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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2005

  • question_answer
    A circular coil of radius 5 cm has 500 turns of a wire. The approximate value of the coefficient of self-induction of the coil will be

    A)  25 mH                                 

    B)  \[25\times {{10}^{-3}}mH\]

    C)  \[50\times {{10}^{-3}}mH\]                       

    D)  \[50\times {{10}^{-3}}mH\]

    Correct Answer: A

    Solution :

    Given:  \[r=5\times {{10}^{-2}}m,\] \[N=500\] \[\therefore \] Self-inductance                                 \[L=\frac{{{\mu }_{0}}\pi {{N}^{2}}r}{2}\]                                 \[=\frac{4\pi \times {{10}^{-7}}}{2}=\pi {{(500)}^{2}}\times 5\times {{10}^{-2}}\]                                 \[=2.5{{\pi }^{2}}\times {{10}^{-3}}\]                                 \[=25\times {{10}^{-3}}H\]                                 \[=25mH\]


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