A 2MeV proton is moving perpendicular to a uniform magnetic field of 2.5 T the force on the proton is (mass of proton $=1.6\times {10}^{-27}\text{kg}$ )

$\frac{1}{2}{\text{mv}}^{\text{2}}=2\left(\text{Mev}\right)$

$\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}=2\times \left(1.6\times {10}^{-18}\right)$

$\frac{1}{2}\times 1.6\times {10}^{-27}{\text{v}}^{\text{2}}=3.2\times {10}^{-13}$

${\text{v}}^2=4\times {10}^{14}$

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

$\text{F}=\text{qVB}=1.6\times {10}^{-19}\times 2\times {10}^7\times 2.5$

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