= Solution
The macroscopic <Maxwell equations>, with free charge and current, are
$$
\nabla\cdot\mathbf D=\rho_f,\quad \nabla\cdot\mathbf B=0,\quad
\nabla\times\mathbf E=-\partial_t\mathbf B,\quad
\nabla\times\mathbf H=\mathbf J_f+\partial_t\mathbf D.
$$
Inside this body $\mathbf D=\epsilon_0\mathbf E$ and $\mathbf B=\mu_0\mathbf H$. The wave <magnetic field> lies along $z$; its sign follows the local wave component. Faraday's and Ampère's laws give
$$
\partial_xE_y=-\mu_0\partial_tH_z,\qquad
\boxed{\mathbf J=(0,-\partial_xH_z-\epsilon_0\partial_tE_y,0).}
$$
Static background fields are excluded from this wave solution. The magnetic <force> <density> is along $x$, with $f_x=\mu_0J_yH_z$. For a steady periodic state, $\overline{H_z\partial_tE_y}=-\overline{E_y\partial_tH_z}$. Therefore
$$
\boxed{\overline f_x=-\partial_x\left(\tfrac12\mu_0\overline{H_z^2}+\tfrac12\epsilon_0\overline{E_y^2}\right).}
$$
The result gives <radiation-force density without Ohm's law>. Integrating through the body gives its <pressure> as the boundary value of the averaged <electromagnetic energy> <density> minus its value deep inside. In both limiting cases below the latter is zero. Take $A$ as the incident <electric field>'s peak amplitude and write $Z_v=\sqrt{\mu_0/\epsilon_0}$.
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