Magnetic Effects Of Current Question 265
Question: A proton of mass m and charge +e is moving in a circular orbit in a magnetic field with energy 1 MeV. What should be the energy of $ \alpha - $ particle (mass = 4m and charge = + 2e), so that it can revolve in the path of same radius [BHU 1997]
Options:
A) 1 MeV
B) 4 MeV
C) 2 MeV
D) 0.5 MeV
Show Answer
Answer:
Correct Answer: A
Solution:
$ r=\frac{\sqrt{2mK}}{qB} $
Þ $ K\propto \frac{q^{2}}{m} $
Þ $ \frac{K_{p}}{{K_{\alpha }}}={{( \frac{q_{p}}{{q_{\alpha }}} )}^{2}}\times \frac{{m_{\alpha }}}{m_{p}} $
Þ $ \frac{1}{{K_{\alpha }}}={{( \frac{q_{p}}{2q_{p}} )}^{2}}\times \frac{4m_{p}}{m_{p}}=1 $
Þ $ {K_{\alpha }}=1,MeV $ .
BETA
