A proton (charge $+e=1.6\times 10^{-19}\text{C}$) moves with speed $v=2.0\times 10^6{\text{m s}}^{-1}$ perpendicular to a uniform magnetic field of magnitude $B=0.80\text{T}$ directed into the page. Find the magnitude of the magnetic force on the proton and state the direction of the force using the right‑hand rule.

Step-by-step solution

1. The magnetic force on a moving charge is given by $F=qvB\sin \theta$. Since the velocity is perpendicular to the field, $\theta =90^{\circ }$ and $\sin \theta =1$.
2. Substitute the values: $F=(1.6\times 10^{-19}\text{C})(2.0\times 10^6{\text{m s}}^{-1})(0.80\text{T})=2.56\times 10^{-13}\text{N}$.
3. For direction, point the thumb of the right hand along the velocity (to the right), fingers into the page (direction of $B$); the palm then faces upward, indicating the force is upward (out of the page).

Concept tested: Magnetic force on a moving charge

Published 6 September 2026 · How this solution was prepared and checked