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Use Faraday's law ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$ and the solenoidal magnetic-field constraint $\nabla \cdot \mathbf{B}=0$. In the nonrelativistic, single-fluid approximation, neglect Hall and other nonideal electromotive terms and use the moving-conductor moving-conductor Ohm law, $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\mathbf{J}/\sigma$. Infinite electrical conductivity with finite current gives $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, henceThis is the ideal magnetohydrodynamic induction equation. For comparison, neglecting displacement current in Ampère-Maxwell equation gives $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$; with uniform finite electrical conductivity it produces magnetic diffusion ${\eta }_m{\nabla }^2\mathbf{B}$, where the magnetic diffusivity is ${\eta }_m=1/({\mu }_0\sigma )$. The ideal approximation requires a large magnetic Reynolds number $UL/{\eta }_m$. Dropping displacement current is useful for this finite-conductivity comparison, but Faraday's law and the ideal Ohm relation already suffice for the ideal induction equation.

Expanding the curl and using the solenoidal magnetic-field constraint gives the material derivative formCombine this with mass conservation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$, to obtainNow parametrize a material curve by a fixed label $a$: $\mathbf{X}(a,t)$ obeys ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$. Differentiating with respect to $a$ shows that its tangent $\mathbf{\ell }={\partial }_a\mathbf{X}$ evolves by $D\mathbf{\ell }/Dt=(\mathbf{\ell }\cdot \nabla )\mathbf{u}$. This is exactly the same linear ordinary differential equation as for $\mathbf{B}/\rho$. Initially parallel tangents remain parallel by uniqueness, with a label-dependent proportionality factor constant along each particle trajectory. Thus magnetic field lines are transported as material curves, wherever the field and fluid flow are smooth and the field is nonzero. This is the field-line form of magnetic flux freezing.

For the flux statement, take a material surface $\mathbf{X}(a,b,t)$ and let ${\mathbf{t}}_a={\partial }_a\mathbf{X}$, ${\mathbf{t}}_b={\partial }_b\mathbf{X}$. Both tangents obey $D{t}_i/Dt=({\partial }_j{u}_i)t_j$. Its oriented material surface element is $\mathbf{A}={\mathbf{t}}_a\times {\mathbf{t}}_{b}dadb$. Differentiating the cross product, rather than assuming its transport rule, givesEquivalently, $D\mathbf{A}/Dt=(\nabla \cdot \mathbf{u})\mathbf{A}-(\nabla \mathbf{u}{)}^T\mathbf{A}$, with $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. Contracting this derived rule with the induction equation gives a pointwise cancellation:Integrating over the fixed material labels therefore proves conservation of flux through an open material surface:The surface need not be closed; its boundary is carried with the fluid. The result follows from material transport, rather than from the zero flux through a closed surface.

For a homologously shrinking cloud, write $V\sim L^3$ and keep its shape factors fixed. Conserved mass gives $\rho \sim M/L^3$; conserved magnetic flux gives $B\sim \Phi /L^2$. Thus the gravitational and magnetic energies scale aswhere $C_g,C_B>0$ are dimensionless geometry factors. Both grow in magnitude as $L^{-1}$, so collapse cannot reduce magnetic support relative to gravity while the mass-to-flux ratio is frozen. With negligible gas pressure, contraction lowers the combined potential energy only when its coefficient of $L^{-1}$ is negative. Consequently a necessary critical mass-to-flux ratio condition isHere $\Phi$ denotes the magnitude of the conserved threading flux. The numerical coefficient depends on geometry and boundary conditions; the scaling argument does not determine it or make the condition sufficient in the presence of other support.

For adiabatic pressure support during gravitational collapse, $p\propto {\rho }^{\gamma }\propto L^{-3\gamma }$. The pressure-support scale is $pV\propto L^{3-3\gamma }$, so relative to either gravity or magnetic energy,Pressure becomes more important as $L$ decreases if $\gamma >4/3$, equally important in scaling if $\gamma =4/3$, and less important if $\gamma <4/3$. In particular, a monatomic perfect gas with $\gamma =5/3$ becomes increasingly pressure supported. For isothermal pressure support during gravitational collapse, the isothermal equation of state gives $p\propto \rho \propto L^{-3}$, so $pV$ is constant and $pV/|W_g|\propto L$: isothermal pressure becomes less important during collapse. The same comparisons hold against magnetic support because its energy has the same $L^{-1}$ scaling as gravity.

Use metric $(+,-,-)$ and the paper's normalization. A static multi-vortex solution is a smooth finite-energy field configuration of Abelian Higgs vortices whose complex Higgs field has winding at spatial infinity and zeros in the plane. In a positive $N$-vortex sector, the zeros have total multiplicity $N$ and the magnetic flux is $2\pi N$; they may be separated or coincident. Finite energy requires $|\varphi |\to 1$, $D_i\varphi \to 0$ and $B=f_{12}\to 0$ at infinity. Then $a_i$ asymptotically compensates the phase gradient, so Stokes' theorem relates flux to the phase winding. The positive Bogomolny vortex equations describe this same-sign sector; negative flux uses the opposite signs and is not covered by simply mixing positive and negative zero multiplicities.

For the critically coupled Abelian Higgs field equations, vary $\varphi$ and $\bar{\varphi }$ independently. After covariant integration by parts,Thus the scalar equation and its complex conjugate areVarying the gauge field gives ${\partial }_{\mu }{f}^{\mu \nu }\delta a_{\nu }$ from the Maxwell term. The matter variation isConsequentlyThese factors use the scalar kinetic coefficient $1/2$; importing the current from a convention with coefficient one would give an incorrect factor two. The current is invariant under gauge transformations, and its conservation follows from the field equations and antisymmetry of $f^{\mu \nu }$.

For the static purely magnetic solutions take temporal gauge $a_0=0$. The scalar equation reduces toSince $[D_1,D_2]=-iB$, applying $D_1-i{D}_2$ to the first Bogomolny vortex equation givesThe second equation sets $B=(1-|\varphi {|}^2)/2$, proving the scalar equation, including at smooth vortex cores without dividing by $\varphi$.

For the spatial gauge equations, put $k_i=\mathrm{Im}(\bar{\varphi }{D}_i\varphi )$. The first equation implies $D_1\varphi =-i{D}_2\varphi$, and ${\partial }_i|\varphi {|}^2=2\mathrm{Re}(\bar{\varphi }{D}_i\varphi )$. ThereforeWith the stated metric, $j^i=-k_i$ and the two static Maxwell equations are ${\partial }_{2}B=k_1$ and ${\partial }_{1}B=-k_2$. Both follow by differentiating $B=(1-|\varphi {|}^2)/2$. The time component is the Gauss law constraint in gauge theory; it is identically satisfied because both the electric field and $j^0$ vanish in this static temporal gauge. HenceAs an energy check, the Bogomolny square completion for an Abelian Higgs vortex gives $E\ge \pi N$ in this normalization, saturated by these equations. On the plane, the Abelian Higgs vortex moduli space admits arbitrary vortex positions, including coincidence; its static energy is $\pi N$, so there is no position-dependent static potential. Their slow dynamical interactions are a different question.

The logarithmic form of a Bogomolny vortex is only local away from zeros. Write $\varphi =e^{u+i\chi }$ there, with $u=\log |\varphi |$. ThenSeparating the real and imaginary parts of $(D_1+i{D}_2)\varphi =0$ yields $u_1-{\chi }_2+a_2=0$ and ${\chi }_1-a_1+u_2=0$. Hencewhere $J(v_1,v_2)=(-v_2,v_1)$ is the positive quarter-turn. Under a gauge transformation, $\chi \mapsto \chi +\omega$ and $\mathbf{a}\mapsto \mathbf{a}+\nabla \omega$, as this relation requires.

At regular contour points, the stipulated orthogonality means $\nabla \chi \cdot \nabla u=0$. Since $J\nabla u$ is also perpendicular to $\nabla u$,Thus, in that gauge, the gauge potential is tangent to contours of $u$, equivalently contours of $|\varphi |$, and is parallel to the phase gradient wherever that gradient is nonzero. This is a gauge-dependent statement about the vector potential, not an assertion that the magnetic field is tangent to these curves, nor that the potential is a pure gradient or divergence-free. The two gradient magnitudes are not fixed by contour orthogonality.

On a phase patch $B={\partial }_1{a}_2-{\partial }_2{a}_1=-\Delta u$. Combining with the second equation gives $\Delta u+(1-e^{2u})/2=0$ away from the cores. At a zero of multiplicity $n_j$, $u\sim n_j\log |x-X_j|$, and the phase has winding $2\pi n_j$; $u$ is not a finite real smooth field there. Distributionally, the full equality isThis is the Taubes equation in the paper's amplitude logarithm, rather than the often-used $\log |\varphi {|}^2=2u$ convention. No single smooth globally defined phase or regular contour description at every zero has been assumed.

A Bogomolny bound expresses a static field energy as nonnegative squares plus a term fixed by the topological charge or boundary data. Setting the squares to zero gives the first-order Bogomolny equations. Their solutions minimize the energy in that sector and satisfy the second-order Euler-Lagrange field equations, although a general stationary solution need not attain the bound.

For one real scalar in one spatial dimension, take the Lagrangian densityA finite-energy field configuration in this static sector must approach scalar-field vacua ${\varphi }_{\pm }$ with $W^{\prime }({\varphi }_{\pm })=0$ at the two ends. Completing the square gives the square completion for a one-dimensional kink:Choose the sign for which the boundary term is nonnegative. ThusTaking the derivative of the equality equation gives ${\varphi }_{xx}=W^{\prime }{W}^{\prime \prime }=d([W^{\prime }{]}^2/2)/d\varphi$, the static Euler-Lagrange field equation. The boundary term is invariant under deformations keeping the asymptotic scalar-field vacua fixed; it is not a contribution from the local shape of the kink.

For a phi-four kink, let $c>0$, $W(\varphi )=c^2\varphi -{\varphi }^3/3$, and select the sector ${\varphi }_{-}=-c$, ${\varphi }_{+}=c$. Then $W^{\prime }=c^2-{\varphi }^2$, and the increasing Bogomolny equation is ${\varphi }_x=c^2-{\varphi }^2$. Separating variables yieldsThe integration constant $X$ is the translational collective coordinate. Directly, ${\varphi }_K^{\prime }=c^2{\mathrm{sech}}^2(c(x-X))$ and its energy density is $c^4{\mathrm{sech}}^4(c(x-X))$, whose integral is $4c^3/3$. The decreasing antikink has the reversed boundary values, profile $-{\varphi }_K$, and the same energy. A Lorentz boost produces the exact uniformly moving kink $c\tanh [c\gamma (x-X_0-vt)]$, with Lorentz factor $\gamma =(1-v^2{)}^{-1/2}$ and energy $\gamma M$.

In two spatial dimensions, choose the Abelian Higgs model at critical coupling, with $D_i={\partial }_i-i{A}_i$, $B={\partial }_1{A}_2-{\partial }_2{A}_1$, and energyThe gauge covariant derivative is used throughout this normalization. Integration by parts, using $[D_1,D_2]=-iB$, gives $\int |D_i\varphi {|}^2=\int |(D_1+i{D}_2)\varphi {|}^2+\int B|\varphi {|}^2$, with a vanishing boundary divergence for the decaying vortex fields. Combining this identity with the magnetic and potential terms gives the Bogomolny square completion for an Abelian Higgs vortex:Finite-energy field configurations have $|\varphi |\to 1$ at infinity, and $D_i\varphi \to 0$ makes the magnetic flux equal to the phase winding $2\pi N$. For $N>0$,These are the Bogomolny vortex equations for an Abelian Higgs vortex. Opposite signs give antivortices and the bound $E\ge \pi |N|$. The coefficient $\pi$ follows from the explicit energy normalization above; other conventions can give $2\pi$. Static solutions of fixed positive vortex number have equal energy independent of their positions, giving the Abelian Higgs vortex moduli space used for slow dynamics.

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

Align the polar axis with the magnetic dipole moment. The magnetic dipole field hasThe magnetic-field-line equation gives $dr/(rd\theta )=2\cot \theta$, hence the dipole magnetic-field line is $r=C{\sin }^2\theta$. Normalize the loaded bundle by its small surface angular radius ${\theta }_{\star }$ at $r=R_{\star }$. Its boundary obeysThe dipolar flux-tube area for one polar cap is consequentlyEquivalently, $BA$ is constant by conservation of magnetic flux and $B\propto r^{-3}$ near the axis. Two equal loaded caps double the total area. The $r^3$ scaling is independent of the normalization; if the loading angle is specified at another radius $r_L$, use $A(r)=\pi {\theta }_L^2{r}^3/r_L$ instead.

For $f(w)=w^M=r^M{e}^{iM\phi }$, the angular momentum operator $L_z=-i\hslash {\partial }_{\phi }$ givesThe probability density is proportional towhose maximum occurs atIn a large disk of area $A=\pi R^2$, the states that fit inside satisfy $M\lesssim qB{R}^2/(2\hslash )$. Thus the degeneracy of each Landau level is asymptoticallywhere ${\Phi }_B=BA$ is the magnetic flux. There is one state per magnetic flux quantum $h/q$, up to boundary corrections.

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

Translation parallel to the wire leaves every point of the loop at the same distance from the wire, so its magnetic flux is constant. Equivalently, the line integrals of the motional field $\mathbf{u}\times \mathbf{B}$ cancel on opposite sides. Hence

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.

Finite energy requires $|\varphi |\to 1$ and $D\varphi \to 0$ on the circle at spatial infinity. Writing $\varphi \sim e^{i\chi }$ there gives $A\sim d\chi$. The vortex number is the winding numberBy Stokes theorem, the magnetic flux is quantized:

For the rotationally symmetric Abelian Higgs vortex ansatz, $A=f(r)d\theta$ and $\varphi =h(r)e^{ik\theta }$ giveAfter the angular integration, the Abelian Higgs model energy becomesCompleting the square in the two pairs of terms givesRegularity at the origin and approach to the vacuum at infinity requireMore precisely, $h(r)=O(r^k)$ and $f(r)=O(r^2)$ near the origin. The boundary term is $k$, soThe bound is saturated exactly when both squares vanish, giving the radial Bogomolny vortex equationsThe asymptotic value $f(\infty )=k$ also makes the flux $2\pi k$, so the ansatz has vortex number $N=k$.

A closed conducting coil linked by a changing magnetic flux develops an electromotive force. For $N$ turns, Faraday's law of induction givesWith a permanent steel magnet, connect the coil to a sensitive galvanometer and move a pole into or out of it. Entry and withdrawal give opposite deflections; exchanging the pole reverses the signs. Holding both magnet and coil stationary gives no sustained induction merely because a magnetic field is present. Moving the coil instead, or changing its orientation, can produce the same necessary flux change.

Earth's magnetic field can serve as the field source. Turn a many-turn coil so that its area normal changes direction relative to the terrestrial field. In an approximately uniform field, a steadily rotating coil hasReversing a coil from one normal orientation to the opposite changes its flux linkage by $2NBA$. The Earth's weak field calls for an appropriate coil area, number of turns and detector sensitivity; it does not require a chemically supplied current in the detecting circuit. Faraday described terrestrial magneto-electric induction in his Second Series.^[ref]

The coil must provide a conducting return path for a circulating current. Its electrical resistance and self-inductance affect the response. Lenz's law fixes the sign: the induced effect opposes the change responsible for it. The moving system supplies mechanical work when current delivers energy or dissipates heat; a permanent magnet is not thereby an inexhaustible fuel.

The coils form a mutual induction apparatus. Changing primary electric current changes secondary magnetic flux linkage ${\Lambda }_2$. Faraday's law of induction givesOn closing the primary circuit, increasing electric current produces a transient secondary electric current and galvanometer deflection. By Lenz's law its direction opposes the increasing magnetic flux. Once primary electric current and core magnetization are steady, the derivative vanishes: the galvanometer returns to zero despite the continuing magnetic field.

On opening the circuit, the decreasing magnetic flux induces a secondary electric current tending to sustain that flux. The galvanometer deflects oppositely, then returns to zero. Making and breaking produce opposite transient deflections; steady primary current produces no sustained induced current.

For fixed linear coils, ${\Lambda }_2=M{I}_1$ and ${\mathcal{E}}_2=-Md{I}_1/dt$, with mutual inductance $M$. Opposite equal changes of primary electric current give opposite equal areas under the induced-electromotive force curves in that model. Peak sizes can differ with switching times. Real iron-core magnetic hysteresis and remanent magnetization can also change the flux excursion, so equality of the actual peaks is not required.

Three standard approaches are piezoelectric, electromagnetic and electrostatic energy harvesting.

In all three cases the output is constrained by the available mechanical input and damping. Nanoscale dimensions alone do not supply an energy advantage.

In the magnetic Aharonov-Bohm effect, coherent electron paths enclose inaccessible magnetic flux even though the field vanishes along the paths. The magnetic vector potential contributes relative phaseRecombination turns this phase into a fringe displacement. For an electron $q=-e$, changing flux by $h/e$ gives one full fringe period. Reversing the loop orientation reverses the phase sign. A gauge change modifies each path's phase convention but leaves the closed-loop phase and observed interference unchanged. The accessible region surrounds excluded flux, so a globally trivial gauge cannot remove that phase.

A shielded toroidal magnet provides a concrete test: electron holography compared waves passing through its hole and outside it. Superconducting and copper coatings confined the field and excluded electrons from the magnet; the measured relative phase tested enclosed flux rather than a classical Lorentz deflection along the paths.^[ref]

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.

Choose the loop normal along $+z$. Rotation about $x$ gives $B_z=B\cos \omega t$, so the linked magnetic flux is ${\Phi }_B=B{a}^2\cos \omega t$. ThereforeThe sign uses the contour orientation associated with $+z$. For the small open gap, this is the corresponding open-circuit terminal emf, with the stated terminal ordering. The tangential induced electric field is nonconservative, so its circulation is not the integral of a single-valued electrostatic potential gradient.

Faraday's law statesfor a stationary loop, with ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{S}$ and consistent contour and area orientations. The minus sign expresses Lenz's law: the induced response opposes the change of magnetic flux. An emf can exist across an open loop even when no conduction current flows.

Faraday's law states that the induced electromotive force equals minus the rate of change of linked magnetic flux:The minus sign expresses Lenz's law: the induced electric current acts to oppose the change in magnetic flux. For a fixed contour this is equivalent to $\oint \mathbf{E}\cdot d\mathbf{\ell }=-d\int \mathbf{B}\cdot d\mathbf{S}/dt$.

Let $\theta =\omega t+{\theta }_0$ be the angle between the coil's area normal and the uniform magnetic flux density. The linked magnetic flux is ${\Phi }_N=NBA\cos \theta$. Faraday's law givesThe sign uses the winding direction associated with the chosen normal; reversing the terminal convention changes it. The magnitude is largest when $\cos \theta =0$: the coil's normal is perpendicular to the magnetic field, so the plane of the coil is parallel to it. At that orientation the instantaneous magnetic flux is zero but its rate of change has maximum magnitude.

Faraday's law of induction states that the induced electromotive force around a material electrical circuit equals the negative rate of change of its linked magnetic flux:The surface orientation and direction around its boundary are related by the right-hand rule. Lenz's law supplies the physical meaning of the minus sign: the induced electric current produces a magnetic field opposing the change in linked magnetic flux. For a stationary contour the local form is $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$; for this moving wire, the magnetic Lorentz force term is also needed in the electromotive force.