A 2 MeV proton is moving perpendicular to a uniform magnetic field of 2.5T. The force on the proton is

$\text{E}=\frac{1}{2}{\text{mv}}^2$

$3.2\times {10}^{-13}=\frac{1}{2}\times 1.67\times {10}^{-27}\times {\text{v}}^{\text{2}}$

$\frac{6.4\times {10}^{-13}}{1.67\times {10}^{-27}}={\text{v}}^{\text{2}}$

${\text{v}}^2=3.83\times {10}^{14}$

$\text{v}\cong 2\times {10}^7\text{m/s}$

$\text{F}=\text{qvB}$

$=\text{1}\text{.602}\times {\text{10}}^{-19}\times 2\times {10}^7\times 2.5$

$=8\times {10}^{-12}\text{N}$