A circular loop of radius $R$ carries a current $I$. The magnetic field at a point on the axis at a distance $x$ from the center is $B=\frac{{\mu }_{0}I{R}^2}{2(R^2+x^2{)}^{3/2}}$. If the loop is replaced by a square loop of the same perimeter carrying the same current, the magnetic field at the same axial distance $x$ is:

Solution

1. The perimeter of the circular loop is $2\pi R$, so the side of the square loop is $$\begin{array}{rl}a& =\frac{2\pi R}{4}\\ & =\frac{\pi R}{2}\end{array}$$.
2. The magnetic field on the axis of a square loop is $B_{sq}=\frac{{\mu }_{0}I{a}^2}{2\pi (x^2+a^2/4)\sqrt{x^2+a^2/2}}$.
3. Comparing the magnetic moments, the circular loop has $m_{circ}=I\pi R^2$ and the square loop has $$\begin{array}{rl}{m}_{sq}& =I{a}^2\\ & =I(\frac{\pi R}{2}{)}^2\\ & =I\frac{{\pi }^2{R}^2}{4}\end{array}$$.
4. Since ${\pi }^2<4\pi$, the magnetic moment of the circular loop is larger.
5. For large $x$, the field is proportional to the magnetic moment, so $B_{circ}>B_{sq}$. Hence the answer is (B).