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VMMC VMMC Medical Solved Paper-2015

  • question_answer
    A particle having charge q and mass m starts moving from the origin under the action of an electric field \[E={{E}_{0}}L\]and \[B={{B}_{0}}I\]with a velocity\[v={{v}_{0}}\hat{j}.\]The speed of particle will becomes \[\frac{\sqrt{5}}{2}{{v}_{0}}\]after a time

    A) \[\frac{m{{v}_{0}}}{qE}\]                                             

    B) \[\frac{m{{v}_{0}}}{2qE}\]

    C) \[\frac{\sqrt{3}m{{v}_{0}}}{2qE}\]                           

    D) \[\frac{\sqrt{5}m{{v}_{0}}}{2qE}\]

    Correct Answer: B

    Solution :

    The E and B are acting along X-axis and v is acting along Y-axis, i.e. perpendicular to both E and B. Therefore, the path of charged particle is a helix with increasing speed. The speed of particle at time t is   \[v=\sqrt{v_{x}^{2}+v_{y}^{2}}\] Here,\[{{v}_{y}}={{v}_{0}},{{v}_{x}}=\frac{dE}{m}t\] and\[v\frac{\sqrt{5}}{2}{{v}_{0}}\] Putting these values in Eq. (i), we get time,\[t=\frac{m{{v}_{0}}}{2qE}\]


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