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BCECE Engineering BCECE Engineering Solved Paper-2003

  • question_answer
    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

    Correct Answer: A

    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.,        \[Bqv=\frac{m{{v}^{2}}}{r}\] or            \[mv=Bqr\] or            \[p=Bqr\] Kinetic energy of proton \[=\frac{{{p}^{2}}}{2m}\]                 or            \[KE=\frac{{{B}^{2}}{{q}^{2}}{{r}^{2}}}{2m}\]                 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.,\[1\,MeV.\]


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