= Solution
The <Biot-Savart law> gives the <complex amplitude> of the field from the real wire as
$$
\mathbf B_{\rm wire}(x,y)=\frac{\mu_0J}{2\pi}\frac{-(y+b)\mathbf e_x+x\mathbf e_y}{x^2+(y+b)^2}.
$$
An oppositely directed image current $-J$ at $(0,b)$ gives
$$
\mathbf B_{\rm image}(x,y)=-\frac{\mu_0J}{2\pi}\frac{-(y-b)\mathbf e_x+x\mathbf e_y}{x^2+(y-b)^2}.
$$
The <method of images> makes the normal oscillating <magnetic field> vanish at $y=0$, as required when the alternating component cannot penetrate a <perfect conductor>. The two tangential components reinforce one another there. Consequently
$$
\boxed{\mathbf B(x,0^-)=C(x)\mathbf e_x,\qquad C(x)=-\frac{\mu_0Jb}{\pi(x^2+b^2)}.}
$$
Only the oscillating field is being excluded. A pre-existing steady <magnetic field> is not erased by the <perfect conductor> limit.
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