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In a vacuum with no charge or current, the Maxwell equations areFor the proposed plane electromagnetic wave, $\nabla \cdot B=0$ givesThe Ampère-Maxwell equation givesand substituting this into Faraday's law gives the dispersion relationThus both fields have transverse polarization relative to $k$, and the corresponding real electric field is

For incidence in the positive $x$-direction, choose the incident fieldsThe perfect conductor requires the tangential electric field to vanish at $x=0$. The reflected wave therefore hasThis is normal reflection of an electromagnetic wave from a perfect conductor. The magnetic field in $x\le 0$ isso its tangential value at the surface is

For the stated magnetic vector potential, part a givesApplying the same identity once more and using the Ampère-Maxwell equation in the magnetostatic limit givesExpanding the Lorentz force density ${\mathbf{F}}_m=\mathbf{J}\times \mathbf{B}$, using $\nabla \varphi ={\mathbf{e}}_{\varphi }/r$ and the vector triple-product identity, separates its poloidal and azimuthal parts:where

By cylindrical symmetry the field is azimuthal and constant on a circle of radius $r$. The Ampère's law for a steady current givessoIn the plane $x=0$ with $y>0$, it points in the $-x$ direction.

Take the wire along the $x$-axis and let $r$ be perpendicular distance from it. Gauss's law and cylindrical symmetry giveFor an observer boosted parallel to the wire,The moving observer sees length contraction and hence line density ${\lambda }^{\prime }=\gamma \lambda$, together with current $I^{\prime }=-\gamma \lambda v$. The transformed magnetic field is exactlywith the direction prescribed by the current. This is Ampère's law for the current seen in the moving frame.

In vacuum, curl Faraday's law and use the Ampère-Maxwell equation:Since $\nabla \cdot \mathbf{B}=0$, the left side is $-{\nabla }^2\mathbf{B}$. Hence

Dot the vacuum Ampère-Maxwell equation with $\mathbf{E}$ and Faraday's law with $\mathbf{B}/{\mu }_0$. The vector identity
$\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot \nabla \times \mathbf{E}-\mathbf{E}\cdot \nabla \times \mathbf{B}$
gives the Poynting theoremThusis the Poynting vector, the electromagnetic energy-flux density.

Take the divergence of the Ampère-Maxwell equation. Since the divergence of a curl is zero, Gauss's law givesThuswhich is charge conservation from Maxwell equations. Integrating over a fixed volume and using the divergence theorem yieldsCharge in $V$ is conserved provided no current crosses $\partial V$. For total charge in all space, the corresponding assumption is sufficient decay of $J$ so that the flux at infinity vanishes.

In empty space the fields are divergence-free. Taking the curl of Faraday's law and usinggivesThe same calculation, starting from the Ampère-Maxwell equation, gives the corresponding equation for $B$. Thus every component satisfies the electromagnetic wave equation with speedIts agreement with the measured speed of light identifies light as an electromagnetic wave.

Dot the Ampère-Maxwell equation with $E$, dot Faraday's law with $B/{\mu }_0$, and use the stated vector identity. The result is the local balance lawHere $S$ is the Poynting vector. Integration over $V$ gives the Poynting theoremThe field energy decreases through outward electromagnetic energy flux and through work done on charges.

The magnetic field is $B=\nabla \times A$. The magnetostatic Ampère-Maxwell equation is $\nabla \times B={\mu }_{0}J$, so the curl of the curl identity givesA gauge transformation $A\mapsto A+\nabla \chi$ does not change $B$. We may therefore choose the Coulomb gauge $\nabla \cdot A=0$ by solving the Poisson equation ${\nabla }^2\chi =-\nabla \cdot A$. It follows component by component in a Cartesian coordinate system thatThe free-space Green function of the Laplacian then givesFor a localized steady current density, conservation of electric charge gives ${\nabla }^{\prime }\cdot J=0$; integration by parts confirms directly that this integral has zero divergence and hence obeys the Coulomb gauge.

Dot the Ampère-Maxwell equation with $\mathbf{E}$ and Faraday's law with $\mathbf{H}$:Adding and using the vector identitygivesThis is the local Poynting theorem before specializing the constitutive relations.

Substitution of the sinusoidal plane waves into Faraday's law givesThe source-free Ampère-Maxwell equation, with $\mathbf{H}=\mathbf{B}/\mu$ and $D_i={\epsilon }_{ij}{E}_j$, givesEliminating $\mathbf{b}=(\mathbf{k}\times \mathbf{e})/\omega$ yields the wave equation for a plane wave in an anisotropic dielectric:

Split the microscopic sources into free and bound parts and defineSubstitution of $\rho ={\rho }_{\mathrm{f}}-\nabla \cdot P$ into Gauss's law gives $\nabla \cdot D={\rho }_{\mathrm{f}}$. Similarly, using $J=J_{\mathrm{f}}+{\partial }_{t}P+\nabla \times M$ in the Ampère-Maxwell equation gives $\nabla \times H=J_{\mathrm{f}}+{\partial }_{t}D$. The homogeneous Maxwell equations are unchanged. Thus

Maxwell equations areTaking the divergence of the Ampère-Maxwell equation and using Gauss's law givesThereforebecause $J$ vanishes on the boundary. Componentwise integration by parts also giveswith the boundary term again zero.

In vacuum, curl Faraday's law and use $\nabla \cdot E=0$ and the Ampère-Maxwell equation to obtainThe plane wave solves this when $\omega =c|k|$ and $k\cdot E_0=0$. Faraday's law gives

For the stated wave, propagation is along $+x$ and polarization is $(0,1,1)/\sqrt{2}$. ThusThe Poynting vector is electromagnetic energy flux, and its average is the wave intensity.