A proton with a kinetic energy of 10 MeV moves at right angles to a uniform field of 5.0 × 10⁻⁵ T…

Step 1. The speed. $K = 10 \times 1.6 \times 10^{-13}\ \text{J} = 1.6 \times 10^{-12}$ J. We want $v$, so $v = \sqrt{\dfrac{2K}{m}} = \sqrt{\dfrac{2 \times 1.6 \times 10^{-12}\ \text{J}}{1.67 \times 10^{-27}\ \text{kg}}} = 4.38 \times 10^{7}\ \text{m/s}$.

Step 2. The radius. The top of $r = \dfrac{mv}{eB}$ is $mv = 1.67 \times 10^{-27}\ \text{kg} \times 4.38 \times 10^{7}\ \text{m/s} = 7.31 \times 10^{-20}\ \text{kg m/s}$. The bottom is $eB = 1.6 \times 10^{-19}\ \text{C} \times 5.0 \times 10^{-5}\ \text{T} = 8.0 \times 10^{-24}\ \text{C T}$. So $r = \dfrac{7.31 \times 10^{-20}\ \text{kg m/s}}{8.0 \times 10^{-24}\ \text{C T}} = 9.14 \times 10^{3}\ \text{m}$, that is $9.14$ km.

Step 3. The period. $T = \dfrac{2\pi m}{eB} = \dfrac{2\pi \times 1.67 \times 10^{-27}\ \text{kg}}{8.0 \times 10^{-24}\ \text{C T}} = 1.31 \times 10^{-3}\ \text{s}$, that is $1.31$ ms.