JEE Main & Advanced JEE Main Paper (Held on 09-4-2019 Morning)

  • question_answer The magnetic field of a plane electromagnetic wave is given by : \[\vec{B}={{B}_{0}}\hat{i}[cos(kz-\omega t)]+{{B}_{1}}\hat{j}\cos (kz+\omega t)\] where \[{{B}_{0}}=3\times {{10}^{-5}}T\] and \[{{B}_{1}}=2\times {{10}^{-6}}T.\] The rms  value of the force experienced by a stationary charge \[Q={{10}^{-4}}C\]at \[z=0\]is closest to :             [JEE Main 9-4-2019 Morning]

    A) \[0.9\text{ }N\]                       

    B) \[0.1\text{ }N\]

    C) \[3\times {{10}^{2}}N\]                  

    D) \[0.6N\]

    Correct Answer: D

    Solution :

    Maximum Electric field E = \[{{\vec{E}}_{0}}=(3\times {{10}^{-5}})c\left( -\hat{j} \right)\] \[{{\vec{E}}_{1}}=(2\times {{10}^{-6}})c\left( -\hat{i} \right)\] Maximum force \[{{\vec{F}}_{net}}=q\vec{E}=qc\left( -3\times {{10}^{-5}}\hat{j}-2\times {{10}^{-6}}\hat{i} \right)\] \[{{\vec{F}}_{0\max }}={{10}^{-4}}\times 3\times {{10}^{8}}\sqrt{{{(3\times {{10}^{-5}})}^{2}}+{{(2\times {{10}^{-6}})}^{2}}}\]\[=0.9N\] \[{{F}_{rms}}=\frac{{{F}_{0}}}{\sqrt{2}}=0.6N\]                     (approx) Option


You need to login to perform this action.
You will be redirected in 3 sec spinner