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12th Class Physics Magnetism Question Bank MCQ - Moving Charges and Magnetism

  • question_answer
    A charged panicle with charge 'q’ enters a region of constant, uniform and mutually orthogonal fields E B, with velocity v perpendicular to both E and B. The particle comes out without any change in magnitude or direction of its velocity. Then:

    A) \[\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{{{E}^{2}}}\]                  

    B) \[\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{{{B}^{2}}}\]

    C) \[\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{E}}{{{E}^{2}}}\]                  

    D) \[\overrightarrow{v}=\frac{\overrightarrow{B}\times \overrightarrow{E}}{{{B}^{2}}}\]

    Correct Answer: B

    Solution :

    (b) \[\overrightarrow{v}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{{{B}^{2}}}\] Force acting on a particle under both electric and magnetic field is \[\overrightarrow{F}=q\overrightarrow{E}+q(\overrightarrow{v}\times \overrightarrow{B})=0\](since neither direction nor magnitude changes)                                  Taking cross product with \[\overrightarrow{B}\] on both sides.                              \[0=q\overrightarrow{E}\times \overrightarrow{B}+q(\overrightarrow{v}\times \overrightarrow{B})\times \overrightarrow{B}\]             \[=q\overrightarrow{E}\times \overrightarrow{B}+q(\overrightarrow{v}.\overrightarrow{B})\overrightarrow{B}-\overrightarrow{v}\,{{B}^{2}}\] \[\overrightarrow{v}=\overrightarrow{E}\times \overrightarrow{B}/{{B}^{2}}(\because \overrightarrow{v}\,.\,\overrightarrow{B}=0)\]


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