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The magnetic rigidity gives the momentum, the bending direction gives the sign of electric charge, and the calorimetric energy then gives the mass. For planar tracks perpendicular to a uniform magnetic field, the Lorentz force and circular motion giveTrace each track away from its production or decay vertex. With the magnetic field into the page, a positive track bends counterclockwise and a negative track bends clockwise. For example, a positive particle initially travelling left accelerates downwards. The known incoming negative pion fixes the sign convention experimentally. A radius alone does not distinguish electric charge signs.

For a calibrated total energy $E$, the relativistic energy-momentum relation givesIf the calorimeter instead supplies only kinetic energy $T$, solve $(T+m{c}^2{)}^2=p^2{c}^2+m^2{c}^4$ to obtainCompare with candidate pion, kaon, and proton masses, together with the electric charge sign and shower behavior. An electron makes an electromagnetic shower; a muon is penetrating; a hadron generally requires a hadronic calorimeter and its own calibration. If either energy escapes or the response is uncalibrated, simply substituting a deposited signal for $E$ need not identify the particle correctly. The absent figure prevents attaching identities to individual drawn tracks, but the requested identification procedure does not require its geometry.

The electric potential and magnetic vector potential represent the fields through $\mathbf{B}=\nabla \times \mathbf{A}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$. They are not unique: a differentiable function $\chi$ giveswithout changing either field. Gauge fixing restricts this freedom to suit the sources, boundaries or symmetry; it does not alter observables.

The Lorenz gauge $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\varphi =0$ is Lorentz covariant. Maxwell's equations then reduce to sourced wave equations,Retarded solutions implement causal radiation conditions. This gauge is useful for radiation and relativistic calculations; residual transformations obey the homogeneous wave equation.^[ref]

The Coulomb gauge $\nabla \cdot \mathbf{A}=0$ instead gives ${\nabla }^2\varphi =-\rho /{ϵ}_0$. Its electric potential is an instantaneous Poisson solution, but the accompanying magnetic vector potential makes the full physical electric field causal. It is useful in separating longitudinal electric fields from transverse radiation. In a source-free vacuum region, suitable radiation boundary conditions permit radiation gauge $\varphi =0$, $\nabla \cdot \mathbf{A}=0$, leaving two transverse plane-wave polarizations. These extra conditions are not a gauge choice applicable unchanged to arbitrary nonzero charge distributions.

An ideal current-carrying wire consists of opposite charge populations, one associated with the stationary lattice and one with drifting carriers. They may cancel in the lattice frame while carrying a net current, as in part (c), up to exchanging the charge signs and current direction. In a carrier-rest description their densities no longer cancel because the two populations undergo different length contractions. The wire then has both electric fields and magnetic fields.

A neutral steady current in one frame generally becomes a charged steady current in another. This illustrates how electric fields and magnetic fields are parts of one relativistic field rather than separately invariant forces. Real resistive wires also have a driving longitudinal electric field and geometry-dependent surface charges; the infinite uniform filament model isolates the transverse fields of a neutral current and does not describe every field of a complete circuit.

Use ideal large plates and neglect edge effects. A Lorentz boost normal to the plates does not contract either transverse plate dimension. Charge is invariant, so charge per plate area remains ${\sigma }^{\prime }=\sigma$. Gauss's law then givesThe plate separation contracts, but the infinite-plate field is independent of separation. It is therefore incorrect to infer a stronger field merely from the shorter gap. The ideal field is parallel to the boost, so there is no induced transverse magnetic field from this boost.

A Hertzian dipole is an ideal localized oscillating source much smaller than its radiation wavelength. Its electric version is an infinitesimal current element; its magnetic version is a small current loop with moment $\mathbf{m}=IA\mathbf{n}$. Such elements model the far-field radiation and radiation resistance of electrically small antennas. The approximation requires source dimensions much smaller than $c/\omega$.

The supplied magnetic field has a quasistatic $[m]/r^3$ term, an induction $[\dot{m}]/(c{r}^2)$ term and a transverse radiation $[\ddot{m}]/(c^{2}r)$ term. The electric field has induction and radiation terms. Brackets mean evaluation at the retarded time $t_r=t-r/c$, not a time average. Near terms mainly exchange stored energy; only the $r^{-1}$ fields give a nonvanishing outgoing power integral as $r\to \infty$.

For a fixed dipole axis the radiation fields satisfy $E_{\varphi }=-c{B}_{\theta }$. The radial Poynting flux is thereforeUsing $\int {\sin }^2\theta d\Omega =8\pi /3$ yieldsFor a general changing dipole direction, replace the squared scalar acceleration by $|\ddot{\mathbf{m}}{|}^2$.

Radiation resistance is the equivalent input resistance whose mean power equals radiated power, not ohmic heating. For a fixed small loop carrying $I(t)=I_0\cos \omega t$, $m=AI$, soThis uses peak current $I_0$; with rms current the power is $I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$.

The satellite loop needs its changing orientation as well as its induced current. Put the spin axis along z, the external field along x, and let $\mathbf{n}(t)=(\cos \omega t,\sin \omega t,0)$. With $A=\pi b^2$ and resistance $R=10^{-4}\Omega$, the resistance-only approximation givesWrite $m_0=B{A}^2\omega /R$. The full magnetic moment isIt has a static component and a rotating component at $2\omega$. Consequently $|\ddot{\mathbf{m}}{|}^2=4m_0^2{\omega }^4$, andHolding the dipole axis fixed while inserting the sinusoidal current would instead give $B^2{A}^4{\omega }^6/(12\pi {ϵ}_0{c}^5{R}^2)$, eight times smaller. That is the fixed-axis-loop result; it does not describe the stated rotating loop. This is frequency doubling of an induction-driven rotating loop; the changing vector direction is essential.

The resistance-only model is conditional: loop inductance is not supplied. If self-inductance L matters, solve $L\dot{I}+RI=BA\omega \sin \omega t$. In the steady state $I_0=BA\omega /\sqrt{R^2+{\omega }^2{L}^2}$. Its phase changes the static moment direction but not the twice-frequency component's magnitude, so the power becomesThe first expression is the requested resistive idealization and an upper estimate within this passive RL correction. Radiation reaction is negligible compared with the stipulated ohmic resistance. The radiated energy comes from rotational kinetic energy: the induced magnetic torque brakes the loop. Most lost rotational energy is Joule heating; a small part escapes as radiation. Maintaining a constant spin would require external mechanical power.

For scale only, take $B=50\mu \mathrm{T}$, a representative near-Earth field rather than a supplied orbital measurement. The terrestrial surface range reported by NOAA is roughly $25$–$65\mu \mathrm{T}$; the actual orbital value depends on position.^[ref] With $b=1\mathrm{m}$ and $\omega =2\pi (500{\mathrm{s}}^{-1})$, the resistance-only model givesInductance would reduce this power. The kilohertz signal is exceptionally weak and a metre-scale receiving antenna is electrically tiny. Ionospheric absorption/reflection and strong natural or instrumental low-frequency backgrounds make ordinary ground reception impractical. This is not a theorem that the ionosphere blocks every low-frequency mode: field-guided propagation and condition-dependent leakage are real, as the VIPER mission's measurements were designed to test.^[ref]

At a separation $r=100\mathrm{km}$, $k_{\mathrm{rad}}r\simeq 2.1$, so the observer is not securely in the far field. The radiation term alone has at most the characteristic magnetic amplitudein this optimistic resistance-only vacuum model. Near and induction fields must be retained for a full prediction; a simple far-field inverse-square power estimate is not an exact received signal here. Another satellite avoids the whole ground ionospheric path, but plasma propagation, receiving sensitivity, bandwidth, orientation, integration time and spacecraft noise remain unspecified. Detection is not forbidden in principle, but routine practical detection is implausible at this scale; a definitive instrument-level yes/no does not follow from the supplied data.

Synchrotron radiation is emitted by relativistic charges accelerated along curved paths. A magnetic field can supply the transverse acceleration while doing no work on the charge; the radiated energy must ultimately be replenished if the orbit energy is to remain constant. Relativistic beaming confines much of the radiation to angles of order $1/\gamma$ about the instantaneous velocity.

For circular motion of curvature radius $\rho$ and $\gamma \gg 1$,The exact classical circular-orbit power contains an additional factor ${\beta }^4$. A distant observer sees short pulses as the beamed direction sweeps past, producing a broad high-frequency spectrum. Radiation is predominantly linearly polarized in the orbital plane and becomes elliptically polarized off it. The critical frequency is a spectral scale, not a monochromatic emission frequency or a hard cutoff.

Storage rings provide intense radiation for spectroscopy and diffraction, while relativistic electrons in astrophysical magnetic fields also produce synchrotron emission. The classical theory relates total power, angular beaming and spectrum to the particle's trajectory; sufficiently strong quantum recoil requires corrections.^[ref]

The Faraday effect rotates the plane of linear polarization when light propagates through a material with a magnetic field component along the ray. In the weak-field transparent regime,with a stated propagation and transverse-basis convention; $V$ is the Verdet constant. For a uniform longitudinal field and sample length $L$, the magnitude is $|VBL|$. A scalar intensity change without a corresponding polarization analysis would not establish this effect.

The microscopic field breaks the equivalence of the two circular polarization modes: the Lorentz force and the Zeeman shifts of optical transitions produce different susceptibilities for their opposite senses of rotation. A linearly polarized wave is an equal-amplitude superposition of these two modes. If their real refractive refractive indices are $n_{+},n_{-}$, their relative wave phase after the sample is $\Delta \varphi =(\omega L/c)(n_{+}-n_{-})$. Recombination rotates the line byEqual transmitted circular amplitudes preserve linear polarization. Unequal absorption produces circular dichroism and generally elliptical polarization as well, so pure rotation is an approximation appropriate away from strong absorption.

A modern motivating experiment sends polarized laser light through a glass rod inside a solenoid, then through a rotatable analyzer onto a photodiode. Scan the analyzer first to calibrate its angular response. Driving the solenoid sinusoidally and recording the detector synchronously separates a small rotation signal from unsynchronized noise; the conversion to angle still requires the analyzer slope and the magnetic-field integral through the rod. A first-party William & Mary laboratory manual specifies this apparatus and measurement procedure.^[ref]

Additional controls are equal-magnitude field reversals, zero-field readings before and after each pair, and a blocked-beam pickup test. The odd-in-field rotation is $[{\theta }_{\mathrm{obs}}(B)-{\theta }_{\mathrm{obs}}(-B)]/2$; a fixed analyzer offset cancels. Heating, stressed-glass birefringence, source drift and field-dependent transmission need separate checks. A single reading near crossed extinction loses the sign because its leading response is quadratic; an analyzer near $45^{\circ }$ gives a linear small-rotation response.

Faraday rotation is nonreciprocal. With the magnetic bias held fixed, retracing the path adds rotation in the same laboratory sense rather than undoing it, unlike reciprocal natural optical activity. This permits an optical isolator with a $45^{\circ }$ Faraday rotator between appropriately oriented polarizers. One must compare the returned polarization in a consistently defined laboratory basis; merely reversing the sign of $\mathbf{B}\cdot \hat{\mathbf{k}}$ while retaining the old ray-basis convention gives an incorrect cancellation.

antenna quantities. radiation resistance represents emitted power by $P_{\mathrm{rad}}=I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$ at the feed. It is not Joule heating in the antenna; a separate loss resistance accounts for that. antenna gain is the directional far-field intensity relative to an isotropic radiator supplied with the same input power:It includes radiation efficiency and equals antenna directivity only for a lossless antenna. antenna effective area is defined by the power available to a matched load, $P_{\mathrm{av}}=A_{\mathrm{eff}}{S}_{\mathrm{inc}}$, for incidence from a specified direction and matched polarization. It need not equal a literal geometric collecting area.

Gain–area relation. To derive the antenna gain-effective-area relation, take a passive reciprocal antenna. Its electromagnetic reciprocity makes the receive pattern proportional to the transmit pattern; write $A_{\mathrm{eff}}(\Omega )=KG(\Omega )$. The normalization follows from thermal equilibrium. In the Rayleigh-Jeans law limit, blackbody radiation radiance per unit frequency per unit solid angle is $2k_{B}T/{\lambda }^2$ for two polarizations. A single antenna polarization couples to one half, so the available incident noise power isThe open-circuit Johnson voltage spectrum of the antenna resistance is $4k_{B}T{R}_{\mathrm{rad}}d\nu$. A matched load receives half the open-circuit voltage, giving available power $k_{B}Td\nu$. In a lossless antenna at thermal equilibrium these powers agree; hence $\int A_{\mathrm{eff}}d\Omega ={\lambda }^2$. Since $\int Gd\Omega =4\pi$,The same efficiency factor multiplies receiving area and gain if reciprocal dissipative losses are included consistently. Impedance mismatch or polarization mismatch reduces delivered power further. The factor of two for thermal polarization modes and the factor of four between open-circuit voltage variance and matched-load power are separate effects.

Normalize the radar beam. Interpret the azimuth as signed about the beam centre at $\varphi =0$, so both sides of the narrow lobe are included. Put $u=\theta -\pi /2$. To leading order in the small angular width, $\sin \theta \simeq 1$ and the angular integrals can be extended over the real line:The Gaussian antenna beam solid angle gives the peak on the positive $x$-axis consequently hasHere $N$ is power per solid angle, equal to $r^2{S}_r$ in the far field. The leading neglected solid-angle correction is relative order ${\alpha }^2$.

Return using the supplied electric-dipole response. The electric-dipole radar cross section of a small conducting sphere determines this return. At target range $D$, the incident intensity is $S_t=C/D^2$. Write its root mean square electric field as $E_t$, with $S_t={ϵ}_{0}c{E}_t^2$. The induced electric dipole is $p_{\mathrm{rms}}=4\pi {ϵ}_0{a}^3{E}_t$. A dipole's angular radiation pattern, normalized by $\int {\sin }^2\chi d\Omega =8\pi /3$, gives $d{P}_{\mathrm{sc}}/d\Omega =(3P_{\mathrm{sc}}/8\pi ){\sin }^2\chi$. The backward direction is perpendicular to the induced dipole, so the supplied total dipole power givesThe received intensity is this angular power divided by $D^2$. Using the monostatic radar equation, or directly multiplying by the receive area, givesThe receiver threshold is a voltage across the matched load, so the detectable power is $V_m^2/R$, without another matching factor:This is the electric-dipole result following from the response data provided. It assumes $ka\ll 1$, far-field propagation on each leg, alignment at the beam maximum and no attenuation or additional receiver losses.

Perfect-conductor qualification. perfect-conductor sphere dipole interference arises because the sphere also excludes the oscillating magnetic field. Solving the magnetic boundary problem gives $\mathbf{m}=-2\pi a^3{\mathbf{H}}_t$ in addition to the electric dipole. The magnetic polarizability of a perfectly conducting sphere follows from its induced surface currents.^[ref] In the backward direction the magnetic contribution has half the electric scattering amplitude and the same sign, so the returned intensity is $9/4$ of the electric-only value. Equivalently, the monostatic radar cross section is ${\sigma }_{\mathrm{back}}=9\pi k^4{a}^6$, rather than $4\pi k^4{a}^6$. Including both leading dipoles therefore givesBoth results have been distinguished because the supplied auxiliary data describe only the electric dipole, while the literal perfect-conductor boundary conditions also generate the magnetic one. The backward interference is a leading-order effect, not a higher-order correction in $ka$.^[ref]

Synchrotron radiation is radiation from relativistic charges undergoing transverse acceleration, for example electrons bent by a magnetic field. Relativistic aberration concentrates the radiation within an angle of order $1/\gamma$ about the instantaneous velocity. An observer sees short pulses as this cone sweeps past; synchrotron pulse compression produces a broad spectrum extending far above the orbital frequency.

For an ultrarelativistic circular trajectory of curvature radius $\rho$, the characteristic scales are^[ref] The synchrotron critical frequency sets the spectral turnover; it is not a sharp upper cutoff. The transverse-acceleration Lienard formula givesThe strong energy dependence is a major loss mechanism in electron storage rings. Radiation observed in the orbital plane is linearly polarized in that plane; away from it the two components generally produce elliptical polarization, with opposite handedness on opposite sides. Accelerator light sources exploit its brightness, broad tunable spectrum and pulsed timing. At low speeds the relativistic beaming and high-frequency extension disappear, approaching ordinary cyclotron radiation.^[ref]

The magnetic vector potential is defined by $\mathbf{B}=\nabla \times \mathbf{A}$. The gauge transformationleaves both $\mathbf{B}$ and $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ unchanged. In magnetostatics choose the Coulomb gauge $\nabla \cdot \mathbf{A}=0$. Then Ampère's law and the curl-of-curl identity giveThe free-space Green function identity ${\nabla }^2|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^{-1}=-4\pi {\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}^{\prime })$ therefore givesThe boundary condition selects the homogeneous solution; the usual decay condition assumes suitably localized sources. An infinitely long static wire needs a reference for its logarithmically divergent potential, although its magnetic field is finite away from the wire.

For time-dependent sources use the Lorenz gauge, $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\varphi =0$, and the outgoing retarded potentialThe scalar potential has the analogous retarded charge-density integral. The two gauges should not be conflated for a general time-dependent source.^[ref]

For the prescribed uncharged wire, $\rho =0$ and $\varphi =0$ is consistent. Translation invariance makes $\mathbf{A}=A_z(r,t)\hat{\mathbf{z}}$ with $\nabla \cdot \mathbf{A}=0$, so this particular solution satisfies both gauge conditions. With $s=z^{\prime }-z$,It is zero for $t<r/c$. For $t>r/c$, set $L=\sqrt{c^2{t}^2-r^2}$. The causal interval is $-L<s<L$, giving the switched infinite-wire retarded potentialDifferentiation gives the fields and Poynting vector, for $r>0$, $t>r/c$:Here $\mathbf{B}={\mu }_0\mathbf{H}$ and $B_{\varphi }=-{\partial }_r{A}_z$. All three fields vanish before arrival. For positive $I$, $\mathbf{E}$ opposes the imposed current and the radial energy flow after switching a line current points radially outward: $(-\hat{\mathbf{z}})\times \hat{\mathbf{\varphi }}=\hat{\mathbf{r}}$.

For the graph, let $u=ct/r$. Normalized to $E_{\ast }={\mu }_{0}Ic/(2\pi r)$, $H_{\ast }=I/(2\pi r)$ and $N_{\ast }={\mu }_0{I}^{2}c/(4{\pi }^2{r}^2)$, the curves areAt the wavefront, $|E_z|$ and $H_{\varphi }$ diverge as $(u-1{)}^{-1/2}$ and $N_r$ as $(u-1{)}^{-1}$. This abrupt line-current wavefront singularity reflects the ideal instantaneous turn-on of a line current; a finite rise time smooths the front. At late times,The vector potential grows logarithmically with time in this infinite-wire idealization even as the local magnetic field approaches its static limit. The divergent front should not be interpreted as a finite-energy pulse from a real instantaneous switching experiment.

General pure state. Use time dependence $e^{i(kz-\omega t)}$. The most general nonzero Jones vector can be writtenThere are four real parameters in the full complex field. Removing overall intensity $A^2$ and the unobservable common phase $\chi$ leaves two real degrees of freedom for pure polarization, $\alpha$ and $\delta$. Their endpoint degeneracies simply reflect the fact that the phase of a vanishing component is irrelevant. A single Jones vector cannot represent partially polarized light.

Three linear polarizers. Define the transmission axis ${\mathbf{a}}_{\theta }=(\cos \theta ,\sin \theta {)}^T$. An ideal linear polarizer has Jones matrix $P_{\theta }={\mathbf{a}}_{\theta }{\mathbf{a}}_{\theta }^T$. For definiteness label $(1,-i{)}^T/\sqrt{2}$ as right-circular in the chosen convention; reversing the handedness convention changes no intensity below. The first polarizer produces $(1,0{)}^T/\sqrt{2}$. The intermediate polarizer and final $y$ analyzer then giveso Malus's law givesThe maximum is $1/8$ at $\theta =\pi /4+m\pi /2$, and the transmission vanishes at $\theta =m\pi /2$. The first polarizer accounts for an initial factor $1/2$; retaining that factor is essential when quoting transmission relative to the incident circularly polarized beam.

Waveplate and quarter-wave matrix. A lossless waveplate has linear birefringence: its orthogonal fast and slow axes have indices $n_f<n_s$. A thickness $L$ produces slow-minus-fast phase retardationResolving the incident Jones vector into the plate axes, advancing the two phases and transforming back gives, up to a common phase,A quarter-wave plate has $\delta =\pi /2$ modulo $2\pi$. With $c_{\theta }=\cos \theta$ and $s_{\theta }=\sin \theta$, its Jones matrix is thereforeIn particular $Q(0)=\mathrm{diag}(1,i)$. Equal incident components along the two axes acquire a relative quarter-cycle phase, converting $45^{\circ }$ linear polarization to circular polarization; an incident component along either principal axis stays linearly polarized.

Half-wave replacement. A half-wave plate has $\delta =\pi$ and henceIt converts the $x$-polarized output of the first polarizer to a linear state at angle $2\theta$. The final $y$ analyzer givesThe maximum is $1/2$, because the ideal half-wave plate loses no light after the first polarizer.

Quarter-wave replacement. The transmitted $y$ amplitude is $(1-i)c_{\theta }{s}_{\theta }/\sqrt{2}$, givingMore generally, the retarder gives $I_{\mathrm{out}}/I_{\mathrm{in}}=\frac{1}{2}{\sin }^2(\delta /2){\sin }^2(2\theta )$. All three requested curves have the same zeros and angular period, but different heights:

Preparing an arbitrary ellipse. Yes: two rotatable quarter-wave plates already suffice. Let the input left-circular state be ${\mathbf{J}}_L=(1,i{)}^T/\sqrt{2}$. Up to an overall phase, any target pure polarization can be parameterized by its polarization ellipse:Here $\beta$ is the major-axis orientation, and $\eta$ sets the signed minor-to-major semiaxis ratio $\tan \eta$. First note directly from $Q(\theta )$ thatChoose the two fast-axis angles asThe first quarter-wave plate produces linear light at $\beta +\eta$. In the axes of the second, its components are $(\cos \eta ,\sin \eta )$; the second gives precisely ${\mathbf{J}}_{\mathrm{target}}$. This proves pure-polarization preparation with two quarter-wave plates, including the linear and circular limiting cases. A half-wave plate can also rotate the axes, but is not required. The phrase about left-handed input is understood as left-circular light in this construction; the explicit vector fixes the convention.

Switchable retardation. The electro-optic Kerr effect induces linear birefringence quadratic in a static transverse electric field. Defining the coefficient $K$ by $\Delta n={\lambda }_{0}K{E}_{\mathrm{dc}}^2$, its phase retardation isThe two eigenpolarizations are linear, parallel and perpendicular to the applied field. Thus an appropriate field produces a switchable quarter-wave plate:The sign of $K$ determines which axis is fast. Switching the field off removes the induced retardation, assuming no residual birefringence.

The Faraday effect instead gives circular birefringence for a longitudinal magnetic field. The two circular eigenmodes acquire different phases, rotating a linear polarization throughwith sign set by the handedness convention and the Verdet constant $V$. Its Jones matrix in the linear basis is a real rotation, up to a common phase. It preserves the ellipticity of a pure state, so a linear input remains linear. A $\pi /2$ phase difference between circular eigenmodes rotates a linear state by $\pi /4$; it does not make it circular. The Kerr device supplies the switchable quarter-wave plate; a Faraday rotator alone does not.

Gauge freedom. The magnetic vector potential satisfies $\mathbf{B}=\nabla \times \mathbf{A}$. The transformationleaves $\mathbf{B}$ unchanged because the curl of a gradient is zero. For time-dependent fields, transform the electric scalar potential as $\varphi \mapsto \varphi -{\partial }_t\chi$ as well, so that $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ is unchanged. The scalar $\chi$ must have the regularity and single-valuedness appropriate to the physical region and its boundary conditions. Local values of $\mathbf{A}$ therefore depend on the gauge transformation.

Gauge and nonzero component. In the standard axisymmetric Coulomb gauge representative for the long-solenoid approximation,The divergence vanishes since $A_{\varphi }$ is independent of the azimuth. Inside, this agrees with the usual symmetric gauge representative $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. The Coulomb gauge condition alone does not force the other two components to vanish; the symmetry, regularity and customary boundary choice select this representative from the remaining gauge freedom.

Radial dependence. Write the interior magnetic field as $B_0\hat{\mathbf{z}}$, where $B_0={\mu }_{0}nI$ for a vacuum solenoid of solenoid turn density $n$. Its exterior magnetic field vanishes in the infinitely long idealization. Applying Stokes theorem to a coaxial circle gives $2\pi r{A}_{\varphi }(r)=\Phi (r)$. Therefore the Coulomb-gauge vector potential of a long solenoid isIt grows linearly inside, is continuous at $R$, and decays as $r^{-1}$ outside. Although its exterior curl is zero, its circulation is the enclosed magnetic flux $\Phi =B_0\pi R^2$. No regular single-valued gauge function removes it throughout the exterior while preserving this circulation.

Electric field of the rotating charged surface. For clarity retain absolute permittivity $ϵ$ and permeability $\mu$ in the cylinder formulae; setting them to unity gives the rationalized-unit convention in the question. If the stated unity means relative material constants in SI, use ${ϵ}_0,{\mu }_0$ instead. Assume prescribed charge densities in the laboratory frame and the nonrelativistic, steady-rotation model.

The axisymmetric surface charge is stationary as a distribution even while the material rotates. A Gaussian cylinder of length $L$ has enclosed charge zero for $r<R$ and $2\pi RL\sigma$ for $r>R$. Gauss's law therefore givesThe radial electric field jumps from zero to $\sigma /ϵ$ at the surface and then decays as $r^{-1}$. Its direction reverses for negative $\sigma$. The angular velocity does not enter this electric-field result for a specified surface charge.

Magnetic field of the volume-charged cylinder. A shell of radius $a$ and thickness $da$ has surface charge $\rho da$, azimuthal surface current $d{K}_{\varphi }=\rho \omega ada$, and the field of an infinite solenoid: $d{B}_z=\mu \rho \omega ada$ inside that shell and zero outside it. At radius $r<R$, only shells with $a>r$ contribute. Thus the magnetic field of a rotating charged solid cylinder isWith no applied background magnetic field, $\mathbf{B}=0$ outside. The profile is parabolic, largest on the axis and zero at the boundary. The sign follows $\rho \omega$. As a check, the current density is $J_{\varphi }=\rho \omega r$, and Ampère's circuital law gives $-d{B}_z/dr=\mu J_{\varphi }$, as required.

Electron impulse when the current is switched off. Choose positive $I$ so the initial magnetic flux is ${\Phi }_i={\mu }_{0}nI\pi R^2$ along $+z$. Faraday's law of induction gives, outside the solenoid,For an electron of electric charge $q=-e$, treat its position as fixed while the short pulse acts. The inductive impulse outside a switched solenoid isConsequently the nonrelativistic final velocity isThe induced electric field circulates in the positive azimuthal direction while a positive flux decreases, so the electron accelerates in the negative direction. Once the pulse has passed, there is no exterior field and it travels along the tangent at the impulse point, rather than continuing in a circular orbit. If the resulting momentum is relativistic, use ${\mathbf{v}}_f=c{\mathbf{p}}_f/\sqrt{m_e^2{c}^2+|{\mathbf{p}}_f{|}^2}$ instead. The sudden-switch result assumes negligible displacement during the pulse and the usual quasistatic electromagnetic induction approximation; “instantaneous” switching does not mean instantaneous propagation of fields to arbitrary distance.

One can check the sign using conserved canonical angular momentum: axial symmetry conserves $p_{\varphi }^{\mathrm{can}}=L_z+qr{A}_{\varphi }$. Initially $L_z=0$, and finally $A_{\varphi }=0$, so $L_z=q{\Phi }_i/(2\pi )=-e{\Phi }_i/(2\pi )$.

Interferometric measurement at constant current. Use a coherent electron beam split into two paths passing on opposite sides of a shielded long solenoid, then recombine them to form interference fringes. Keep the current constant during the passage and suppress leakage fields on the paths. In the Aharonov-Bohm effect, the phase difference due to the magnetic vector potential is the closed-loop integralThe overall sign reverses if the path orientation is reversed. The displacement in fringe periods is $\Delta \phi /(2\pi )$, so the measured fringe shift is linear in $I$ and repeats whenA gauge transformation adds a gradient whose closed-loop integral vanishes for a single-valued $\chi$. Endpoint changes in electron wavefunction phase cancel in the interference measurement. The measured Aharonov-Bohm phase is therefore gauge invariant: it determines the flux, or equivalently the circulation of $\mathbf{A}$. In the chosen Coulomb gauge this implies $A_{\varphi }(d)={\mu }_{0}nI{R}^2/(2d)$, but the experiment does not select or measure a unique local vector potential independently of gauge.

Far field of the rotating sphere. Interpret uniform charge as uniform volume charge of total $Q$, and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. The magnetic dipole moment of a rigidly rotating charge distribution isIts axial component is $m_z=(\rho \omega /2)\int r_{\perp }^2{d}^{3}r$. For a solid sphere, $\int r_{\perp }^2{d}^{3}r=8\pi R^5/15$ and $\rho =3Q/(4\pi R^3)$, soThe magnetic field of a rotating charged spherical body is a static magnetic dipole field at large distances:In particular, $|\mathbf{B}|={\mu }_0|\mathbf{m}|\sqrt{1+3{\cos }^2\vartheta }/(4\pi r^3)$. Both the angular mean of the magnitude and its root mean square therefore scale as $\boxed{r^{-3}}$; the latter is $B_{\mathrm{rms}}=\sqrt{2}{\mu }_0|\mathbf{m}|/(4\pi r^3)$. The angular average of the vector is zero, which is different from average strength. Steady rotation leaves the charge and current distributions stationary, so this is magnetostatics, not a radiative $r^{-1}$ field. If uniform charge instead means a thin spherical surface, replace the dipole moment by $Q{R}^2\mathbf{\omega }/3$; the angular pattern and $r^{-3}$ scaling are unchanged.

For a dipole along $z$, axial symmetry removes dependence on azimuth $\varphi$. In the Lorenz gauge, its magnetic vector potential points along $z$, so it has spherical components $A_r=A_z\cos \theta$, $A_{\theta }=-A_z\sin \theta$ and $A_{\varphi }=0$. The scalar potential is also independent of $\varphi$. Thus $E_{\varphi }=0$, while taking $\nabla \times \mathbf{A}$ leaves only an azimuthal magnetic field:The generally nonzero components are $E_r,E_{\theta },B_{\varphi }$. This statement includes the full dipole near field. In the leading $1/r$ radiation zone term, $E_r$ also vanishes; its nonzero full-field contributions fall faster than $1/r$. Confusing those two levels of approximation would incorrectly suggest four exact zeros everywhere.

The consistent Liénard–Wiechert potential obeys $\mathbf{A}=\mathbf{u}\phi /c^2$. To first order in $\mathbf{\beta }=\mathbf{u}/c$, its large-distance form isTerms varying as $1/r^2$ give near fields and are negligible for the radiated power at large $r$. Keep the first-order velocity term in $\phi$: although small, its retarded-time derivative contributes a leading acceleration field. To the required order, $\partial t_r/\partial t\simeq 1$ and $\nabla t_r\simeq -\mathbf{n}/c$. Holding the observation direction fixed at leading order givesThus the outgoing electric field is transverse:Here $\mathbf{a}=d\mathbf{u}/d{t}_r$. The magnetic field follows from $\nabla \times \mathbf{A}$, or from the outgoing plane electromagnetic wave relation:The minus sign is essential. Both fields fall as $1/r$, are mutually orthogonal and obey $|\mathbf{E}|=c|\mathbf{B}|$. The corresponding Liénard-Wiechert field also contains nonradiative terms, omitted in this leading far-field calculation.

Write a nonzero vacuum plane electromagnetic wave with wavevector $\mathbf{k}=(\omega /c)\mathbf{s}$, where $\mathbf{s}$ is a unit propagation direction. Its propagation Lorentz four-vector is $K^{\mu }=(\omega /c,\mathbf{k})$, or, up to a scalar normalization in any one frame, $n^{\mu }=(1,\mathbf{s})$. It is null because $K_{\mu }{K}^{\mu }=0$.

Choose coordinates $x^{\mu }=(ct,\mathbf{x})$ and Minkowski metric ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$. With the supplied three-dimensional boost formulas, a consistent convention assigns the printed array to upper-index components,The index conventions for the electromagnetic field tensor then require $F_{0i}=+E_i/c$ and $F_{ij}=F^{ij}$ when both indices are lowered. The printed lower-index label therefore cannot be kept together with its electric signs and a forward propagation vector in this convention.

This is not merely an arbitrary overall sign. For a valid forward wave with $\mathbf{s}=\hat{\mathbf{z}}$, $\mathbf{E}=E_0\hat{\mathbf{x}}$ and $\mathbf{B}=(E_0/c)\hat{\mathbf{y}}$, multiplying the printed array, unchanged, by $n^{\mu }=(1,0,0,1)$ gives an $x$ component $-2E_0/c$, rather than zero. One consistent repair is to retain the printed array as $F^{\mu \nu }$ and contract with $n_{\mu }=(1,-\mathbf{s})$. Equivalently, lower both tensor indices before forming the contraction asked for in the question.

Using that corrected lower-index tensor givesTo establish both electric and magnetic transversality, include the dual electromagnetic field tensor. Choose ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$; lowering both indices then givesThe source-free plane-wave Maxwell equations are precisely the pair of covariant conditionsTheir temporal components give $\mathbf{k}\cdot \mathbf{E}=0$ and $\mathbf{k}\cdot \mathbf{B}=0$. Their spatial components give the associated amplitude and orientation relations, $\mathbf{E}/c=-\mathbf{s}\times \mathbf{B}$ and $\mathbf{B}=\mathbf{s}\times \mathbf{E}/c$.