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BITSAT 2021

Exam: BITSAT
Year: 2021
Subject: Physics
Chapter: Magnetic Effects of Current

BITSAT 2021 · Physics · Magnetic Effects of Current

A proton moving with a velocity 3 × 10^ 5 m / s enters a magnetic...

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}

)

Options

A $2 \mathrm{ cm}$
B $0.5 \mathrm{ cm}$
C $0.02 \mathrm{ cm}$
D $1.25 \mathrm{ cm}$

Correct answer

B. $0.5 \mathrm{ cm}$

Step-by-step 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}

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