JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    The electric field of a plane electromagnetic wave is given by \[\vec{E}={{E}_{0}}\frac{\hat{i}+\hat{j}}{\sqrt{2}}\,\cos \,(kz+\omega t)\] At t = 0, a positively charged particle is at the point \[(x,y,z)\,=\,\left( 0,0,\,\frac{\pi }{k} \right)\]. If its instantaneous velocity at (t = 0) is\[{{v}_{0}}\,\hat{k},\], the force acting on it due to the wave is: [JEE MAIN Held on 07-01-2020 Evening]

    A) Antiparallel to \[\frac{\hat{i}+\hat{j}}{\sqrt{2}}\]

    B) Zero

    C) Parallel to \[\frac{\hat{i}+\hat{j}}{\sqrt{2}}\]

    D) Parallel to \[\hat{k}\]

    Correct Answer: A

    Solution :

    [a]  \[\vec{E}\] at t = 0 at \[z=\pi k\] is given by
    \[\vec{E}=\frac{{{E}_{0}}}{\sqrt{2}}\,(\hat{i}+\hat{j})\,cos\,[\pi ]\,=-\,\frac{{{E}_{0}}}{\sqrt{2}}\,(\hat{i}+\hat{j})\]
    \[\Rightarrow \] Force due to electric field, \[{{\vec{F}}_{E}}\] is antiparallel to \[\frac{\hat{i}+\hat{j}}{\sqrt{2}}\].
    \[{{\vec{F}}_{mag}}=q(\vec{v}\times \vec{B})\]
    \[\vec{B}\,\,(at\,t=0,\,z\,=\pi k)\]is \[\frac{{{B}_{0}}}{\sqrt{2}}\,(-\hat{i}+\hat{j})\]
    \[\Rightarrow \,\,\,{{\vec{F}}_{mag}}\,=q\,{{v}_{0}}\hat{k}\times \frac{{{B}_{0}}}{\sqrt{2}}\,(-\hat{i}+\hat{j})\]which is antiparallel
    to \[\frac{(\hat{i}+\hat{j})}{\sqrt{2}}\]
    \[\Rightarrow \,\,{{\vec{F}}_{net}}={{\vec{F}}_{E}}+{{\vec{F}}_{B}}\] is Antiparallel to \[\frac{\hat{i}+\hat{j}}{\sqrt{2}}\]

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