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. 25 ke V
• B. 50 ke V
• C. 200 ke V
• D. 100 ke V

Answer 1

Answer: (D)

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,
$E_K=\frac{1}{2}m{v}^2=\frac{1}{2}Bqvr$
$=Bq\frac{r}{2}\cdot \frac{Bqr}{m}=\frac{B^2{q}^2{r}^2}{2m}$
For deuteron, $E_1=\frac{B^2{q}^2{r}^2}{2\times 2m}$
For proton, $E_2=\frac{B^2{q}^2{r}^2}{2m}$
$\frac{E_1}{E_2}=\frac{1}{2}$
$\Rightarrow \frac{50keV}{E_2}=\frac{1}{2}$
$\Rightarrow E_2=100keV.$