Magnetic Effects Of Current Question 282
Question: An electron and a proton enter region of uniform magnetic field in a direction at right angles to the field with the same kinetic energy. They describe circular paths of radius $ r_{e} $ and $ r_{p} $ respectively. Then [Manipal MEE 1995]
Options:
A) $ r_{e}=r_{p} $
B) $ r_{e}<r_{p} $
C) $ r_{e}>r_{p} $
D) $ r_{e} $ may be less than or greater than $ r_{p} $ depending on the direction of the magnetic field
Show Answer
Answer:
Correct Answer: B
Solution:
$ r=\frac{\sqrt{2mK}}{qB}i.e.\ \ r\propto \frac{\sqrt{m}}{q} $ Here kinetic energy K and B are same.
$ \therefore \ \frac{r_{e}}{r_{p}}=\sqrt{\frac{m_{e}}{m_{p}}}\times \frac{q_{p}}{q_{e}}\Rightarrow \frac{r_{e}}{r_{p}}=\sqrt{\frac{m_{e}}{m_{p}}}\ \ (\because \ q_{e}=q_{p}) $ Since me < mp, therefore re < rp
BETA
