VIT Engineering VIT Engineering Solved Paper-2011

  • question_answer
    A particle carrying a charge equal to 100 times the charge on an electron is rotating per second in a circular path of radius 0.8 m. The value of the magnetic field produced at the centre will be (\[{{\mu }_{0}}\]= permeability for vacuum)

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

    B)  \[{{10}^{-17}}{{\mu }_{0}}\]

    C)  \[{{10}^{-6}}{{\mu }_{0}}\]

    D)  \[{{10}^{-7}}{{\mu }_{0}}\]

    Correct Answer: B

    Solution :

    \[t=\frac{q}{t}=100\times e\] \[{{B}_{centre}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\pi i}{r}\] \[=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\pi \times 100e}{r}\] \[=\frac{{{\mu }_{0}}\times 200\times 1.6\times {{10}^{-19}}}{4\times 0.8}\] \[={{10}^{17}}{{\mu }_{0}}\]


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