= Solution
The <solenoidal magnetic-field constraint> implies
$$
\nabla_\perp\cdot\delta\mathbf B_\perp=-\partial_z\delta B_\parallel.
$$
With $\delta B/B_0=O(\epsilon)$ and $\partial_z=O(\epsilon\nabla_\perp)$, the right-hand side is $O(\epsilon^2k_\perp B_0)$. Thus the leading, order-$\epsilon$ perpendicular field is <divergence-free>. On a simply connected perpendicular patch, or for the nonzero perpendicular <Fourier modes> of a periodic domain, it has a <stream function>. Choose its normalization by
$$
\delta B_x=-\frac{B_0}{v_A}\partial_y\Psi,\qquad\delta B_y=\frac{B_0}{v_A}\partial_x\Psi.
$$
Consequently the leading <reduced electron magnetohydrodynamics> representation is
$$
\boxed{\frac{\delta\mathbf B}{B_0}=\frac1{v_A}\hat{\mathbf z}\times\nabla_\perp\Psi+\hat{\mathbf z}\frac{\delta B_\parallel}{B_0}+O(\epsilon^2).}
$$
The error denotes an order-$\epsilon^2$ field relative to $B_0$. The parallel perturbation is an independent scalar and is not removed by leading perpendicular solenoidality.
For clarity, the representation is asymptotic rather than an exact decomposition of an arbitrary three-dimensional <solenoidal> field. If $\delta B_\parallel$ depends on $z$, exact solenoidality can be restored by adding $\nabla_\perp\chi$ to $\delta\mathbf B_\perp$, where $\nabla_\perp^2\chi=-\partial_z\delta B_\parallel$. Its amplitude is $O(\epsilon^2B_0)$, so it lies beyond the retained field order. A perpendicular harmonic component can be absorbed into the guide field or fixed separately by <boundary conditions>.
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