RAJASTHAN PMT Rajasthan - PMT Solved Paper-2009

  • question_answer
    A particle of charge \[-16\,\times {{10}^{-18}}C\] moving with velocity \[10\,m{{s}^{-1}}\] along the \[x\]-axis enters a region where a magnetic field of induction B is along the y-axis and an electric field of magnitude \[\text{1}{{\text{0}}^{\text{4}}}\text{V/m}\] is along the negative z-axis. If the charged particle continues moving along the \[x\]-axis, the magnitude of B is

    A)  \[{{10}^{16}}\,Wb/{{m}^{2}}\]

    B)  \[{{10}^{5}}\,Wb/{{m}^{2}}\]

    C)  \[{{10}^{3}}\,Wb/{{m}^{2}}\]                    

    D)         \[{{10}^{-3}}\,Wb/{{m}^{2}}\]

    Correct Answer: C

    Solution :

    As particle is moving without deviation, therefore                 \[Eq=Bqv\] \[\therefore \]  \[B=\frac{E}{v}=\frac{{{10}^{4}}}{10}\]                                 \[={{10}^{3}}Wb/{{m}^{2}}\]


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