An equivalent way of looking at loosely coupled coils is to consider the magnetic moment , this is a convenient measure of the "strength" of a magnet, and can be used with bar magnets and coils, obviously more turns, more amps, or more area make a magnet "stronger" and magnetic moment is a way of capturing that "strength".
For a coil of wire of
n turns, carrying
I amps, with an area of
S sq metres, then the magnetic moment calculation is simply \$\mu\$ =
n I S a nice easy representation! The moment is a vector quantity, and it points along the centreline of a circular coil (or in general is othogonal to the area, i.e. the "directed area"). The units are A·m
2 .
We can use magnetic moment , for example , to describe how strong a transmitter we can make with our mains cable.
First consider a single wire, make it 3.14m long, and let it carry 1 amp, if we wind this into a horizontal loop 100mm across, we have an area of 0.0078 and ten turns so \$\mu\$ = n I S = 10 x 1 x 0.0078 = 0.078 A·m
2 .
What if we only make 1 turn that is 1m diameter? then it is n I S = 1 x 1 x 0.78 = 0.78 A·m
2 So winding a fixed length cable into many small loops is better (part of your original question)
What if we take the ten turn, 100mm loop , and flop half of it over, so it is two loops, side by side, each of 5 turns, so one loop has \$\mu\$ = .0039 A·m
2 pointing up , and the other half has 0.0039 pointing down, the net effect is zero in the far field. So you should be able to work out what happens with figure-8 loops or serpentine loops yourself and drawing pictures to determine the effective directed area.
Let's now consider something approximating a power cable, say we have a 3.14m long cable where the active and neutral are spaced by 10mm and there is no twist (a bit like the old 300ohm TV cable), so the area is 3.14 x 0.01=.0314 and n I S = 1 x 1 x 0.0314 = 0.0314 A·m
2. If the cable is laying flat, then the moment is directed upwards.
Lets twist this cable so its one twist per 400mm, i.e. one twist per 100mm, So one 100mm segment has an average moment pointing up, the next segment points left, then down, then right, so in the far-field the net magnetic moment vanishes for each of the 31 100mm segments, so just the remainder of 40mm actually radiates, so an area of 0.01 x 0.04 =.0004 sqm or \$\mu\$ = .0004 A·m
2 .
--- let's get back to your case ----
Assume the power cable feeds a 240VA load from 240V, so the active carries one amp of load current and the neutral 1 amp. In actual practice you may have a line filter in your amplifier, the capacitors in this will carry say 1mA on the active to earth, and none on the neutral to earth, (they are the same potential) and 1mA flows back on the earth line, so if we consider the power cable is say 3.14m long in a straight line, and twisted, its much the same problem as above, but you have 1.001A on A, 1.00A on N and 0.001. So the net magnetic moment in the far-field is miniscule as the directed areas cancel out.
Let's take this cable and make 10 loops 100mm across, we can calculate the moment again of this as \$\mu\$ = n I S = 10 x 0 x 0.0078 = 0.0000000 A·m
2 .(That's because the net current is 0 = I
A + I
N +I
E = 1.001-1.000-0.001)
More realistically some of that 1mA of earth current might flow through the chassis, even if all of did divert through the chassis, the moment would then be \$\mu\$ = n I S = 10 x 0.001 x 0.0078 = 0.000078 A·m
2.
Once you've got the magnetic moment calculated, i.e. your transmitter strength, you can determine the magnetic field applied outside your "leaky" transformer, numerically this will be a handful of nanotesla's for a magnetic moment of 1 A m
2 .
(See about the middle of this page:
https://en.wikipedia.org/wiki/Magnetic_moment )
So the whole signal chain from mains current to your transformer is milli x nano x micro = vanishingly small.
Compare this to the earth leakage current of ~ 1mA due to filter caps , which if applied across a 1ohm resistance between various chassis parts yields a millivolt of noise , if you are real sloppy with earthing your mikes, and see even a tenth of this across your mike input it will be very loud!
So earth leakage is likely to be a stronger noise source, and this won't change if you cut the power cables shorter.