JEE Main & Advanced Physics Magnetic Effects of Current / करंट का चुंबकीय प्रभाव Question Bank Mock Test - Magnetics

  • question_answer
    An electric field acts along positive x-axis. A charged particle of charge q and mass m is released from origin and moves with velocity\[\vec{v}={{v}_{0}}\hat{j}\] under the action of electric' field and magnetic field, \[\vec{B}={{B}_{0}}\hat{i}\]. The velocity of particle becomes 2\[{{v}_{0}}\] after time\[\frac{\sqrt{3}m{{v}_{0}}}{\sqrt{2}q{{E}_{0}}}\]. Find the electric field.

    A) \[\frac{\sqrt{2}}{\sqrt{3}}{{E}_{0}}\hat{i}\]      

    B) \[\frac{\sqrt{3}}{\sqrt{2}}{{E}_{0}}\hat{i}\]

    C) \[\sqrt{3}{{E}_{0}}\hat{i}\]      

    D) \[\sqrt{2}{{E}_{0}}\hat{i}\]

    Correct Answer: D

    Solution :

    [d] Path of particle is helix with increasing pitch \[v={{({{v}_{x}}^{2}+{{v}_{y}}^{2}+{{v}_{z}}^{2})}^{1/2}}\] Here \[{{v}_{x}}^{2}={{\left( \frac{qE}{m}t \right)}^{2}}\]and \[{{v}_{y}}^{2}+{{v}_{z}}^{2}={{v}_{0}}^{2}\] Also \[v=2{{v}_{0}}\] \[{{(2{{v}_{0}})}^{2}}=\frac{{{q}^{2}}{{E}^{2}}{{t}^{2}}}{{{m}^{2}}}+{{v}_{0}}^{2}\] \[t=\frac{\sqrt{3}m{{v}_{0}}}{qE}=\frac{\sqrt{3}m{{v}_{0}}}{\sqrt{2}q{{E}_{0}}}given\] \[\vec{E}=\sqrt{2}{{E}_{0}}\hat{i}\]


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