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A charged particle in a magnetic field is one of the most reliable sources of JEE questions: find the radius, the time period, the pitch of a helix, or the energy from a cyclotron. All of these come from one force law and one idea. This post works through each with numbers you can check.
Why does a charge move in a circle in a magnetic field?
The magnetic force on a charge q moving with velocity v in a field B is F = q(v × B). It is always perpendicular to the velocity, so it changes the direction of motion but never the speed. A force of constant size that always points sideways is exactly what circular motion needs.
The NCERT Class 12 chapter Moving Charges and Magnetism sets the magnetic force equal to the centripetal force: mv²/r = qvB, which gives
radius r = mv/(qB)
angular frequency ω = qB/m, and frequency ν = qB/(2πm)
NCERT then points out that ν "is independent of the velocity or energy", and that this "has important application in the design of a cyclotron". A faster particle moves on a bigger circle, so it takes the same time to go round.
The JEE (Main) 2026 syllabus lists "force on a moving charge in uniform magnetic and electric fields" in Unit 13. The cyclotron itself is not named there, so treat its details as useful practice rather than a listed topic.
Worked example 1: radius and frequency of a proton
A proton (m = 1.67 × 10⁻²⁷ kg, q = 1.60 × 10⁻¹⁹ C) moves at 2.0 × 10⁶ m/s perpendicular to a field of 0.50 T.
r = mv/(qB) = (1.67 × 10⁻²⁷ × 2.0 × 10⁶) / (1.60 × 10⁻¹⁹ × 0.50) = 0.042 m, about 4.2 cm.
ν = qB/(2πm) = (1.60 × 10⁻¹⁹ × 0.50) / (2π × 1.67 × 10⁻²⁷) = 7.6 × 10⁶ Hz, about 7.6 MHz.
Period T = 1/ν = 1.3 × 10⁻⁷ s.
Double the speed and the radius doubles to 8.4 cm, but ν stays at 7.6 MHz. The NIST values for e and the proton mass give the same answers to two figures.
What happens when the velocity is not perpendicular to B?
Split the velocity into a part along B and a part across it. The part across B gives the circle as before. The part along B feels no force, so it carries the particle steadily along the field. The path is a helix. NCERT defines the pitch as the distance moved along the field in one rotation: p = v∥T = 2πmv∥/(qB).
Worked example 2. The same proton moves at 2.0 × 10⁶ m/s at 30° to the 0.50 T field.
v⊥ = 2.0 × 10⁶ × sin 30° = 1.0 × 10⁶ m/s, so the helix radius is 2.1 cm.
v∥ = 2.0 × 10⁶ × cos 30° = 1.73 × 10⁶ m/s, so the pitch is 1.73 × 10⁶ × 1.31 × 10⁻⁷ = 0.23 m.
A common mistake is using the full speed in the radius formula. Only v⊥ goes into r, and only v∥ goes into the pitch.
How does a cyclotron accelerate particles?
A magnetic field cannot speed a particle up, so a cyclotron uses an electric field for that and a magnetic field to bring the particle back. Lawrence and Livingston described the idea in Physical Review in 1932: two hollow semicircular electrodes ("dees") sit in a uniform magnetic field, with an oscillating voltage across the gap between them. Each time the ion crosses the gap the field has reversed, so it gains energy again. Because the time for each half-circle does not depend on speed, the ion stays in step and spirals outward.
Their paper reports a beam of protons at 1,220,000 electron-volts from a tube whose maximum applied voltage was only 4000 volts, using a magnet with pole faces 11 inches across. Ernest Lawrence received the 1939 Nobel Prize in Physics "for the invention and development of the cyclotron and for results obtained with it". Our profile of Ernest Lawrence tells that story.
Worked example 3. A cyclotron has dees of radius R = 0.50 m in a field of 1.5 T. What is the maximum kinetic energy of a proton?
At the edge, v = qBR/m, so K = q²B²R²/(2m) = (1.60 × 10⁻¹⁹ × 1.5 × 0.50)² / (2 × 1.67 × 10⁻²⁷) = 4.3 × 10⁻¹² J, about 27 MeV. The oscillator must run at qB/(2πm), about 23 MHz. Note that K does not depend on the gap voltage: a smaller voltage only means more turns.
Common mistakes
Saying the period depends on speed. It does not, as long as the speed is well below that of light.
Using the gap voltage to find the final energy. The final energy is set by B and R.
Getting the direction of circling wrong for negative charges. Find the force for a positive charge, then reverse it for an electron.
Mixing up eV and J. Divide by 1.60 × 10⁻¹⁹ to convert J to eV.
A practice routine
Derive r and ν from mv²/r = qvB without looking.
Solve one proton, one electron and one alpha-particle problem at the same B and compare the radii.
Solve two helix problems, writing v⊥ and v∥ before anything else.
Solve one cyclotron energy problem and check that the answer does not change when you change the gap voltage.
When your answer disagrees with the key, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can see whether the slip was in v⊥, the units or the formula.
Start with worked example 2: redo it at 60° and check that the radius grows while the pitch shrinks.
