Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    The ratio of the magnetic field at the centre of a current carrying coil of radius a and at a distance x from centre of the coil on its axial line is a

    A)  \[\frac{1}{\sqrt{2}}\]                                    

    B)  \[\sqrt{2}\]

    C)  \[\frac{1}{2\sqrt{2}}\]                                  

    D)  \[2\sqrt{2}\]

    Correct Answer: D

    Solution :

    Magnetic field at the centre      \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi nI}{a}\]       Magnetic field at a distance x from centre                 \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi nI{{a}^{2}}}{{{({{a}^{2}}+{{x}^{2}})}^{3/2}}}\] at            \[x=a\]                 \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\,\frac{2\pi nI}{{{2}^{3/2}}a}\]                 \[\frac{{{B}_{1}}}{{{B}_{2}}}={{2}^{3/2}}=2\sqrt{2}\]


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