Question

A proton of mass $m$ and charge $q$ is moving in a plane with kinetic energy $E$. If there exists a uniform magnetic field $B$, perpendicular to the plane of the motion the portion will move in a circular path of radius

• A. $\frac{2Em}{qB}$
• B. $\frac{\sqrt{2Em}}{qB}$
• C. $\frac{\sqrt{Em}}{2qB}$
• D. $\sqrt{\frac{2Eq}{mB}}$

Answer 1

The correct option is B $\frac{\sqrt{2Em}}{qB}$

Since the proton moves in a circular path due to magnetic field, the centripetal force is provided to the proton by the same.

$⟹\frac{m{v}^2}{R}=qvB$

$⟹R=\frac{mv}{qB}$

Also $E=\frac{1}{2}m{v}^2$

$⟹v=\sqrt{\frac{2E}{m}}$

Thus $R=\frac{\sqrt{2Em}}{qB}$