A thin, rigid insulating material ring of radius $R$ is uniformly charged with $Q$. What will be the magnitude and direction of the magnetic field at the center of the ring if we rotate the ring with an angular velocity $\omega$ about some symmetry axis lying in its plane?
For the small arc segment $\Delta s$ marked in Figure 1, the charge $\Delta q = (\Delta s / 2 R \pi) Q$ is distributed.

![](https://cdn.mathpix.com/cropped/2024_05_02_9f939c6e8ba8f6f7c97ag-1.jpg?height=694&width=569&top_left_y=222&top_left_x=759)

Figure 1

This represents a current of strength

$$
I=\frac{\Delta q}{T}=\frac{\Delta q \omega}{2 \pi}=\frac{\omega \Delta s \cdot Q}{4 R \pi^{2}}
$$

Considering the symmetric segment $\Delta s^{\prime}$ about the axis, a current of strength $J=2 I$ flows in the elementary spherical cap at a distance $r$ from the center. Calculate the magnetic field produced by such a circular current at the center of the ring. A segment of length $\Delta l$ in a conductor through which a current $J$ flows produces a magnetic field

$$
\Delta H^{\prime}=\frac{1}{4 \pi} \cdot J \cdot \frac{\Delta l \sin \alpha}{R^{2}}
$$

at a point a distance $R$ away, with the direction of the field being perpendicular to both the segment $\Delta l$ and the direction of $R$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_9f939c6e8ba8f6f7c97ag-1.jpg?height=610&width=466&top_left_y=1446&top_left_x=816)

Figure 2

$\alpha$ is the angle between $\Delta l$ and the direction of $R$, which in this case is $90^{\circ}$. The total magnetic field of the circular current is the sum of the fields from each segment. The components of the magnetic field perpendicular to the axis of the circular current cancel each other out, while the components parallel to the axis, $\Delta H^{\prime} \cdot \sin \varphi$, add up. Since the sum of the segments $\Delta l$ is the circumference of the circle,

$$
\Delta H=\frac{1}{4 \pi} \cdot J \cdot \frac{2 l \pi}{R^{2}} \cdot \sin \varphi
$$

It remains to sum the magnetic fields of the circular currents. Substituting $t=R \sin \varphi, J$, and $\Delta s=R$, and taking the limit as $\Delta l \rightarrow 0$, we get:

$$
H=\frac{\omega Q}{R \cdot 4 \pi^{2}} \int_{0}^{\pi} \sin ^{2} \varphi \mathrm{d} \varphi=\frac{\omega Q}{8 \pi R}
$$

The magnetic field strength $H$ is parallel to the axis of rotation, and for positive $Q$, its direction is the same as the direction from which the rotation is positive.