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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2019
  • Medium
  • MCQ

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

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Answer & Solution

Correct Answer

(C) \(\frac {4m}{\pi }\)

Step-by-step Solution

Detailed explanation

\(m = I{l^2}\)         \(2\pi r = 4l\)

\({{{\text{m}}^\prime } = \frac{{I4{\ell ^2}}}{\pi }}\)      \({{\text{r}} = \frac{{2\ell }}{\pi }}\)

\({{{\text{m}}^\prime } = \frac{{I4{\ell ^2}}}{\pi }}\)      \({\pi {r^2} = \frac{{\pi 4{\ell ^2}}}{{{\pi ^2}}} = \frac{{4{\ell ^2}}}{\pi }}\)  

\({{{\text{m}}^\prime } = \frac{4}{\pi }\,{\text{m}}}\)

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