BCECE Engineering BCECE Engineering Solved Paper-2008

  • question_answer
    A beam of light travelling along \[x-\]axis is described by the electric field\[{{E}_{y}}=(600\,V{{m}^{-1}})\sin \omega (t-x/c)\] 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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