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. $\frac{3m}{\pi }$
• B. $\frac{4m}{\pi }$
• C. $\frac{2m}{\pi }$
• D. $\frac{m}{\pi }$

Answer 1

Answer: (B)

$\frac{4m}{\pi }$

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