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Write the jump from side $-$ to side $+$ as $[\mathbf{F}]={\mathbf{F}}_{+}-{\mathbf{F}}_{-}$. A thin rectangular loop in the static Faraday's law and a thin pillbox in Gauss's law for magnetism giveThus the tangential electric field and normal magnetic field are continuous in the plane's rest frame.

For the boost $\mathbf{v}=v\mathbf{n}$, the Lorentz transformation of electromagnetic fields givesThereforeMoreover $[\mathbf{E}]$ is purely normal, so $[{\mathbf{B}}_t^{\prime }]=\gamma [{\mathbf{B}}_t]$. Consequently the transformed tangential boundary condition isUsing the magnetic-field jump across a surface current, $\mathbf{n}\times [\mathbf{B}]={\mu }_0\mathbf{K}$, this can equivalently be writtenThe nonzero jump is consistent with Faraday's law because the magnetic-field discontinuity itself moves through Albert's frame.

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 moving circuit, Faraday's law states that the electromotive forceequals $-d{\Phi }_B/dt$, where ${\Phi }_B={\int }_S\mathbf{B}\cdot d\mathbf{S}$ is the magnetic flux through any spanning surface. Choose the upward normal. The horizontal square hasso Ohm's law gives the induced currentwith positive current taken counterclockwise from above.

On the edge $x=a$, $d\mathbf{\ell }=dy{\mathbf{e}}_y$. The magnetic force on a current-carrying wire is $d\mathbf{F}=I,d\mathbf{\ell }\times \mathbf{B}$, henceThe ratio of the horizontal to vertical magnitudes is $2|z|/a$, so the directed force makes angle ${\tan }^{-1}(2z/a)$ with the vertical, with the sign fixing the horizontal side.

The horizontal forces on opposite edges cancel, while all four vertical contributions add:This is an electromagnetic drag force opposing the motion, as required by Lenz's law. Including gravity givesAt the terminal velocity, $\ddot{z}=0$, and the constant downward speed has magnitude

The source-free Maxwell equations in a linear medium areHere the constitutive relations are $\mathbf{D}=\epsilon \mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$.

Substitution of the sinusoidal plane waves into the two divergence equations gives the transversality of an electromagnetic plane wave,Faraday's law and the Ampère-Maxwell equation give, respectively,Taking $\mathbf{k}\times$ the first equation and using transversality yieldsThus a nonzero plane electromagnetic wave in a linear medium must satisfyIn particular, ${\mathbf{E}}_0$, ${\mathbf{B}}_0$, and $\mathbf{k}$ are mutually orthogonal, with $|{\mathbf{B}}_0|=\sqrt{\mu \epsilon }|{\mathbf{E}}_0|$.

Take the interface to be $x=0$ and the plane of incidence to be the $xy$-plane. The incident, reflected, and transmitted wavevectors may be writtenThe phase matching at a planar wave interface requires one common angular frequency and one common tangential wavevector:Since $k_{\pm }=\omega \sqrt{{\epsilon }_{\pm }\mu }=n_{\pm }\omega /c$, this gives the Snell law for electromagnetic waves,

For the final polarization, take the signed electric amplitudes along $\hat{\mathbf{z}}$. The dielectric-interface boundary conditions require continuity of tangential $\mathbf{E}$ and $\mathbf{H}$. Sincetheir $z$- and $y$-components giveandEliminating $E_I$ yieldsUsing ${\theta }_R={\theta }_I$ and Snell's law,Therefore the reflected-to-transmitted transverse-electric amplitude ratio iswith the polarization directions chosen so that the displayed signed factor is nonnegative; otherwise its absolute value gives the magnitude ratio.

For a time-independent field, Faraday's law gives $\nabla \times \mathbf{E}=0$. Integrating around a thin rectangular loop crossing $x=0$ and shrinking its height gives continuity of the tangential component:A thin Gaussian pillbox and $\nabla \cdot \mathbf{D}=0$ give continuity of the normal electric displacement field:These are the electrostatic cases of the dielectric-interface boundary conditions.

In a region with no current, the Ampère-Maxwell equation and Faraday's law areThereforewhere the last line uses the divergence and curl of a cross product. Thus Poynting's theorem takes the formThe vector $\mathbf{S}$ is the Poynting vector.

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

The Maxwell equation $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$, integrated over a fixed spanning surface $S$, gives by Stokes theoremThus the electromotive force and magnetic flux obey Faraday's law

At time $t$, let the centre of the loop have coordinate $Y=d+vt$. With normal $+{\mathbf{e}}_x$, the magnetic flux isFaraday's law and Ohm's law therefore give a current of magnitudeThe original flux points into the plane and decreases in magnitude as the loop recedes. By Lenz's law, the induced current reinforces the into-plane field: it is clockwise when viewed from the $+x$ side.

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.

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.

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:

Faraday's law isChoose the positive circulation to be anticlockwise as viewed from $+z$. Since ${\Phi }_B=\pi B{r}^2$,

If $\dot{r}>0$, the current is clockwise and its field points in the $-z$ direction, reducing the field inside the loop. Since $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$, the force due to the imposed field is radially inward. If $\dot{r}<0$, the current is anticlockwise, its field points in the $+z$ direction and increases the field inside, while the force is radially outward. In both cases the magnetic force opposes the radial motion, in accordance with Lenz's law.

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.

The source-free Maxwell equations in a linear medium arewith $D=ϵE$ and $B=\mu H$.

Take complex plane wavesThe divergence equations giveand Faraday's law givesAmpère's law givesSubstituting the expression for $B_0$ and using
$k\times (k\times E_0)=-|k{|}^2{E}_0$ yieldsHence the plane electromagnetic wave in a linear medium hasand equivalently

Faraday's law is $\mathcal{E}={\oint }_C(E+v\times B)\cdot dl=-d{\Phi }_B/dt$. Here $I=-L(B\dot{x}+x\dot{B})/R$ (sign relative to the chosen circuit orientation), and the bar force is $ILB$. For $B=B_0/(1+t)$ this gives the displayed equation with $\alpha =B_0^2{L}^2/(mR)$. At $\alpha =1$, the independent solutions are $u$ and $u{e}^{1/u}$, $u=1+t$; the data give $x=x_{0}u{e}^{1/u-1}$. Thus $\dot{x}\to x_0/e$ and total mechanical work is $m{x}_0^2/(2e^2)$. The current circulates to reinforce the decaying $+z$ flux, and its force drives increasing area, both opposing the flux decrease as Lenz's law requires.