JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Evening)

  • question_answer
    The energy associated with electric field is (\[{{U}_{E}}\]) and with magnetic field is (\[{{U}_{B}}\]) for an electromagnetic wave in free space. Then: [JEE Main Online Paper (Held On 09-Jan-2019 Evening]

    A) \[{{U}_{E}}<{{U}_{B}}\]                               

    B) \[{{U}_{E}}=\,\frac{{{U}_{B}}}{2}\]

    C) \[{{U}_{E}}={{U}_{B}}\]      

    D)               \[{{U}_{E}}>{{U}_{B}}\]

    Correct Answer: C

    Solution :

    Energy density in electric field \[{{U}_{E}}=\frac{1}{2}\,{{\varepsilon }_{0}}{{E}^{2}}\] Energy density in magnetic field \[{{U}_{B}}=\frac{1}{2}\,\frac{{{B}^{2}}}{{{\mu }_{0}}}\] \[\frac{{{U}_{E}}}{{{U}_{B}}}\,=\,\frac{{{\mu }_{0}}{{\varepsilon }_{0}}{{E}^{2}}}{{{B}^{2}}}\,=\,\frac{{{C}^{2}}}{{{C}^{2}}}\,=\,1\] \[C=\frac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}and\frac{E}{B}=C\]


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