An electron enters a uniform magnetic field of magnitude $B=0.30\text{T}$ with speed $v=1.5\times 10^6{\text{m s}}^{-1}$ at an angle of $45^{\circ }$ to the field direction. Find the radius of the circular component of its trajectory.

Step-by-step solution

1. Decompose the velocity into components parallel and perpendicular to $\vec{B}$: $v_{\perp }=v\sin 45^{\circ }=\frac{v}{\sqrt{2}}$.
2. Use the magnetic force for the perpendicular component: $F=e{v}_{\perp }B$.
3. Equate magnetic force to the required centripetal force: $e{v}_{\perp }B=\frac{m{v}_{\perp }^2}{r}$.
4. Cancel $v_{\perp }$ and solve for $r$: $r=\frac{m{v}_{\perp }}{eB}$.
5. Insert $m=9.11\times 10^{-31}\text{kg}$, $e=1.60\times 10^{-19}\text{C}$, $v_{\perp }=\frac{1.5\times 10^6}{\sqrt{2}}$, and $B=0.30\text{T}$.
6. Calculate $r=\frac{9.11\times 10^{-31}\times 1.06\times 10^6}{1.60\times 10^{-19}\times 0.30}=2.4\times 10^{-3}\text{m}$.

Concept tested: Radius of charged particle in magnetic field

Published 9 September 2026 · How this solution was prepared and checked