CBSE 2024 · Region 5 · Set 1 · Q25 · 3 marks
An electron moving with a velocity $\displaystyle \vec{v}=\left(1.0 \times 10^{7} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}+\left(0.5 \times 10^{7} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{j}}$ enters a region of uniform magnetic field $\displaystyle \overrightarrow{\mathrm{B}}=(0.5 \mathrm{mT}) \hat{\mathrm{j}}$. Find the radius of the circular path described by it. While rotating; does the electron trace a linear path too? If so, calculate the linear distance covered by it during the period of one revolution.
Marking-scheme solution
Radius of circular path:
$\displaystyle r=\frac{m v_{x}}{e B}$
$\displaystyle r=\frac{9.1 \times 10^{-31} \times 1 \times 10^{7}}{1.6 \times 10^{-19} \times 0.5 \times 10^{-3}}$
$\displaystyle =11.38 \times 10^{-2} \mathrm{~m}$
Yes, it traces a linear path too.
Linear distance during period of one revolution:
$\displaystyle y=\frac{2 \pi m}{e B} \times v_{y}$
$\displaystyle =\frac{2 \times \pi \times 9.1 \times 10^{-31} \times 0.5 \times 10^{7}}{1.6 \times 10^{-19} \times 0.5 \times 10^{-3}}$
$\displaystyle =0.357 \mathrm{~m}=0.36 \mathrm{~m}$
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.