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Electric Field

A charge part...

Question

A charge particle of mass $m$ and charge $q$ entered int magnetic field $B$ normally after accelerating by potential difference $V$ . Calculate radius of its circular path.

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Solution

Kinetic energy of the particle $K . E = q V$
So, $m v = \sqrt{2 m K . E} = \sqrt{2 m q V}$
Radius of circular path $R = \frac{m v}{q B}$
$⟹ R = \frac{\sqrt{2 m q V}}{q B}$

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