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JEE Main & Advanced JEE Main Online Paper (Held On 08 April 2018)

  • question_answer
    The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the loop is\[{{B}_{1}}\]. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre of the loop is \[{{B}_{2}}.\] The ratio \[\frac{{{B}_{1}}}{{{B}_{2}}}\] is: [JEE Main Online 08-04-2018]

    A)  \[\sqrt{2}\]                             

    B)  \[\frac{1}{\sqrt{2}}\]

    C)  2                                

    D)  \[\sqrt{3}\]

    Correct Answer: A

    Solution :

    \[I(\pi {{r}^{2}})=M,2M=I\pi {{(r)}^{2}}\]                                                 \[r=\sqrt{2}r\] \[{{B}_{1}}=\frac{{{\mu }_{o}}I}{2r}\] \[{{B}_{2}}=\frac{{{\mu }_{o}}I}{2\sqrt{2}r}\] \[\therefore \frac{{{B}_{1}}}{{{B}_{2}}}=\frac{\sqrt{2}}{1}\]


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