A charge particle of charge $q$ and mass $m$ is accelerated through a potential diff. $V\, volts$. It enters a region of orthogonal magnetic field $B$. Then radius of its circular path will be
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$\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{m}(\mathrm{KE})}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{m}(\mathrm{qv})}}{\mathrm{qB}}$
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