Maxwell equations

The Maxwell equations are the field equations of classical electromagnetism. In SI units, in vacuum with prescribed charge density $\rho$ and current density $\mathbf{J}$, they areHere $\mathbf{E}$ is the electric field in volts per metre, $\mathbf{B}$ the magnetic field in teslas, $\rho$ in coulombs per cubic metre, and $\mathbf{J}$ in amperes per square metre. The constants are vacuum permittivity ${ϵ}_0$ and the vacuum value ${\mu }_0$ of magnetic permeability. “Vacuum” specifies the field laws without a material response; “source-free” additionally means $\rho =0$ and $\mathbf{J}=0$. Matter described by averaged fields instead requires Maxwell equations in matter and constitutive relations.

The first equation, Gauss's law, makes electric charge a source of electric flux. Gauss's law for magnetism sets the net magnetic flux through a closed surface to zero. The two curl equations, Faraday's law and the Ampère-Maxwell equation, couple changes of the electric field and magnetic field. Appropriate initial conditions and boundary conditions select a physical solution; the two divergence equations constrain the initial fields. ^[ref].

An approximation that keeps slow electromagnetic time variation while treating propagation across characteristic size $L$ as rapid compared with the variation time $1/\omega$. The dimensionless parameter $\omega L/c$ measures that separation. Which terms of the Maxwell equations can be dropped still depends on the physical scaling; small propagation delay alone does not justify deleting both displacement current and electromagnetic induction. For a slowly driven parallel-plate capacitor, a nearly uniform gap electric field can coexist with the magnetic field generated by its changing electric flux.

Vacuum Maxwell equations without sources are invariant under $E^{\prime }=E\cos \alpha +cB\sin \alpha$, $c{B}^{\prime }=cB\cos \alpha -E\sin \alpha$. At $\alpha =\pi /2$ this is $E^{\prime }=cB$, $B^{\prime }=-E/c$. Electric-conductor boundary conditions are not invariant under arbitrary duality rotations; the transformed fields must be tested separately against the wall conditions.

Use the negative curvature convention defined in curvature-sign convention in Killing derivative identities, and $C_{ab}=C^c{}_{acb}$. Covariant derivatives of a vector obey $[{\nabla }_b,{\nabla }^a]A^b=-C^a{}_b{A}^b$. Hence the flat-space equivalent expressions $\square A^a-{\partial }_b{\partial }^a{A}^b$ and $\square A^a-{\partial }^a{\partial }_b{A}^b$ acquire a difference $+C^a{}_b{A}^b$ under the comma-to-semicolon prescription. Covariantizing $F_{ab}=2{\nabla }_{[a}{A}_{b]}$ and its Maxwell equations gives $\square A^a-{\nabla }^a({\nabla }_b{A}^b)+C^a{}_b{A}^b={\mu }_0{J}^a$. This form respects gauge invariance. Dropping the Ricci term is generally not gauge invariant. The two expressions coincide in a Ricci-flat vacuum spacetime; vanishing matter alone with nonzero cosmological constant does not imply Ricci flatness.

The magnetic nonrelativistic limit neglects displacement current and uses ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$, $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$, $\nabla \cdot \mathbf{B}=0$. Under a Galilean transformation ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{v}t$, $t^{\prime }=t$, its consistent field transformations are ${\mathbf{B}}^{\prime }=\mathbf{B}$, ${\mathbf{J}}^{\prime }=\mathbf{J}$ and ${\mathbf{E}}^{\prime }=\mathbf{E}+\mathbf{v}\times \mathbf{B}$. Indeed ${\partial }_{t^{\prime }}={\partial }_t+\mathbf{v}\cdot \nabla$ and $\nabla \times (\mathbf{v}\times \mathbf{B})=-(\mathbf{v}\cdot \nabla )\mathbf{B}$, giving covariance of the induction equation. The full charge-current transformation has ${\mathbf{J}}^{\prime }=\mathbf{J}-{\rho }_e\mathbf{v}$; neglecting its convective correction belongs to the magnetic quasineutral approximation, not exact relativistic electrodynamics. With ${\mathbf{u}}^{\prime }=\mathbf{u}-\mathbf{v}$, the combination $\mathbf{E}+\mathbf{u}\times \mathbf{B}$ is invariant. Vanishing electric field in the fluid rest frame therefore gives the ideal magnetohydrodynamic induction equation.

Gauss's law for magnetism is $\nabla \cdot \mathbf{B}=0$. Its integral form says that the net magnetic flux through every closed surface vanishes, so the normal magnetic field is continuous across an ordinary interface.

Let a smooth magnetic field with axisymmetry have axial value $B_0(z)$ and no line singularity on its axis. Gauss's law for magnetism gives ${\partial }_r(r{B}_r)=-r{\partial }_z{B}_z$. Since $B_z(r,z)=B_0(z)+O(r^2)$, integration from the regular axis gives the displayed expansion. A falling axial field therefore implies outward field lines above a positive-source loop. In a current-free region, Ampère's law also gives $B_z(r,z)=B_0(z)-r^2{B}_0^{\prime \prime }(z)/4+O(r^4)$, which follows by equating ${\partial }_z{B}_r={\partial }_r{B}_z$.

For a smooth axisymmetric magnetic field, write $B_z(\rho ,z)=B_0(z)+O({\rho }^2)$. The absence of magnetic monopoles gives ${\rho }^{-1}{\partial }_{\rho }(\rho B_{\rho })=-B_0^{\prime }(z)+O({\rho }^2)$. Integrating from the axis and excluding a singular $1/\rho$ term proves the displayed relation. An axial field decreasing along $+z$ has an outward radial component, even when the field points purely axially on the axis itself.

For the electromagnetic two-form $F$, the source-free Maxwell equations are $dF=0$ and $d(\ast F)=0$. The first is the Bianchi identity and the second is the dynamical equation.

In the two-component spinor convention with primed symmetric gauge curvature self-dual, an Abelian Maxwell field is anti-self-dual when its primed symmetric gauge curvature two-component spinor vanishes. For a locally exact field $F=dA$, the Bianchi identity gives $dF=0$; the duality relation then also gives $d\ast F=0$. Thus it obeys the source-free Maxwell equations. In complexified Lorentzian signature the Hodge star eigenvalues are $\pm i$; the two-component spinor condition specifies which component is meant without imposing a Euclidean eigenvalue on that signature.

// Destination: general-relativity.bigb

In macroscopic matter with free charge density $\rho$ and free current density $J$,The electric displacement $D$ and magnetic field strength $H$ encode the material response.

These are the Maxwell equations after separating free and bound material sources.

Constitutive relations specify how matter responds to fields. A homogeneous isotropic linear electromagnetic medium obeys $\mathbf{D}=\epsilon \mathbf{E}$ and $\mathbf{B}=\mu \mathbf{H}$.

Electromagnetic permeability relates magnetic flux density to magnetic field strength through $\mathbf{B}=\mu \mathbf{H}$ in an isotropic linear medium.

Vacuum permeability is the magnetic permeability appearing in the vacuum Maxwell equations. In SI, it relates vacuum magnetic field strength to magnetic flux density through $\mathbf{B}={\mu }_0\mathbf{H}$ and obeys ${\mu }_0{\epsilon }_0{c}^2=1$, where $c$ is the speed of light and ${\epsilon }_0$ the vacuum permittivity.

A linear anisotropic medium has constitutive relations $D_i={\epsilon }_{ij}{E}_j$ and $B_i={\mu }_{ij}{H}_j$. Symmetric, time-independent tensors make the field-energy density a quadratic form.

Here permittivity is a tensor, so the electric displacement need not be parallel to the electric field.

Birefringence is the existence of different optical propagation constants for different eigenpolarizations in an anisotropic material. It provides the propagation-mode separation needed for adiabatic optical following in a twisted nematic.

Two different refractive indices can produce two displaced images, as in the calcite double-image experiment, or a phase retardation between overlapping components, as in a waveplate. Spatial separation concerns where energy flux travels; phase retardation concerns the relative wave phase. A specimen need not show two visibly separated images to have birefringence.

A known retarder inserted in a polarization microscope to compare phase delay with an unknown specimen. Parallel slow directions add signed retardations; a slow direction paired with a fast one subtracts them. General relative angles require composing polarization transformations rather than scalar addition.

A quartz retarder used to identify specimen fast and slow directions through changes of interference colour or compensation. Its own retardation and orientation must be specified before unique final colours can be predicted.

For a lossless uniaxial dielectric, $n_{\mathrm{ex}}^{-2}={\cos }^2\vartheta /n_o^2+{\sin }^2\vartheta /n_e^2$, where $\vartheta$ is the angle between the wave vector and optic axis. Along the axis $n_{\mathrm{ex}}=n_o$, while perpendicular propagation gives $n_{\mathrm{ex}}=n_e$.

Colour produced when wavelength-dependent interference selectively transmits or reflects part of white light. Between crossed polarizers, a uniform birefringent plate gives a factor ${\sin }^2(\pi \Gamma /\lambda )$, where $\Gamma$ is its optical path difference. The observed colour also depends on illumination and the spectral response of the observer.

A birefringent plate with retardation $\Gamma =t\Delta n$ gives $I/I_P={\sin }^2(2\alpha ){\sin }^2(\pi \Gamma /\lambda )$. At non-extinction azimuths, wavelengths satisfying $\Gamma =m\lambda$ are removed by a crossed analyser. At a principal-axis azimuth every wavelength is extinguished.

Birefringence produced by subwavelength structure rather than solely by intrinsic crystalline anisotropy. For ideal dielectric lamellae, continuity of tangential electric field gives ${ϵ}_{\parallel }=f{ϵ}_1+(1-f){ϵ}_2$, while continuity of normal electric displacement field gives ${ϵ}_{\perp }^{-1}=f/{ϵ}_1+(1-f)/{ϵ}_2$. The unequal effective permittivities give different principal refractive indices.

Linear birefringence means that two orthogonal linear polarizations have different refractive indices. Their different propagation phases produce phase retardation. In a lossless waveplate, with fast-axis angle $\theta$ and slow-minus-fast retardation $\delta$, the Jones matrix, up to a common phase, isusing time dependence $e^{-i\omega t}$. Circular birefringence instead has circular eigenpolarizations.

Place a clear calcite specimen on a transparency bearing printed words on an overhead projector. The projected readout is two copies of the lettering. Put a rotatable linear polarizer after the specimen: suitable orientations suppress one copy or the other. These are the actual apparatus and observations in the University of Iowa demonstration.^[ref]

For a useful control, remove the specimen while keeping the transparency, projector and screen fixed: the doubling should disappear. Replace it, hold it stationary and rotate only the polarizer; complementary brightness changes distinguish two linear polarizations from an accidental duplicate transparency. These are suggested controls, rather than additional reported measurements. Ideally the two copies obey complementary Malus's law curves, $I_o(\alpha )=I_{o,\max }{\cos }^2\alpha$ and $I_e(\alpha )=I_{e,\max }{\sin }^2\alpha$, with angle $\alpha$ measured from the exiting ordinary wave's linear polarization. Unequal maxima do not invalidate the test. Finite polarizer extinction, poor focus and specimen defects limit visibility; image separation alone does not determine either refractive index without the specimen geometry.

A transparent, nonmagnetic linear anisotropic dielectric is uniaxial when two principal permittivities agree. With the optic axis along $z$, its permittivity tensor has the displayed diagonal matrix form. Here $n_o$ and $n_e$ are principal refractive indices at a specified frequency; $n_e$ is not the extraordinary wave's index in every direction.^[ref]

The symmetry direction of a uniaxial optical medium. For a wavevector along it, the two refractive indices coincide: there is no birefringence for that propagation direction, even though $n_e\ne n_o$.

The ordinary wave's electric field is perpendicular to the plane containing the wavevector and optic axis. Its refractive index $n_o$ is direction independent at fixed frequency.

The extraordinary wave's electric displacement field lies in that plane and is perpendicular to the wavevector; its electric field need not be. For angle $\theta$ between the wavevector and optic axis,Thus $n_{\mathrm{ex}}(0)=n_o$ and $n_{\mathrm{ex}}(\pi /2)=n_e$.^[ref]

A short derivation starts from the existing plane wave in an anisotropic dielectric equation. Put $\mathbf{k}=(k_x,0,k_z)$ and $k_0=\omega /c$, with angular frequency $\omega$ and vacuum speed of light $c$. The $y$ component gives the ordinary wave relation $k_x^2+k_z^2=n_o^2{k}_0^2$. The determinant of the $x,z$ system givesSubstituting $k_x=n_{\mathrm{ex}}{k}_0\sin \theta$ and $k_z=n_{\mathrm{ex}}{k}_0\cos \theta$ yields the displayed directional refractive index.

The extraordinary wave's Poynting vector can differ in direction from its wavevector. The latter specifies wave phase propagation, while the former specifies energy flux. This angular separation is walk-off.^[ref]

In the lossless, nondispersive model above, take the gradient of $\omega =c\sqrt{k_x^2/n_e^2+k_z^2/n_o^2}$ to obtain the group velocity, parallel to the Poynting vector. If its angle to the optic axis is $\varphi$, this gives $\tan \varphi =(n_o^2/n_e^2)\tan \theta$. Consequently the extraordinary wave generally carries energy away from its wavevector direction. An internal wavevector normal to a slab face therefore does not guarantee a normal energy path. Birefringent walk-off vanishes at $\theta =0$ and $\theta =\pi /2$; the latter direction still has unequal refractive indices. This is a useful waveplate geometry: overlapping components can acquire phase retardation without spatial splitting. A thin specimen, insufficient imaging resolution, or excitation of only one linear polarization can also hide a double image, for different reasons.

For a lossless nonmagnetic uniaxial optical medium, the electric displacement field of a plane electromagnetic wave is perpendicular to its wavevector. The magnetic field is perpendicular to the plane containing the electric field and wavevector, while the Poynting vector is perpendicular to the electric field in that same plane. Thus the angle between $\mathbf{E}$ and $\mathbf{D}$ equals the walk-off angle between energy flow and phase propagation. At normal incidence on a parallel slab of thickness $L$, a lateral separation $\Delta$ of the ordinary and extraordinary exiting rays gives ${\alpha }_e=\arctan (|\Delta |/L)$, while ${\alpha }_o=0$. The output rays can be parallel even though their internal energy paths differ.

A waveplate uses birefringence to change polarization of an electromagnetic wave through phase retardation. Its fast axis has the smaller refractive index, hence the larger phase velocity; its slow axis has the larger index. Consider normal incidence with the optic axis in the plate face, so both components follow the same straight path. A linear polarization at $45^{\circ }$ to these axes excites equal wave amplitudes.

A complementary experiment uses two sheet polarizers with a cellophane sheet between them. First cross the polarizers and remove the sheet: the field is dark. Insert and rotate the sheet about the beam: light returns at intermediate orientations, but extinction recurs at two perpendicular orientations. The sheet has changed the polarization of an electromagnetic wave, without requiring a resolved double image. With white light, different wavelengths can produce different transmitted colors.^[ref]. Brightness alone does not establish circular polarization: an ideal rotating linear polarizer also gives orientation-independent light intensity for unpolarized light.

A detected lower-path photon with vertical polarization evolves under an angle-dependent Jones matrix. For fast-axis angle $\eta$, its horizontal amplitude is $(1-i)\sin \eta \cos \eta$ and its vertical amplitude is ${\sin }^2\eta +i{\cos }^2\eta$. A later vertical-path probability therefore depends on the plate orientation and on whether the earlier lower-path event has been conditioned on.

The relative phase accumulated by orthogonal eigenpolarizations is $\Delta =2\pi L(n_2-n_1)/{\lambda }_0$. A quarter-wave plate has relative phase $\pi /2$ modulo $2\pi$, while a half-wave plate has $\pi$ modulo $2\pi$.

For the overlapping monochromatic wave components, let $d$ be the thickness and ${\lambda }_0$ the vacuum wavelength. Define $\Delta n=n_s-n_f>0$ using the slow and fast refractive indices. With complex amplitude convention $e^{i(kz-\omega t)}$, the accumulated slow-minus-fast phase difference isThis is the optical path difference $\Delta nd$ expressed as a wave phase. It is a signed accumulation convention; with the opposite time convention the sign reverses. The outgoing polarization of an electromagnetic wave depends on $\delta$ modulo $2\pi$. Here the positive slow-minus-fast $\Delta n$ differs from the signed principal-index $n_e-n_o$ used in the parent birefringence entry when $n_e<n_o$.

For an original worked example, stipulate $\Delta n=0.0100$ at ${\lambda }_0=600\mathrm{nm}$ and a uniform thickness $d=15.0\mu \mathrm{m}$. These are design assumptions, not measured material data. The optical path difference is $150\mathrm{nm}$, soIt is a quarter-wave plate. At the same wavelength and refractive indices, doubling $d$ to $30.0\mu \mathrm{m}$ gives $\delta =\pi$, a half-wave plate. A $1.0\mu \mathrm{m}$ thickness increase changes $\delta$ by $\pi /30$, or $6^{\circ }$; retaining the same birefringence at $500\mathrm{nm}$ would instead give $\delta =3\pi /5$ for the $15.0\mu \mathrm{m}$ plate. Real refractive indices also vary with wavelength, so this last calculation explicitly neglects optical dispersion.

For these equal transmitted wave amplitudes, orthogonal projection onto a polarizer crossed with the initial $45^{\circ }$ linear polarization gives a simple independent readout. After removing a common wave phase, the component complex amplitudes are proportional to $(1,e^{i\delta })/\sqrt{2}$; orthogonal projection onto $(1,-1)/\sqrt{2}$ gives $(1-e^{i\delta })/2$. Squaring its modulus yields $I_{\perp }/I_{\mathrm{in}}={\sin }^2(\delta /2)$: zero at $d=0$, one half at $15\mu \mathrm{m}$, and one at $30\mu \mathrm{m}$. This assumes ideal polarizers, negligible specular reflection and wave absorption, overlapping components and stable relative wave phase. It measures a periodic function of phase retardation, so one brightness measurement cannot determine the thickness uniquely.

A zero-order quarter-wave plate satisfies ${\omega }_{0}d\Delta n({\omega }_0)/c=\pi /2$. Its retardance error at nearby angular frequency is $ϵ=\omega d\Delta n(\omega )/c-\pi /2$. Taylor expansion gives the title expression, including material dispersion. Neglecting that dispersion yields $ϵ=(\pi /2)\Delta \omega /{\omega }_0$. A half-wave plate made from twice the same material thickness has exactly twice this phase error. Common propagation phase can be removed from its Jones matrix without changing polarization.

A quarter-wave plate gives a quarter-cycle phase retardation. Equal transmitted wave amplitudes initially in linear polarization at $45^{\circ }$ become circular polarization; unequal wave amplitudes generally give elliptical polarization. Reversing the sign reverses the traversal of the same circle.

Take a half-wave plate with fast axis at $15^{\circ }$ to a linearly polarized input, followed by a quarter-wave plate at a further $60^{\circ }$, so the second axis is at $75^{\circ }$ to the input. With shared quarter-wave retardance error $ϵ$ and half-wave error $2ϵ$, use $Q=\mathrm{diag}(1,i{e}^{iϵ})$ and $H=\mathrm{diag}(1,-e^{2iϵ})$. Rotating each matrix into the input axes and multiplying gives $\mathbf{J}(0)=e^{i75^{\circ }}(1,-i{)}^{\mathsf{T}}/\sqrt{2}$ and the title expansion. Its first-order change is common phase only, so the output remains circular to first order. This input-specific local compensation does not prove a frequency-independent quarter-wave operator for arbitrary input across a finite spectral band.

Suppose the unknown ideal elements are a linear polarizer and a quarter-wave plate, and the two unknown sources are unpolarized and circularly polarized. Test both element orders while rotating their relative axes and monitoring output intensity. Only circular input followed by the quarter-wave plate and then the polarizer gives complete modulation: the plate makes linear light and the analyzer obeys Malus's law. Polarizer-first output is constant half-intensity because the following lossless plate changes polarization but not intensity. Unpolarized input is constant in either order. This identifies both elements and both sources without comparing source brightnesses.

Two rotatable ideal quarter-wave plates can transform a fixed circular polarization into any pure polarization represented by a Jones vector. Use time dependence $e^{-i\omega t}$, $Q(\theta )=R(-\theta )\mathrm{diag}(1,i)R(\theta )$ and input $J_L=(1,i{)}^T/\sqrt{2}$. Every target polarization ellipse can be written, up to a common phase, as $R(-\beta )(\cos \eta ,i\sin \eta {)}^T$, with signed ellipticity angle $-\pi /4\le \eta \le \pi /4$. The first quarter-wave plate at ${\theta }_1=\beta +\eta +\pi /4$ produces linear polarization at $\beta +\eta$. The second, at ${\theta }_2=\beta$, therefore produces the required target. The two angles supply the two independent parameters of pure polarization.

For a reciprocal ideal quarter-wave plate followed by a mirror at normal incidence, use the same fixed transverse laboratory axes for the forward and returning Jones vectors. If the axes coincide with the plate's fast and slow axes, each pass has Jones matrix $J=\mathrm{diag}(1,i)$ up to a common phase factor. A polarization-independent mirror has scalar reflection coefficient $r_m$, so the returned Jones vector is $r_m{J}^2\mathbf{e}=r_m\mathrm{diag}(1,-1)\mathbf{e}$. The plate acts twice as a half-wave plate, rather than undoing its phase retardation.

For input linear polarization at angle $\alpha$ to the fast axis, the returned line is at $-\alpha$ modulo $\pi$. At $\alpha =45^{\circ }$ it is perpendicular to the original line. An axis-aligned input instead returns along the same line. General elliptical polarization requires its full input Jones vector; no unique returned state follows from the word “polarized” alone. The propagation direction reverses, so comparing descriptions of circular polarization must also specify the viewing convention.

Take a linear polarizer $P=\mathrm{diag}(1,0)$ and a reciprocal quarter-wave plate $Q=2^{-1/2}\left(\begin{array}{cc}1& i\\ i& 1\end{array}\right)$ in fixed physical transverse axes. A near-normal isotropic mirror has $M=-I_2$ up to global phase. If the source/detector side encounters P before Q, the round-trip matrix is $PQMQP=-P{Q}^{2}P=0$. Reversing the filter gives $QPMPQ=-QPQ$, which transmits half an initially unpolarized intensity. The result is double-pass rejection, not optical nonreciprocity; changing direction without changing the input polarization does not preserve the same physical input state.

A half-wave plate reverses the relative sign of the two components. For initial linear polarization at angle $\alpha$ to the fast axis, the outgoing line is at $-\alpha$ modulo $\pi$. The $45^{\circ }$ input therefore rotates by $90^{\circ }$.

Circular birefringence means different refractive indices for opposite circular polarizations. In a transparent isotropic medium, their relative wave phase changes rotate the direction of transmitted linear polarization; optical activity gives the solution measurement and convention.

The rotatory power is the rotation of transmitted linear polarization per unit path length, in $\mathrm{rad}{\mathrm{m}}^{-1}$. With equally transmitted circular eigenmodes $({\mathbf{e}}_1\mp i{\mathbf{e}}_2)/\sqrt{2}$ and time convention $e^{-i\omega t}$, their accumulated phase difference gives $\theta =({\varphi }_{+}-{\varphi }_{-})/2$. For uniform circular birefringence, $\rho =\pi (n_{+}-n_{-})/{\lambda }_0$. Generic linear birefringence need not admit one rotation angle common to all incident polarization directions. In a weak longitudinal Faraday effect, $\rho =V{B}_{\parallel }$, with signed Verdet constant $V$.

The Ampère-Maxwell equationrelates magnetic circulation to conduction and displacement current.

An Amperian loop is a closed curve used in the integral form of Ampère's circuital law or the Ampère-Maxwell equation.

In vacuum magnetostatics, the Ampère-Maxwell equation reduces to $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$. Applying Stokes theorem gives ${\oint }_C\mathbf{B}\cdot d\mathbf{\ell }={\mu }_0{I}_{\mathrm{enc}}$, where $I_{\mathrm{enc}}$ is the electric current crossing an oriented spanning surface. Time-dependent electric fields require the additional displacement current term in the Ampère-Maxwell equation.

In vacuum, displacement current is the term ${\mathbf{J}}_D={ϵ}_0{\partial }_t\mathbf{E}$ in the Ampère-Maxwell equation. It has the units of current density, but need not describe electric charge crossing the region. In Maxwell equations in matter, the corresponding term is ${\partial }_t\mathbf{D}$, with $\mathbf{D}$ the electric displacement field.

Consider a slowly charging parallel-plate capacitor with vacuum gap, plate area $A$, and charge $Q(t)$ on its positive plate. Neglect fringing and assume the charging timescale is long compared with the light-travel time across the apparatus. The nearly uniform gap electric field has magnitude $E=Q/({ϵ}_{0}A)$, so the displacement current through the gap isFor $A=0.010{\mathrm{m}}^2$ and charging electric current $I=1.0\mathrm{mA}$, this gives $dE/dt\simeq 1.13\times 10^{10}\mathrm{V}{\mathrm{m}}^{-1}{\mathrm{s}}^{-1}$ and $J_D=0.10A/{\mathrm{m}}^2$. No conducting electric current crosses the gap. A spanning surface cutting the wire and another passing between the plates nevertheless give the same magnetic circulation when the full Ampère-Maxwell equation is used. ^[ref].

The local consistency check is short. Without displacement current, taking the divergence of $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}$ would force $\nabla \cdot \mathbf{J}=0$, allowing no local accumulation of electric charge. With displacement current, the divergence of a curl is zero and Gauss's law giveThis is precisely the charge continuity equation. Electric charge conservation fixes the required divergence of the added term; it does not by itself exclude other divergence-free additions. The specific Maxwell equations also yield the electromagnetic wave equation, linking this consistency correction to propagation at the speed of light.

The time derivative of the electric displacement field enters the macroscopic Ampère-Maxwell equation alongside free electric current density. In vacuum, ${\mathbf{J}}_D={ϵ}_0{\partial }_t\mathbf{E}$. It is not transport of free charges through the vacuum gap of a capacitor; its surface integral nevertheless supplies the term needed for compatibility with the continuity equation.

For a circular air-filled parallel-plate capacitor with voltage phasor $V_0{e}^{-i\omega t}$, uniform gap $d$ and radius $a$, the Ampère-Maxwell equation on a circular loop of radius $s<a$ gives the displayed magnetic field. The enclosed displacement current is proportional to $s^2$, whereas the circulation is $2\pi s{B}_{\varphi }$. Consequently its amplitude grows linearly from zero on the axis. This assumes negligible fringing fields and a quasistatic approximation $\omega a/c\ll 1$.

If a thin capacitor interior is modeled with axially uniform axisymmetric fields, Maxwell's equations give $r^{-1}(r{E}_z^{\prime }{)}^{\prime }+(\omega /c{)}^2{E}_z=0$. The solution regular at the axis is proportional to $J_0(\omega r/c)$. Normalizing to a central sinusoidal voltage gives $E_z=-(V_0/L)J_0(\omega r/c)\sin \omega t$ in this ideal interior model. It can change sign across the radius when propagation matters. Feed, edge and radiation boundary conditions are needed for a full driven-device solution.

For a uniform ohmic material under harmonic forcing, the magnitudes obey $|J_{\mathrm{cond}}|/|J_{\mathrm{disp}}|=\sigma /(\omega ϵ)$. A large ratio supports a local conduction-dominated approximation; it does not remove the displacement term from Maxwell's exact continuity relation.

A spanning surface that cuts a charging wire carries free current $I$. A surface passing through the capacitor gap instead carries displacement flux $\int {\partial }_t\mathbf{D}\cdot d\mathbf{A}=\dot{Q}=I$. Their common boundary has the same magnetic circulation only when both terms are included in Ampère's law.

Faraday's law of induction relates a changing magnetic field to circulation of the electric field. For a fixed oriented surface $S$ with boundary $C=\partial S$, choose the positive direction around $C$ by the right-hand rule: the fingers follow $C$ while the thumb follows the chosen unit normal. The fixed-loop electromagnetic electromotive force isThe final derivative of magnetic flux is an ordinary time derivative, with the surface held fixed. Reversing both the unit normal and the boundary direction reverses both signs. The loop may be an imagined path: the field law does not require a wire or an electric current. ^[ref].

For a computed example, a fixed single-turn loop has area $A=0.020{\mathrm{m}}^2$ and out-of-page unit normal. Prescribe an external uniform magnetic field in that direction with $B(t)=0.10\mathrm{T}+(0.30T/s)t$ for $0\le t\le 1.0\mathrm{s}$. ThenIn a quasistatic resistive model with electrical resistance $R=2.0\Omega$, neglect the loop's self-inductance and its own contribution to changing magnetic flux. Ohm's law givesThe magnetic flux rises from $0.0020\mathrm{Wb}$ to $0.0080\mathrm{Wb}$ while the induced electromotive force stays constant: it measures the rate of change, rather than the amount of magnetic flux. The negative sign means clockwise electric current; its magnetic field through the loop points into the page, as explained by Lenz's law. The toy model describes the interval of steady ramping, excluding switching transients at its ends.

For a moving conducting loop $C(t)$, the electromagnetic electromotive force instead includes the Lorentz force contribution from the local velocity $\mathbf{v}$ of each conductor segment:This form assumes a thin-wire model with a well-defined closed contour and spanning oriented surface that move compatibly, a magnetic field satisfying Gauss's law for magnetism, and conductor speeds small compared with the speed of light. Ideal sliding contacts such as those in motional electromotive force can be treated piecewise. Arbitrary changes of electrical connections require checking the actual contour and conductor velocities; differentiating the flux through an arbitrary chosen moving path is insufficient. The velocity here describes conductor motion, not the charge carriers' drift along the wire. ^[ref].

A short derivation of the fixed-loop integral statement applies Stokes theorem to the local curl equation and differentiates under the fixed surface integral. For the moving loop, the magnetic flux transport for a moving circuit adds the swept-area term $-{\oint }_C(\mathbf{v}\times \mathbf{B})\cdot d\mathbf{l}$ to $d{\Phi }_B/dt$. Combining these two identities gives the moving-loop statement. This derives the integral forms from the local field law; it is not a proof of that empirical law from mathematics alone.

The useful unity is geometric: changing the magnetic field, the enclosed area, or the orientation can change magnetic flux and hence produce an electromotive force. A coil's multiple turns are accounted for by magnetic flux linkage, rather than by multiplying a single-turn flux without checking which flux each turn receives. ^[ref].

Faraday's report read on 24 November 1831 describes opposite momentary responses when a neighboring circuit is connected and disconnected, with no continuing response while its steady electric current is maintained. These are concrete experimental roots of the changing-flux idea. The report predates the modern Maxwell equations notation used above. ^[ref].

Choose the positive loop normal and its positive circulation by the right-hand rule. A spatially uniform external magnetic field of fixed direction and signed magnitude $B_0\cos \omega t$, at angle $\theta$ to that normal, gives magnetic flux $\Phi =\pi r^2{B}_0\cos \theta \cos \omega t$. Neglecting the loop's self-inductance, Faraday's law of induction and Ohm's law give the displayed electric current. At a quarter-cycle the external field is zero but its normal component is decreasing most rapidly, so the current produces a field in the direction of the initial normal flux, as required by Lenz's law. With appreciable self-inductance the circuit instead obeys $L\dot{I}+RI=-{\dot{\Phi }}_{\mathrm{ext}}$.

Michael Faraday's induction ring separates two questions: can one electrical circuit affect another without an electrical connection, and does that effect continue under steady conditions? The surviving apparatus uses insulated copper windings around an iron ring; the Royal Institution dates the successful experiment to 29 August 1831.^[ref]

Faraday connected one winding group to a battery, the other to a galvanometer. Making battery contact deflected the needle; maintaining contact let it return to rest; breaking contact reversed the response. Reversing the battery reversed the deflections. His separate cylinder comparison enhanced induction with iron, whereas replacing it with copper gave no enhancement over the windings alone.^[ref]

The contrast between a switching response and a quiet steady state makes mutual induction tangible. A steady primary electric current can maintain magnetic flux without sustaining a secondary electromotive force. Changes in magnetic flux linkage matter, as expressed by Faraday's law of induction. This is why a driven electrical transformer uses repeated changes in magnetic flux to supply secondary power.

The galvanometer readout of an induction pulse also separates the measured motion from the inferred electric current. Its moment of inertia and damping mean that a deflection is not an instantaneous trace of electric current. The following coupled RL induction model calculates currents in a stipulated circuit, rather than fitting Faraday's needle motion or reconstructing his switching times.

Take two electrical circuits containing coupled inductors. Let their constant self-inductances be $L_1,L_2>0$, mutual inductance $M$, and positive electrical resistances $R_1,R_2$. Choose current directions so that the off-diagonal entries of the inductance matrix are $+M$. Assume a linear, quasistatic model, with no capacitive effects or core hysteresis. The magnetic flux linkages are ${\lambda }_1=L_1{i}_1+M{i}_2$ and ${\lambda }_2=M{i}_1+L_2{i}_2$. Combining their time derivatives with resistive electric potential differences givesThe secondary is a closed, undriven electrical circuit. Require $M^2<L_1{L}_2$, making $\mathbf{L}$ a positive-definite matrix. The chosen current convention fixes the signs; reversing the secondary's reference direction changes the sign assigned to $M$ and $i_2$ without changing the physical circuit.

There is an essential switching assumption: the primary loop remains closed when $V$ becomes zero. An ideal voltage source then imposes zero electric potential difference, allowing the current to decay. Literally opening an inductive circuit requires additional physics to determine the interruption, including possible capacitance and an electric arc. This closed-return model explains transient induction while keeping that distinction from Faraday's historical making and breaking of contact explicit.

For an exactly soluble example, set $L_1=L_2=L$, $R_1=R_2=R$, and $0<M<L$. Start with zero electric currents, and apply a constant electric potential difference $V_0>0$ at $t=0$. The sums $s=i_1+i_2$ and $d=i_1-i_2$ obey separate first-order linear differential equations:Their time constants are ${\tau }_{+}=(L+M)/R$ and ${\tau }_{-}=(L-M)/R$. Thus, for $t\ge 0$,Since ${\tau }_{-}<{\tau }_{+}$, $i_2(t)<0$ for $t>0$, but $i_2(t)\to 0$ as $t\to \infty$. The primary approaches $V_0/R$. The secondary response opposes the rise of linked magnetic flux, in accordance with Lenz's law. Removing mutual inductance by setting $M=0$ makes the two exponentials equal and gives zero secondary electric current.

For stipulated $L=0.1\mathrm{H}$, $M=0.06\mathrm{H}$, $R=10\Omega$ and $V_0=1\mathrm{V}$, the two time constants are $16\mathrm{ms}$ and $4\mathrm{ms}$. Setting ${\dot{i}}_2=0$ gives its largest negative magnitude atThe continuing primary electric current eventually reaches $100\mathrm{mA}$ while the secondary returns to zero. These numbers belong to the stated model, not the 1831 apparatus.

Apply $V_0$ for $0\le t<T$ and then set the drive to zero while retaining the closed primary return. By superposition, each pulse-response electric current is its equal-loop RL induction step response at $t$ minus that response at $t-T$, with the step response defined as zero before its onset. All currents remain continuous at $T$; the drive changes their derivatives.

For the displayed pulse, the secondary responds negatively as the primary builds and positively as it decays. Between these transients, the primary is almost steady and the secondary is almost quiet. The energy balance of coupled RL circuits accounts for the entire process, including the energy remaining in the magnetic field when the drive changes.

The magnetic energy of this constant-inductance model isThe positive-definite matrix condition makes $W\ge 0$. Multiplying the two circuit equations by $i_1$ and $i_2$, then adding, gives the displayed conservation of energy statement: input power $V{i}_1$ supplies stored magnetic energy and Joule heating. With zero drive, $\dot{W}\le 0$. After the drive becomes zero, the stored magnetic energy supplies the remaining Joule heating.

A pickup coil links magnetic flux from a source and develops an electromotive force when that linked magnetic flux changes.

Lenz's law expresses the minus sign in Faraday's law of induction: the induced electromotive force drives a passive resistive loop's electric current in the direction whose own magnetic field opposes the imposed change in magnetic flux. It opposes the change of flux, not invariably the external magnetic field. Choose an oriented surface and its positive boundary direction first, find the sign of $d{\Phi }_{\mathrm{ext}}/dt$, and then use the right-hand rule to translate the signed electric current into a physical direction. ^[ref].

In the fixed-loop example in Faraday's law of induction, out-of-page magnetic flux increases at $0.0060Wb/s$. With counterclockwise positive, $I=-0.0030\mathrm{A}$ is clockwise, producing an in-page magnetic field through the loop. Reverse the ramp to $B(t)=0.40\mathrm{T}-(0.30T/s)t$ on the same one-second interval: now $\mathcal{E}=+0.0060\mathrm{V}$ and $I=+0.0030\mathrm{A}$. The counterclockwise electric current produces an out-of-page magnetic field, supporting the decreasing external magnetic flux. A steady out-of-page magnetic field with fixed loop geometry gives no induced electromotive force in either direction.

For the increasing-area example in motional electromotive force, clockwise electric current runs down the right-hand rod. Its magnetic force on a current-carrying wire points left, opposing the imposed motion to the right. Writing $I_{\mathrm{mag}}=B\ell v/R$ for the current magnitude, the power required to maintain constant speed isThus the sign agrees with conservation of energy: the external drive supplies the Joule heating. The magnetic forces do no work on individual charges; mechanical work by the drive supplies the electrical energy. This also explains the usefulness of electromagnetic braking of a sliding rod and the terminal speed of a falling conducting loop. ^[ref].

The sign of induced electromotive force alone does not determine an arbitrary circuit's instantaneous electric current. Retaining constant self-inductance $L$ gives $LdI/dt+RI=-d{\Phi }_{\mathrm{ext}}/dt$, so a pre-existing electric current can persist while the induced drive opposes its change. The distinction matters in the lossless limit of magnetic loop braking. The resistive toy examples use $L=0$ as an approximation, not a general consequence of Lenz's law.

If a loop of resistance $R$ has magnetic flux ${\Phi }_B=-kz$, then its induced current is $I=k\dot{z}/R$. When its magnetic force is $-kI$ in the vertical direction, its equation under gravity isso its downward terminal speed is $Rmg/k^2$.

For an external loop flux $Kz$, positive current orientation and upward Lorentz force $KI$, including self-inductance gives $\mathcal{L}\dot{I}+RI=-K\dot{z}$ and $m\ddot{z}=-mg+KI$. At zero resistance total linked flux is conserved. If initially $I=0$ at $z=z_0$, then $I=-K(z-z_0)/\mathcal{L}$ and the mechanical motion is an undamped oscillator about $z_0-mg\mathcal{L}/K^2$. The finite-resistance terminal speed tends to zero as $R\to 0$, but neglecting self-inductance is a singular approximation in that limit. A lossless loop does not dispose of energy through Joule heating.

Gauss's law relates the electric field on a closed surface to the net electric charge inside it. In SI units, with outward unit normal $\mathbf{n}$,Here $\rho$ describes a volume charge density, and ${\epsilon }_0$ is the vacuum permittivity; for ideal point or surface charges, use the enclosed electric charge directly rather than an ordinary volume-density integral. The left side is electric flux: the surface integral of the outward normal component of the electric field. Field directed outward contributes positively; field directed inward contributes negatively. The equality uses net signed electric charge, so positive and negative enclosed charges can cancel. The Gaussian surface is an imagined accounting boundary, not necessarily a material object. See OpenStax, explaining Gauss's law.

Where the electric field and charge density are smooth, the local statement is $\nabla \cdot \mathbf{E}=\rho /{\epsilon }_0$. The divergence theorem connects this divergence to the surface integral. This is the satisfying connection between local sources and global accounting: a surface can register how much net electric charge it encloses without revealing where that charge is located within it. A point charge requires care with the singular field at its position; the integral formulation remains useful with a boundary that avoids the charge.

The law fixes total electric flux, rather than every value of the electric field. In an uncharged region, a uniform electric field $\mathbf{E}=(E_0,0,0)$ crosses a cube of side $\ell$. Its left and right faces contribute $-E_0{\ell }^2$ and $+E_0{\ell }^2$; the other four faces contribute zero. Thus the net electric flux is zero while the electric field is nonzero everywhere. Charges outside a Gaussian surface can change the field on it without changing its net electric flux. Conversely, fields cannot generally be reconstructed from enclosed electric charge alone. Electrostatics also requires the electric field to be the negative gradient of an electric potential, with appropriate boundary information. This distinction is emphasized in Feynman lectures II.5, application of Gauss' law.

Spherical symmetry supplies the extra information needed for a particularly simple calculation. An isolated spherically symmetric charge distribution in vacuum produces a radial electric field, constant in magnitude on each concentric sphere. Its electric flux is therefore $4\pi r^2{E}_r(r)$, givingOutside the distribution, doubling $r$ quadruples the surface area and divides the radial electric field by four, while the electric flux stays $Q/{\epsilon }_0$. The inverse-square dependence is an area-dilution result. The electric field of a uniformly charged ball and electric field of a uniformly charged spherical shell show that the same total charge can give different interior fields. These symmetry arguments and their limits are developed in OpenStax, applying Gauss's law.