Question: Q. 5. A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, acting perpendicular to their paths. Which of them will move in a circular path with higher frequency?
R [Delhi, O.D. 2018]
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Solution:
Ans. Electron.
(No explanation need to be given. If a student only writes the formula for frequency of charged particle i.e., $v_{c} \propto \frac{q}{m}$ award $1 / 2$ mark)
[CBSE Marking Scheme 2018]
Detailed Answer :
Electron moves in circular path with higher frequency.
$$ \begin{align*} \frac{m g^{2}}{r} & =q v B, r=\frac{m v}{q B} \ \omega & =\frac{v}{r}=\frac{q B}{m} \ \omega & =2 \pi f \Rightarrow \frac{q B}{m}=2 \pi f \Rightarrow f \propto \frac{1}{m} \end{align*} $$
Since $m_{e}<m_{p}$, therefore $f_{e}>f_{p}$
Thus, electron moves in circular path with higher frequency.
