JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
              A plane electromagnetic wave having a frequency v = 23.9 GHz propagates along the positive z-direction in free space. The peak value of the electric field is 60 V/m. Which among the following is the acceptable magnetic field component in the electromagnetic wave? [JEE Main 12-4-2019 Afternoon]

    A) \[\vec{B}=2\times {{10}^{7}}\sin (0.5\times {{10}^{3}}z+1.5\times {{10}^{11}}t)\hat{i}\]

    B) \[\vec{B}=2\times {{10}^{-7}}\sin (1.5\times {{10}^{2}}x+0.5\times {{10}^{11}}t)\hat{j}\]

    C) \[\vec{B}=2\times {{10}^{-7}}\sin (0.5\times {{10}^{3}}z-1.5\times {{10}^{11}}t)\hat{i}\]

    D) \[\vec{B}=60\sin (0.5\times {{10}^{3}}x+1.5\times {{10}^{11}}t)\hat{k}\]

    Correct Answer: C

    Solution :

    Magnetic field when electromagnetic wave propagates in +z direction \[B={{B}_{0}}\sin (kz-\omega t)\] where \[{{B}_{0}}=\frac{60}{3\times {{10}^{8}}}=2\times {{10}^{-7}}\] \[k=\frac{2\pi }{\lambda }=0.5\times {{10}^{3}}\] \[\omega =2\pi f=1.5\times {{10}^{11}}\]


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