RAJASTHAN PMT Rajasthan - PMT Solved Paper-2003

  • question_answer
    A particle having charge 100 times that of an electron is revolving in a circular path of radius 0.8 m at the rate of 1 revolution/second. The magnetic field produced at the centre of orbit is

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

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

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

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

    Correct Answer: D

    Solution :

    Current due to circulating electron, \[i=\frac{q}{T}=nq\] \[=(1\,rev/\sec )\times 100\,e\] \[=100\times 1.6\times {{10}^{-19}}amp\] \[=1.6\times {{10}^{17}}amp\] \[\therefore \]Magnetic field at centre, \[B=\frac{{{\mu }_{0}}i}{2r}\] \[\frac{={{\mu }_{0}}\times 1.6\times {{10}^{-17}}}{2\times 0.8}\] \[-{{10}^{17}}{{\mu }_{0}}\] (Where \[\mu {{  }_{0}}=\] permeability of free space.)


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