= Solution
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 define
$$
E_*\equiv\frac{\eta}{2\pi\epsilon_0d},\qquad
a_y=0,\quad a_z=-E_*.
$$
In both cases A moves at $+v\widehat{\mathbf x}$, hence at B
$$
\mathbf E_A=-\gamma E_*\widehat{\mathbf z},\qquad
\mathbf B_A=+\frac{\gamma vE_*}{c^2}\widehat{\mathbf y}.
$$
For a filament with <linear charge density> $\lambda_B$ and convective <electric current> $I_B$, the <Lorentz force> per laboratory length is
$$
\mathbf f_B=\lambda_B\mathbf E_A+I_B\widehat{\mathbf x}\times\mathbf B_A.
$$
In Case 1, $\lambda_B=\gamma\eta$ and $I_B=\gamma\eta v$. Electric repulsion and magnetic attraction give
$$
\mathbf f_{B,1}=-\gamma^2\eta E_*\widehat{\mathbf z}
+\gamma^2\beta^2\eta E_*\widehat{\mathbf z},
$$
so
$$
\boxed{\mathbf f_{B,1}=-\frac{\eta^2}{2\pi\epsilon_0d}\widehat{\mathbf z}.}
$$
The cancellation uses $\gamma^2(1-\beta^2)=1$. In Case 2, B is stationary, $\lambda_B=\eta$ and $I_B=0$, giving
$$
\boxed{\mathbf f_{B,2}=-\frac{\gamma\eta^2}{2\pi\epsilon_0d}\widehat{\mathbf z}.}
$$
Both results have units $\mathrm{N\,m^{-1}}$ 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.
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