= Solution
Keep the index convention of <8/b>: $A^\mu=(A_0,\mathbf A)$, so $A_\mu=(A_0,-\mathbf A)$ and the spatial <gauge covariant derivative> is $\nabla+ie\mathbf A$. The physical three-current is the spatial part of $J^\mu$:
$$
\boxed{\mathbf j=ie(\varphi^*\nabla\varphi-\varphi\nabla\varphi^*)-2e^2|\varphi|^2\mathbf A.}
$$
Equivalently, writing $\varphi=R e^{i\theta}$ gives $\mathbf j=-2eR^2(\nabla\theta+e\mathbf A)$. Under the <gauge transformation> in <8/a>, $\theta\mapsto\theta+\alpha$ and $\mathbf A\mapsto\mathbf A-\nabla\alpha/e$, so this expression is manifestly invariant. Switching to a convention with the opposite spatial gauge sign reverses the phase-gradient term, not the screening term proportional to $e^2$.
For static fields the spatial <Ampère-Maxwell equation> is $\nabla\times\mathbf B=\mathbf j$, while $\nabla\cdot\mathbf B=0$. Taking its curl yields the general magnetic-field equation
$$
\boxed{\nabla^2\mathbf B=-\nabla\times\mathbf j
=-2ie\,\nabla\varphi^*\times\nabla\varphi
+2e^2\nabla|\varphi|^2\times\mathbf A
+2e^2|\varphi|^2\mathbf B.}
$$
No uniform-condensate approximation has yet been used. For a varying scalar amplitude the gradient terms must be kept; simply replacing $|\varphi|^2$ by a constant before deriving the current would conceal their origin. The homogeneous-condensate limit supplies the <London screening> equation in <8/d>.
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