A proton of speed $v=2.0\times 10^6{\text{m s}}^{-1}$ enters a uniform magnetic field $B=0.20\text{T}$ at an angle $\theta =30^{\circ }$ to the field direction. Find the radius of the circular component of its helical trajectory.

Step-by-step solution

1. The component of velocity perpendicular to the field is $v_{\perp }=v\sin \theta$.
2. Calculate $v_{\perp }=(2.0\times 10^6)\sin 30^{\circ }=1.0\times 10^6{\text{m s}}^{-1}$.
3. Use the magnetic radius formula $r=\frac{m{v}_{\perp }}{qB}$, where $m=1.67\times 10^{-27}\text{kg}$ and $q=1.60\times 10^{-19}\text{C}$ for a proton.
4. Insert the numbers: $r=\frac{(1.67\times 10^{-27})(1.0\times 10^6)}{(1.60\times 10^{-19})(0.20)}$.
5. Simplify numerator: $1.67\times 10^{-21}$; denominator: $3.20\times 10^{-20}$.
6. Divide to obtain $r=5.22\times 10^{-2}\text{m}$; check arithmetic, the correct division yields $r=2.78\times 10^{-2}\text{m}$.

Concept tested: Radius of motion of a charged particle in a magnetic field

Published 11 September 2026 · How this solution was prepared and checked