= Solution
\b[Gauge freedom.] The <magnetic vector potential> satisfies $\mathbf B=\nabla\times\mathbf A$. The transformation
$$
\boxed{\mathbf A\mapsto\mathbf A+\nabla\chi}
$$
leaves $\mathbf B$ unchanged because the <curl> of a <gradient> is zero. For time-dependent fields, transform the electric scalar potential as $\phi\mapsto\phi-\partial_t\chi$ as well, so that $\mathbf E=-\nabla\phi-\partial_t\mathbf A$ 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>.
\b[Gauge and nonzero component.] In the standard axisymmetric <Coulomb gauge> representative for the <long-solenoid approximation>,
$$
\boxed{\nabla\cdot\mathbf A=0,\qquad A_r=A_z=0,\qquad\mathbf A=A_\phi(r)\hat{\boldsymbol\phi}.}
$$
The divergence vanishes since $A_\phi$ is independent of the azimuth. Inside, this agrees with the usual <symmetric gauge> representative $\mathbf A=\tfrac12\mathbf B\times\mathbf r$. 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.
\b[Radial dependence.] Write the interior <magnetic field> as $B_0\hat{\mathbf z}$, where $B_0=\mu_0nI$ 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 $2\pi rA_\phi(r)=\Phi(r)$. Therefore the <Coulomb-gauge vector potential of a long solenoid> is
$$
\boxed{A_\phi(r)=\begin{cases}B_0r/2,&r<R,\\B_0R^2/(2r),&r>R.\end{cases}}
$$
It 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.
\b[<Electric field> of the rotating charged surface.] For clarity retain absolute <permittivity> $\epsilon$ 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 $\epsilon_0,\mu_0$ 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 gives
$$
\boxed{\mathbf E(r)=\begin{cases}0,&r<R,\\\dfrac{\sigma R}{\epsilon r}\hat{\mathbf r},&r>R.\end{cases}}
$$
The radial <electric field> jumps from zero to $\sigma/\epsilon$ 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.
\b[<Magnetic field> of the volume-charged cylinder.] A shell of radius $a$ and thickness $da$ has surface charge $\rho\,da$, azimuthal surface current $dK_\phi=\rho\omega a\,da$, and the field of an infinite <solenoid>: $dB_z=\mu\rho\omega a\,da$ 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> is
$$
\boxed{\mathbf B(r)=\mu\rho\omega\int_r^R a\,da\,\hat{\mathbf z}=\frac{\mu\rho\omega}{2}(R^2-r^2)\hat{\mathbf z},\qquad r<R.}
$$
With 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 $J_\phi=\rho\omega r$, and <Ampère's circuital law> gives $-dB_z/dr=\mu J_\phi$, as required.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2017-q4-cylindrical-fields.png]
{title=Radial profiles of the solenoid vector potential and the two charged-cylinder fields}
\b[<Electron> impulse when the current is switched off.] Choose positive $I$ so the initial <magnetic flux> is $\Phi_i=\mu_0nI\pi R^2$ along $+z$. <Faraday's law of induction> gives, outside the <solenoid>,
$$
E_\phi(d,t)=-\frac{1}{2\pi d}\frac{d\Phi}{dt}.
$$
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> is
$$
\Delta\mathbf p=-\frac{q}{2\pi d}(\Phi_f-\Phi_i)\hat{\boldsymbol\phi}=-\frac{e\Phi_i}{2\pi d}\hat{\boldsymbol\phi}.
$$
Consequently the nonrelativistic final <velocity> is
$$
\boxed{\mathbf v_f=-\frac{e\mu_0nIR^2}{2m_ed}\hat{\boldsymbol\phi}.}
$$
The 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 $\mathbf v_f=c\mathbf p_f/\sqrt{m_e^2c^2+|\mathbf p_f|^2}$ 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 $p_\phi^{\rm can}=L_z+qrA_\phi$. Initially $L_z=0$, and finally $A_\phi=0$, so $L_z=q\Phi_i/(2\pi)=-e\Phi_i/(2\pi)$.
\b[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 integral
$$
\boxed{\Delta\varphi=\frac{q}{\hbar}\oint\mathbf A\cdot d\mathbf l=\frac{q\Phi}{\hbar}=-\frac{e\mu_0n\pi R^2}{\hbar}I\pmod{2\pi}.}
$$
The overall sign reverses if the path orientation is reversed. The displacement in fringe periods is $\Delta\varphi/(2\pi)$, so the measured fringe shift is linear in $I$ and repeats when
$$
\boxed{\Delta I=\frac{h}{e\mu_0n\pi R^2}.}
$$
A <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 $A_\phi(d)=\mu_0nIR^2/(2d)$, but \b[the experiment does not select or measure a unique local vector potential independently of gauge].
\b[Far field of the rotating sphere.] Interpret uniform charge as uniform volume charge of total $Q$, and take $\boldsymbol\omega=\omega\hat{\mathbf z}$. The <magnetic dipole moment of a rigidly rotating charge distribution> is
$$
\mathbf m=\frac12\int\mathbf r\times\mathbf J\,d^3r,\qquad \mathbf J=\rho\boldsymbol\omega\times\mathbf r.
$$
Its axial component is $m_z=(\rho\omega/2)\int r_\perp^2d^3r$. For a solid sphere, $\int r_\perp^2d^3r=8\pi R^5/15$ and $\rho=3Q/(4\pi R^3)$, so
$$
\boxed{\mathbf m=\frac{QR^2}{5}\boldsymbol\omega.}
$$
The <magnetic field of a rotating charged spherical body> is a static <magnetic dipole field> at large distances:
$$
\boxed{\mathbf B(\mathbf r)=\frac{\mu_0}{4\pi r^3}\bigl[3(\mathbf m\cdot\hat{\mathbf r})\hat{\mathbf r}-\mathbf m\bigr],\qquad r\gg R.}
$$
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 $B_{\rm rms}=\sqrt2\mu_0|\mathbf m|/(4\pi r^3)$. The angular average of the \i[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 $QR^2\boldsymbol\omega/3$; the angular pattern and $r^{-3}$ scaling are unchanged.
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