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Haryana PMT Haryana PMT Solved Paper-2007

  • question_answer
    A beam of light travelling along x-axis is described by the electric field \[{{\text{E}}_{\text{y}}}\text{=(600V}{{\text{m}}^{\text{-1}}}\text{)sin }\!\!\omega\!\!\text{ (t-x/c)}\] then maximum magnetic force on a charge.\[\text{q=}{{\text{2}}^{\text{e}}}\text{,}\] 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,      \[{{E}_{0}}=600V{{m}^{-1}}\]                 \[c=3\times {{10}^{8}}m/s\] \[\therefore \]  \[{{B}_{0}}=\frac{600}{3\times {{10}^{8}}}\]                 \[=2\times {{10}^{-6}}T\] 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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