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Question

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

zero

$\frac{mE^3}{qB^2}$

$\frac{mE^2}{qB^3}$

none of the above

Solution

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