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In magnetostatics, an ideal infinitely long solenoid is a cylindrical sheet of circumferential surface current density $K=nI$. It has negligible winding advance, no magnetic core, and no imposed external magnetic field. Choose positive electric current to produce magnetic field along $+z$. In this ideal model the interior magnetic field is uniform and axial, $B={\mu }_{0}nI$, and the exterior magnetic field is zero; ${\mu }_0$ is the vacuum permeability. For a finite, closely wound air-core coil, the same result is an approximation well inside the coil, several radii from either end. Air is treated as having the magnetic permeability of vacuum. ^[ref].

For a short proof, the translational and rotational symmetries of the ideal cylindrical current sheet, together with the Biot-Savart law, give an axial magnetic field. Ampère's law on rectangles wholly outside makes that axial exterior value independent of radius; the condition of no imposed magnetic field makes it zero at infinity and hence everywhere outside. Rectangles wholly inside similarly show that the interior value is independent of radius. Now choose an Amperian loop with axial length $s$, one long side inside and the other outside. The transverse sides are perpendicular to the magnetic field, so their contributions to the line integral vanish. The loop crosses $ns$ turns of the winding, givingHere $s$ is the integration-path length, not the whole coil length $\ell$. An Amperian loop surrounding both sides of the winding encloses equal and opposite electric currents; zero net circulation alone would not prove zero magnetic field at every exterior point.

For the $1000$-turn, $0.50\mathrm{m}$ coil with electric current $I=0.50\mathrm{A}$, the long-solenoid approximation givesThis illustrative calculation uses vacuum permeability ${\mu }_0\approx 1.256637\times 10^{-6}H/m$, a rounded value, not an exact constant. Since the 2019 redefinition of SI, vacuum permeability is determined experimentally.^[ref] Doubling $I$ doubles the ideal magnetic field; doubling $N$ at fixed $\ell$ does the same. Increasing $\ell$ while keeping $N$ fixed decreases $n$ and hence the ideal magnetic field.

For this radially layered coaxial cable, defineUse the solution regular at the axis with bounded exterior electric potential, so that the empty interior and exterior have the constant electric potentials fixed by their adjacent conducting cylinders.

There is no free charge in either electrical insulator or on their interface. Gauss's law and continuity of normal electric displacement field give $D_{\rho }=C/\rho$ in both layers. The electric potential difference fixes the constant becauseThe electric potential, including the empty regions bounded by the conducting cylinders, is thereforeBoth inner and outer Dirichlet boundary conditions hold, and the electric potential is continuous at the material interface, whereTaking the negative gradient of the electric potential yieldsThe radial electric field generally jumps at $\rho =c$, but $ϵE_{\rho }(c^{-})={ϵ}^{\prime }{E}_{\rho }(c^{+})$, as required by the absence of free surface charge density there. The conducting cylinders carry opposite linear charge densities $\lambda =2\pi C$ and $-\lambda$.

For an Amperian loop of radius $\rho$, the enclosed electric current is zero, $I$, or zero according as $\rho <a$, $a<\rho <b$, or $\rho >b$. The magnetostatic Ampère law therefore gives $H_{\phi }=I/(2\pi \rho )$ in the gap, and zero outside. Using the two magnetic permeabilities,The tangential magnetic field strength is continuous at the material interface, while the tangential magnetic field need not be. At each ideal conducting surface and at the material interface, the displayed electric field and magnetic field formulas specify one-sided limits; a single interface value is unnecessary when the limiting values differ.

Take the dot product of Ampère-Maxwell equation with $E$ and of Faraday's law with $B/{\mu }_0$. The divergence of a cross product identitythen gives the local Poynting theoremIntegration and the divergence theorem produce the stated equation. Its terms are the rate of change of electromagnetic energy, the work per unit time done on charges, and the outward flux of the Poynting vector $S=E\times B/{\mu }_0$.

Let the plate separation be $d$ and their radius be $R$, so $A=\pi R^2$. Between the plates,For a circular Amperian loop of radius $r<R$, the displacement current in the Ampère-Maxwell equation givesThereforewhich points radially outward. Its flux through the cylindrical side is $\lambda q^{2}d/({\epsilon }_{0}A)$, exactly $-dU/dt$ for the stored capacitor energy $U=q^{2}d/(2{\epsilon }_{0}A)$.

For a thin current-carrying cylindrical shell, a coaxial Amperian loop encloses zero current inside and the entire current outside. HenceMore generally $B(r)=0$ for $r<R$ and $B(r)={\mu }_0{I}_0/(2\pi r)$ for $r>R$. The tangential field discontinuity is supported by the surface current.

Let the wire lie along $z$, and observe at perpendicular distance $r$. In the Biot-Savart law, the azimuthal contributions from line elements add with magnitudeIndependently, cylindrical symmetry makes the field constant and azimuthal on a circular Amperian loop, so Ampère's law gives $2\pi rB={\mu }_0{I}_0$. Both yieldwith direction fixed by the right-hand rule.

For a charging parallel-plate capacitor of area $A$, neglect fringing. The field in the gap is $E=Q/({ϵ}_{0}A)$, so the displacement current through the gap isAn Amperian loop around the feed wire can bound a surface cutting the wire, which has conduction current $I$, or a bulging surface passing through the dielectric gap, which has no conduction current. Including displacement current makes both surfaces give the same magnetic circulation. It is also required by charge conservation: the divergence of the Ampère-Maxwell equation and Gauss's law yields $\nabla \cdot \mathbf{J}+\partial \rho /\partial t=0$, the continuity equation.

Let the wire radius be $a$, and take its electric current along $+z$. Cylindrical symmetry makes the magnetic flux density azimuthal. For a circular Amperian loop of radius $r$, Ampère's law gives $2\pi rB={\mu }_0{I}_{\mathrm{enclosed}}$. Uniform current density means $I_{\mathrm{enclosed}}=I{r}^2/a^2$ inside and $I$ outside. ConsequentlyIts direction follows the right-hand rule. The magnitude rises linearly from zero to ${\mu }_{0}I/(2\pi a)$ at the surface, is continuous there, and then falls as $1/r$.

Outside the wire, the Amperian loop encloses the entire electric current, soIts direction is again azimuthal, fixed by the right-hand rule. The inside and outside formulas join continuously at the maximum $B(a)={\mu }_{0}I/(2\pi a)$. The outer branch decays as $1/r$.

For a steady uniform current density in a nonmagnetic wire, the enclosed electric current at radius $r<a$ is $I_{\mathrm{enc}}=I{r}^2/a^2$. A circular Amperian loop and Ampère's law then giveThe magnetic field circles the wire according to the right-hand rule relative to the current direction. Its magnitude increases linearly from zero on the axis. The uniform-current assumption excludes skin effect.

In magnetostatics, Ampère's circuital law in vacuum isIt is especially useful when symmetry makes the tangential field constant on an Amperian loop. It gives $B={\mu }_{0}I/(2\pi r)$ for an infinite straight wire, $B\simeq {\mu }_{0}nI$ inside a long solenoid, and $B\simeq \mu NI/(2\pi r)$ in an ideal toroid. The long-solenoid and toroidal expressions neglect fringing.

For time-dependent fields, displacement current is essential:In matter, the corresponding macroscopic form is $\oint \mathbf{H}\cdot d\mathbf{\ell }=I_{\mathrm{free}}+d\int \mathbf{D}\cdot d\mathbf{A}/dt$. The enclosed electric current alone is sufficient only in the magnetostatic limit.

Take the electric current along $+z$, and use cylindrical coordinates. A uniform current in a straight homogeneous wire hasThe local Ohm's law $\mathbf{J}=\sigma \mathbf{E}$ gives the longitudinal electric fieldFor a circular Amperian loop of radius $r\le a$, the enclosed electric current is $I_{\mathrm{enc}}=I{r}^2/a^2$. The magnetostatic Ampère law gives $2\pi r{H}_{\varphi }=I_{\mathrm{enc}}$, henceSince $\hat{\mathbf{z}}\times \hat{\mathbf{\varphi }}=-\hat{\mathbf{r}}$, the Poynting vector isIt is zero at the centre and points radially inward elsewhere. Its units are $\mathrm{W}{\mathrm{m}}^{-2}$. These are the interior fields in the long-wire approximation, neglecting end and contact effects.

Let the unit normal $\mathbf{n}$ point from material 1 to material 2, and let their relative magnetic permeabilities be ${\mu }_1,{\mu }_2$, so ${\mathbf{B}}_i={\mu }_0{\mu }_i{\mathbf{H}}_i$ for linear isotropic media. Apply Gauss's law for magnetism to an infinitesimal Gaussian pillbox straddling the interface. The side contribution vanishes as its height tends to zero, leavingThe normal magnetic flux density is continuous.

Next use a narrow rectangular Amperian loop crossing the interface. The two short-side contributions vanish in the zero-height limit. The magnetostatic Ampère law for magnetic field intensity giveswhere ${\mathbf{K}}_f$ is any free surface current density. Thus, if there is no free surface current density on the interface, the tangential magnetic field intensity is continuous. In that case the component relations areThe distinction between free and bound electric currents matters: bound currents due to magnetization are already incorporated in the definition of magnetic field intensity. A change of magnetic permeability alone does not make the tangential magnetic flux density continuous.