Adva van szabályos háromszög, négyzet, szabályos ötszög, szabályos hatszög és kör alakú áramvezető. Mindegyiknek a területe egyenlő, mindegyikben ugyanolyan erősségú áram folyik. Melyiknek a középpontjában a legnagyobb az áram által létrehozott mágneses tér intenzitása?

Given are regular triangular, square, regular pentagonal, regular hexagonal, and circular conductors. Each has the same area, and the same strength current flows through each. Which one has the highest magnetic field intensity at its center due to the current?
First, let's express the magnetic field strength created by the current flowing in one side of a regular $n$-sided polygonal conductor at the center of the polygon.

![](https://cdn.mathpix.com/cropped/2024_05_02_4ae9d09066758ed2331eg-1.jpg?height=526&width=398&top_left_y=279&top_left_x=858)

According to the Biot-Savart law, the contribution of an element of the side is

$$
d H=k \frac{I \cdot d l \cdot \sin \vartheta}{r^{2}}
$$

its direction is perpendicular to the plane of the polygon, where $k$ is a constant depending on the unit choice. The total magnetic field strength created by the entire side is

$$
H_{0}=k I \int_{-l}^{+l} \frac{\sin \vartheta d l}{r^{2}}
$$

Since $\sin \vartheta=\cos \varphi, d l=r \cos \varphi d \varphi$ and $r=\frac{R}{\cos \varphi}, H_{0}$ can be written as:

$$
H_{0}=k I \int_{-\varphi_{0}}^{+\varphi_{0}} \frac{\cos \varphi}{R} d \varphi=\frac{2 k I}{R} \sin \varphi_{0}
$$

For an $n$-sided polygon, $\varphi=\pi / n$, and the direction of the magnetic field strength created by each side is the same, so the total magnetic field strength at the center of the polygon is

$$
H=\frac{2 n k I}{R} \sin (\pi / n)
$$

The area of the regular $n$-sided polygon is

$$
\begin{aligned}
T=n R l & =n R^{2} \operatorname{tg}(\pi / n), \quad \text { hence } \\
R & =\sqrt{\frac{T}{n \operatorname{tg}(\pi / n)}}
\end{aligned}
$$

Using this, the magnetic field strength is

$$
H=\frac{2 n k I}{\sqrt{T}} \sqrt{n \operatorname{tg}(\pi / n) \sin ^{2}(\pi / n)}=\frac{2 k I}{\sqrt{T}} \pi^{3 / 2} \sqrt{\left(\frac{n}{\pi}\right)^{3} \frac{\sin ^{3}(\pi / n)}{\cos (\pi / n)}}
$$

For given values of $n$, calculating the value of $H$, we find that the magnetic field strength at the center of the triangle is the largest $\left(\approx 11.85 \cdot \frac{k I}{\sqrt{T}}\right)$, and it decreases monotonically with increasing number of sides. (The numerical value of the coefficient is approximately 11.34 for a square, 11.20 for a regular pentagon, 11.17 for a regular hexagon, etc.) For a circle, we take the limit as $n \rightarrow \infty$, which, using the relation $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$, gives $H=\frac{2 k I}{\sqrt{T}} \pi^{3 / 2} \approx 11.12 \frac{k I}{\sqrt{T}}$.

Gál Péter (Bp., Fazekas M. Gyak. Gimn., IV. o. t)

Note. The differences are very small (within 6%), consistent with the fact that at a greater distance, any current loop can be replaced by a dipole, whose strength depends only on the current and the area of the loop.