= 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:
$$
\boxed{\begin{aligned}
E_x&=0,&B_x&=0,\\
E_y&=\gamma a_y+b_y,&B_y&=-\frac{\gamma v}{c^2}a_z,\\
E_z&=\gamma a_z+b_z,&B_z&=\frac{\gamma v}{c^2}a_y.
\end{aligned}}
$$
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 $O'$, A is stationary while B moves with <velocity> $-v\widehat{\mathbf x}$. Thus A contributes its rest electrostatic field, and B contributes the boosted electric field and the opposite-sign convective magnetic field:
$$
\boxed{\begin{aligned}
E'_x&=0,&B'_x&=0,\\
E'_y&=a_y+\gamma b_y,&B'_y&=+\frac{\gamma v}{c^2}b_z,\\
E'_z&=a_z+\gamma b_z,&B'_z&=-\frac{\gamma v}{c^2}b_y.
\end{aligned}}
$$
The densities in this frame are $\lambda'_A=\eta$, $\lambda'_B=\gamma\eta$, with $I'_A=0$ and $I'_B=-\gamma\eta v$. Substituting the laboratory result into part (a) gives the same fields; for instance $E'_y=\gamma[(\gamma a_y+b_y)-v\gamma va_y/c^2]=a_y+\gamma b_y$. The magnetic contribution attributed to A disappears in its own frame, but B's contribution does not.
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