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A proton or electron fired straight across a magnetic field goes round in a circle. Fire it at an angle and it corkscrews along the field. Exam questions on this helix look varied, but almost all of them come down to one split of the velocity. This post sets out that split, works through electron examples with real constants, and shows a pitch shortcut that saves time.
Where is this in the syllabus?
Section 4.3, Motion in a Magnetic Field, of the NCERT Class 12 Physics chapter Moving Charges and Magnetism (Reprint 2026-27) derives the helix and defines its pitch. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list "force on a moving charge in uniform magnetic and electric fields" in Unit 13.
Why is the path a helix?
NCERT's reasoning has two parts. The magnetic force qv × B is always perpendicular to the velocity, so it does no work and the speed never changes. Then split the velocity relative to B:
v⊥ = v sin θ, across the field. The force acts on this part like a centripetal force, giving a circle.
v∥ = v cos θ, along the field. The field has no effect on it, so the particle drifts steadily along B.
Circle plus steady drift is a helix. The formulas follow from NCERT's equations 4.5 and 4.6:
Radius: r = m v⊥ / (qB)
Period: T = 2πm / (qB), which does not depend on speed
Pitch, the distance moved along B in one turn: p = v∥ T = 2πm v∥ / (qB)
Worked example 1: an electron at 60°
An electron with 100 eV of kinetic energy enters a uniform 1.0 mT field at 60° to the field. Use the NIST values for the electron mass, 9.109 × 10⁻³¹ kg, and the elementary charge, 1.602 × 10⁻¹⁹ C.
Speed: K = 100 × 1.602 × 10⁻¹⁹ = 1.602 × 10⁻¹⁷ J. v = √(2K/m) = 5.93 × 10⁶ m/s.
Components: v⊥ = v sin 60° = 5.14 × 10⁶ m/s; v∥ = v cos 60° = 2.97 × 10⁶ m/s.
Radius: r = (9.109 × 10⁻³¹ × 5.14 × 10⁶) / (1.602 × 10⁻¹⁹ × 1.0 × 10⁻³) = 0.0292 m, about 2.9 cm.
Period: T = 2π × 9.109 × 10⁻³¹ / (1.602 × 10⁻¹⁹ × 1.0 × 10⁻³) = 3.57 × 10⁻⁸ s, or 35.7 ns.
Pitch: p = 2.97 × 10⁶ × 3.57 × 10⁻⁸ = 0.106 m, about 10.6 cm.
If the field region is 0.50 m long, the electron makes 0.50 ÷ 0.106 ≈ 4.7 turns before leaving it.
Is there a shortcut for pitch and radius?
Divide the two formulas. Everything except the angle cancels:
p / r = 2π v∥ / v⊥ = 2π cot θ
Check with example 1: 2π cot 60° = 2π ÷ 1.732 = 3.63, and 0.106 ÷ 0.0292 = 3.63. The ratio does not depend on the mass, the charge, the speed or the field.
Worked example 2. At what angle does the pitch equal the radius? Set 2π cot θ = 1, so tan θ = 2π and θ ≈ 81°. The particle must enter almost perpendicular to the field, which makes sense: an angle close to 90° leaves little velocity along B.
How can a magnetic field focus a beam?
Worked example 3. Electrons accelerated through 500 V leave a point on the axis of a long solenoid, spreading at small angles to the field. Find the field that brings them back to a point on a screen 0.20 m away.
Speed: v = √(2 × 500 × 1.602 × 10⁻¹⁹ / 9.109 × 10⁻³¹) = 1.33 × 10⁷ m/s.
Every electron has the same period, T = 2πm/(eB), whatever its angle. After one period each is back on the axis line it started from.
For small angles cos θ ≈ 1, so all of them travel about the same distance along the axis in that time: p ≈ 2πmv/(eB).
Set p = 0.20 m: B = 2π × 9.109 × 10⁻³¹ × 1.33 × 10⁷ / (1.602 × 10⁻¹⁹ × 0.20) ≈ 2.4 × 10⁻³ T.
The beam spreads, then reconverges after each pitch. This is the idea behind magnetic focusing, and a favourite JEE Advanced set-up.
Which mistakes cost marks?
Using the full speed in r. Only v⊥ belongs there.
Swapping sin and cos. If θ is given from the field, v∥ = v cos θ. If the question gives the angle from the plane perpendicular to B, the roles swap.
Thinking the period depends on angle or speed. T = 2πm/(qB) only.
Confusing pitch with the helix diameter. Pitch is measured along the field.
Forgetting the sign of the charge. It reverses the sense of rotation, not the radius or pitch.
Using these formulas when E is along B. Then v∥ changes and the pitch grows or shrinks turn by turn.
The cyclotron and charged-particle motion guide covers circles and the cyclotron, and the right-hand rule post covers the direction of the force.
A practice routine
Redo example 1 for a proton with the same energy, and confirm that p/r is unchanged.
Find the angle at which the pitch is twice the radius.
Solve three past-paper helix questions, writing v⊥ and v∥ before anything else.
Sketch a helix and mark r, p and the direction of B, then check the sense of rotation for a positive and a negative charge.
When a helix will not form in your head, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than to read.
Start with the proton version of example 1 now, and predict the answer before you calculate it.
