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A circular loop of radius $R$ carrying electric current $I$ has, on its axis,where the sign of $\hat{z}$ is fixed by the right-hand rule. At its centre, $B(0)={\mu }_{0}I/(2R)$.

Put the loop in the $xy$ plane, centered at the origin, with positive electric current counterclockwise when viewed from $+z$. The right-hand rule then gives the central magnetic field along $+z$. Parameterize source positions by the angle $\varphi$ and observe at $(0,0,z)$:Here $R$ without bold type is the loop radius, whereas $\mathbf{R}$ is the source-to-observer displacement. Their cross product isThe denominator $|\mathbf{R}{|}^3=(R^2+z^2{)}^{3/2}$ is constant around the loop. The integrals of sine and cosine over a full turn vanish, while the axial contribution adds to $2\pi R^2$. Substitution into the Biot-Savart law proves the on-axis formula. Symmetry cancels transverse components; it does not cancel the axial components.

For an original one-turn example in the same vacuum approximation, choose $R=0.04\mathrm{m}$ and $I=2\mathrm{A}$. With ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$,At $z=0.03\mathrm{m}$, the source-to-observer distance is $0.05\mathrm{m}$, so $B(z)/B(0)=64/125=0.512$ and $B(z)\approx 16.08\mu \mathrm{T}$. At $z=-0.03\mathrm{m}$, the magnetic field has the same magnitude and still points along $+z$. The point has crossed the loop plane, but the loop's electric current has not reversed. Reversing the electric current reverses the magnetic field on both sides.

For $|z|\ll R$, a Taylor expansion gives $B(z)/B(0)=1-3z^2/(2R^2)+O(z^4/R^4)$: the center is a smooth maximum, with zero first derivative. For $|z|\gg R$,This is the axial magnetic dipole field of magnetic dipole moment $\mathbf{m}=I\pi R^2\hat{\mathbf{z}}$.^[ref] A compact closed loop has an inverse-cube axial far field, in contrast to the inverse-distance field of the infinite straight wire. The SI units of the exact loop formula are ${\mu }_{0}I{R}^2/(R^2+z^2{)}^{3/2}\sim {\mu }_{0}I/\text{length}$, hence tesla.

For $N$ coincident turns carrying the same electric current, the superposition principle multiplies the ideal result by $N$. A real coil has radial width and axial length: if those are not negligible, integrate or sum over its individual turns. This is why a single measured coil diameter and a turn count may not determine the full spatial magnetic field precisely. A collection of loops becomes a solenoid; integrating their axial contributions gives the on-axis magnetic field of a finite solenoid.

Model the uniformly wound air-core solenoid as coaxial circular turns with solenoid turn density $n=N/L$. Put its ends at $x^{\prime }=0,L$ and an internal axial point at $x\in [0,L]$. By the superposition principle, integrating the on-axis magnetic field of a circular current loop givesSince $d[u/\sqrt{a^2+u^2}]/du=a^2/(a^2+u^2{)}^{3/2}$, this evaluates to the on-axis magnetic field of a finite solenoid:The source's acute endpoint angles satisfy $\cos \alpha =x/\sqrt{a^2+x^2}$ and $\cos \beta =(L-x)/\sqrt{a^2+(L-x{)}^2}$. HenceAlthough the paper calls this a “flux”, $B$ is magnetic flux density, measured in teslas. Magnetic flux is an surface integral ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{A}$, measured in webers, and cannot be assigned to a point. For example, two differently sized small disks at the same point have the same limiting $B$ but different magnetic fluxes, approximately $B$ times their normal areas. The displayed plus sign uses the internal-point angle convention shown in the source. Outside the coil, the integral expression still works, but one must retain signed axial projections rather than adding two acute-angle cosines.

At an end, $x=0$, the on-axis magnetic field of a finite solenoid isAt its centre, $x=L/2$, it isThe ratio is thereforeFor a long solenoid, $B_{\mathrm{centre}}\simeq {\mu }_{0}NI/L$ and $B_{\mathrm{end}}\simeq {\mu }_{0}NI/(2L)$. For a very short solenoid, the ratio tends to one, providing a separate check: its centre and end become nearly coincident compared with its radius.

A solenoid is a coil of electrical conductor, usually wire wound into a helix around a circular cylinder. The same electric current $I$ flows through its successive turns. Each turn contributes a magnetic field; the superposition principle makes their axial contributions reinforce inside the coil. Curl the fingers of your right hand in the direction of the electric current around the turns: the thumb gives the interior magnetic field direction by the right-hand rule. Throughout this branch, $N$ is the total number of turns, $\ell$ is the coil length, and $n=N/\ell$ is the solenoid turn density. ^[ref].

The attraction of a long solenoid is a controllable, nearly uniform magnetic field over an interior working region: change the electric current to tune its strength, or reverse the electric current to reverse its direction. This makes it a useful setting for studying the Lorentz force, measuring magnetic flux, and constructing an inductor. Its geometry turns the many contributions from individual turns into the simple long-solenoid approximation. A real finite coil has end fields and leakage; its exterior magnetic field is not zero. The on-axis magnetic field of a finite solenoid below quantifies that difference.

Historically, the coil helped connect magnetic fields to electric currents. In his memoir on the mathematical theory of electrodynamic phenomena, Ampère modeled the interactions of magnets using the ends of solenoids in place of microscopic magnetic constituents. This was a route toward explaining magnetism by electric currents, rather than a modern microscopic account of magnetization. ^[ref]. The historical setting is documented by the ^[ref].

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.