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RAJASTHAN PMT Rajasthan - PMT Solved Paper-1995

  • question_answer
    Radius of a current carrying circular coil is R at a distance\[x(x>>R)\] on the axis of circular coil the magnetic field produced B is proportional to:

    A) \[B\propto \frac{1}{{{x}^{3/2}}}\]                            

    B)        \[B\propto \frac{1}{{{x}^{2}}}\]                

    C)        \[B\propto \frac{1}{{{x}^{3}}}\]

    D)        \[B\propto \frac{1}{{{x}^{-1/2}}}\]

    Correct Answer: C

    Solution :

    Intensity of magnetic field \[B=\frac{{{\mu }_{0}}\,Ni\,{{a}^{2}}}{2{{({{a}^{2}}+{{x}^{2}})}^{3/2}}}\] a = radius of coil if \[x>>a\]                 \[\therefore \]  \[B\propto \frac{1}{{{x}^{3}}}\]


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