The Wikipedia article on ground plane says
In addition, a ground plane under printed circuit traces [the paths that the circuit currents take] can reduce crosstalk between adjacent traces. When two traces run parallel, an electrical signal in one can be coupled into the other through electromagnetic induction by magnetic field lines from one linking the other; this is called crosstalk. When a ground plane layer is present underneath, it forms a transmission line with the trace. The oppositely-directed return currents flow through the ground plane directly beneath the trace. This confines most of the electromagnetic fields to the area near the trace and consequently reduces crosstalk.
but unfortunately there are no citations. Page 5 of this document says the same thing ("Return Current Flow is directly below the signal trace") and gives a formula that the magnitude of the return current at point $x$ on the ground plane is proportional to $\cos(\theta)/r$, where $r$ is the distance between point $x$ and the circuit trace and $\theta$ is the angle that point $x$ makes relative to the vertical. But the document gives no explanation justifying this formula.
My naive intuition is that the return current would want to avoid the circuit trace, since opposite currents repel, and so it would spread out and increase crosstalk. Why does the return current instead concentrate below the circuit trace?
Answer
I just asked a professor this question, and he thinks that the concentration of return current below the circuit trace is actually just an AC effect. The rapidly oscillating electric and magnetic fields induce a fictitious image current on the other side of the conductor, and the real and image wires together act as a transmission line. But this only works if the fields oscillate fast enough - in the DC limit, the magnetostatic repulsion between the trace and the return current does indeed cause the return current to spread out far away from the circuit trace.
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