Question

A deutron of kinetic energy $50\, keV$ is describing a circular orbit of radius $0.5 \,m$, in a plane perpendicular to magnetic field $B$. The kinetic energy of a proton that describes circular orbit of radius $0.5 \,m$ in the same plane with the same magnetic field is

• A. 200 keV
• B. 50 keV
• C. 100 keV
• D. 25 keV

Answer 1

Answer: (C)

In this case magnetic force provides necessary
centripetal force $i e, q v B=\frac{m v^{2}}{r}$
Radius of path $r=\frac{m v}{B q}=\frac{\sqrt{2 m E}}{q B}$
$r=\frac{\sqrt{2 m E}}{B q}=\frac{\sqrt{2 m_{1} E_{1}}}{B q}$
or $ E_{1}=\frac{m E}{m_{1}}$
$=\frac{\left(2 m_{1}\right)}{m_{1}} \times 50\, keV \left[\because m=2 m_{1}\right]$
$=100 \,keV$