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+### Statement
+
+$9.2.20.$ [Insert the problem statement]
+
+### Solution
+
+### Problem
+
+$9.2.20.$ A flat, thin disk of radius $R$ and thickness $h$ was cut from magnetized iron. The plane of the disk is perpendicular to the direction of magnetization. The magnetic moment per unit volume of the iron is $M$. Determine the magnetic field induction on the axis of the disk at a distance $l$ from its center.
+
+### Solution
+
+The magnetic moment per unit volume of a magnet $M$ is called its magnetization.
+
+Generally, the magnetic properties of magnets are determined by the so-called magnetization currents. $M = const$, which means magnetization currents flow only on the surface of the disk (all molecular currents inside the volume cancel each other out).
+
+To find these currents, we use the circulation theorem for the vector $\vec{M}$:
+$$\oint \vec{M} d\vec{l} = I$$
+
+where $I$ is the algebraic sum of the magnetization currents enclosed by a closed contour of length $l$.
+
+Consider a contour enclosing the end surface of the cylinder. In this case, $\vec{M} \perp \vec{l}$,
+$$I = \oint \vec{M} d\vec{l} = \vec{M} \cdot \vec{l} = 0 \quad (1)$$
+
+Now consider a contour enclosing the lateral surface of the cylinder. In this case, $\vec{M} \parallel \vec{l}$,
+$$I = \oint \vec{M} d\vec{h} = \vec{M} \cdot \vec{h} = Mh \quad (2)$$
+
+Thus, the magnetic field induction of our disk is equivalent to the field on the axis of a ring of radius $R$ carrying a current $I$, which is equal to
+$$B(l) = \frac{\mu_0 I R^2}{2(R^2 + l^2)^{3/2}} \quad (3)$$
+
+This formula was already obtained in problem 9.2.10.
+
+Substituting (2) into (3), we get
+$$B = \frac{\mu_0 M R^2 h}{2(R^2 + l^2)^{3/2}}$$
+
+### Answer
+$$B = \frac{\mu_0 M R^2 h}{2(R^2 + l^2)^{3/2}}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]