Solution

A proton is accelerated by a potential difference of $5 \times 10^{6}$ volts. It then enters perpendicularly in a uniform magnetic field $B = 1.0\,\mathrm{Wb\,m^{-2}}$. Find the radius of curvature of the path of the proton. $m_{p} = 1.67 \times 10^{-27}\,\mathrm{kg}$, $e = 1.6 \times 10^{-19}\,\mathrm{C}$.