Gujarati
Hindi
4.Moving Charges and Magnetism
medium

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

A

$\sqrt {\frac{{Vm}}{{2q{B^2}}}} $

B

${\frac{{2Vm}}{{q{B^2}}}}$

C

$\sqrt {\frac{{2Vm}}{q}} \left( {\frac{1}{B}} \right)$

D

$\sqrt {\frac{{Vm}}{q}} \left( {\frac{1}{B}} \right)$

Solution

$\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{m}(\mathrm{KE})}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{m}(\mathrm{qv})}}{\mathrm{qB}}$

Standard 12
Physics

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