Analyse the interaction between charged particles and uniform magnetic fields, including: acceleration, perpendicular to velocity F = qv x B, circular motion of a charged particle moving perpendicular to a uniform magnetic field

What this dot point is asking

NESA wants you to use the Lorentz force law $\vec{F} = q \vec{v} \times \vec{B}$, recognise that the magnetic force is always perpendicular to the velocity (so does no work and changes only the direction of motion), and apply Newton's second law in the form $qvB = mv^2/r$ to extract the radius and period of circular motion. You should also use the right-hand rule fluently to find the direction of the force.

The answer

The Lorentz force

A particle of charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a force:

The magnitude is:

where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. Key features:

• If $\vec{v}$ is parallel or antiparallel to $\vec{B}$ ($\theta = 0$ or $180°$), the force is zero. The particle moves in a straight line.
• If $\vec{v}$ is perpendicular to $\vec{B}$ ($\theta = 90°$), the force has its maximum magnitude $F = qvB$ and points perpendicular to both.
• The direction is given by the right-hand rule (with a sign flip for negative charges).

Units: tesla (T), where $1$ T $= 1$ N/(A m).

Why the magnetic force does no work

Because $\vec{F} \perp \vec{v}$, the dot product $\vec{F} \cdot d\vec{s} = \vec{F} \cdot \vec{v} \, dt$ is always zero. The work done over any displacement is zero, so the kinetic energy and hence the speed are constant. The magnetic force can change a particle's direction but never its speed.

This is the deepest difference between electric and magnetic forces: an electric field can accelerate a charge (change its speed); a magnetic field can only steer it.

Circular motion in a uniform field

When a charged particle moves perpendicular to a uniform magnetic field, the constant-magnitude force perpendicular to $\vec{v}$ provides exactly the centripetal force needed for circular motion. Setting magnetic = centripetal:

Solving for the radius:

Solving for the period (using $v = 2 \pi r / T$):

The period depends only on $m/q$ and $B$, not on the speed of the particle. This is the principle behind the cyclotron: particles of all speeds (below relativistic limits) orbit with the same period in a given field.

The frequency $f = 1/T = qB / (2 \pi m)$ is called the cyclotron frequency.

Direction: the right-hand rule

For a positive charge:

1. Point the fingers of your right hand in the direction of $\vec{v}$.
2. Curl them toward $\vec{B}$ (through the smaller angle).
3. Your thumb points in the direction of $\vec{F}$.

Equivalently (the "slap" rule): flat right hand, fingers along $\vec{B}$, thumb along $\vec{v}$, palm pushes in the direction of $\vec{F}$.

For a negative charge (such as an electron), the force is in the opposite direction to the rule above. Either reverse the rule by using your left hand, or apply the right-hand rule for the positive case and then flip.

The result is that positive charges and negative charges in the same field, moving the same way, orbit in opposite senses.

Worked example: mass spectrometer

A singly-ionised carbon-12 atom is accelerated from rest through 500 V, then enters a uniform magnetic field of $0.20$ T perpendicular to its velocity. Find the radius of its circular path. ($m = 1.99 \times 10^{-26}$ kg, $q = 1.60 \times 10^{-19}$ C.)

Find the speed first. Energy conservation in the accelerator:

$qV = \frac{1}{2} m v^2$
$v = \sqrt{\frac{2 q V}{m}} = \sqrt{\frac{2 \times 1.60 \times 10^{-19} \times 500}{1.99 \times 10^{-26}}}$
$v = \sqrt{8.04 \times 10^9} = 8.97 \times 10^4$ m/s.

Radius in the magnetic field:

$r = \frac{m v}{q B} = \frac{1.99 \times 10^{-26} \times 8.97 \times 10^4}{1.60 \times 10^{-19} \times 0.20} = 5.6 \times 10^{-2}$ m = 5.6 cm.

This is a mass spectrometer: different isotopes (different $m$) yield different radii at the detector, separating them by mass.

Try it: Lorentz force calculator for radius, period, and speed of a charge in a uniform magnetic field.

Examples in context

Example 1. Mass spectrometer at the Australian Synchrotron. A singly-ionised iron-56 ion ($m = 56 \times 1.66 \times 10^{-27} = 9.30 \times 10^{-26} \text{ kg}$) is accelerated to $v = 2.0 \times 10^5 \text{ m/s}$ then enters a uniform magnetic field $B = 0.50 \text{ T}$ perpendicular to its velocity. The radius of the circular path is $r = m v / (q B) = 9.30 \times 10^{-26} \times 2.0 \times 10^5 / (1.6 \times 10^{-19} \times 0.50) = 0.232 \text{ m}$. A nickel-58 ion of the same speed traces $r = 9.62 \times 10^{-26} \times 2.0 \times 10^5 / (1.6 \times 10^{-19} \times 0.50) = 0.241 \text{ m}$. The $9 \text{ mm}$ separation lets a detector array resolve the two isotopes.

Example 2. Electron path bending in the Australian Synchrotron booster ring. Electrons at $E = 3.0 \text{ GeV}$ are highly relativistic, but we can illustrate the principle with a non-relativistic estimate of bending radius for a beam test electron at $v = 5.0 \times 10^7 \text{ m/s}$ in a dipole field $B = 1.4 \text{ T}$. Radius $r = m v / (e B) = 9.11 \times 10^{-31} \times 5.0 \times 10^7 / (1.6 \times 10^{-19} \times 1.4) = 2.03 \times 10^{-4} \text{ m} = 0.20 \text{ mm}$. For the actual relativistic case at $\gamma \approx 5870$, the effective mass is $\gamma m$, giving the design radius of $\sim 7 \text{ m}$. The same physics scales up by the Lorentz factor.

Try this

Q1. Write the equation for the force on a charge $q$ moving with velocity $\vec{v}$ in magnetic field $\vec{B}$ and state the direction of the force when $\vec{v} \perp \vec{B}$. [2 marks]

• Cue. $\vec{F} = q \vec{v} \times \vec{B}$, magnitude $qvB \sin\theta$; perpendicular to both (right-hand rule for positive charge).

Q2. A proton enters a $0.20 \text{ T}$ field perpendicular to its velocity at $4.0 \times 10^6 \text{ m/s}$. Calculate the radius and period of its circular motion. [4 marks]

• Cue. $r = m_p v / (e B) = 1.67 \times 10^{-27} \times 4.0 \times 10^6 / (1.6 \times 10^{-19} \times 0.20) = 0.209 \text{ m}$. $T = 2\pi m / (e B) = 3.28 \times 10^{-7} \text{ s}$.

Q3. A velocity selector uses crossed $\vec{E}$ and $\vec{B}$ fields, with $E = 5000 \text{ V/m}$ and $B = 0.020 \text{ T}$. (a) Show that only particles with $v = E/B$ pass through undeflected. (b) Calculate this selected speed. (c) Explain what happens to slower-moving charged particles. [2+1+3 marks]

• Cue. (a) $qE = qvB$ requires $v = E/B$. (b) $v = 5000 / 0.02 = 2.5 \times 10^5 \text{ m/s}$. (c) Magnetic force smaller than electric force, so they deflect toward the plate of opposite sign.