Source: High school physics (Chinese)
Problem Sets:
Problem
A uniform magnetic field of magnitude $B = 1.5$ T points from east to west. A proton with kinetic energy $E = 5.0\times 10^{6}$ eV moves vertically downward through this field. The proton charge is $e = 1.6\times 10^{-19}$ C and the proton mass is $m = 1.7\times 10^{-27}$ kg. Find the magnitude and direction of the magnetic force on the proton.
Find the proton speed from its kinetic energy (non-relativistic since $E \ll m c^2 \approx 938$ MeV):
$$E = \tfrac{1}{2}m v^2 \implies v = \sqrt{\frac{2E}{m}}.$$Convert $E = 5.0\times 10^{6}\text{ eV} = (5.0\times 10^{6})(1.6\times 10^{-19})\text{ J} = 8.0\times 10^{-13}$ J.
$$v = \sqrt{\frac{2\times 8.0\times 10^{-13}}{1.7\times 10^{-27}}} = \sqrt{9.41\times 10^{14}} \approx 3.07\times 10^{7}\text{ m/s}.$$Since $\vec{v}$ (downward) is perpendicular to $\vec{B}$ (westward), $\sin\theta = 1$:
$$F = evB = (1.6\times 10^{-19})(3.07\times 10^{7})(1.5) \approx 7.4\times 10^{-12}\text{ N}.$$Direction: with $\vec{v} = -v\,\hat{z}$ (down) and $\vec{B} = -B\,\hat{x}$ (west), $\vec{v}\times\vec{B} = vB\,\hat{y}$ (north). The proton is positive, so $\vec{F}$ points north.