Manipal Medical Manipal Medical Solved Paper-2014

  • question_answer
    A particle having charge 100 times that of an electron is revolving in a circular path of radius 0.8 m with one rotation per second. Magnetic field produced at centre of particle is

    A)  \[{{10}^{-17}}{{\mu }_{0}}\]                          

    B) \[{{10}^{-11}}{{\mu }_{0}}\]

    C) \[{{10}^{-7}}{{\mu }_{0}}\]                             

    D) \[{{10}^{-3}}{{\mu }_{0}}\]

    Correct Answer: A

    Solution :

    Here charge on particle \[q=ne=100e=100\times 1.6\times {{10}^{-19}}C\]               \[=1.6\times {{10}^{-17}}C\] Radius of circular path r = 0.8 m Time period T = 1 rotation/s The current associated with the particle is \[i=\frac{ch\arg e}{time}\]             \[=\frac{1.6\times {{10}^{-17}}}{1}=1.6\times {{10}^{-17}}A\] Magnetic field produced at the center of the coil is given by \[B={{\mu }_{0}}\times \frac{i}{2r}\]  \[={{\mu }_{0}}\times \frac{1.6\times {{10}^{-17}}}{2\times 0.8}\]   \[={{10}^{-17}}{{\mu }_{0}}\]


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