• # 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 $\frac{{{R}_{\alpha }}}{{{R}_{p}}}=\frac{{{e}_{P}}}{{{e}_{\alpha }}}=\frac{e}{2e}=\frac{1}{2}$ $\therefore$ $\frac{{{R}_{\alpha }}}{{{R}_{p}}}=\frac{1}{2}$.

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