Q. A deutron of kinetic energy 50 keV is describing a circular orbit of radius 0.5m, is plane perpendicular to magnetic field $\overrightarrow{B}.$ The kinetic energy of a proton that describes a circular orbit of radius 0.5 m in the same plane with the same magnetic field $\overrightarrow{B}$ is
A) 200 keV
B) 50 keV
C) 100 keV
D) 25 keV
Explanation
(c) $r = \frac{\sqrt{2mE}}{Bq} = \frac{\sqrt{2m_{1}E_{1}}}{Bq}$
Or $E_{1} = \frac{mE}{m_{1}} = \frac{(2m_{1})}{m_{1}} \times 50\ $keV = 100 keV
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.