Question

A deuteron of kinetic energy $50keV$ is describing a circular orbit of radius $0.5$ metre in a plane perpendicular to the 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. $25keV$
• B. $50keV$
• C. $200keV$
• D. $100keV$

Answer 1

Answer: (D)

$100keV$

For a charged particle orbiting in a circular path in a magnetic field
$\frac{m{v}^2}{r}=Bqv\Rightarrow v=\frac{Bqr}{m}$
or, $m{v}^2=Bqvr$
Also,
$E_K=\frac{1}{2}m{v}^2=\frac{1}{2}Bqvr=Bq\frac{r}{2}.\frac{Bqr}{m}=\frac{B^2{q}^2{r}^2}{2m}$
For deuteron, $E_1=\frac{B^2{q}^2{r}^2}{2\times 2m}$
For proton, $E_2=\frac{B^2{q}^2{r}^2}{2m}$
$\frac{E_1}{E_2}=\frac{1}{2}\Rightarrow \frac{50keV}{E_2}=\frac{1}{2}\Rightarrow E_2=100keV$.