Given, Kinetic energy of proton, K=2MeV=2×106eV =2×106×1.6×10−19J =3.2×10−13J Magnetic field, B=2.5T and mass of proton, m=1.67×10−27kg Now, kinetic energy, K=
1
2
mv2 .......(i) Substitute given values in Eq. (i) ∴3.2×10−3=
1
2
×1.67×10−27×v2 or v2=
2×3.2×10−13
1.67×10−27
or v2=3.83×1014 or v=1.95×107m∕s Force applied on proton in a uniform magnetic field is given by F=qvB‌sin‌θ Here, θ=90°,(∵ Proton is moving perpendicular to a uniform magnetic field). ∴F=qvB[∵sin‌90°=1] or F=1.6×10−19×1.95×107×2.5 or F=7.8×10−12N or F≈8×10−12N