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  4. A proton of energy 2 MeV is moving in a circular path in a magnetic field. What should be the energy of a deuteron, so that it also describes circular path of radius equal to that of the proton?

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Q. A proton of energy $2\, MeV$ is moving in a circular path in a magnetic field. What should be the energy of a deuteron, so that it also describes circular path of radius equal to that of the proton?

Bihar CECEBihar CECE 2003Moving Charges and Magnetism

Solution:

When a charged particle moves in a circular path in a magnetic field, then magnetic force provides the centripetal force to the particle.
i.e., $B q v=\frac{m v^{2}}{r}$
or $m v=B q r$
or $p=B q r$
Kinetic energy of proton $=\frac{p^{2}}{2 m}$
or $KE =\frac{B^{2} q^{2} r^{2}}{2 m}$
or $KE \propto \frac{1}{m}$
(for same circular path).
As mass of deuteron is twice that of proton, hence it should have half the energy of proton, i.e., $1\,MeV.$