Single Choice

A deuteron of kinetic energy $$50\ keV$$ 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$$25\ keV$$
B$$50\ keV$$
C$$200\ keV$$
D$$100\ keV$$
Correct Answer

Solution

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


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