A proton with charge $+e=1.6\times 10^{-19},\text{C}$ moves at a speed of $4.0\times 10^5,{\text{m,s}}^{-1}$ perpendicular to a uniform magnetic field of magnitude $0.30,\text{T}$ directed towards the observer (along the $+z$–axis). Calculate the magnitude and direction of the magnetic force on the proton.

Step-by-step solution

1. Identify the velocity vector $vecv$ as $+x$–directed and the magnetic field $vecB$ as $+z$–directed.
2. Write the Lorentz force law for a positive charge: $\vec{F}=q,\vec{v}\times \vec{B}$.
3. Compute the magnitude: $F=qvB=1.6\times 10^{-19},\text{C}\times 4.0\times 10^5,{\text{m,s}}^{-1}\times 0.30,\text{T}=1.92\times 10^{-14},\text{N}$.
4. Use the right–hand rule: point fingers along $+x$, curl them towards $+z$; the thumb points in the $-y$ direction, so $vecF$ is along $-y$.

Concept tested: Magnetic force on moving charge

Published 19 September 2026 · How this solution was prepared and checked