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

Answer 1

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

The speed of particle just before entering into magnetic field is $v_0$.
$\therefore qV=\frac{1}{2}m{v}_0^2$
$\Rightarrow v_0=\sqrt{\left(\frac{2qV}{m}\right)}$

Radius of the circular path is given by $r=\frac{m{v}_0}{qB}$

$\therefore r=\frac{m}{qB}\sqrt{\left(\frac{2qV}{m}\right)}=\frac{1}{B}\sqrt{\left(\frac{2mV}{q}\right)}$