Magnetic Effects Of Current Question 273
Question: A deutron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to magnetic field $ \overrightarrow{B} $ . The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same $ \overrightarrow{B} $ is [CBSE PMT 1991]
Options:
A) 25 keV
B) 50 keV
C) 200 keV
D) 100 keV
Show Answer
Answer:
Correct Answer: D
Solution:
$ r=\frac{\sqrt{2mK}}{qB}\Rightarrow K\propto \frac{q^{2}}{m} $
$ \Rightarrow \frac{K_{p}}{K_{d}}={{( \frac{q_{p}}{q_{d}} )}^{2}}\times \frac{m_{d}}{m_{p}}={{( \frac{1}{1} )}^{2}}\times \frac{2}{1}=\frac{2}{1} $
$ \Rightarrow k_{p}=2\times 50=100\ keV. $
BETA
