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JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Morning)

  • question_answer
    The electric fields of two plane electromagnetic plane waves in vacuum are given by \[{{\vec{E}}_{1}}={{E}_{0}}\hat{j}\cos \left( \omega t-kx \right)\] and \[{{\vec{E}}_{2}}={{E}_{0}}\hat{k}\cos \left( \omega t-ky \right)\] At t = 0, a particle of chcarge q is at origin with a velocity \[\vec{v}=0.8c\hat{j}\] (c is the speed of light in vaccum). The instantaneous force experienced by the particle is:     [JEE MAIN Held on 09-01-2020 Morning]

    A) \[{{E}_{0}}q\left( 0.4\hat{i}-3\hat{j}+0.8\hat{k} \right)\]

    B) \[{{E}_{0}}q\left( -\,0.8\hat{i}+\hat{j}+\hat{k} \right)\]

    C) \[{{E}_{0}}q\left( 0.8\hat{i}+\hat{j}+0.2\hat{k} \right)\]

    D) \[{{E}_{0}}q\left( 0.8\hat{i}-\hat{j}+0.4\hat{k} \right)\]

    Correct Answer: C

    Solution :

    [c] \[\vec{F}=q\vec{E}+q\left( \vec{V}\times \vec{B} \right)\] \[{{\vec{B}}_{1}}=\frac{{{E}_{0}}}{c}\hat{k}\cos \left( \omega t-kx \right)\And {{\vec{B}}_{2}}=\frac{{{E}_{0}}}{c}\hat{i}\cos \left( \omega t-ky \right)\] \[\vec{F}=q\left( {{{\vec{E}}}_{1}}+{{{\vec{E}}}_{2}} \right)+q\vec{v}\times \left( {{{\vec{B}}}_{1}}\times {{{\vec{B}}}_{2}} \right)\] If t = 0 and x = y = 0 \[\vec{F}=q{{E}_{0}}\left( 0.8\hat{i}+\hat{j}+0.2\hat{k} \right)\]


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