9.2.19∗. a. Inside a large square circuit with current, many square micro-circuits with current are evenly distributed. The magnetic moment of each micro-circuit is $M_0$. Prove that at a distance much greater than the distance between the micro-contours, the induction of their magnetic field coincides with the induction of the magnetic field of a large contour, the magnetic moment of which is $nM_0$, where n is the number of micro-contours inside the large contour.
b. A thin square plate with dimensions $a × a × h (h ≪ a)$ is magnetized in a direction perpendicular to its plane. Magnetic field induction in the center of plate $B$. Determine the magnetic moment of the unit volume of the plate substance.
Solution
К задаче $9.2.19$
a) let current flow in each micro circuit $I = \frac{M_0}{\Delta S}$
This means that at large distances we can assume that we are dealing with a frame with the same current, the magnetic moment of such a frame (assuming the distribution of contours as in the problem $9.2.18$):
$$M = n \Delta S = nM_0$$
b)Find the equivalent current flowing along the edges of the plate: $$MhS=IS$$ $$I = Mh$$ Here, the field in the center is represented as a superposition of four fields from the frame contour. A field from one frame: $$dB_1 = \frac{\mu_0 Idlsin\varphi}{4\pi r^2}$$ $$dB_1 = \frac{\mu_0 I\frac{a/2d\varphi}{sin^2\varphi} sin\varphi}{4\pi \frac{(a/2)^2}{sin^2\varphi}}=\frac{\mu_0 Isin\varphi d\varphi}{4\pi a/2}$$ $$B_1 = 2\int_{0}^{\pi/4}\frac{\mu_0 Isin\varphi d\varphi}{4\pi a/2}$$ $$B_1 = \frac{\mu_0 I}{4\pi a/2}\cdot 2 cos(\pi/4)$$ From total 4: $$B = 4B_1 = \frac{2\sqrt{2}\mu_0I}{\pi a}$$ Substitute the expression for the current and get the answer: $$M = \frac{B\pi a}{\sqrt{8}\mu_0 h}$$
