Question

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

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

Answer 1

Answer: (D)

For a charged particle orbiting in a circular path in a magnetic field
$\frac{m v^{2}}{r}=B v q$
$ \Rightarrow v=\frac{B q r}{m} $
$m v^{2}=B q v r $
$E_{k}=\frac{1}{2} m v^{2}=\frac{1}{2} B q v r=B q \frac{r}{2} \cdot \frac{B q r}{m}=\frac{B^{2} q^{2} r^{2}}{2 m}$
For deuteron, $E_{1}=\frac{B^{2} q^{2} \times r^{2}}{2 \times 2 m}$
For proton, $E_{2}=\frac{B^{2} q^{2} r^{2}}{2 m}$
$\frac{E_{1}}{E_{2}}=\frac{1}{2} $
$\Rightarrow \frac{50 keV }{E_{2}}=\frac{1}{2}$
$ \Rightarrow E_{2}=100 \,keV$