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:

IMATH_39
IMATH_40
$v = \sqrt{8.04 \times 10^9} = 8.97 \times 10^4$ 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.

Common traps

Forgetting the $\sin \theta$. $F = qvB$ is only the magnitude when $\vec{v} \perp \vec{B}$. In other cases use $F = qvB \sin \theta$.

Using $F = qE$ for a magnetic-field problem. $F = qE$ is the electric-field force. The magnetic-field force is $F = qvB \sin \theta$. They look similar; mix them up and you lose easy marks.

Applying the right-hand rule to a negative charge without reversing. Always note whether the charge is positive or negative and flip the direction for an electron.

Saying the magnetic force changes the speed. It changes the direction only. Speed and kinetic energy are constant.

Forgetting that $T$ is independent of $v$. The period depends only on $m/(qB)$. Faster particles orbit on larger circles in the same time, not faster.

Mixing up the radius formula. $r = mv/(qB)$, not $r = qB/(mv)$ or $r = mvB/q$.

In one sentence

A charged particle moving perpendicular to a uniform magnetic field experiences a force $F = qvB$ perpendicular to its velocity, which acts as a centripetal force and produces circular motion of radius $r = mv/(qB)$ and period $T = 2 \pi m / (qB)$ at constant speed.