AMU Medical AMU Solved Paper-1997

  • question_answer An\[\alpha \]particle and a proton having same momentum enter into a region of uniform magnetic field and move in circular paths. The ratio of the radii of curvature of their paths,\[\frac{{{r}_{\alpha }}}{{{r}_{p}}}\], in the field is       

    A)  \[\frac{1}{2}\]                 

    B)  \[\frac{1}{4}\]

    C)  1                  

    D)  4

    Correct Answer: A

    Solution :

    : \[\alpha -\]particle \[{{=}_{2}}H{{e}^{4}}\] \[\therefore \]charge\[=2e\] Proton\[{{=}_{1}}{{H}^{1}}\] \[\therefore \] charge\[=e\] Momentum\[={{m}_{\alpha }}{{v}_{\alpha }}={{m}_{p}}{{v}_{p}}\] For circular motion in magnetic field, \[Bev=\frac{m{{v}^{2}}}{R}\] or \[BeR=mv\] \[\therefore \]\[B{{e}_{\alpha }}{{R}_{\alpha }}=B{{e}_{p}}{{R}_{p}}\]or \[\frac{{{R}_{\alpha }}}{{{R}_{p}}}=\frac{{{e}_{P}}}{{{e}_{\alpha }}}=\frac{e}{2e}=\frac{1}{2}\] \[\therefore \] \[\frac{{{R}_{\alpha }}}{{{R}_{p}}}=\frac{1}{2}\].


You need to login to perform this action.
You will be redirected in 3 sec spinner