A proton of mass $1.67\times10^{-27}\, kg$ and charge $1.6\times10^{-19}\, C$ is projected with a speed of $2\times10^6\, m/s$ at an angle of $60^o$ to the $X-$ axis. If a uniform magnetic field of $0.104\, tesla$ is applied along the $Y-$ axis, the path of the proton is
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By using $r=\frac{m v \sin \theta}{q B}$
$\Rightarrow r=\frac{1.67 \times 15^{27} \times 2 \times 10^{6} \times \sin 30^{\circ}}{1.6 \times 10^{-19} \times 0.104}$
$=0.1 m$
and it's time period
$T=\frac{2 \pi m}{q B}=\frac{2 \times \pi \times 9.1 \times 10^{-31}}{1.6 \times 10^{-19} \times 0.104}$
$=2 \pi \times 10^{-7} \mathrm{sec}$
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