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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 39

  • question_answer
    The magnetic field at the centre of a circular current carrying coil of radius r is\[{{B}_{c}}\]. The magnetic field on its axis at a distance r from the centre is\[{{B}_{a}}\]. The value of \[{{B}_{c}}:{{B}_{a}}\] will be:

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

    B)        \[1:2\sqrt{2}\]

    C) \[2\sqrt{2}:1\] 

    D)        \[\sqrt{2}:1\]

    Correct Answer: C

    Solution :

    [c] \[\frac{{{B}_{c}}}{{{B}_{a}}}={{\left( \frac{{{R}^{2}}+{{x}^{2}}}{{{R}^{2}}} \right)}^{3/2}}={{\left( \frac{{{R}^{2}}+{{R}^{2}}}{{{R}^{2}}} \right)}^{3/2}}=2\sqrt{2}\]           


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