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.

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    <h3 id="back-link"><a href="/#9.2">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $9.2.16.$ 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.
    </p>

    <h3>Solution</h3>
    <p>
        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!

    <h4>Answer</h4>
    <p>
        $$B = \frac{\mu_0 M}{2\pi h^3}$$
    </p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
      Andrei Yersh<br>
    </p>


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