A proton of charge $+e$ and mass $1.67\times 10^{-27},\text{kg}$ moves with speed $4.0\times 10^6,\text{m/s}$ making an angle of $70^{\circ }$ with a uniform magnetic field of magnitude $0.15,\text{T}$. What is the radius of the circular component of its helical trajectory?

Step-by-step solution

1. Find the component of velocity perpendicular to the field: $v_{\perp }=vsin\theta =(4.0\times 10^6,\text{m/s})sin70^{\circ }$. Calculate $sin70^{\circ }\approx 0.9397$, giving $v_{\perp }\approx 3.7588\times 10^6,\text{m/s}$.
2. Use the radius formula for circular motion in a magnetic field: $r=\frac{m{v}_{\perp }}{qB}$. Substituting $m=1.67\times 10^{-27},\text{kg}$, $q=1.6\times 10^{-19},\text{C}$, $B=0.15,\text{T}$, and $v_{\perp }=3.7588\times 10^6,\text{m/s}$.
3. Compute numerator: $m{v}_{\perp }=(1.67\times 10^{-27})(3.7588\times 10^6)\approx 6.279\times 10^{-21},\text{kg⋅m/s}$.
4. Compute denominator: $qB=(1.6\times 10^{-19})(0.15)=2.4\times 10^{-20},\text{C⋅T}$.
5. Divide: $r=\frac{6.279\times 10^{-21}}{2.4\times 10^{-20}}\approx 0.2616,\text{m}$.
6. Round to three significant figures: $r\approx 0.262,\text{m}$.

Concept tested: Helical trajectory radius for moving charge

Published 21 September 2026 · How this solution was prepared and checked