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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
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The correct answer is:
$$ rac{4 m^2 E^2}{q B^3}$
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,
$\begin{aligned} & v=\frac{E}{B} \\ \therefore \quad L & =\frac{4 m^2 E^2}{q B^3}\end{aligned}$
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