JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Morning)

  • question_answer
    If the magnetic field of a plane electromagnetic wave is given by (the speed of light \[=~\,3~\times ~{{10}^{8}}m/s)\] \[B\,=\,100\times {{10}^{-6}}\,\sin \,\left[ 2\pi \times 2\times {{10}^{15}}\left( t-\frac{x}{c} \right) \right]\]then the maximum electric field associated with it is - [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

    A) \[4.5\times {{10}^{4}}\,N/C\]               

    B) \[4\times {{10}^{4}}\,N/C\]

    C) \[6\times {{10}^{4}}\,N/C\]      

    D)   \[3\times {{10}^{4}}\,N/C\]

    Correct Answer: D

    Solution :

    \[B=100\times {{10}^{-6}}\,\,\sin \left[ 2\pi \times 2\times {{10}^{15}}\left( t-\frac{x}{c} \right) \right]\]Maximum magnetic field \[=\text{ }{{B}_{0}}=100\times {{10}^{-}}^{6}\] Tesla \[{{E}_{0}}=c{{B}_{0}}\] \[{{E}_{0}}=3\times {{10}^{8}}\times 100\times {{10}^{-}}^{6}\] \[=\,\,3\times {{10}^{4}}\,N/C\]


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