Question

A charged particle of mass m and charge $q$ is accelerated through a potential difference of $V$ volts.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

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

Answer 1

The correct option is C $\sqrt{\left(\frac{2Vm}{q}\right)}\left(\frac{1}{B}\right)$

Kinetic energy of the charge $K=\frac{1}{2}m{v}^2=qV$

Radius of circular path $r=\frac{mv}{Bq}$

Or $r=\sqrt{\frac{m^2{v}^2}{B^2{q}^2}}=\sqrt{\frac{2Km}{q^2}}\left(\frac{1}{B}\right)$

$⟹$ $r=\left(\sqrt{\frac{2qVm}{q^2}}\right)\left(\frac{1}{B}\right)$ $=\left(\sqrt{\frac{2Vm}{q}}\right)\left(\frac{1}{B}\right)$