= Solution
Let $\delta=\sqrt{2\eta/\omega}$ and $q=(1+i)/\delta$. Within the <magnetic skin layer>, differentiation across the layer dominates differentiation along the wall by the factor $b/\delta$. The leading <resistive induction equation> is therefore
$$
i\omega B_x=\eta\partial_y^2B_x,\qquad B_x(x,0)=C(x),\qquad B_x\longrightarrow0\quad(y/\delta\longrightarrow\infty).
$$
Because $\eta q^2=i\omega$ and $\operatorname{Re}q>0$, its decaying solution is
$$
\boxed{B_x=C(x)e^{-qy},\qquad B_z=0.}
$$
The <solenoidal magnetic-field constraint> determines the first smaller component,
$$
B_y=\frac{C'(x)}q e^{-qy},\qquad \frac{B_y}{B_x}=O(\delta/b).
$$
Corrections to $B_x$ from longitudinal diffusion have relative order $(\delta/b)^2$. Thus the zero normal boundary field from the <perfect conductor> calculation is a leading-order condition, not an additional exact condition at finite $\eta$. A smaller exterior-field correction matches $B_y$.
To leading order the <electric current density> is $j_z=qCe^{-qy}/\mu_0$. For two real harmonic fields, the period average of their product is half the real part of the product of one amplitude with the conjugate of the other. Thus the <mean Lorentz force in a magnetic skin layer> is
$$
\overline{\mathbf F}=\frac1{2\mu_0}\operatorname{Re}\{(\nabla\times\mathbf B)\times\mathbf B^*\}.
$$
Its normal component is
$$
\boxed{\overline F_y=\frac{C(x)^2}{2\mu_0\delta}e^{-2y/\delta},\qquad \overline{\mathbf F}=\overline F_y\mathbf e_y\quad\hbox{to leading order}.}
$$
The first possible tangential contribution from the smaller $B_y$ vanishes: it is $-CC'e^{-2y/\delta}\operatorname{Re}(q/q^*)/(2\mu_0)$, and $q/q^*=i$. This cancellation matters in computing the tangential streaming flow.
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