JEE Main & Advanced Sample Paper JEE Main - Mock Test - 5

  • question_answer
    The rms value of the electric field of the light coming from the sun is \[720N{{C}^{-1}}\]. The average total energy density of the electromagnetic wave is

    A) \[3.3\times {{10}^{-3}}J{{m}^{-3}}\]

    B) \[4.58\times {{10}^{-6}}J{{m}^{-3}}\]

    C) \[6.37\times {{10}^{-9}}J{{m}^{-3}}\]

    D) \[81.35\times {{10}^{-12}}J{{m}^{-3}}\]

    Correct Answer: B

    Solution :

    [b]: Total average energy density of electro- magnetic wave is \[<u>=\frac{1}{2}{{\varepsilon }_{0}}E_{rms}^{2}+\frac{1}{2{{\mu }_{0}}}B_{rms}^{2}\] \[=\frac{1}{2}{{\varepsilon }_{0}}E_{rms}^{2}+\frac{1}{2{{\mu }_{0}}}\left( \frac{E_{rms}^{2}}{{{c}^{2}}} \right)\left( \because {{B}_{rms}}=\frac{{{E}_{rms}}}{c} \right)\] \[=\frac{1}{2}{{\varepsilon }_{0}}E_{rms}^{2}+\frac{1}{2{{\mu }_{0}}}E_{rms}^{2}{{\varepsilon }_{0}}{{\mu }_{0}}\] \[=\frac{1}{2}{{\varepsilon }_{0}}E_{rms}^{2}+\frac{1}{2}{{\varepsilon }_{0}}E_{rms}^{2}={{\varepsilon }_{0}}E_{rms}^{2}\] \[=(8.85\times {{10}^{-}}^{12})\times {{(720)}^{2}}=4.58\times {{10}^{-}}^{6}J\text{ }{{m}^{-}}^{3}.\]


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