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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2000

  • question_answer
    A charged particle travels along a straight line with a speed v in a region where both electric field\[\overrightarrow{E}\]and magnetic fields\[\overrightarrow{B}\]are present. It follows that:

    A) \[|\overrightarrow{E}|=v|\overrightarrow{B}|\]and the two fields are parallel

    B) \[|\overrightarrow{E}|=v|\overrightarrow{B}|\]and the two fields are perpendicular

    C) \[|\overrightarrow{B}|=v|\overrightarrow{E}|\]and the two fields are parallel

    D) \[|\overrightarrow{B}|=v|\overrightarrow{E}|\]and the two fields are perpendicular

    E)  \[|\overrightarrow{E}|=|\overrightarrow{B}|\]and the two fields are perpendicular

    Correct Answer: B

    Solution :

    Force due to magnetic field is \[F=qvB\text{ }sin\theta \] when         \[\theta =90{}^\circ \]      \[F=qvB\]                       ...(1) Force due to electric field E is      \[F=qE\]                        ...(2) Equating Eqs. (1) and (2), we get                 \[|\overrightarrow{E}|=v|\overrightarrow{B}|\]


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