Question

A proton of energy $2MeV$ is moving perpendicular to uniform magnetic field of $2.5T$. The force on the proton is : ($M_p=1.6\times {10}^{-27}kg$ and $q_p=e=1.6\times {10}^{-19}C$ )

• A. $2.5\times {10}^{-16}N$
• B. $8\times {10}^{-11}N$
• C. $2.5\times {10}^{-11}N$
• D. $8\times {10}^{-12}N$

Answer 1

The correct option is D $8\times {10}^{-12}N$

$F=qvB$
$\frac{m{v}^2}{2}=K_e$
$v=\sqrt{\frac{2K_e}{m}}$
$\therefore f=q\times \sqrt{\frac{2K_e}{m}}\times B$
=$1.6\times {10}^{-19}\times \sqrt{\frac{2\times 2\times {10}^6\times 1.6\times {10}^{-19}}{1.6\times {10}^{-27}}}\times 2.5$
=$8\times {10}^{-12}N$