= Uniformly magnetized cylinder
{title2=$\mathbf M=M\widehat{\mathbf z},\quad\mathbf K_{\rm b}=M\widehat{\boldsymbol\phi}$}
Prescribe constant <magnetization> $\mathbf M=M\widehat{\mathbf z}$, with $M>0$, inside a <circular cylinder> of <radius> $R$ and <length> $L$, surrounded by zero <magnetization>. In <cylindrical coordinates>, the outward <unit normal> on the curved <surface> is $\widehat{\mathbf r}$, so
$$
\mathbf J_{\rm b}=\mathbf0\quad\text{in the interior},\qquad
\mathbf K_{\rm b}=M\widehat{\mathbf z}\times\widehat{\mathbf r}
=M\widehat{\boldsymbol\phi}\quad\text{on the side}.
$$
The endcaps have outward <unit normals> $+\widehat{\mathbf z}$ and $-\widehat{\mathbf z}$. Their <cross products> with <magnetization> vanish, so $\mathbf K_{\rm b}=0$ on each endcap. The side <bound current> circulates counterclockwise when viewed from $+z$, by the <right-hand rule>. Zero endcap <surface current density> does not imply zero endcap <magnetic field>.
\Image[/magnetized-cylinder-bound-current-geometry.png]
{title=A uniformly magnetized <circular cylinder> is equivalent to a circumferential <bound current> sheet}
{description=Original prescribed-<magnetization> geometry. Blue arrows represent $\mathbf M$; orange arrows represent side <surface current density>, not independent wires. The end view looks from $+z$. The endcap <unit normals> are parallel or antiparallel to $\mathbf M$, so neither endcap carries <bound current>. The drawing specifies source geometry, not a uniform finite-cylinder <magnetic field>.}
For an original example, let $M=800\,\mathrm{A/m}$, $R=0.020\,\mathrm m$, and $L=0.10\,\mathrm m$. The <volume> and total <magnetic dipole moment> are
$$
V=\pi R^2L=4\pi\times10^{-5}\,\mathrm{m^3},\qquad
\mathbf m=MV\widehat{\mathbf z}
=0.032\pi\,\mathrm{A\,m^2}\widehat{\mathbf z}
\approx0.100531\,\mathrm{A\,m^2}\widehat{\mathbf z}.
$$
Check the result independently using the equivalent <bound current>: a side strip of axial <length> $dz$ forms a <current loop> with $dI=M\,dz$ and <magnetic dipole moment> $d\mathbf m=\pi R^2M\,dz\,\widehat{\mathbf z}$. Its <integral> over $L$ gives the same $M\pi R^2L\widehat{\mathbf z}$. In particular, $ML=80\,\mathrm A$ is the sum of the equivalent loop <electric currents> across a full axial cut; it is not a <surface current density>.
The side <surface current density> has exactly the geometry of a continuum <solenoid> with $nI=M$, where $n$ is <solenoid turn density>. In the infinite-cylinder idealization with no applied <magnetic field>, the <long-solenoid approximation> therefore becomes
$$
\mathbf B_{\rm in}=\mu_0M\widehat{\mathbf z},\qquad
\mathbf B_{\rm out}=\mathbf0,\qquad\mathbf H=\mathbf0.
$$
The <limit> is useful well away from the ends of a sufficiently long <circular cylinder>. A finite <circular cylinder> instead has fringing <magnetic field> and a <demagnetizing field>. For this prescribed uniform <magnetization>, integrate the <on-axis magnetic field of a circular current loop>, or substitute $nI=M$ in the <on-axis magnetic field of a finite solenoid>, to obtain the central <magnetic field>:
$$
B_z(0)=\mu_0M\frac{L}{\sqrt{L^2+4R^2}}.
$$
Our example has $L/(2R)=2.5$ and $B_z(0)/(\mu_0M)=5/\sqrt{29}\approx0.92848$, rather than unity. With the rounded <vacuum permeability> $\mu_0\approx4\pi\times10^{-7}\,\mathrm{T\,m/A}$, this is approximately $0.9334\,\mathrm{mT}$, compared with the infinite-cylinder value $1.0053\,\mathrm{mT}$. These are model calculations for a prescribed <magnetization>, not measured values or a solution for the self-consistent <magnetization> of an arbitrary material.
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