Originally Posted by Joe Skubinski
You may have covered this but I don't recall... What cable topology are you using when making these measurements?
The cable topology per se is unimportant to the discussion.
One source, two single ended outputs, one amp, two single ended inputs, two 3 prong line cords, two coaxial IC's. Source output Z zero (for clarity of discussion), amp input z 10K.
When the source pushes say, 1 volt at 20 hz in the right IC, 100 uAmp flows in the IC center conductor. Current always requires a return path, so how does the 100 uAmp get back to the source? By the path of least resistance.
Examine the paths that are available...you have two IC shields as return paths, and the safety grounds via the duplex outlet.
Without the safety ground, the 100 uA will split between the IC grounds, 50 uA in the right, 50 uA in the left. With a massive safety ground, the bulk of the return current will flow through the ground, NOT the IC shield.
As long as return current flows through the safety ground, the actual loop will be picking up any and all time varying magnetic fields. Faraday's law of induction will cause a voltage to be generated around that loop, and that voltage will appear across the 10k input resistance of the amp.
Let's recap the salient point:
The return current does NOT for the most part, flow back to the source via the outer shield of the coax that the signal came in on. For a shield to actually work, ALL the return current MUST flow back on the shield.
For a balanced pair to "shield", in this case by centroid proximity, all the current in one wire MUST return via the other wire.
This is the standard conceptual understanding within the EMC field.