Question

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

Answer 1

Radius of the charge in a perpendicular magnetic field is $r=\frac{\sqrt{2mK}}{qB}\Rightarrow K\propto \frac{q^2}{m}$
Ratio of kinetic energy for proton and deutron is
$\Rightarrow \frac{K_p}{K_d}={\left(\frac{q_p}{q_d}\right)}^2\times \frac{m_d}{m_p}={\left(\frac{1}{1}\right)}^2\times \frac{2}{1}=\frac{2}{1}$
$\Rightarrow K_p=2\times 50=100\text{keV}$