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Charged particles in magnetic fields (SACE Stage 2 Physics)

Syllabus dot point

“Analyse the circular motion of a charged particle in a uniform magnetic field, relating radius to mass, charge, speed and field strength.”

9 min read

Quick answer

A charge entering a uniform magnetic field at right angles moves in a circle, because the magnetic force is always perpendicular to the velocity and so acts as a centripetal force. Setting qvB=mv2rqvB = \dfrac{mv^2}{r} gives the radius r=mvqBr = \dfrac{mv}{qB}. Heavier or faster particles curve in larger circles; stronger fields or larger charges curve them tighter.

Jump to a section
  1. What this dot point is asking
  2. Why the path is a circle
  3. The radius of the path
  4. Applications: the mass spectrometer
  5. The cyclotron
  6. How SACE assesses this
  7. Exam-style questions

What this dot point is asking

You need to analyse the circular motion of a charged particle in a uniform magnetic field and relate the radius to the particle's mass, charge, speed and the field.

Radius of an electron's path

An electron moving at 3.0×106 m s−13.0\times10^6\ \text{m s}^{-1} enters a 0.020 T0.020\ \text{T} field perpendicularly. This example finds the radius of its circular path, using electron mass m=9.11×10−31 kgm = 9.11\times10^{-31}\ \text{kg} and charge q=1.60×10−19 Cq = 1.60\times10^{-19}\ \text{C}.

Step 1: set up the force balance to obtain the radius formula

The magnetic force qvBqvB acts perpendicular to the velocity at every point, so it provides the centripetal force mv2/rmv^2/r needed for circular motion. Setting these equal and solving for the radius:

qvB=mv2r⇒r=mvqB.qvB = \frac{mv^2}{r} \quad \Rightarrow \quad r = \frac{mv}{qB}.

Step 2: substitute the values and compute the radius

Inserting the electron's mass, speed, charge and the field strength into the formula:

r=mvqB=(9.11×10−31)(3.0×106)(1.60×10−19)(0.020)=2.73×10−243.20×10−21=8.5×10−4 m≈0.85 mm.r = \frac{mv}{qB} = \frac{(9.11\times10^{-31})(3.0\times10^6)}{(1.60\times10^{-19})(0.020)} = \frac{2.73\times10^{-24}}{3.20\times10^{-21}} = 8.5\times10^{-4}\ \text{m} \approx 0.85\ \text{mm}.

Final answer: the electron follows a circular arc of radius approximately 0.85 mm0.85\ \text{mm}.

Why the path is a circle

The magnetic force F=qvBF = qvB on a charge moving perpendicular to the field is always at right angles to the velocity. A force that is always perpendicular to motion changes only the direction, not the speed - exactly the condition for uniform circular motion. So the particle traces a circle at constant speed.

Magnetic force as centripetal force

qvB=mv2rqvB = \frac{mv^2}{r}

The magnetic force supplies the centripetal force. The magnetic force does no work, so the speed and kinetic energy stay constant throughout.

The radius of the path

Rearranging the balance of forces:

Radius of circular motion

r=mvqBr = \frac{mv}{qB}

rr is the radius of the circular path (m), mm the particle's mass (kg), vv its speed (m s−1\text{m s}^{-1}), qq its charge (C), and BB the flux density (T).

So the radius:

  • increases with mass and speed (more momentum is harder to bend),
  • decreases with charge and field strength (a stronger push bends it tighter).

The period of the circular motion, T=2πmqBT = \dfrac{2\pi m}{qB}, is independent of speed: faster particles travel larger circles in the same time. This is the principle that lets a cyclotron use a fixed-frequency accelerating voltage.

Applications: the mass spectrometer

The radius equation is the basis of the mass spectrometer. Ions of known charge and speed (often selected by a velocity selector) enter a uniform field. Since r=mvqBr = \dfrac{mv}{qB}, ions of different mass-to-charge ratio land at different radii, separating them so their masses can be measured. This is how isotopes are identified and their abundances measured. On a graph of radius against speed for ions of the same charge in the same field, the gradient is mqB\dfrac{m}{qB}, so a steeper line means a larger mass.

If a charge enters the field at an angle (not perpendicular), the component of velocity along the field is unaffected and the perpendicular component drives circular motion, so the overall path is a helix.

The cyclotron

A cyclotron accelerates charged particles to high energy using the speed-independent period. The particle spirals outward through two D-shaped electrodes, gaining energy from an alternating voltage applied across the gap each time it crosses. Because T=2πmqBT = \dfrac{2\pi m}{qB} does not depend on the speed or radius, the accelerating voltage can alternate at a single fixed frequency and stay synchronised with the particle no matter how fast it is moving. The radius grows as the speed grows, but the time for each half-circle stays the same, which is exactly why the design works.

How SACE assesses this

SACE Stage 2 questions on this dot point typically give a particle's mass, charge, speed and the field, then ask for the radius using r=mvqBr = \dfrac{mv}{qB}, or rearrange it to find the speed or field. A second common style supplies a graph of radius against speed and asks you to interpret the gradient mqB\dfrac{m}{qB} to compare masses or charges. "Show that" parts asking you to verify a cyclotron period use T=2πmqBT = \dfrac{2\pi m}{qB} directly. Quote the formula, substitute in consistent SI units (convert any radius given in centimetres to metres), and rearrange before substituting numbers rather than after.

Common errors
Thinking the field changes the speed
The magnetic force does no work; speed and kinetic energy are constant. Only the direction changes.
Confusing which way the circle curves
The direction depends on the sign of the charge; electrons and positive ions curve in opposite senses for the same field and velocity.
Forgetting period is speed-independent
T=2πmqBT = \dfrac{2\pi m}{qB} does not contain vv; a faster particle just takes a bigger circle in the same time.
Mixing up the radius dependences
Larger mass or speed gives a larger radius; larger charge or field gives a smaller (tighter) radius.

Exam-style questions

Questions in the style of SACE Board exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

SACE 2023-style3 marks
An electron enters a uniform magnetic field with its velocity perpendicular to the field and undergoes uniform circular motion. The field has magnitude 0.0230 T0.0230\ \text{T} and the radius of the path is 1.50 cm1.50\ \text{cm}. Determine the speed of the electron.
Show worked answer →

The magnetic force supplies the centripetal force, so r=mvqBr = \dfrac{mv}{qB}.

Rearrange for speed: v=qBrmv = \dfrac{qBr}{m}, with q=1.60×10−19 Cq = 1.60\times10^{-19}\ \text{C}, m=9.11×10−31 kgm = 9.11\times10^{-31}\ \text{kg} and r=0.0150 mr = 0.0150\ \text{m}.

v=(1.60×10−19)(0.0230)(0.0150)9.11×10−31=5.52×10−239.11×10−31=6.06×107 m s−1.v = \dfrac{(1.60\times10^{-19})(0.0230)(0.0150)}{9.11\times10^{-31}} = \dfrac{5.52\times10^{-23}}{9.11\times10^{-31}} = 6.06\times10^7\ \text{m s}^{-1}.

1 mark for the rearrangement, 1 mark for converting the radius to metres, 1 mark for the answer of about 6.1×107 m s−16.1\times10^7\ \text{m s}^{-1}.

SACE 2025-style2 marks
A cyclotron accelerates a helium-3 ion of mass 5.02×10−27 kg5.02\times10^{-27}\ \text{kg} and charge 3.20×10−19 C3.20\times10^{-19}\ \text{C} in a magnetic field of 1.08 T1.08\ \text{T}. Show that the period of the circular motion is 9.13×10−8 s9.13\times10^{-8}\ \text{s}.
Show worked answer →

The period of circular motion is independent of speed and radius: T=2πmqBT = \dfrac{2\pi m}{qB}.

T=2π(5.02×10−27)(3.20×10−19)(1.08)=3.154×10−263.456×10−19=9.13×10−8 s.T = \dfrac{2\pi(5.02\times10^{-27})}{(3.20\times10^{-19})(1.08)} = \dfrac{3.154\times10^{-26}}{3.456\times10^{-19}} = 9.13\times10^{-8}\ \text{s}.

1 mark for the formula T=2πmqBT = \dfrac{2\pi m}{qB}, 1 mark for the substitution and answer. The period not depending on radius is the key cyclotron principle.

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Sources & how we know this

Based on the official SACE Board subject outline. Not yet individually reviewed by a person; always check SACE Board and your school's materials for current rules.

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