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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Afternoon)

  • question_answer
    A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be : [JEE Main 10-4-2019 Afternoon]

    A) \[\frac{3m}{\pi }\]                  

    B)   \[\frac{4m}{\pi }\]

    C) \[\frac{2m}{\pi }\]                              

    D) \[\frac{m}{\pi }\]

    Correct Answer: B

    Solution :

    \[m=NIA=1\times I\times {{a}^{2}}\]here a = side of square Now, \[4a=2\pi r\] \[r=\frac{2a}{\pi }\] For circular loop \[m'=1\times I\times \pi {{r}^{2}}\] \[=1\times I\times \pi \times {{\left( \frac{2a}{\pi } \right)}^{2}}\] \[m'=\frac{4m}{\pi }\]


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