A positively charged particle $q$ of mass $m$ is passed through a velocity selector. It moves horizontally righward without deviation along the line $y=\frac{2 m v}{q B}$ with a speed $v$. The electric field is vertically downwards and magnetic field is into the plane of paper. Now, the electric field is. switched off at $t=0$. The angular momentum of the charged particle about origin $O$ at $t=\frac{\pi m}{q B}$ is

• A

  $\frac{m E^2}{q B^3}$
• Correct answer
  B

  $$rac{4 m^2 E^2}{q B^3}$
• C

  Zero
• D

  $\frac{m E^3}{q B^2}$

Solution

Radius of the circle, $r=\frac{m v}{B q}$ $\therefore$ Equation of line, $y=\frac{2 m v}{B q}=2 R$ Time pariod, $T=\frac{2 \pi m}{B q}$ $t=\frac{\pi m}{B q}=\frac{T}{2}$ $\therefore$ Angular momentum of charged particle about origin $\begin{aligned} L & =m v y+m v y=2 m v y \\ & =2 m v \times 2 R \\ & =4 m v \times \frac{m v}{B q}=\frac{4 m^2 v^2}{B q}\end{aligned}$ From velocity selector, $\begin{aligned} & v=\frac{E}{B} \\ \therefore \quad L & =\frac{4 m^2 E^2}{q B^3}\end{aligned}$

Hence, the correct answer is option: 'B'.