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

Given data

Magnetic field $B=2.5 T$

Charge of the proton q=$1.6\times {10}^{-19}$

Mass of the proton m =$1.6\times {10}^{-27}$

Kinetic energy $=2 Mev$

Concepts used moving charges and magnetism

Explanation

We know that

$\frac{1}{2}m{v}^2=2MeV$

$½\times 1.6\times {10}^{-27}{v}^2=2\times {10}^6\times 1.6\times {10}^{-19}$

By solving we get

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

Force $F=Bqv$

$2.5\times 1.6\times {10}^{-19}\times 2\times {10}^7$

$F=8\times {10}^{-12}$ N