Question

A $2\text{MeV}$ proton is moving perpendicular to a uniform magnetic field of $2.5\text{T}.$ The magnetic force on the proton is-
[Take mass of the proton, $m=1.6\times {10}^{-27}\text{kg}$]

• A. $0.8\times {10}^{-9}\text{N}$
• B. $0.8\times {10}^{-10}\text{N}$
• C. $8\times {10}^{-11}\text{N}$
• D. $8\times {10}^{-12}\text{N}$

Answer 1

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

Given, $B=2.5\text{T};E=2\text{MeV}$

Magnetic force on a charge particle is,

$F=Bqv\sin \theta .......\left(1\right)$

$E=\frac{p^2}{2m}\Rightarrow p=\sqrt{2mE}$

$\Rightarrow v=\frac{\sqrt{2mE}}{m}=\sqrt{\frac{2E}{m}}$

$=\sqrt{\frac{2\times 2\times {10}^6\times 1.6\times {10}^{-19}}{1.6\times {10}^{-27}}}$

$\therefore v=2\times {10}^7{\text{ms}}^{-1}$

Now, from (1) we get,

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

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.