Magnetic fields and the Lorentz force on charges: VCE Physics Unit 3

What this dot point is asking

VCAA wants you to describe magnetic fields around magnets, current-carrying wires and solenoids, apply the right-hand rule to find field and force directions, and apply $F = qvB$ to a charged particle moving perpendicular to a uniform field (including the resulting circular motion).

The answer

The magnetic field model

A magnetic field $B$ is a vector at each point in space, measured in tesla (T). Field lines run from the north pole of a magnet to its south pole externally (and from south to north inside the magnet, forming closed loops). They never start or end on a charge; magnetic monopoles do not exist.

Field shapes

Bar magnet

Field lines emerge from the north pole, curve around, and enter the south pole. The field is strongest near the poles.

Straight current-carrying wire

The field forms concentric circles around the wire. The direction is given by the right-hand grip rule: thumb in the direction of conventional current, fingers curl in the direction of $B$.

Solenoid

A long coil of wire carrying current. Inside the solenoid the field is uniform and parallel to the axis (like a bar magnet's interior). Outside, the field resembles that of a bar magnet, with one end acting as north and the other as south. Apply the right-hand grip rule with fingers curling in the direction of current to find which end is north.

The strength of a solenoid's field can be increased by increasing the current, increasing the turns per unit length, or inserting a ferromagnetic core.

Force on a moving charge

A charged particle moving with velocity $v$ through a magnetic field $B$ experiences a force:

where $\theta$ is the angle between $v$ and $B$. For $v$ perpendicular to $B$ (the VCE case), $F = qvB$.

The direction is given by the right-hand slap rule for a positive charge:

• Fingers point in the direction of velocity $v$.
• Curl them toward the field $B$.
• The thumb points in the direction of the force $F$.

For a negative charge, the force is in the opposite direction.

If $v$ is parallel or anti-parallel to $B$, the force is zero. The magnetic force never does work on a charged particle (it is always perpendicular to $v$); it changes the direction of motion, not the speed.

Circular motion of a charged particle in a uniform $B$

When a charged particle moves perpendicular to a uniform $B$, the constant magnitude of $F = qvB$ supplies the centripetal force:

So the radius of the circular path is:

and the period is:

The period is independent of speed. Faster particles travel in larger circles at the same period.

If $v$ has a component along $B$ as well as across it, the path is a helix (the parallel component is unaffected by the field; the perpendicular component drives circular motion).

Comparing the three field models

Examples in context

Example 1. Australian Synchrotron storage-ring dipole magnets. The Australian Synchrotron at Clayton uses dipole bending magnets with field $B = 1.3$ T to steer $3$ GeV electrons around the $216$ m circumference storage ring. Electron momentum is $p = E/c = 3 \times 10^9 \times 1.6 \times 10^{-19}/(3 \times 10^8) = 1.6 \times 10^{-18}$ kg m s$^{-1}$. Radius of curvature is $r = p/(qB) = 1.6 \times 10^{-18}/(1.6 \times 10^{-19} \times 1.3) = 7.7$ m. The right-hand rule gives the force direction: with $\vec{v}$ tangent to the orbit and $\vec{B}$ vertical, $\vec{F} = q\vec{v} \times \vec{B}$ points horizontally inward, providing the centripetal force.

Example 2. Mass-spectrometer separation of forensic samples at Victoria Police. A Victoria Police forensic mass spectrometer accelerates singly charged ions through a $5000$ V potential, giving each ion kinetic energy $qV = 8 \times 10^{-16}$ J. A uranium-238 ion ($m = 3.95 \times 10^{-25}$ kg) reaches speed $v = \sqrt{2qV/m} = \sqrt{4.05 \times 10^9} = 6.36 \times 10^4$ m s$^{-1}$. In a $0.5$ T field, radius of circular motion is $r = mv/(qB) = 3.95 \times 10^{-25} \times 6.36 \times 10^4/(1.6 \times 10^{-19} \times 0.5) = 0.314$ m. Mass-235 ions follow a slightly smaller radius, allowing isotope separation at the detector.

Try this

Q1. State the magnitude and direction of the magnetic force on a charged particle moving perpendicular to a magnetic field. [2 marks]

• Cue. $F = qvB$; direction given by right-hand rule (fingers from $\vec{v}$ to $\vec{B}$, thumb gives $\vec{F}$ for positive charge).

Q2. A proton travels at $2.0 \times 10^6$ m s$^{-1}$ perpendicular to a $0.40$ T field. Calculate (a) the force on the proton, and (b) the radius of its circular path. [4 marks]

• Cue. (a) $F = qvB = 1.6 \times 10^{-19} \times 2 \times 10^6 \times 0.4 = 1.28 \times 10^{-13}$ N. (b) $r = mv/(qB) = 1.67 \times 10^{-27} \times 2 \times 10^6/(1.6 \times 10^{-19} \times 0.4) = 0.052$ m.

Q3. Refer to the Synchrotron dipole magnets. (a) Calculate the magnetic force on an electron of momentum $1.6 \times 10^{-18}$ kg m s$^{-1}$ in a $1.3$ T field. (b) Determine the orbit radius. (c) Use the right-hand rule to identify the force direction relative to the orbit. [2+2+2 marks]

• Cue. (a) $F = qvB \approx qcB = 6.24 \times 10^{-11}$ N for ultrarelativistic electron. (b) $r = 7.7$ m. (c) Toward the orbit centre, providing centripetal force.