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

Question

Dr. Abubacker · Accepted Answer

(b) $r = \ \sqrt{\frac{2m_{1}{E_{k}}_{1}}{Bq_{1}}}$ =$\sqrt{\frac{2m_{2}{E_{k}}_{2}}{Bq_{2}}}$
Or ${E_{k}}_{2} = \frac{m_{1}}{m_{2}}\frac{q_{2}}{q_{1}}\ {E_{k}}_{1}$
$= \frac{2m}{m} 	imes \frac{q}{q} 	imes 50$ keV = 100 keV

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

Explanation

Reviewed by Dr. Abubacker

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