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 $$B = \frac{\mu_0 M R^2 h}{2(R^2 + l^2)^{3/2}}$$