Past exam of the optics and electrodynamics course of the University of Cambridge / 2018 / Section B / B4 / Opposite currents

Keep wire 1's Electrons moving at $+v$, but let wire 2's Electrons move at $-v$, so $I_2=-I_1$. The positive charges remain lab-stationary and still feel zero external electrostatic force in the lab. The appropriate rest frame for wire 2's Electrons now moves at $-v$.

In this frame, wire 1's total line charge isIt produces an electric field pointing toward wire 1. The stationary test Electron is negatively charged, so its electric force points away from wire 1. Transforming back again divides the individual transverse force by $\gamma$, and multiplying by $\lambda /e$ givesThe changed sign arises from the changed transformed charge imbalance, without needing a magnetic-force calculation. More generally the signed result is $F_x/\ell =-{\mu }_0{I}_1{I}_2/(2\pi d)$ for the chosen outward radial direction.