 Solution

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.