Question

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

• A. $\sqrt{\frac{V m}{q B^{2}}}$
• B. $\frac{2 V m}{q B^{2}}$
• C. $\sqrt{\frac{2 V m}{q}} \cdot\left(\frac{1}{B}\right)$
• D. $\sqrt{\frac{V m}{q}} \cdot\left(\frac{1}{B}\right)$

Answer 1

Answer: (C)

The speed of particle just before entering into magnetic field is $v_{0}$.
$\therefore q V=\frac{1}{2} m v_{0}^{2}$
$\Rightarrow v_{0}=\sqrt{\left(\frac{2 q V}{m}\right)}$
$\therefore r=\frac{m v_{0}}{q B}=\frac{m}{q B} \sqrt{\left(\frac{2 q V}{m}\right)}$
$=\frac{1}{B} \sqrt{\left(\frac{2 m V}{q}\right)}$