CET Karnataka Medical CET - Karnataka Medical Solved Paper-2001

  • 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 the centre of particle will be:

    A)  10-17\[{{\mu }_{0}}\]

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

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

    D)  \[{{10}^{-3}}{{\mu }_{0}}\]

    Correct Answer: A

    Solution :

     Charge on the particle =100C \[=100\times 1.6\times {{10}^{-}}^{19}=1.6\times {{10}^{-}}^{17}C\] Radius of circular path r = 0-8 m Time period t = 1 rotation/sec The current in the particle is \[i=\frac{ch\arg e}{time}=\frac{1.6\times {{10}^{-17}}}{1}=1.6\times {{10}^{-17}}\]amp The centre of the coil is given by \[B=\frac{{{\mu }_{0}}i}{r}=\frac{{{\mu }_{0}}\times 1.6\times {{10}^{-17}}}{2\times 0.8}={{10}^{-17}}{{\mu }_{0}}\]


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