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)

100 keV

Answer 2

For a charged particle orbiting in a circular
path in a magnetic field
$\frac{ m v^2}{r } = Bvq \Rightarrow v= \frac{ Bqr}{m }$
$mv^2= Bqvr$
$E_k = \frac{1}{2} m v^2 = \frac{1}{2}Bqvr = B q \frac{r}{2}\cdot \frac{Bqr}{m} = \frac{B^2q^2r^2}{2m}$
For deuteron, $E_1= \frac{ B^2q^2\times r^2 }{ 2 \times 2m}$
For proton, $E_2= \frac{ B^2q^2r^2 }{ 2m}$
$\frac{ E_1}{E_2} = \frac{1}{2} \Rightarrow \frac{50\,keV}{E_{2}}=\frac{1}{2} \Rightarrow E_2 = 100 \,keV$