Magnetic Effects Of Current Question 267
Question: A proton and an $ \alpha - $ particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 $ \mu $ sec to make 5 revolutions, then the periodic time for the $ \alpha - $ particle would be [MP PET 1993]
Options:
A) 50 $ \mu $ sec
B) 25 $ \mu $ sec
C) 10 $ \mu $ sec
D) 5 $ \mu $ sec
Show Answer
Answer:
Correct Answer: C
Solution:
Time period of proton $ T_{p}=\frac{25}{5}=5\mu ,\sec $ By using $ T=\frac{2\pi ,m}{qB} $
$ \Rightarrow $ $ \frac{{T_{\alpha }}}{T_{p}}=\frac{{m_{\alpha }}}{m_{p}}\times \frac{q_{p}}{{q_{\alpha }}} $ $ =\frac{4m_{p}}{m_{p}}\times \frac{q_{p}}{2q_{p}} $
$ \Rightarrow , $ $ {T_{\alpha }}=2T_{p}=10\mu \sec . $
BETA
