BCECE Engineering BCECE Engineering Solved Paper-2010

  • question_answer
    In hydrogen atom the electron is making \[6.6\,\times {{10}^{15}}\,rev/s\] around the nucleus of radius \[0.53\,\overset{\text{o}}{\mathop{\text{A}}}\,\]. The magnetic field produced at the centre of the orbit is nearly

    A)  \[0.12\,Wb/{{m}^{2}}\]

    B)  \[1.2\,Wb/{{m}^{2}}\]

    C)  \[12\,Wb/{{m}^{2}}\]   

    D)         \[120\,Wb/{{m}^{2}}\]

    Correct Answer: C

    Solution :

    Magnetic field produced at the centre \[B=\frac{{{\mu }_{0}}i}{2r}\] \[=\frac{{{\mu }_{0}}}{2r}\frac{q}{T}=\frac{{{\mu }_{0}}}{2r}ef\]                 \[\therefore \]  \[B=\frac{4\pi \times {{10}^{-7}}}{2\times 0.53\times {{10}^{-10}}}\times 1.6\times {{10}^{-19}}\]                                 \[=\frac{2\pi }{0.53}\times 1.6\times 6.6\times {{10}^{-1}}\]                                 \[\simeq 12Wb/{{m}^{2}}\]


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