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Write $m_0={\mu }_0\rho$ and treat the imposed mean magnetic field as uniform on the forcing scale. In first-order smoothing, retain terms linear in the fluctuating fields in their equations, while keeping their quadratic correlation when calculating the mean-field electromotive force. The inertial fluctuation, nonlinear induction term and quadratic fluctuating Lorentz force density are neglected in the steady response. The linear magnetic force isAbsorb the gradient into the total pressure. The resulting equations arewith $\nabla \cdot \mathbf{u}=\nabla \cdot \mathbf{b}=0$. Taking the divergence of the first equation gives ${\nabla }^{2}P=0$ for solenoidal forcing. In a periodic or homogeneous Fourier calculation its nonconstant part therefore vanishes.

For each Fourier mode with wavevector $\mathbf{q}$, $|\mathbf{q}|=k$, define $\ell =\mathbf{B}\cdot \mathbf{q}$ andSolving the two coupled linear equations gives the monochromatic coupled magnetic and velocity responseThe magnetic coupling suppresses modes with large projection of their wavevector along the imposed field.

To calculate the mean-field electromotive force, group each pair of opposite wavevectors into a real mode and average its spatial phase. Put $\mathbf{n}=\mathbf{q}/k$, and define its forcing correlationsFor any solenoidal real mode $\mathbf{v}=\mathbf{a}\cos (\mathbf{q}\cdot \mathbf{x})+\mathbf{c}\sin (\mathbf{q}\cdot \mathbf{x})$, its two amplitude vectors are perpendicular to $\mathbf{q}$. Direct multiplication givesAlso, integration by parts gives $⟨(L\mathbf{g})\times (L\mathbf{f})⟩=-{\ell }^2⟨\mathbf{f}\times \mathbf{g}⟩$, where $L=\mathbf{B}\cdot \nabla$. Expanding the response products therefore yieldsSumming over independent real modes, or integrating over their angular spectrum, is the required general answer for $⟨\mathbf{u}\times \mathbf{b}⟩$.

At zero field, $\mathcal{E}(0)={\sum }_{\mathbf{q}}{\mathbf{C}}_{\mathbf{q}}/(\nu \eta k^4)$. Thus the stated zero-field condition removes the unweighted sum of the cross correlations. For anisotropic correlated forcing it does not necessarily remove their differently weighted sum at nonzero field. The displayed general formula retains that allowed contribution. It disappears if cross forcing is absent mode by mode, or under the jointly isotropic statistics assumed for the final part.

The helicity content can also be expressed directly using the actual fluctuating fields. Define their kinetic helicity $H_u(\mathbf{q})=⟨{\mathbf{u}}_{\mathbf{q}}\cdot \nabla \times {\mathbf{u}}_{\mathbf{q}}⟩$ and current helicity $H_b(\mathbf{q})=⟨{\mathbf{b}}_{\mathbf{q}}\cdot \nabla \times {\mathbf{b}}_{\mathbf{q}}⟩$. Substitute $\mathbf{f}=\nu k^2\mathbf{u}-L\mathbf{b}/m_0$ and $\mathbf{g}=\eta k^2\mathbf{b}-L\mathbf{u}$ into their cross correlation. This gives the exact identityIt shows the competing kinetic and magnetic contributions without introducing a spurious singularity when the factor on the left vanishes. For a monochromatic field, ${\mathbf{a}}_b=\nabla \times \mathbf{b}/k^2$ is a magnetic vector potential, so $H_b=k^2⟨{\mathbf{a}}_b\cdot \mathbf{b}⟩$: current helicity is $k^2$ times magnetic helicity. In the isotropic weak-field limit, with zero-field fluctuations ${\mathbf{u}}_0,{\mathbf{b}}_0$, the scalar alpha effect isThis identifies both signs and their different viscous and magnetic relaxation times.

For jointly isotropic forcing let $h_f,h_g$ now denote the total forcing helicities, and set $S_h=\nu h_g/m_0-\eta h_f$. The cross-forcing contribution is even in $\mathbf{B}$, whereas isotropy requires any vector depending only on $\mathbf{B}$ to be parallel to it and odd under reversal. Hence that contribution vanishes. A natural symmetric alpha tensor isOnly its longitudinal coefficient contributes to $\mathcal{E}={\alpha }_{\parallel }\mathbf{B}$. Introduce $h=|B|/[k\sqrt{m_0\nu \eta }]$ and $A_0=\nu \eta k^4$. Angular integration givesTherefore the strong-field quenching of an isotropic electromotive force isprovided $S_h\ne 0$. If the two helicity contributions cancel exactly, this electromotive force vanishes.

There is a tensor qualification to the printed scaling claim. The natural transverse coefficient instead contains $\frac{1}{2}{\int }_0^1(1-{\mu }^2)(1+h^2{\mu }^2{)}^{-2}d\mu \sim \pi /(8h)$ and scales as $|B{|}^{-1}$. The relation ${\mathcal{E}}_i={\alpha }_{ij}{B}_j$ by itself leaves transverse components undetermined. One may choose the effective representative ${\alpha }_{ij}={\alpha }_{\parallel }{\delta }_{ij}$, for which the requested $|B{|}^{-3}$ scaling holds. It is the observable longitudinal response that is fixed, rather than every component of a unique field-dependent tensor.

For a thin wire carrying current $I$, the given magnetic vector potential reduces to $\mathbf{A}=({\mu }_{0}I/4\pi ){\oint }_{C}d{\mathbf{r}}^{\prime }/|\mathbf{r}-{\mathbf{r}}^{\prime }|$. Taking its curl and using ${\nabla }_{\mathbf{r}}(1/|\mathbf{r}-{\mathbf{r}}^{\prime }|)=-(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$ gives the Biot-Savart lawThis is exactly the sign convention in the printed form using ${\mathbf{r}}^{\prime }-\mathbf{r}$.

Let $\mathbf{v}=(\mathbf{r}-{\mathbf{r}}^{\prime })/|\mathbf{r}-{\mathbf{r}}^{\prime }{|}^3$. Away from the wire, $\nabla \cdot \mathbf{v}=0$. The supplied divergence and curl of a cross product therefore gives $\nabla \times (d{\mathbf{r}}^{\prime }\times \mathbf{v})=-(d{\mathbf{r}}^{\prime }\cdot {\nabla }_{\mathbf{r}})\mathbf{v}$. But ${\nabla }_{\mathbf{r}}\mathbf{v}=-{\nabla }_{{\mathbf{r}}^{\prime }}\mathbf{v}$, so the integrand is the total differential of $\mathbf{v}$ along the source loop. Its closed-loop integral is zero. Thus $\nabla \times \mathbf{B}=0$ at every point outside $C$, as demanded by the current-free magnetostatic Ampère's law.

For $r^{\prime }=f(\theta )>0$, $z^{\prime }=0$, and counterclockwise current, $d{\mathbf{r}}^{\prime }=(f^{\prime }{\mathbf{e}}_r+f{\mathbf{e}}_{\theta })d\theta$. At the origin, $\mathbf{v}=-{\mathbf{e}}_r/f^2$, soFor the ellipse expressed relative to a focus, $1/f=(1-e\cos \theta )/\ell$. The cosine integrates to zero, givingHere $\ell$ is the semilatus rectum. Reversing the current reverses the field direction.

The product of the magnetic field with the velocity in this question is the three-dimensional cross product. Fix ${ϵ}_{123}=1$ and write $K^i{}_j={ϵ}^i{}_{kj}{B}^k$, so $K\mathbf{v}=\mathbf{B}\times \mathbf{v}$. In coordinates $y^0=t$, $y^i=x^i$ on ${\mathbb{R}}^3\times \mathbb{R}$, prescribe an affine connection by the following Christoffel symbols:with all other Christoffel symbols zero. These are smooth globally in the given Cartesian coordinates. The lower-index symmetry makes this a torsion-free connection. It is the geometrization of the Lorentz force by an affine connection; no condition on the curl of $\mathbf{E}$ or the divergence of $\mathbf{B}$ is required for this construction.

An affinely parametrized geodesic, with primes denoting differentiation with respect to $\lambda$, satisfiesOn the branch $t^{\prime }=c\ne 0$, use $t$ itself as an affine parameter. Dividing the spatial equation by $c^2$ givesConversely, each solution of this equation makes $\lambda =t$, $y(t)=(t,\mathbf{x}(t))$ an affinely parametrized geodesic of the constructed affine connection. Therefore its image is also an unparametrized geodesic. A general change of parameter adds a term proportional to the tangent in the geodesic equation, leaving the curve unchanged. The branch $t^{\prime }=0$ is not a trajectory with time as parameter.

For the metric realization, assume $\mathbf{E}=-\nabla U$ and $\nabla \cdot \mathbf{B}=0$. Introduce the differential formThen $dF=(\nabla \cdot \mathbf{B})d{x}^1\wedge d{x}^2\wedge d{x}^3=0$. The global Poincaré lemma on the contractible space ${\mathbb{R}}^3$ supplies a one-form $A=A_{i}d{x}^i$ with $dA=F$, equivalently a magnetic vector potential with $\nabla \times \mathbf{A}=\mathbf{B}$. An explicit radial-gauge potential for a divergence-free magnetic field isThis is the radial homotopy formula for a closed two-form and is smooth even at the origin. For a constant magnetic field it gives $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.

Consider the Eisenhart-Duval lift with the Lorentzian metricThe independent one-forms $d{x}^1,d{x}^2,d{x}^3,dt,du-2A_{i}d{x}^i-Udt$ exhibit three positive directions and a two-dimensional block with one positive and one negative direction. Thus the metric is nondegenerate, of signature $(4,1)$. Its coefficients do not depend on $u$, and $g_{uu}=0$, so $\xi ={\partial }_u$ is a null Killing vector field. Indeed $g_{Au}$ are constant, so ${\Gamma }^A{}_{Bu}=0$ for the Levi-Civita connection and $\xi$ is parallel.

The geodesic Lagrangian for an affine parameter $\lambda$ isThe cyclic coordinate $u$ gives the conserved quantity $p_u=\partial L/\partial u^{\prime }=t^{\prime }$. Work at a nonzero value of $p_u$ and rescale the affine parameter to set $t^{\prime }=1$. This is the essential step in the null Kaluza-Klein reduction of a stationary force: one fixes the momentum along the null isometry and projects its geodesics, rather than dividing by $g_{uu}$.

The spatial Euler-Lagrange equations, before setting $t^{\prime }=1$, areSince $t^{\prime \prime }=0$ and $F_{ij}{v}^j=-(\mathbf{B}\times \mathbf{v}{)}_i$, their reduction isEquivalently, after taking $t$ as the affine parameter, the term $\dot{u}$ is a total derivative and the reduced Lagrangian is $L_{\mathrm{red}}=\frac{1}{2}|\dot{\mathbf{x}}{|}^2-2\mathbf{A}\cdot \dot{\mathbf{x}}-U$. Its Euler-Lagrange equations give the same sign and factor two.

There is also an explicit converse using null geodesics. For any physical trajectory, setThis makes its five-dimensional tangent null. The conserved momentum of the cyclic coordinate $t$ becomesThe quantity in parentheses is conserved, because its derivative is $\dot{\mathbf{x}}\cdot (2\mathbf{B}\times \dot{\mathbf{x}})=0$. Hence the $t$ equation, as well as the spatial and $u$ Euler-Lagrange equations, is satisfied. Every trajectory admits a null geodesic lift with $p_u=1$, and every such lift projects to the required trajectory.

Finally, a time-independent gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ is absorbed by $u\mapsto u+2\chi$. The one-form $du-2A_{i}d{x}^i-Udt$ and the five-dimensional metric are unchanged. These gauge transformations of an Eisenhart-Duval lift alter the reduced Lagrangian only by the total derivative $-2d\chi /dt$.

The MHD induction equation can be written in material form asIts $y$ component immediately givesFor the Cartesian magnetic flux function, the $x,z$ components of the MHD induction equation, or the $y$ component of the magnetic vector potential equation, giveThe additive function of time in $\psi$ has no effect on $\mathbf{B}$. Choose this gauge so that $C=0$. Then the flux label is materially conserved:This is magnetic flux freezing in the two-dimensional geometry.

To prove the absence of axial motion, construct an invariant solution with $u_y=0$. Along each Lagrangian trajectory, $\psi$ is fixed, and the assumed divergence givesThus $B_y$ remains a function of $\psi$ and time alone. Its gradient stays parallel to $\nabla \psi$, so the $y$ component of the Lorentz force remains zero. The axial momentum equation isbecause $(B_x{\partial }_x+B_z{\partial }_z)\psi =0$. With $u_y=0$ initially, it remains zero. Hence neither axial force nor axial motion is generated. This flux-surface preservation during magnetic-tube expansion argument assumes the smooth ideal evolution for which the initial-value problem is unique; the particular in-plane rising motion need not be calculated.

Choose an origin near the bounded loop and write ${\mathbf{r}}^{\prime }$ for the source position. The magnetic vector potential and its far-field multipole expansion arewith the loop held fixed as $r=|\mathbf{r}|\to \infty$. The first term integrates to zero since $C$ is closed. Define its oriented vector area byThe second equality is Stokes theorem, applied componentwise to any oriented spanning surface. In particular the vector area is independent of the choice of spanning surface. From ${\oint }_{C}d(r_i^{\prime }{r}_j^{\prime })=0$, the matrix ${\oint }_C{r}_i^{\prime }d{r}_j^{\prime }$ is antisymmetric, which gives ${\oint }_C(\mathbf{r}\cdot {\mathbf{r}}^{\prime })d{\mathbf{r}}^{\prime }=\mathbf{S}\times \mathbf{r}$. HenceThe orientation follows the current by the right-hand rule. For $r\ne 0$, taking the curl of this leading magnetic vector potential gives the magnetic dipole fieldFor a planar loop, $\mathbf{S}$ is the signed area times the unit normal vector. For a nonplanar loop it is the oriented vector area, rather than the scalar area of an arbitrarily chosen surface.

The free-space Green function of the Laplacian for the Poisson equation gives the electric potentialFor $r=|\mathbf{x}|\gg R$, its electric multipole expansion begins withDefine the total electric charge and electric dipole moment byThen

The analogous solution for the magnetic vector potential isIts apparent $r^{-1}$ coefficient is ${\mu }_0(4\pi {)}^{-1}\int \mathbf{J}{d}^3{x}^{\prime }$. For each component, the supplied identity and the divergence-free vector field condition giveThe divergence theorem turns this into a surface integral outside the compact support of the current density, where $\mathbf{J}=0$. Hence $\int \mathbf{J}{d}^3{x}^{\prime }=0$, so the $1/r$ term vanishes.

Because $u^0=c,dt/d\tau =\gamma c$ is constant, $dt/d\tau =\gamma$. The coordinate speed along the circle is consequentlysoFor the convention in the question, take $qB>0$, so $\omega >0$ and $\Delta \tau =2\pi /\omega$.

On the circular solution, the free part of the relativistic charged-particle action isIn the gauge $A_y=Bx$, the interaction is the closed line integralThe orbit is clockwise for $qB>0$, so $\oint xdy=-\pi R^2$. ThusCombining the two terms and using ${\gamma }^2{v}^2=c^2({\gamma }^2-1)$ gives the charged-particle action over one relativistic cyclotron orbitAlthough the intermediate magnetic line integral depends on the chosen magnetic vector potential, the closed-orbit result is gauge invariant.

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

Vanishing net charge density does not require vanishing current. Positive and negative charge carriers can cancel in charge density while their oppositely directed motions add to a nonzero current. Charge conservation only requiresfor magnetostatics this becomes $\nabla \cdot J=0$.

Because $\nabla \cdot B=0$, one may introduce a magnetic vector potential withIt is not unique: $A+\nabla \chi$ gives the same field for any scalar $\chi$.

For the stated current,so it is consistent with stationary charge conservation. Direct calculation givesThus $J$ is a Beltrami field. For $\lambda \ne 0$, chooseThen $\nabla \cdot B=0$ and $\nabla \times B={\mu }_{0}J$. A convenient Coulomb-gauge potential issince $\nabla \times A=B$ and $\nabla \cdot A=0$. Gradient gauge terms may of course be added.

If $\lambda =0$, the current is the constant field $J=(0,J_0,0)$. The formulas involving $1/\lambda$ do not apply; one valid choice is

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.

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]

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.

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.