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 proton 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

Answer: (B)

$\frac{\sqrt{2Em}}{qB}$

Kinetic energy of proton $E=\frac{1}{2}m{v}^2$

$⟹$ $mv=\sqrt{2mE}$ .........(1)

Let radius of the circular path in which the proton revolves be $r$

Using $Bqr=mv$

$Bqr=\sqrt{2mE}$ (using (1))

$⟹$ $r=\frac{\sqrt{2mE}}{qB}$