KinexiaLEARN PHYSICS

Magnetic Force — Cyclotron Proton (Worked Example)

Medium
Problem

A proton moves at 1.0 × 10⁷ m/s perpendicular to a 1.5 T magnetic field in a cyclotron. What magnetic force acts on it? (q = 1.602 × 10⁻¹⁹ C)

Given

  • q = 1.602 × 10⁻¹⁹ C

  • v = 1.0 × 10⁷ m/s

  • B = 1.5 T

  • perpendicular

Find

The magnetic force F.

Principle

F = qvB (perpendicular): the field bends the proton into a circular path — the cyclotron's whole trick.

Formula

Solution

  1. Write the formula: F=qvBF = qvB

  2. Substitute: F=(1.602×10−19)(1.0×107)(1.5)F = (1.602 \times 10^{-19})(1.0 \times 10^7)(1.5)

  3. Multiply: F=2.40×10−12 NF = 2.40 \times 10^{-12}\ \mathrm{N}

2.40 × 10⁻¹² N, perpendicular to v and B

Unit check

C × m/s × T = C × m/s × N·s/(C·m) = N — correct.

What this means physically

2.40 × 10⁻¹² N — tiny, but on a near-massless proton it produces enormous centripetal acceleration, whirling it around to gain energy each lap.

Common mistake

Forgetting the perpendicular condition. F = qvB needs v ⊥ B; at other angles use F = qvB sinθ.