A proton of speed $3.0\times 10^6,\text{m/s}$ enters a uniform magnetic field of magnitude $0.15,\text{T}$ at an angle of $45^{\circ }$ to the field direction. Find the radius of the circular component of its helical trajectory.

Step-by-step solution

1. Use $r=\frac{mv}{qBsin\theta }$ for the component of velocity perpendicular to $B$
2. Take $m=1.6726\times 10^{-27},\text{kg}$, $v=3.0\times 10^6,\text{m/s}$, $q=1.602\times 10^{-19},\text{C}$
3. Compute $sin45^{circ}=0.7071$
4. Calculate numerator $mv=5.0178\times 10^{-21},\text{kg⋅m/s}$
5. Compute denominator $qBsin\theta =1.602\times 10^{-19}\times 0.15\times 0.7071=1.698\times 10^{-20},\text{C⋅T}$
6. Divide: $r=5.0178\times 10^{-21}/1.698\times 10^{-20}\approx 0.2955,\text{m}$

Concept tested: Radius of charged particle in magnetic field

Published 13 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked