$9.2.16.$ Determine the magnetic field induction on the axis of a circuit whose magnetic moment is $M$, at large distances $h$ in cases where the circuit is a circle, square, or regular triangle.
For a circle, the magnetic moment $$ M = \pi IR^2 $$ Magnetic field at the ring axis (see 9.2.10) $$ B = \frac{\mu_0 I R^2}{2(R^2+h^2)^{3/2}} = \frac{\mu_0 M}{2\pi (R^2 + h^2)^{3/2}} \approx \frac{\mu_0 M}{2\pi h^3} $$ For large distances the shape of the contour is obviously not important, then $$ B = \frac{\mu_0 M}{2\pi h^3} $$ For the unbelievers, let us prove this: let us have a regular $n$ polygon, described around a circle of radius $r$, the side of the polygon $a$, its magnetic moment $$ M = \frac{1}{2}narI $$ Let's find the projection of the induction of the side onto the axis: $$ B_i = \int dB = \int\limits_0^a\frac{\mu_0 I dx}{4\pi (h^2+r^2)} \cdot \frac{r}{\sqrt{h^2+r^2}} $$ $$ B_i = \frac{\mu_0 Ir}{4\pi}\int\limits_0^a \frac{dx}{(h^2 + r^2)^{3/2}} $$ in approximation $r\ll h$ we get $$ B_i \approx \frac{\mu_0 Ir}{4\pi}\int\limits_0^a \frac{dx}{h^3} = \frac{\mu_0 Ira}{4\pi h^3} $$ Summarizing $B_i$, I get: $$ B = \frac{\mu_0 Iran}{4\pi h^3} = \frac{\mu_0 2M}{4\pi h^3} \Rightarrow \boxed{B= \frac{\mu_0 M}{2\pi h^3}} $$ And you didn't believe it!
$$B = \frac{\mu_0 M}{2\pi h^3}$$