= Solution
The information in the original question is insufficient to determine a unique spin-induced change of <charge density>. It specifies ideal dielectric filaments with a prescribed excess <linear charge density>, but not the <charge carrier> species, their <spin> state, <charge carrier mobility>, binding potential or relaxation mechanism. A classical line charge has no spin variable at all. The following conditional conclusions are supported by the fields already calculated.
For a carrier moving with its filament, the local <magnetic field> in its instantaneous <rest frame> is, for the transverse fields here,
$$
\mathbf B_{\rm rest,\perp}=\gamma_u\left(\mathbf B-\frac{\mathbf u\times\mathbf E}{c^2}\right),
$$
where $\mathbf u$ is that carrier's laboratory <velocity>. Exclude its filament's singular self-field and first retain only the prescribed straight convective motion.
In Case 1 there is a common <rest frame> in which both filaments and the excess charges are stationary and the interfilament <magnetic field> vanishes. Equivalently, in the laboratory $\mathbf B_A=\mathbf v\times\mathbf E_A/c^2$, so B's co-moving charges have $\mathbf B_{A,\rm rest}=0$. Thus uniform common translation alone creates no extra mutual rest-field spin splitting. This does not exclude intrinsic <spin-orbit coupling> of electrons moving within their atomic binding potentials; that material physics was never specified.
In Case 2 there is relative motion and no frame in which both source distributions are stationary. Put $E_*=\eta/(2\pi\epsilon_0d)$ as in part (c). At the respective filament axes, the external fields experienced in each carrier distribution's own rest frame are
$$
\boxed{\mathbf B_{A\to B}^{(B\,\rm rest)}
=\frac{\gamma vE_*}{c^2}\widehat{\mathbf y},\qquad
\mathbf B_{B\to A}^{(A\,\rm rest)}
=\frac{\gamma vE_*}{c^2}\widehat{\mathbf y}.}
$$
For the first expression B is at rest in $O$; for the second A is at rest in $O'$ and the field comes from B moving at $-v\widehat{\mathbf x}$. For a carrier with <magnetic dipole moment> $\boldsymbol\mu_s$, the interaction $H_s=-\boldsymbol\mu_s\cdot\mathbf B_{\rm rest}$ distinguishes opposite transverse spin orientations. A field gradient could exert a <spin>-dependent <force> $\nabla(\boldsymbol\mu_s\cdot\mathbf B_{\rm rest})$ on mobile carriers; <spin relaxation> could instead produce a <spin polarization>. These are possible responses, not consequences of charge density alone. In B's frame the displayed interaction is directly a magnetic-moment coupling to the moving source's field; treating the moving carrier's electrostatic orbital term in another frame also requires consistent binding and acceleration dynamics, including any applicable <Thomas precession>.
For example, if $\eta>0$ and mobile electron spins can relax, the lower magnetic-energy orientation has the electron's magnetic moment toward $+\widehat{\mathbf y}$ and its <spin> toward $-\widehat{\mathbf y}$. Oppositely oriented moments would experience opposite gradient forces. The sign changes if the source charge changes sign. None of these statements fixes a net charge redistribution without a confinement and response model, and the stipulated zero transverse size supplies no resolved “sides” on which charge could accumulate.
\b[Therefore Case 1 has no new mutual spin effect from common bulk motion alone; Case 2 permits motion-induced spin-dependent interactions, but neither case determines an additional charge-density profile from the supplied data.] In particular, the factor $\gamma$ relating proper and laboratory <linear charge densities> in part (b) follows from <length contraction> and is not an effect of <spin-orbit coupling>. If the charges remain pinned in an ideal dielectric, their imposed density remains the prescribed density in either case.
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