Q87 / 125 · BITSAT 2021 · Physics

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 $=10^{8} \mathrm{C} / \mathrm{kg}$ )

  • A
    $2 \mathrm{~cm}$
  • Correct answer
    B
    $0.5 \mathrm{~cm}$
  • C
    $0.02 \mathrm{~cm}$
  • D
    $1.25 \mathrm{~cm}$

Solution

$r=\frac{m v \sin \theta}{\mathrm{B} e}=\frac{3 \times 10^{5} \sin 30^{\circ}}{0.3 \times 10^{8}}$ $\frac{3 \times 10^{5} \times \frac{1}{2}}{3 \times 10^{7}}=0.5 \times 10^{-2}$ $m=0.5 \mathrm{~cm}$

Hence, the correct answer is option: 'B'.

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