Is r = mv/qB on the HSC data sheet?

No. The 2019 data sheet gives F = qv⊥B = qvB sin θ and you are expected to know Fc = mv^2/r from Module 5. Set qvB = mv^2/r and rearrange: r = mv/qB. Show those two lines whenever you use it.

What is the difference between the effect of an electric field and a magnetic field on a charge?

An electric field pushes on any charge, along the field lines, does work and changes its speed; fired across the field, it follows a parabola. A magnetic field pushes only on a moving charge, at right angles to its velocity, does no work and leaves the speed unchanged; fired across the field, it follows a circle.

Do you ignore gravity for charged particles in fields?

For electrons, protons and ions in the fields used in HSC questions, yes: the electric or magnetic force is many orders of magnitude bigger than the weight. Questions about heavier charged objects, like oil drops, are the exception, and they will make gravity part of the problem.

Which hand rule should I use for the force on a moving charge?

Any correct right-hand rule works. With the right-hand palm rule, fingers point along B, the thumb along the velocity of a positive charge, and the palm pushes in the direction of the force. For a negative charge, reverse the result.

What is an electron-volt?

A unit of energy: the energy a charge of 1.602 × 10^−19 C (one electron's worth) gains or loses moving through a potential difference of 1 V. 1 eV = 1.602 × 10^−19 J.

A proton and an electron are each accelerated from rest through the same 500 V potential difference. Which statement is correct?

They gain the same kinetic energy, and the electron ends up much faster.

The proton gains more kinetic energy because it has more mass.

The electron gains more kinetic energy because it has less mass.

HSC Physics · Module 6 · Electromagnetism · 13 min read

Charged particles in electric and magnetic fields

Electrons in an old TV tube, protons in a cyclotron, ions in a mass spectrometer: all of them are steered by electric and magnetic fields. In HSC Physics the two fields do very different jobs. An electric field pushes a charge along the field lines and changes its speed. A magnetic field pushes a moving charge sideways and only changes its direction.

This guide covers the field between parallel plates, the force and work on a charge, parabolic paths that work just like projectiles, the magnetic force F = qvB sin θ and circular paths with r = mv/qB, and how the electric, magnetic and gravitational fields compare. Then a quick quiz and a full worked example.

What you need to know

Investigate charged particles in uniform electric fields, deriving and analysing the interaction quantitatively, including E = V/d between oppositely charged parallel plates, F_net = ma and F = qE for the field accelerating the particle, and the work the field does on the charge, W = qV = qEd, with K = ½mv².
Model the path of a charge in an electric field qualitatively and quantitatively, and compare it with the path of a projectile under gravity.
Analyse how uniform magnetic fields act on charged particles, including their acceleration at right angles to the field and the magnetic force on each charge, F = qv_⊥B = qvB sin θ.
Compare how a magnetic field affects a moving charge with how electric fields act on charges, and with other cases of uniform circular motion.

The equations (from the 2019 HSC data sheet unless marked):

E = V/dField strength between parallel plates (V m⁻¹, the same as N C⁻¹). d is the plate separation in metres.

F = qEForce on charge q. Positive charges are pushed along the field; negative charges against it.

W = qV W = qEdWork done by the field on the charge, which becomes kinetic energy: qV = ½mv² from rest.

K = ½mv²Kinetic energy, to turn the work done into a speed.

F = qv_⊥B = qvB sin θMagnetic force on a moving charge; θ is the angle between v and B.

r = mv/qB not on the data sheetRadius of the circular path in a magnetic field, from qvB = mv²/r. Derive it.

Dot points: PH-M6-01, PH-M6-02, PH-M6-03, PH-M6-04 (2017 Physics syllabus, HSC 2026 and 2027), in our own words.

Mini lesson

The uniform field between parallel plates

Connect two parallel metal plates to a voltage V and the field between them is uniform: the same size and direction everywhere (except near the edges). It points from the positive plate to the negative plate, and its strength is E = V/d.

A charge q in the field feels a constant force F = qE: along the field for a positive charge, against it for a negative one. With F = ma, the acceleration is a = qE/m. For electrons and protons, gravity is negligible: in a field of 8.0 × 10⁴ V m⁻¹, the electric force on an electron is about 10¹⁵ times its weight.

The field E points from + to − and is the same everywhere between the plates. A positive charge is pushed along the field with F = qE.

Work, energy and the electron-volt

When a charge moves across the plates, the field does work on it: W = qV, or W = qEd for a distance d along the field. That work becomes kinetic energy. For a particle starting from rest, qV = ½mv², so v = √(2qV/m).

The work depends only on the voltage crossed, not on the gap. Pull the plates twice as far apart at the same voltage and E halves, but the charge travels twice as far, so W = qEd is unchanged.

Energies this small are often given in electron-volts: 1 eV is the energy one electron gains crossing 1 V, which is 1.602 × 10⁻¹⁹ J. An electron accelerated through 500 V gains 500 eV = 8.0 × 10⁻¹⁷ J and reaches about 1.3 × 10⁷ m s⁻¹.

Paths between plates: projectiles again

Fire a charge into the field at right angles to it and the motion splits exactly like a projectile. Across the plates there is a constant force, so a constant acceleration a = qE/m. Along the plates there is no force, so the velocity in that direction stays the same. The result is a parabola, bending towards the plate of opposite sign.

Same as a projectile: constant velocity along the plates, uniform acceleration across them; use s = ut + ½at² and v = u + at in each direction, linked by the time.
Different: the acceleration is qE/m, not g. It depends on the charge and the mass, can point any way the field points, and for an electron in a typical plate field (such as 8.0 × 10⁴ V m⁻¹) it is about 10¹⁵ times bigger than g.
Time between the plates = plate length ÷ entry speed, because the speed along the plates doesn't change.

Inside the plates the electron has a constant force towards the positive plate, so it follows a parabola. Outside the field there is no force, so it goes straight.

The magnetic force on a moving charge

A magnetic field only pushes on a charge that is moving, and only if it moves across the field: F = qvB sin θ, where θ is the angle between v and B. A charge moving along the field lines (θ = 0°) feels no force; one moving at right angles (θ = 90°) feels the most, qvB.

The force is at right angles to both v and B. Find the direction with the right-hand palm rule: fingers along B, thumb along the velocity of a positive charge, and the palm pushes in the direction of the force. For an electron or any negative charge, reverse the answer.

Because the force is always at right angles to the velocity, a magnetic field does no work on a charge. It changes the direction of motion but never the speed or the kinetic energy.

Circular paths: r = mv/qB

A charge moving at right angles to a uniform magnetic field feels a force of constant size that is always at right angles to its velocity. That is exactly the condition for uniform circular motion, so the magnetic force is the centripetal force:

qvB = mv²/r → r = mv/qB

B into the page (×). At every point the force qvB points to the centre, at 90° to v, so the charge goes round a circle of radius r = mv/qB at constant speed.

Heavier or faster particles curve less (bigger r); more charge or a stronger field curves them more (smaller r). The direction they curve tells you the sign of the charge. The time for one lap, T = 2πr/v = 2πm/qB, doesn't depend on the speed, which is the idea behind the cyclotron.

Comparing electric, magnetic and gravitational fields

Gravitational field: acts on mass, always attracts, gives every object the same acceleration g. Fired sideways, an object follows a parabola.
Electric field: acts on charge, moving or not, along (or against) the field lines. It does work and changes the speed. Fired across the field, a charge follows a parabola, like a projectile.
Magnetic field: acts only on a moving charge, at right angles to both v and B. It does no work, so the speed is constant. Fired across the field, a charge follows a circle, like any other uniform circular motion (a car on a curve, a satellite in orbit).

Common mistakes

Using the plate separation in millimetres (or the voltage in kV) in E = V/d.Convert to metres and volts first: 400 V across 5.0 mm is 400 ÷ 0.0050 = 8.0 × 10⁴ V m⁻¹.
Saying a magnetic field speeds up or slows down a charged particle.The magnetic force is always at 90° to the velocity, so it does no work. Only the direction changes.
Using the right-hand rule for an electron without reversing it.Hand rules give the force on a positive charge. For electrons, find that direction, then reverse it.
Drawing a curved arc (part of a circle) for a charge between parallel plates.A constant force in one direction gives a parabola, like a projectile. Circles come from magnetic fields.
Mixing electron-volts and joules in qV = ½mv².1 eV = 1.602 × 10⁻¹⁹ J. Put energies into kinematics equations in joules.
Using the wire formulas F = lIB sin θ or B = μ₀I/2πr for a single moving particle.For one moving charge use F = qvB sin θ. The wire formulas are for currents in conductors.

Worked example

In a mass-analysis experiment, a proton starts from rest and is accelerated through a potential difference of 1.8 kV. It then passes through a small hole into a region of uniform magnetic field of strength 0.12 T, directed into the page. It enters moving to the right, at right angles to the field.

Calculate the radius of the proton's path in the magnetic field, and state which way it curves.

(5 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. Through about 2 kV a proton gains about 3 × 10⁻¹⁶ J, so v ≈ √(2 × 3 × 10⁻¹⁶ ÷ 1.7 × 10⁻²⁷) ≈ 6 × 10⁵ m s⁻¹. Then r = mv/qB ≈ (1.7 × 10⁻²⁷ × 6 × 10⁵) ÷ (1.6 × 10⁻¹⁹ × 0.12) ≈ 0.05 m. I predict a radius of a few centimetres, around 5 cm.

K · Knowns

I sketch the two stages: the accelerating plates, then the field region, and put every value on it. The voltage is in kilovolts, so I convert it to volts. The proton's charge and mass come from the data sheet.

V = 1.8 kV = 1800 V, starting from rest (u = 0)
B = 0.12 T, into the page, θ = 90°
q = +1.602 × 10⁻¹⁹ C, m_p = 1.673 × 10⁻²⁷ kg

Stage 1: accelerated from rest through V = 1.8 kV. Stage 2: enters B = 0.12 T (into the page) at right angles. Unknown: the radius r.

N · Not known

The question wants a radius in metres and a direction of curving. To get r I first need the speed the proton has when it enters the field.

r = ? m (on the way: v = ? m s⁻¹)

O · Organise

Stage 1 is energy: the work done by the field becomes kinetic energy. Stage 2 is circular motion: the magnetic force is the centripetal force. I write both in symbols first.

qV = ½mv² → v = √(2qV/m)
qvB = mv²/r → r = mv/qB

W · Work it out

I find the kinetic energy, then the speed, keeping the full calculator value of v for the radius.

qV = 1.602 × 10⁻¹⁹ × 1800 = 2.8836 × 10⁻¹⁶ J
v = √(2 × 2.8836 × 10⁻¹⁶ ÷ 1.673 × 10⁻²⁷) = 5.8713 × 10⁵ m s⁻¹
r = (1.673 × 10⁻²⁷ × 5.8713 × 10⁵) ÷ (1.602 × 10⁻¹⁹ × 0.12)
r = 9.8227 × 10⁻²² ÷ 1.9224 × 10⁻²⁰ = 0.05110 m

U · Units

kg × m s⁻¹ ÷ (C × T) comes out in metres, which is right for a radius. The data are to two significant figures, so I give two.

r ≈ 0.051 m (5.1 cm)

D · Direction

The proton is positive. Right-hand palm rule at the moment it enters: fingers into the page (B), thumb to the right (v), palm pushes towards the top of the page. So the force, and the centre of the circle, is above the entry point, and the proton curves upwards: anticlockwise as drawn. As a sense check, 5.9 × 10⁵ m s⁻¹ is only 0.2% of c, so the ordinary ½mv² is fine.

Curves towards the top of the page (anticlockwise, seen with B into the page)

The proton moves in a circle of radius about 0.051 m (5.1 cm), curving towards the top of the page (anticlockwise as viewed with the field into the page).

Check against the prediction. I predicted a few centimetres, around 5 cm. The answer, 5.1 cm, matches.

HSC advice

What markers keep saying

Recognise that a charge fired across a uniform electric field moves like a projectile and follows a parabola, not a circle. Know that at a fixed voltage, moving the plates apart weakens the field (E = V/d) while the work done across the whole gap (W = qV) stays the same.NESA marking feedback, 2020, summarised
A particle that starts from rest has zero initial kinetic energy, so the work done by the field equals its final kinetic energy. When comparing an electron and a proton in the same field, the forces are equal in size but opposite in direction; the electron's acceleration is far larger because its mass is far smaller.NESA marking feedback, 2021, summarised
Convert carefully between electron-volts and joules. When explaining tracks in a magnetic field, describe the features (circular shape, which way it curves, how big the radius is) and back each one up with r = mv/qB.NESA marking feedback, 2022, summarised
The radius in a magnetic field depends on charge as well as mass (and on speed and field strength). Don't explain a difference in radius by mass alone.NESA marking feedback, 2019, summarised
Use F = qvB for a single moving charge, not formulas for current-carrying wires, and give a direction for any vector you calculate.NESA marking feedback, 2023, summarised
Convert millimetres to metres before substituting, and carry answers from earlier parts into later parts of the same question.NESA marking feedback, 2024, summarised

Summaries in our own words. Read NESA's own feedback with the exam packs: 2019 · 2020 · 2021 · 2022 · 2023 · 2024.

How often it is examined

Charged particles in electric and magnetic fields questions were worth 49 marks across the seven HSC Physics exams from 2019 to 2025 (about 7.0 marks a paper), and appeared in 7 of 7 papers.

Marks per paper (Module 6, Charged Particles, Conductors and Electric and Magnetic Fields). Source: CortexHSC's own tagging of the NESA HSC Physics exams 2019–2025, each question counted under its main topic, so treat the numbers as close, not exact.

FAQ

Is r = mv/qB on the HSC data sheet?: No. The 2019 data sheet gives F = qv_⊥B = qvB sin θ and you are expected to know F_c = mv²/r from Module 5. Set qvB = mv²/r and rearrange: r = mv/qB. Show those two lines whenever you use it.
What is the difference between the effect of an electric field and a magnetic field on a charge?: An electric field pushes on any charge, along the field lines, does work and changes its speed; fired across the field, it follows a parabola. A magnetic field pushes only on a moving charge, at right angles to its velocity, does no work and leaves the speed unchanged; fired across the field, it follows a circle.
Do you ignore gravity for charged particles in fields?: For electrons, protons and ions in the fields used in HSC questions, yes: the electric or magnetic force is many orders of magnitude bigger than the weight. Questions about heavier charged objects, like oil drops, are the exception, and they will make gravity part of the problem.
Which hand rule should I use for the force on a moving charge?: Any correct right-hand rule works. With the right-hand palm rule, fingers point along B, the thumb along the velocity of a positive charge, and the palm pushes in the direction of the force. For a negative charge, reverse the result.
What is an electron-volt?: A unit of energy: the energy a charge of 1.602 × 10⁻¹⁹ C (one electron's worth) gains or loses moving through a potential difference of 1 V. 1 eV = 1.602 × 10⁻¹⁹ J.

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