JEE Main & Advanced Sample Paper JEE Main - Mock Test - 41

  • question_answer
    A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance \[4.0\text{ }m\]from the source is

    A) \[64.7\text{ }V/m\]

    B) \[57.8\text{ }V/m\]      

    C) \[56.72\text{ }V/m\] 

    D) \[54.77\text{ }V/m\]

    Correct Answer: D

    Solution :

    [d] Intensity of EM wave is given by \[I=\frac{P}{4\pi {{R}^{2}}}={{v}_{av.}}c=\frac{1}{2}{{\varepsilon }_{0}}E_{0}^{2}\times c\] \[\Rightarrow \,\,\,\,\,\,\,{{E}_{0}}=\sqrt{\frac{P}{2\pi {{R}^{2}}{{\varepsilon }_{0}}C}}\] \[=\sqrt{\frac{800}{2\times 3.14\times {{(4)}^{2}}\times 8.85\times {{10}^{-12}}\times 3\times {{10}^{8}}}}\] \[=54.77\frac{V}{m}\]


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