Question

A $2MeV$ proton is moving perpendicular to a uniform magnetic field of $2.5T$. What is the force on the proton?

• A. $2.5\times {10}^{-10}N$
• B. $7.8\times {10}^{-12}N$
• C. $2.5\times {10}^{-11}N$
• D. $7.6\times {10}^{-11}N$

Answer 1

The correct option is B

$7.8\times {10}^{-12}N$

Step 1: Given Data

The energy of the proton, $KE=2MeV=2\times {10}^{6}eV$

Magnetic field $B=2.5T$

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

Mass of a proton $m=1.67\times {10}^{-27}kg$

The angle between the velocity and magnetic field $\theta =90°$

Step 2: Calculate the Velocity

We know that the kinetic energy of a particle is given as,

$KE=\frac{1}{2}m{v}^2=2\times {10}^{6}eV$

$\Rightarrow v^2=\frac{2\times 2\times {10}^6\times 1.6\times {10}^{-19}}{1.67\times {10}^{-27}}$

$\Rightarrow v=1.957\times {10}^{7}m/s$

Step 3: Calculate the Force

We know that the force on a particle moving in a magnetic field is given as,

$\overrightarrow{F}=q\left(\overrightarrow{v}\times \overrightarrow{B}\right)$

$\Rightarrow F=qvB\sin \theta$

Upon substituting the values we get,

$F=1.6\times {10}^{-19}\times 1.957\times {10}^7\times 2.5\times \sin 90°$

$=7.8\times {10}^{-12}N$

Hence, the correct answer is option (B).