Apply the relationships for the magnetic force on a moving charge F = q v B sin(theta) and on a current-carrying conductor F = B I L sin(theta), including the right-hand rule, circular motion of charged particles in uniform magnetic fields, and forces between parallel conductors

What this dot point is asking

QCAA wants you to apply the two magnetic force relationships ($\vec{F} = q \vec{v} \times \vec{B}$ for moving charges, $\vec{F} = I \vec{L} \times \vec{B}$ for current-carrying conductors), use the right-hand rule to determine the direction of the force in 3D, derive the radius and period of a charged particle's circular motion in a uniform field, and compute the force per unit length between two parallel conductors. This dot point underpins IA1 short response on velocity selectors and mass spectrometers, and feeds the EA Paper 2 derivations on motor and generator action.

The answer

Magnetic force on a moving charge

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

where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. The force is perpendicular to both $\vec{v}$ and $\vec{B}$, given in direction by $\vec{F} = q \vec{v} \times \vec{B}$.

Key consequences:

• A charge moving parallel to $\vec{B}$ ($\theta = 0°$ or $180°$) experiences zero force.
• A charge moving perpendicular to $\vec{B}$ ($\theta = 90°$) experiences the maximum force $F = q v B$.
• The force is always perpendicular to the velocity. The magnetic force does no work on the charge. The speed is unchanged; only the direction changes.

Right-hand rule for a moving positive charge

Point your right hand's fingers in the direction of $\vec{v}$, curl them toward $\vec{B}$; the thumb points in the direction of $\vec{F}$ on a positive charge. For a negative charge (electron), reverse the direction.

Equivalent: point the right palm so that fingers extend along $\vec{B}$ and the thumb along $\vec{v}$; the force on a positive charge pushes out of the palm.

Circular motion of a charged particle

A charge moving perpendicular to a uniform field experiences a constant-magnitude force always perpendicular to its velocity. The trajectory is a circle. Setting magnetic force equal to centripetal force:

The period of the circular motion:

Importantly, $T$ is independent of $v$: a fast and a slow charge of the same mass and charge complete the same circle in the same time. This is the basis of the cyclotron, where an oscillating electric field at a fixed frequency accelerates charged particles to high speeds.

If the velocity has a component parallel to $\vec{B}$, that component is unaffected and the path becomes a helix.

Try it: Lorentz force calculator. Enter charge, speed, field and angle to get force magnitude, radius and period.

Magnetic force on a current-carrying conductor

A wire of length $L$ carrying current $I$ in a magnetic field $\vec{B}$ experiences a force:

where $\theta$ is the angle between the current direction and $\vec{B}$. The direction is given by $\vec{F} = I \vec{L} \times \vec{B}$ (right-hand rule with fingers along the current, curled toward $\vec{B}$, thumb gives $\vec{F}$).

This is exactly the per-charge force $qvB$ summed over all charge carriers in the wire. The two relationships ($F = qvB$ and $F = BIL$) are the same physics in two presentations.

Force between parallel conductors

Two long parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ exert a force per unit length on each other:

with $\mu_0 / (2 \pi) = 2.0 \times 10^{-7}$ T m / A.

• Currents in the same direction attract.
• Currents in opposite directions repel.

The qualitative reasoning: wire 1 produces a field at wire 2 that wraps around it (right-hand grip rule). Wire 2 sits in this field carrying its own current, so $\vec{F} = I \vec{L} \times \vec{B}$ acts on it. Working out the cross product gives the same-direction-attract result.

This force was historically used to define the ampere.

Combining electric and magnetic fields: the velocity selector

A charged particle passing through perpendicular $\vec{E}$ and $\vec{B}$ fields experiences forces $qE$ and $qvB$. These can be made to cancel for a specific velocity:

Particles with this exact speed pass straight through; others are deflected. Velocity selectors are the entry stage of a mass spectrometer.

How this appears in IA1 and IA2

IA1 data test. Expect a velocity-selector or mass-spectrometer geometry with diagrams and a question on the radius of curvature, or a current-balance stimulus measuring the force between two parallel conductors. Right-hand-rule direction questions are routine.

IA2 student experiment. A common IA2 design measures the force on a current-carrying conductor between the poles of a permanent magnet as a function of current (or wire length), and extracts the field strength $B$ from the slope. Strong reports linearise $F$ vs $I$ and report $B$ with uncertainty.

Examples in context

Example 1. A Cairns light-rail induction motor carries a $300 \text{ A}$ stator current segment $0.40 \text{ m}$ long in a $0.6 \text{ T}$ rotor field. The force on the segment is $F = BIL\sin\theta = 0.6 \times 300 \times 0.40 = 72 \text{ N}$ (for $\theta = 90^\circ$), providing the torque that drives the rotor. The QCAA Unit 3 dot-point relationship $F = BIL$ is the heart of every electric-traction calculation across Queensland Rail's network.

Example 2. ANSTO Mt Cotton uses a magnetic mass spectrometer to verify satellite-grade rare-earth purity. A singly-ionised $^{60}\text{Ni}$ ion ($m = 9.95 \times 10^{-26} \text{ kg}$) at $1.0 \times 10^5 \text{ m s}^{-1}$ in $B = 0.20 \text{ T}$ moves on a circle of radius $r = mv/(qB) = 0.0311 \text{ m}$. Higher-mass isotopes deflect on larger circles, separating impurities. QCAA EA Unit 3 stems on charges in B-fields use exactly this cyclotron-radius formula.

Try this

Q1. State the equation for the force on a charge moving in a magnetic field. [2 marks]

• Cue. $F = qvB\sin\theta$.

Q2. A $5.0 \text{ A}$ current carrying conductor of length $0.30 \text{ m}$ sits perpendicular to a $0.40 \text{ T}$ field. Calculate the force and determine its direction using the right-hand rule for current to north and field to east. [3 marks]

• Cue. $F = BIL = 0.60 \text{ N}$; direction vertical (up).

Q3. A proton ($m = 1.67 \times 10^{-27} \text{ kg}$, $q = 1.6 \times 10^{-19} \text{ C}$) at $3.0 \times 10^6 \text{ m s}^{-1}$ enters a $0.20 \text{ T}$ field perpendicularly. (a) Calculate the radius of the circular path. (b) Determine the period of motion. (c) Discuss why electrons of the same speed circle in the opposite sense. [3+2+2 marks; ISMG: Analysis and interpretation, Knowledge and conceptual understanding]

• Cue. (a) $r = mv/(qB) = 0.156 \text{ m}$; (b) $T = 2\pi m/(qB) = 3.27 \times 10^{-7} \text{ s}$; (c) opposite charge reverses force direction.