A question about white noise in a single ended data acquisition system
Consider a standard 33 rpm turntable, with 2 output coax cables and that 5th wire connected to the metal base and the tonearm.

Why is this 5th wire needed? to provide a path for interference coupled across the power transformer used in DC supply for the turntable rotation motor.

How much current to handle?

Assume the power transformer has 3.16 cm by 3.16 cm common area between primary and secondary, with 1mm insulation. Assume relative permittitivity is 5. The capacitance is

C = Eo * Er * Area/Distance

C ~~ 50 pF/meter * 0.0316meter * 0.0316 meter / 0.001 

C= 50pF/meter * 0.001 meter^2 / 0.001 meter = 50pF 

Assume the power transformer is badly imbalanced, so we simply model the displacement current (the capacitive charge/discharge current) as

I = C * dV/dT = 50pF * d(160 voltPeak * sin( 2 * pi * 60Hz))

I = 50pF * 160 * 377 radian/second ~~ 50pF * 60,000 ~~ 3 microAmpers at 60Hz.

Your situation is probably similar. You have 3 microAmps exploring all possible paths to get back home. Draw some sketches of these paths.

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Here is +10dB SNR white-noise, riding atop +1/-1 volt squarewave