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NEET AIPMT SOLVED PAPER MAINS 2010

  • question_answer
    The electric field of an electromagnetic wave in free space is given by \[\overset{\to }{\mathop{\mathbf{E}}}\,=10\cos ({{10}^{7}}t+kx)\mathbf{\hat{j}}v/m,\] where t and x are in seconds and metres respectively. It can be inferred that               
    (1) The wavelength \[\lambda \]is 188.4 m.
    (2) The wave number k is 0.33 rad/m.
    (3) The wave amplitude is 10 V/m.
    (4) The wave is propagating along. + x direction
    Which one of the following pairs of statements is correct ?

    A) (3) and (4)    

    B) (1) and (2)         

    C) (2) and (3)    

    D) (1) and (3)

    Correct Answer: D

    Solution :

    The electric field of electromagnetic wave \[\overset{\to }{\mathop{\mathbf{E}}}\,=10\cos ({{10}^{7}}t\pm kx)\,\mathbf{j}\] Amplitude =10 V/m \[\because \]     \[c=\frac{\omega }{k}\] \[\therefore \]  \[3\times {{10}^{8}}=\frac{{{10}^{7}}}{k}\] or\[k=\frac{1}{30}\] or\[\frac{2\pi }{\lambda }=\frac{1}{30}\]or\[\lambda =188.4\,\text{m}\] So, (1) and (3) option are correct.


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