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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer
    The electric field of a plane electromagnetic wave is given by \[\vec{E}={{E}_{0}}\hat{i}\cos (kz)cos(\omega t)\] The corresponding magnetic field \[\vec{B}\]is then given by: [JEE Main 10-4-2019 Morning]

    A) \[\vec{B}=\frac{{{E}_{0}}}{C}\hat{j}\sin (kz)cos(\omega t)\]

    B) \[\vec{B}=\frac{{{E}_{0}}}{C}\hat{j}\sin (kz)sin(\omega t)\]

    C) \[\vec{B}=\frac{{{E}_{0}}}{C}\hat{k}\sin (kz)cos(\omega t)\]

    D) \[\vec{B}=\frac{{{E}_{0}}}{C}\hat{j}\sin (kz)sin(\omega t)\]

    Correct Answer: B

    Solution :

    \[\because \]\[\vec{E}\times \vec{B}||\vec{v}\] Given that wave is propagating along positive z-axis and \[\vec{E}\]along positive x-axis. Hence \[\vec{B}\]along y-axis. From Maxwell equation\[\vec{\nabla }\times \vec{E}=-\frac{\partial B}{\partial t}\] i.e.\[\frac{\partial E}{\partial Z}=-\frac{\partial B}{dt}\]and\[{{B}_{0}}=\frac{{{E}_{0}}}{C}\]


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