JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Morning)

  • question_answer
    A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, \[\overrightarrow{E}=6.3\text{ }\overrightarrow{j}\text{ }V/m\]. The corresponding magnetic field B, at that point will be [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

    A) \[18.9\times {{10}^{-8}}\text{ }\widehat{k}T\]

    B) \[2.1\times {{10}^{-8}}\text{ }\widehat{k}T\]

    C) \[18.9\times {{10}^{8}}\text{ }\widehat{k}T\]   

    D) \[6.3\times {{10}^{-8}}\text{ }\widehat{k}T\]

    Correct Answer: B

    Solution :

    \[\overrightarrow{E}\,\,\,=\,\,\,6.3\,\,\overrightarrow{j}\,\,V/m\] \[\because \,\,\,\,\,\left| \overrightarrow{E} \right|\,\,\,\,\,=\,\,\,\,\left| \overrightarrow{B} \right|\,\,\,.\,\,\,C\] \[\Rightarrow \,\,\,\,\left| \overrightarrow{B} \right|\,\,\,\,=\,\,\,\frac{6.3}{3\times {{10}^{8}}}\,\,\,\,=\,\,\,2.1\times {{10}^{-8}}T\] Direction of wave is in direction of \[\overrightarrow{E}\text{ }\times \text{ }\overrightarrow{B}\] So, option is correct


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