RAJASTHAN ­ PET Rajasthan PET Solved Paper-2007

  • question_answer
    A beam of light travelling along z-axis is described by the electric field \[{{E}_{y}}=(600\,V/m)\sin \omega \left( t-\frac{x}{c} \right)\] then maximum magnetic force on a charge\[q=2e\] moving along y-axis with a speed of \[3.0\times {{10}^{7}}m/s\]is\[(e=1.6\times {{10}^{-19}}C)\]

    A)  \[19.2\times {{10}^{-17}}N\]    

    B)  \[1.92\times {{10}^{-17}}N\]

    C)  0.992 N         

    D)  None of these

    Correct Answer: B

    Solution :

     Maximum magnetic field is given by \[{{B}_{0}}=\frac{{{E}_{0}}}{c}\] Here,      \[{{E}_{0}}=600V/m\] \[c=3\times {{10}^{8}}m/s\] \[\therefore \] \[{{B}_{0}}=\frac{600}{3\times {{10}^{8}}}\] \[=2\times {{10}^{-6}}T\] Maximum magnetic force imposed on given charge is \[{{F}_{m}}=qv{{B}_{0}}\] \[=2ev{{B}_{0}}\] \[=2\times 1.6\times {{10}^{-19}}\times 3\times {{10}^{7}}\times 2\times {{10}^{-6}}\] \[{{F}_{m}}=1.92\times {{10}^{-17}}N\]


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