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Consider lab-neutral parallel wires with stationary positive line density $+\lambda$ and electron line density $-\lambda$. Let wire 1's electrons move at $v_1$ along the wire and wire 2's at $v_2$. The conventional currents are $I_j=-\lambda v_j$. In the rest frame of wire 2's electrons, wire 1's transformed total linear charge density is ${\lambda }_1^{\prime }={\gamma }_2[0-v_2{I}_1/c^2]={\gamma }_2\lambda v_1{v}_2/c^2$. Its radial electric field is ${\lambda }_1^{\prime }/(2\pi {ϵ}_{0}d)$.

A wire-2 electron at rest there feels only an electric force. Transforming its transverse force back to the lab divides it by ${\gamma }_2$, and multiplying by the lab electron count per length $\lambda /e$ giveswith positive radial direction away from wire 1. The lab-stationary positive charges feel zero net electric force because wire 1 is neutral in their frame. Equal current directions attract; opposite directions repel. Transforming force on each carrier and then counting carriers in the specified frame avoids an incorrect extra length-contraction factor in force per unit length.

A force per unit length describes a distributed force on a filament and has units $\mathrm{N}{\mathrm{m}}^{-1}$. For a charged filament with linear charge density $\lambda$ and longitudinal current $I$, the external Lorentz force is $\mathbf{f}=\lambda \mathbf{E}+I\hat{\mathbf{t}}\times \mathbf{B}$. Length, density and force must all refer to the same measurement frame. An infinite uniform filament generally has a finite force density but no finite total force.

Put wire 1 on the $z$ axis and wire 2 at $x=d$, with positive $x$ pointing away from wire 1. Let each wire have stationary positive linear charge density $+\lambda$ and moving Electron line density $-\lambda$ in the lab. For equal currents, both Electron populations have drift velocity $+v\hat{\mathbf{z}}$, so the signed conventional current is $I=-\lambda v$.

First consider wire 2's positive charges in their own rest frame, which is the lab. Wire 1 has zero total line charge there, so its external electrostatic field is zero. The positive-charge force per length is therefore $\boxed{f_{+}=0}$.

Next use the Electron rest frame $S^{\prime }$ moving at $+v\hat{\mathbf{z}}$. Wire 1's ion line density becomes ${\lambda }_{+}^{\prime }=\gamma \lambda$. Its Electron density becomes ${\lambda }_{-}^{\prime }=-\lambda /\gamma$: the Electron rest density is lower than its lab density. The transformed total isEquivalently this follows from the four-current transformation ${\lambda }_1^{\prime }=\gamma ({\lambda }_{1,\mathrm{net}}-v{I}_1/c^2)$, with ${\lambda }_{1,\mathrm{net}}=0$ and $I_1=-\lambda v$. By the electrostatic field of a line charge,A wire-2 Electron is at rest in $S^{\prime }$ and feels only the electric force, $F_{e,x}^{\prime }=-e{\lambda }_1^{\prime }/(2\pi {ϵ}_{0}d)$, toward wire 1.

Apply the supplied transverse relativistic force transformation with the Electron's lab velocity $\mathbf{u}=\mathbf{v}$. Its denominator is $\gamma (1-v^2/c^2)=1/\gamma$, so $F_{e,x}^{\prime }=\gamma F_{e,x}$, or $F_{e,x}=F_{e,x}^{\prime }/\gamma$. There are $\lambda /e$ Electrons per unit lab length. ThusAdding $f_{+}=0$ gives exactly the attractive magnetic result. This electrostatic-frame derivation of parallel-wire forces uses electrostatics in each test carrier's own rest frame; no magnetic force acts on that stationary test carrier there. The other wire can still carry current in that frame. Transform the force per Electron first and then count Electrons per lab length: transforming a force per unit length as if it were a particle force would introduce an incorrect extra contraction factor.

Write $a_y=E_{A,y}^0$, $a_z=E_{A,z}^0$, $b_y=E_{B,y}^0$ and $b_z=E_{B,z}^0$, evaluated at the same transverse coordinates $(y,z)$. A superscript $0$ labels each filament's own rest frame, rather than a single common frame in all cases. The filaments are infinite and uniform along $x$, so the change of longitudinal coordinate under a Lorentz boost creates no extra dependence in these functions. After the acceleration transients have passed, the magnetic field of each filament vanishes in its own rest frame.

For a filament moving at $+v\hat{\mathbf{x}}$, use the inverse of part (a): its transverse electric field in the laboratory is $\gamma {\mathbf{E}}^0$ and its magnetic field is $\mathbf{v}\times \mathbf{E}/c^2$. Superposition for the two co-moving filaments therefore gives, in $O$,In the common co-moving frame $O^{\prime }$ both filaments are at rest, soFor example, $B_y^{\prime }=\gamma [-\gamma v(a_z+b_z)/c^2+v\gamma (a_z+b_z)/c^2]=0$, a useful sign check. The supplied $\eta$ is the proper linear charge density. Thus each laboratory density is $\gamma \eta$ and each laboratory electric current is $\gamma \eta v$, consistently with length contraction and the four-current transformation. In $O^{\prime }$ each density is $\eta$ and both currents vanish.

Retain $a_y,a_z,b_y,b_z$ as defined in Case 1. In the laboratory, A moves and B is stationary. Transform A's fields and add B's unchanged electrostatic field:The laboratory linear charge densities are ${\lambda }_A=\gamma \eta$ and ${\lambda }_B=\eta$, with electric currents $I_A=\gamma \eta v$ and $I_B=0$.

In $O^{\prime }$, A is stationary while B moves with velocity $-v\hat{\mathbf{x}}$. Thus A contributes its rest electrostatic field, and B contributes the boosted electric field and the opposite-sign convective magnetic field:The densities in this frame are ${\lambda }_A^{\prime }=\eta$, ${\lambda }_B^{\prime }=\gamma \eta$, with $I_A^{\prime }=0$ and $I_B^{\prime }=-\gamma \eta v$. Substituting the laboratory result into part (a) gives the same fields; for instance $E_y^{\prime }=\gamma [(\gamma a_y+b_y)-v\gamma v{a}_y/c^2]=a_y+\gamma b_y$. The magnetic contribution attributed to A disappears in its own frame, but B's contribution does not.

The infinitely long filaments have no finite total force; the meaningful result is force per unit length in the laboratory. Evaluate only A's field at B, excluding B's singular self-field. B is at $y=0,z=-d/2$, so defineIn both cases A moves at $+v\hat{\mathbf{x}}$, hence at BFor a filament with linear charge density ${\lambda }_B$ and convective electric current $I_B$, the Lorentz force per laboratory length isIn Case 1, ${\lambda }_B=\gamma \eta$ and $I_B=\gamma \eta v$. Electric repulsion and magnetic attraction givesoThe cancellation uses ${\gamma }^2(1-{\beta }^2)=1$. In Case 2, B is stationary, ${\lambda }_B=\eta$ and $I_B=0$, givingBoth results have units $\mathrm{N}{\mathrm{m}}^{-1}$ and are repulsive, away from A, irrespective of the common sign of $\eta$. In Case 1 the velocity independence of force per laboratory length does not violate the transverse force transformation: a material segment of proper length ${\ell }_0$ occupies laboratory length ${\ell }_0/\gamma$, so its force is the rest-frame force divided by $\gamma$. The calculation interprets the stated $\eta$ as each filament's final proper density; holding a laboratory density fixed during acceleration would be a different preparation.

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,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 ${\mathbf{B}}_{A,\mathrm{rest}}=0$. 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 $E_{\ast }=\eta /(2\pi {ϵ}_{0}d)$ as in part (c). At the respective filament axes, the external fields experienced in each carrier distribution's own rest frame areFor the first expression B is at rest in $O$; for the second A is at rest in $O^{\prime }$ and the field comes from B moving at $-v\hat{\mathbf{x}}$. For a carrier with magnetic dipole moment ${\mathbf{\mu }}_s$, the interaction $H_s=-{\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}}$ distinguishes opposite transverse spin orientations. A field gradient could exert a spin-dependent force $\nabla ({\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}})$ 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 $+\hat{\mathbf{y}}$ and its spin toward $-\hat{\mathbf{y}}$. 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.

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.