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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