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?

Question

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? (A) 1 MeV (B) 2 MeV (C) 4 MeV (D) 0.5 MeV

Tardigrade Admin · Accepted Answer

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.,

Bq v = \frac{m v ^{2}}{r}

or

m v = Bq r

or

p = Bq r

Kinetic energy of proton

= \frac{p ^{2}}{2 m}

or

K E = \frac{B ^{2} q ^{2} r ^{2}}{2 m}

or

K E \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 M e V .

Q. A proton of energy 2MeV 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?

1 MeV

2 MeV

4 MeV

0.5 MeV

Q. A proton of energy $2 M e V$ 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?