= Solution
Transform the <electromagnetic field tensor> by $F'^{\mu\nu}=\Lambda^\mu{}_{\alpha}\Lambda^\nu{}_{\beta}F^{\alpha\beta}$. For example,
$$
F'^{02}=\gamma F^{02}-\gamma(v/c)F^{12}
=-\frac\gamma c(E_y-vB_z).
$$
Reading off the other components in the same way yields \b[the transverse field transformation]
$$
\boxed{\begin{aligned}
E'_y&=\gamma(E_y-vB_z),&E'_z&=\gamma(E_z+vB_y),\\
B'_y&=\gamma(B_y+vE_z/c^2),&B'_z&=\gamma(B_z-vE_y/c^2).
\end{aligned}}
$$
The longitudinal components remain $E'_x=B'_x=0$. These signs correspond to the stated boost in which $S'$ moves with velocity $v$ relative to $S$.
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