B)	$r = \frac{m v}{B q} = \frac{p}{B q}$	Proportional to momentum
Orbit period	$T = \frac{2 π m}{B q}$	Independent of velocity
Cyclotron frequency	$f_{c} = \frac{B q}{2 π m}$	Matches alternating voltage
Velocity selector speed	$v = \frac{E}{B}$	Independent of $m$ , $q$
Max cyclotron KE	$E_{k (ma x)} = \frac{( B q r _{ma x} ) ^{2}}{2 m}$	Depends on maximum radius

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 · 23
Calculate cyclotron frequency
2022 · 22
Find radius of electron orbit
2021 · 12
Explain velocity selector operation

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.