JEE Main & Advanced Sample Paper JEE Main - Mock Test - 2

  • question_answer
    In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of \[2\times {{10}^{10}}\] Hz and amplitude 54 V \[{{\text{m}}^{-1}}\].

    A) The amplitude of oscillating magnetic field will be \[18\times {{10}^{-7}}\text{Wb}\,{{\text{m}}^{-2}}\].

    B) The amplitude of oscillating magnetic field will be \[18\times {{10}^{-8}}\text{Wb}\,{{\text{m}}^{-2}}\].

    C) The wavelength of electromagnetic wave is 1.5 m.

    D) The wavelength of electromagnetic wave is 1.5 cm.

    Correct Answer: D

    Solution :

    [d]: \[{{B}_{0}}=\frac{{{E}_{0}}}{c}=\frac{54}{3\times {{10}^{8}}}=18\times {{10}^{-8}}T\] \[\lambda =\frac{c}{\upsilon }=\frac{3\times {{10}^{8}}}{2\times {{10}^{10}}}=1.5\times {{10}^{-2}}m=1.5cm\]


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