SACE Board SACE 2025-style practice: A sigma particle of charge 1.6010^-19\ C and speed 6.5310^5\ m s^-1 enters a uniform magnetic field of 0.172\ T perpendicular to the field. Calculate the magnitude of the magnetic force on the particle as it enters the field.

The magnetic force is F = qvBsinθ. The particle enters perpendicular to the field, so θ = 90^ and sinθ = 1. $F = qvB = (1.6010^-19)(6.5310^5)(0.172) = 1.8010^-14\ N.$ 1 mark for F = qvB, 1 mark for the answer of about 1.8010^-14\ N. This force is the centripetal force that makes the particle follow a circular path.

§-Syllabus dot point

SAPhysicsSyllabus dot point

Why does a magnetic field only exert a force on a charge that is moving, and what determines the direction of that force?

Calculate the magnetic force on a moving charge and determine its direction using the right-hand rule.

How a magnetic field exerts a force on a moving charge, the equation $F = qvB\sin\theta$ , and using the right-hand rule to find the direction, with worked examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Why the charge must be moving
Direction: the right-hand rule
Circular motion in a magnetic field
Units
How SACE assesses this

What this dot point is asking

You need to calculate the magnetic force on a moving charge and determine its direction using a hand rule.

Why the charge must be moving

A stationary charge in a magnetic field feels no magnetic force. Only when the charge moves does the field push on it, and only the component of velocity perpendicular to the field contributes.

Direction: the right-hand rule

The force is perpendicular to both the velocity and the magnetic field; it cannot point along either. For a positive charge, point the fingers of your right hand along the velocity $v$ , curl them toward the field $B$ (or use the flat-hand "slap" rule: fingers along $B$ , thumb along $v$ , palm pushes in the force direction). The force is along your thumb/palm.

Circular motion in a magnetic field

Because the magnetic force is always perpendicular to the velocity, it never changes the speed; it only changes direction. It does no work on the charge. A charge entering a uniform field perpendicular to it therefore moves in a circle, with the magnetic force providing the centripetal force (this is developed in the "charged particles in magnetic fields" dot point). If the charge enters at an angle, the velocity component along the field is unchanged and the path becomes a helix.

Units

The tesla is defined so that $1\ \text{T} = 1\ \text{N A}^{-1}\text{m}^{-1}$ , or equivalently $1\ \text{N}$ per coulomb-metre-per-second. A 1 T field is strong; the Earth's field is around $5\times10^{-5}\ \text{T}$ .

How SACE assesses this

SACE Stage 2 questions here split into a calculation type and an explanation type. The calculation gives a charge, speed and field and asks for the force using $F = qvB\sin\theta$ , usually with the velocity perpendicular to the field so $\sin\theta = 1$ . The explanation type asks why a charge entering a uniform field perpendicular to it undergoes uniform circular motion; the expected answer is that the force is always perpendicular to the velocity, so it does no work and changes only the direction (constant speed), making it a centripetal force. For the calculation, quote the formula, note the angle, and substitute in SI units; for the explanation, the two marking points are usually "force always perpendicular to velocity" and "acts as a centripetal force, giving uniform circular motion". Reverse the hand-rule direction for negative charges such as electrons.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20252 marksA sigma particle of charge

1.60\times10^{-19}\ \text{C}

and speed

6.53\times10^5\ \text{m s}^{-1}

enters a uniform magnetic field of

0.172\ \text{T}

perpendicular to the field. Calculate the magnitude of the magnetic force on the particle as it enters the field.

Show worked answer →

The magnetic force is $F = qvB\sin\theta$ . The particle enters perpendicular to the field, so $\theta = 90^\circ$ and $\sin\theta = 1$ .

F = qvB = (1.60\times10^{-19})(6.53\times10^5)(0.172) = 1.80\times10^{-14}\ \text{N}.

1 mark for

F = qvB

, 1 mark for the answer of about

1.80\times10^{-14}\ \text{N}

. This force is the centripetal force that makes the particle follow a circular path.

SACE 20242 marksAn electron undergoes uniform circular motion in a uniform magnetic field directed into the page, perpendicular to its velocity. Explain why the electron undergoes uniform circular motion in the field.

Show worked answer →

The magnetic force $F = qvB\sin\theta$ on the electron has a constant magnitude because the speed and the perpendicular field are constant.

The force always acts perpendicular to the velocity, so it does no work; the speed (and kinetic energy) stays constant.

A constant-magnitude force that is always perpendicular to the velocity is a centripetal force: it continually changes the direction of motion while keeping the speed constant, which is uniform circular motion. 1 mark for stating the force is always perpendicular to velocity (changing direction, not speed), 1 mark for identifying it as a centripetal force giving uniform circular motion.