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JEE Main & Advanced Physics EM Waves Question Bank Self Evaluation Test - Electromagnetic Waves

  • question_answer
    The magnetic field in the plane electromagnetic field is given by: \[{{B}_{y}}=2\times {{10}^{-7}}\sin (0.5\times {{10}^{3}}z+1.5\times {{10}^{11}}t)T\] The expression for the electric field may be given by

    A) \[{{E}_{y}}=2\times {{10}^{-7}}\sin (0.5\times {{10}^{3}}z+1.5\times {{10}^{11}}t)V/m\]

    B) \[{{E}_{x}}=2\times {{10}^{-7}}\sin (0.5\times {{10}^{3}}z+1.5\times {{10}^{11}}t)V/m\]

    C) \[{{E}_{y}}=60\sin (0.5\times {{10}^{3}}z+1.5\times {{10}^{11}}t)V/m\]

    D) \[{{E}_{x}}=60\sin (0.5\times {{10}^{3}}z+1.5\times {{10}^{11}}t)V/m\]

    Correct Answer: D

    Solution :

    [d] \[{{B}_{y}}=2\times {{10}^{-7}}\,\sin \,(0.5\times {{10}^{3}}z+1.5\times {{10}^{11}}t)T\] The electric vector is perpendicular to B as well as direction of propagation of electromagnetic wave. Therefore \[{{E}_{x}}\] has to be taken. Further, \[{{E}_{0}}={{B}_{0}}\times c\] \[=2\times {{10}^{-7}}\times 3\times {{10}^{8}}\,V/m\] \[{{E}_{0}}=2\times {{10}^{-7}}\times 3\times {{10}^{8}}=60\,V/m\] \[\therefore \] The corresponding value of the electric field is \[Ex=60\,\sin \,(0.5\times {{10}^{3}}z+1.5\times {{10}^{11}}t)V/m\]


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