RAJASTHAN PMT Rajasthan - PMT Solved Paper-2002

  • question_answer
    An electron is moving in magnetic field of \[16\,wb/{{m}^{2}}\]. Find the radium of its path:

    A)  \[4.47\times {{10}^{11}}\,per\,\sec \]                   

    B)  \[{{10}^{11}}\,per\,\sec \]

    C)  \[5.3\times {{10}^{-11}}\,per\,\sec \]                   

    D)  \[{{10}^{14}}\,per\,\sec \]

    Correct Answer: A

    Solution :

                Frequency \[f=\frac{Bq}{2\pi m}\] \[\begin{align}   & =\frac{16\times 1.6\times {{10}^{-19}}}{2\pi \times 9.1\times {{10}^{-31}}} \\  & =4.47\times {{10}^{+11}} \\ \end{align}\]Per sec


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