Why does a charged particle follow a circular path in a uniform magnetic field, and what sets the radius?
Analyse the circular motion of a charged particle in a uniform magnetic field, relating radius to mass, charge, speed and field strength.
Why a charge moving across a magnetic field travels in a circle, the radius equation r = mv/(qB), and its application to mass spectrometers, with worked examples.
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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.
Why the path is a circle
The magnetic force 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.
The radius of the path
Rearranging the balance of forces:
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, , is independent of speed - faster particles travel larger circles in the same time.
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 , 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 measured.
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.
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.
2023 SACE Stage 23 marksAn electron enters a uniform magnetic field with its velocity perpendicular to the field and undergoes uniform circular motion. The magnitude of the magnetic field is 0.0230 T and the radius of the circular path is 1.50 cm. Determine the speed of the electron.Show worked answer →
For a charged particle in a magnetic field, the magnetic force provides the centripetal force, giving r = m v / (q B).
Rearrange for the speed: v = q B r / m.
Use the electron charge q = 1.60 x 10^-19 C, mass m = 9.11 x 10^-31 kg, and r = 1.50 cm = 0.0150 m.
v = (1.60 x 10^-19)(0.0230)(0.0150) / (9.11 x 10^-31) = (5.52 x 10^-23) / (9.11 x 10^-31) = 6.06 x 10^7 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 x 10^7 m s-1.
2024 SACE Stage 23 marksThree particles A, B, and C undergo circular motion in the same magnetic field. Each particle has the same charge. A graph shows how the radius of the circular path varies with speed, with A having the steepest line and C the shallowest. Use the graph to determine which particle has the greatest mass. Justify your answer.Show worked answer →
From r = m v / (q B), rearranged as r = (m / (q B)) v, the radius is proportional to the speed and the gradient of an r against v graph is m / (q B).
Since all three particles have the same charge q and travel in the same field B, a larger gradient means a larger mass.
Particle A has the steepest line, so it has the largest gradient and therefore the greatest mass.
1 mark for stating gradient = m / (q B), 1 mark for recognising that with q and B constant the gradient is proportional to mass, 1 mark for selecting particle A (the steepest line).
2025 SACE Stage 22 marksA cyclotron accelerates a helium-3 ion of mass 5.02 x 10^-27 kg and charge 3.20 x 10^-19 C. The radius is 0.45 m and the magnetic field is 1.08 T. Show that the period of the circular motion of the helium-3 ion is 9.13 x 10^-8 s.Show worked answer →
The period of circular motion of a charged particle in a magnetic field is independent of speed and radius: T = 2 pi m / (q B).
Substitute the values: T = 2 pi (5.02 x 10^-27) / [(3.20 x 10^-19)(1.08)].
Numerator = 2 pi (5.02 x 10^-27) = 3.154 x 10^-26.
Denominator = (3.20 x 10^-19)(1.08) = 3.456 x 10^-19.
T = 3.154 x 10^-26 / 3.456 x 10^-19 = 9.13 x 10^-8 s.
1 mark for the formula T = 2 pi m / (q B), 1 mark for the substitution and answer. The period does not depend on the cyclotron radius, which is the key cyclotron principle.