Q 12-04-205JEE MainJEE Main 2019 (10 Apr, Shift 2)Easy
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:
Answer: (A) $\dfrac{4m}{\pi}$
Square of side $s$: $m = Is^2$. The same wire as a circle: $2\pi r = 4s$, so $r = \dfrac{2s}{\pi}$ and the area is $\pi r^2 = \dfrac{4s^2}{\pi}$.
$$m' = I\cdot\frac{4s^2}{\pi} = \frac{4m}{\pi}$$
Solution by Sreeraj P, M.Sc Physics