Question

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:

• A. \(\frac{m}{\pi}\)
• B. \(\frac{3m}{\pi}\)
• C. \(\frac{2m}{\pi}\)
• D. \(\frac{4m}{\pi}\)

Hint:

The same wire length is used; magnetic moment is I times area, so compare the areas of the square and the circle of equal perimeter.

Answer 1

The Correct Option is D

Let the wire length be L. Square: side \(a=\frac L4\), area \(a^2=\frac{L^2}{16}\), so \(m=I\frac{L^2}{16}\).
Circle: \(2\pi r=L\Rightarrow r=\frac{L}{2\pi}\), area \(=\pi r^2=\frac{L^2}{4\pi}\), so \(m'=I\frac{L^2}{4\pi}\).
\(\frac{m'}{m}=\frac{16}{4\pi}=\frac4\pi\Rightarrow m'=\frac{4m}{\pi}\).
Answer: (4) \(\frac{4m}{\pi}\)