= Solution
For the <ordinary plasma wave>, choose $\mathbf k=k\hat{\mathbf x}$ and $\delta\mathbf E=E_z\hat{\mathbf z}$. The equations in <1/a> give $u_{xs}=u_{ys}=0$ and
$$
u_{zs}=\frac{iq_s}{m_s\omega}E_z.
$$
The driven particle motion is entirely parallel to $\mathbf B_0$: the parallel <Lorentz force> has no magnetic contribution. Nevertheless, \b[the <ordinary plasma wave> is electromagnetic], since $\delta\mathbf E\perp\mathbf k$ and the <Faraday law> gives
$$
\boxed{\delta\mathbf B=-\frac{ck}{\omega}E_z\hat{\mathbf y}\ne0.}
$$
An <electrostatic plasma wave> instead has $\delta\mathbf E\parallel\mathbf k$ and zero perturbed <magnetic field>. Alignment with the equilibrium <magnetic field> does not make a wave electrostatic. At the exact <plasma wave cutoff> $k=0$, this propagating-wave distinction becomes degenerate.
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