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JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Evening)

  • question_answer
    A 27 mW laser beam has a cross-sectional area of\[10\text{ }m{{m}^{2}}\]. The magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space\[{{\varepsilon }_{0}}=9\times {{10}^{-12}}\] SI units, speed of light \[c=3\times {{10}^{8}}m/s\]] [JEE  Main Online Paper (Held on 11-jan-2019 Evening)]

    A) 2 kV/m                                     

    B) 0.7 kV/m

    C) 1 kV/m                         

    D)   1.4 kV/m

    Correct Answer: D

    Solution :

    Intensity of electromagnetic wave is given by \[I=\frac{Power\,(P)}{Area\,(A)}=\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}c\] \[E=\sqrt{\frac{2P}{A{{\varepsilon }_{0}}c}}=\sqrt{\frac{2\times 27\times {{10}^{-3}}}{{{10}^{-5}}\times 9\times {{10}^{-12}}\times 3\times {{10}^{8}}}}\] \[=\sqrt{2}\times {{10}^{3}}V/m=1.4V/m\]


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