DUMET Medical DUMET Medical Solved Paper-2007

  • question_answer
    A beam of light travelling along x-axis is described by the electric field \[{{E}_{y}}=600\text{ }V{{m}^{-1}}\text{ }sin~\,\omega \,\,(t-x/c)\]then maximum magnetic force on a charge\[q=2e\], moving along .y-axis with a speed of \[3.0\times {{10}^{7}}m{{s}^{-1}}\] is \[(e=1.6\times {{10}^{-19}}C)\]

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

    B)  \[1.92\times {{10}^{-17}}N\]

    C)  \[0.192\,\,N\]

    D)  none of these

    Correct Answer: B

    Solution :

     Maximum magnetic field is given by \[{{B}_{0}}=\frac{{{E}_{0}}}{c}\] Here, \[{{E}_{0}}=600\,\,V{{m}^{-1}}\] \[c=3\times {{10}^{8}}m/s\] \[\therefore \] \[{{B}_{0}}=\frac{600}{3\times {{10}^{8}}}\] \[=2\times {{10}^{-6}}T\] Maximum magnetic force imposed on given charge is \[{{F}_{m}}=qv{{B}_{0}}=2ev{{B}_{0}}\] \[=2\times 1.6\times {{10}^{-19}}\times 3\times {{10}^{7}}\times 2\times {{10}^{-6}}\] \[=1.92\times {{10}^{-17}}N\]


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