A proton with speed $2.5\times 10^6,\text{m/s}$ enters a uniform magnetic field of magnitude $0.25,\text{T}$ making an angle of $70^{\circ }$ with the field direction. Find the radius of the circular component of its helical path.

Step-by-step solution

1. Determine the component of velocity perpendicular to the field: $v_{\perp }=vsin\theta =2.5\times 10^6\times sin70^{\circ }$.
2. Compute $sin70^{\circ }\approx 0.9397$, giving $v_{\perp }\approx 2.349\times 10^6,\text{m/s}$.
3. Use the radius formula for circular motion in a magnetic field: $r=\frac{m{v}_{p}erp}{qB}$.
4. Insert $m=1.6726\times 10^{-27},\text{kg}$ (proton mass), $q=1.602\times 10^{-19},\text{C}$, $B=0.25,\text{T}$ and $v_{\perp }=2.349\times 10^6,\text{m/s}$.
5. Compute numerator: $m{v}_{\perp }=1.6726\times 10^{-27}\times 2.349\times 10^6\approx 3.93\times 10^{-21}$.
6. Compute denominator: $qB=1.602\times 10^{-19}\times 0.25=4.005\times 10^{-20}$.
7. Divide: $3.93\times 10^{-21}/4.005\times 10^{-20}\approx 0.0981,\text{m}$.
8. Round to two significant figures: $0.10,\text{m}$.

Concept tested: radius of helical motion in magnetic field

Published 15 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked