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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 18

  • question_answer
    The electric field (in \[\text{N}{{\text{C}}^{-1}}\]) in an electromagnetic wave is given by \[E=50\sin \omega \left( t-\frac{x}{c} \right)\]. The energy stored in a cylinder of cross- section \[10\text{c}{{\text{m}}^{2}}\]and length 100 cm along the x-axis will be

    A) \[5.5\times {{60}^{-12}}J\]      

    B) \[1.1\times {{10}^{-11}}J\]

    C) \[2.2\times {{10}^{-11}}J\]      

    D) \[1.65\times {{10}^{-11}}J\]

    Correct Answer: B

    Solution :

    [b]: Energy contained in a cylinder U = average energy density\[\times \]volume \[=\frac{1}{2}{{\varepsilon }_{0}}E_{0}^{2}\times Al\] \[=\frac{1}{2}\times (8.85\times {{10}^{-12}})\times {{(50)}^{2}}\times (10\times {{10}^{-4}})\times 1\] \[=1.1\times {{10}^{-11}}J\]


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