A charged particle of mass $m$ and charge $q$ is accelerated through a potential difference of $V$. It enters a region of uniform magnetic field $B$, which is directed perpendicular to the direction of motion of the particle. The particle will move on a circular path of radius given by -

The correct option is C $\sqrt{\frac{2Vm}{q}}\times \left(\frac{1}{B}\right)$
When a charged particle enters perpendicular to a uniform magnetic field, then it follows a circular path.

Its radius is given by,

$R=\frac{mv}{qB}=\frac{\sqrt{2mKE}}{qB}...\left(1\right)$

The given charged particle $q$ is accelerated through a potential difference of $V$.

$\Rightarrow KE=qV...\left(2\right)$

From equations $\left(1\right)$ and $\left(2\right)$,

$R=\frac{\sqrt{2mqV}}{qB}$

$\therefore R=\sqrt{\frac{2Vm}{q}}\times \frac{1}{B}$

Hence, option $\left(C\right)$ is the correct answer.

Why this Question? This question helps you to understand the concept of motion of a charged particle under the influence of uniform magnetic field.