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A proton moving with a velocity $3 \times 10^{5} \mathrm{~m} / \mathrm{s}$ enters a magnetic field of $0.3$ tesla at an angle of $30^{\circ}$ with the field. The radius of curvature of its path will be $(\mathrm{e} / \mathrm{m}$ for proton $=10^{8} \mathrm{C} / \mathrm{kg}$ )
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The correct answer is:
$0.5 \mathrm{~cm}$
$r=\frac{m v \sin \theta}{\mathrm{B} e}=\frac{3 \times 10^{5} \sin 30^{\circ}}{0.3 \times 10^{8}}$
$\frac{3 \times 10^{5} \times \frac{1}{2}}{3 \times 10^{7}}=0.5 \times 10^{-2}$
$m=0.5 \mathrm{~cm}$
$\frac{3 \times 10^{5} \times \frac{1}{2}}{3 \times 10^{7}}=0.5 \times 10^{-2}$
$m=0.5 \mathrm{~cm}$
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