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Near the axis of a solenoid, the magnetic vector potential $A_{\theta }=\rho B(z)/2$ givesThe radial fringe component enforces the solenoidal magnetic-field constraint. A sharp axial step in $B$ must retain this fringe as a distributional field; a purely axial field that jumps to zero violates Gauss's law for magnetism. The hard-edge model is a convenient limit for near-axis trajectories, not a literal monopole sheet.

For a long vacuum-core or air-core solenoid with fixed geometry and linear magnetic response, neglecting end effects gives self-inductanceHere $N$ is the total number of turns, $n=N/\ell$ is the solenoid turn density, and $A$ is the interior cross-sectional area. The reason for the square of $N$ is physical: more turns produce more magnetic field, and more turns link the resulting magnetic flux. Approximately equal magnetic flux $\Phi \approx BA$ through each turn gives magnetic flux linkage $\Lambda \approx N\Phi \approx {\mu }_0{N}^{2}AI/\ell$; division by the electric current yields $L$. ^[ref].

For the same $N=1000$, $\ell =0.50\mathrm{m}$, and $a=0.025\mathrm{m}$, the area is $A=\pi (0.025{)}^2{\mathrm{m}}^2\approx 1.9635\times 10^{-3}{\mathrm{m}}^2$. The long-coil estimate isAt $I=0.50\mathrm{A}$, each turn links approximately $2.4674\times 10^{-6}\mathrm{Wb}$ and the total magnetic flux linkage is approximately $2.4674\times 10^{-3}\mathrm{Wb}$. Their ratio to $I$ is the stated self-inductance. A real finite coil need not give precisely these values: the axial centre correction above is not by itself a correction formula for inductance, which depends on magnetic flux across all the turns.

At fixed $A$ and $\ell$, doubling $N$ quadruples $L$. At fixed solenoid turn density $n$ and area $A$, doubling $\ell$ doubles both $N$ and $L$. These are different changes to the geometry, so there is no contradiction between $L\propto 1/\ell$ at fixed $N$ and $L\propto \ell$ at fixed $n$. A magnetic core requires its own magnetic permeability model; nonlinear magnetization can make a constant-$L$ description inadequate.

The solenoid turn density $n$ counts turns per unit axial length in a uniformly wound solenoid. The total turn count $N$ is dimensionless, whereas $n$ has units ${\mathrm{m}}^{-1}$. A $1000$-turn coil of length $0.50\mathrm{m}$ has $n=2000{\mathrm{m}}^{-1}$; an axial interval of length $s=0.010\mathrm{m}$ then contains $ns=20$ turns. The effective circumferential surface current density in the closely spaced circular-turn model is $K=nI$, with units $A/m$. ^[ref].

For a single-layer helix, the axial advance per turn is approximately $1/n$. The circular-turn model neglects this advance and the associated axial component of electric current, as well as the detailed magnetic field close to individual wires. The number of turns around the whole coil is always $N=n\ell$ here; it must not be substituted for $n$ in $B={\mu }_{0}nI$.

The magnetic energy stored in a linear solenoid of constant self-inductance $L$ isin the long-solenoid approximation. The first formula is an inductor relation; the second identifies its storage with magnetic energy in the interior magnetic field, under the vacuum-core approximation. This agreement joins the electrical circuit description to the spatial magnetic field description. The expression $B^2/(2{\mu }_0)$ is the vacuum magnetic energy density. ^[ref].

A short derivation starts from the terminal electric potential difference $v_L=LdI/dt$, with positive electric current entering the terminal at higher potential. The absorbed power is $v_{L}I$, and integrating from zero electric current givesThe result describes recoverable storage in the inductor; Joule heating in the wire is a separate expenditure of energy. ^[ref].

For the long-coil example, $L\approx 4.9348\mathrm{mH}$ and $I=0.50\mathrm{A}$ give $U\approx 6.1685\times 10^{-4}\mathrm{J}=0.61685\mathrm{mJ}$. Substituting the ideal $B\approx 1.256637\mathrm{mT}$ and volume $A\ell \approx 9.8175\times 10^{-4}{\mathrm{m}}^3$ into $B^{2}A\ell /(2{\mu }_0)$ gives the same value. Doubling electric current quadruples stored magnetic energy. In a real coil, reducing electric current can return this energy to an electrical circuit, or release it as Joule heating in a closed RL circuit.

Prescribe constant magnetization $\mathbf{M}=M\hat{\mathbf{z}}$, 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 $\hat{\mathbf{r}}$, soThe endcaps have outward unit normals $+\hat{\mathbf{z}}$ and $-\hat{\mathbf{z}}$. Their cross products with magnetization vanish, so ${\mathbf{K}}_{\mathrm{b}}=0$ 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.

For an original example, let $M=800A/m$, $R=0.020\mathrm{m}$, and $L=0.10\mathrm{m}$. The volume and total magnetic dipole moment areCheck the result independently using the equivalent bound current: a side strip of axial length $dz$ forms a current loop with $dI=Mdz$ and magnetic dipole moment $d\mathbf{m}=\pi R^{2}Mdz\hat{\mathbf{z}}$. Its integral over $L$ gives the same $M\pi R^{2}L\hat{\mathbf{z}}$. 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 becomesThe 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:Our example has $L/(2R)=2.5$ and $B_z(0)/({\mu }_{0}M)=5/\sqrt{29}\approx 0.92848$, rather than unity. With the rounded vacuum permeability ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$, 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.

A long array of equal co-oriented vortex rings, spaced by $l\ll r$, approximates an azimuthal vortex sheet on a cylinder of radius $r$. The circulation per unit axial length is $\Gamma /l$. The conditions $\nabla \cdot \mathbf{u}=0$ and $\nabla \times \mathbf{u}=\mathbf{\omega }$ are mathematically analogous to the magnetostatic equations for a solenoid. Away from its ends, an array long compared with $r$ produces nearly uniform axial flow $\Gamma /l$ inside and weak flow outside. A finite array has return flow outside connecting its two ends. The uniform-core approximation fails close to individual vortex cores and near the ends, and the field requires a stated circulation orientation and background flow.