Note: "Ground loop" in this question refers to a ground loop caused by linked magnetic fields. (Sometimes ground loop refers to slightly different scenarios.)
Recently I've been looking into ground loops a lot since I've encountered them in circuits I've set up. I've been trying to analyse ground loops using circuit theory to understand why they happen and how to fix them, however my analysis produces results opposite to what is suggested by Wikipedia (as of 14/09/2021) and many pages on the internet. Where am I wrong?
Below is a schematic of a typical ground loop situation, with mesh currents \$i_s\$ and \$i_g\$ circulating around their respective loops. \$R_L\$ models the input impedance to some amplifier/oscilloscope/multimeter/etc. We can assume that the contents of the two boxes are unaffected by any EMI in the environment (maybe they connect off via twisted pairs far, far away). The signal that will end up getting amplified/read will be \$i_s R_L\$.
In the above diagram: the top wire is the signal wire, the middle wire is the signal cable's ground shield, and the bottom wire is the ground connection via the mains wiring.
Each bit of conductor has some resistance, modelled as \$R_1\$ and \$R_2\$. We assume that \$R_L\gg R_1\$ and \$R_L\gg R_2\$ (the amplifier has a high input impedance, say a few megaohms – way higher than any resistance the wires might have). The loops of wire will also have some self-inductance – but we will model that directly using Faraday's law later on. Note that I didn't include any resistance in the top signal wire – this is because it is negligible compared to \$R_L\$ which it is in series with.
Finally, we assume that the area of the loop for \$i_s\$ is negligible compared to the area for \$i_g\$ (you could imagine that the signal and ground wires overlap exactly, or are a twisted pair).
Now we can solve for the loop currents using mesh analysis. For the ground loop, we will just apply Faraday's law directly rather than trying to include the parasitic mutual and self inductances into the model so that we can use KVL. For the signal loop, there is no linked flux per our previous assumption, which means Faraday's law just simplifies to KVL around the loop:
$$
v_s-i_sR_L-(i_s-i_g)R_1=0\tag{1}
$$
For the ground loop, we assume that it links a magnetic flux caused by EMI in the environment and by the loop's own magnetic field, which is proportional to the current flowing around it (this is the loop's self inductance). By Faraday's law, we get an EMF around the loop equal to:
$$
\oint_\text{clockwise ground loop}\mathbf{E}\cdot d\mathbf{l}=-\frac{d\Phi_\text{total}}{dt}=-\frac{d}{dt}(\Phi_\text{noise}+Li_g)=-\frac{d\Phi_\text{noise}}{dt}-L\frac{di_g}{dt}
$$
We will call the noise term \$v_n\$ – it's sort of like a voltage source is distributed around the loop. Thus:
$$
\oint_\text{clockwise ground loop}\mathbf{E}\cdot d\mathbf{l}=v_n-L\frac{di_g}{dt}
$$
Ohm's law can be stated more generally as:
$$
\int_\text{resistor}\mathbf{E}\cdot d\mathbf{l}=iR
$$
Thus, we can split the closed loop integral into integrals over each resistance. This gives:
$$
(i_g-i_s)R_1+i_gR_2=v_n-L\frac{di_g}{dt}\tag{2}
$$
Now we have the two loop equations, equation 1 and equation 2. There's a derivative in there, which is a bit annoying (makes the whole thing a first order ODE in \$i_g\$). However, we can just take this to the Laplace domain to turn it into an algebra problem. (If you're not used to the Laplace transform, \$sL\$ is essentially the same as \$j\omega L\$ used for phasors.)
$$
V_s-I_sR_L-(I_s-I_g)R_1=0\tag{1a}
$$
$$
(I_g-I_s)R_1+I_gR_2=V_n-sLI_g\tag{2a}
$$
Solving equation 2a for \$I_g\$ gives us:
$$
I_g = \frac{V_n+I_sR_1}{R_1+R_2+sL}
$$
Substituting this into equation 1a gives us:
$$
\begin{align*}
V_s-I_sR_L-I_sR_1+I_gR_1&=0\\
V_s-I_sR_L-I_sR_1+\left(\frac{V_n+I_sR_1}{R_1+R_2+sL}\right)R_1&=0\\
V_s-I_sR_L-I_sR_1+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n+\left(\frac{R_1}{R_1+R_2+sL}\right)I_sR_1&=0
\end{align*}
$$
$$
\begin{align*}
V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n&=I_sR_L+I_sR_1-\left(\frac{R_1}{R_1+R_2+sL}\right)I_sR_1\\
V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n&=\left(R_L+R_1-\left(\frac{R_1}{R_1+R_2+sL}\right)R_1\right)I_s
\end{align*}
$$
Therefore:
$$
I_s=\frac{V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n}{R_L+R_1-\left(\frac{R_1}{R_1+R_2+sL}\right)R_1}
$$
The signal that we read will be equal to \$I_sR_L\$. Thus, we multiply by \$R_L\$ to get:
$$
V_L=I_sR_L=\frac{V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n}{R_L+R_1-\left(\frac{R_1}{R_1+R_2+sL}\right)R_1}\cdot R_L
$$
We can apply some algebra to finesse the denominator into the following:
$$
V_L=\frac{V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n}{R_L+\frac{(R_1)(R_2+sL)}{R_1+R_2+sL}}\cdot R_L
$$
The fraction in the denominator is just the impedance of \$R_1\$ in parallel with \$R_2+sL\$, which can be notated as:
$$
V_L=\frac{V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n}{R_L+(R_1)||(R_2+sL)}\cdot R_L\tag{3}
$$
Now, since we assumed that \$R_L\gg R_1\$ and \$R_L\gg R_2\$, the denominator is approximately equal to \$R_L\$, which gives:
$$
V_L\approx\frac{V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n}{R_L}\cdot R_L
$$
This gives our final expression for \$V_L\$:
$$
V_L=V_s+\left(\frac{R_1}{R_1+R_2+sL}\right)V_n
$$
But this doesn't match what Wikipedia says about ground loops. The above equation implies that increasing \$R_1\$ will make the noise worse, not better. However, the Wikipedia page suggests putting a 10 ohm resistor on the ground shield of the signal cable to mitigate ground loops, but this equation suggests that doing that should increase the noise seen at the load!
Furthermore, we can take the limit as \$R_1\rightarrow\infty\$, which models the case of an open circuit. To do this, we should use equation 3 (which is before we approximated that \$R_L\gg R_1\$). Taking equation 3 to the limit of \$R_1\rightarrow\infty\$ gives us:
$$
V_L=(V_s+V_n)\frac{R_L}{R_L+R_2+sL}
$$
This implies that you would see the full noise of the loop, except scaled down by a voltage divider (which would have approximately unity gain if \$R_L\$ was very large). And this even makes sense if you look at the circuit without the middle ground path in it!!! However, this contradicts common practice (according to the internet), where a "ground lift" is often used to mitigate ground loops by open circuiting the ground shield of the signal wire.
Furthermore, if you open circuit \$R_2\$, you get:
$$
V_L=V_s\frac{R_L}{R_L+R_1}
$$
which is similar to before, except with a slightly different voltage divider and no noise. If \$R_L\$ is very large, we get \$V_L=V_s\$.
What's going on here? Why does this analysis seem to indicate that much of the internet is wrong about how to mitigate ground loops? In fact, I could even argue that the above result makes sense. People say that putting resistance in there prevents the ground loop currents from flowing. But who cares about current! We care about voltage (if the input impedance of the load is very high, it is basically just a voltmeter). While increasing \$R_1\$ makes it harder for current to flow overall, it increases the coupling between the loops for \$I_s\$ and \$I_g\$.
Help! Either there is an error in my calculations (which I doubt would affect the final answer, since the final answer makes intuitive sense to some degree). More likely is that I am missing something important in the model. A core assumption is wrong.
Best Answer
Note: When I use the term "ground lift" I am referring to when the ground of one end of a signal cable is disconnected. The term applies to other scenarios too, such as a ground lift switch that isolates the primary and secondary windings of an isolation transformer or other types of isolators, or a (probably dangerous) cheater plug.
I have come to the conclusion that my analysis was correct given the assumptions. I was able to reproduce the core predictions of my analysis in an experiment (see my description of it here), which would suggest that the assumptions are reasonably accurate (at least in my case).
What some get wrong
Ground loop noise from electromagnetic induction is not necessarily related to the current. Electromagnetic induction means that there are effectively parasitic voltage sources across the signal line, ground shield, and ground wiring. Even if no current is flowing, these will still create voltage differences.
In most circuits, increasing a resistance will increase the voltage drop across it, even though the current goes down (the simplest example is a voltage divider - this can be extended to other circuits using Thevenin's theorem). Changing the resistance of parts of the ground loop (e.g., by inserting \$10\Omega\$ resistor, or making its resistance \$\infty\$ by open circuiting the shield) will simply change how the noise voltages are distributed, but will not remove them.
Ground lifts with single-ended connections
Applying a ground lift to a cable shield is like taking the shield resistance to infinity. It causes all of the noise drop to occur across that gap - allowing the grounds on either end to float at completely different voltages, which will result in lots of noise added across the line. It will also prevent ground loop currents from flowing.
Whether this causes the noise seen at the receiver to improve overall depends on where most of the noise was coming from in the first place. If the noise was predominantly coming from the voltage across the shield, then it will make it worse. On the other hand, if the noise was primarily due to ground loop currents creating \$IR\$ drops within sensitive signal circuitry in the transmitter or receiver, then it will make it better. I suspect the former is more likely, since you can avoid the latter with good circuit design by ensuring that the signal ground is not placed in the path of ground loop currents (and it would tend to require larger trace resistances, high gain circuitry, or low signal levels). These parasitic \$IR\$ drops would also be problems for balanced lines.
However, even if internal \$IR\$ drops were the dominant factor, a ground lift could never completely reject the noise, since opening the ground shield puts the noise voltages directly in series with the signal (referring to the symbols in my question: \$V_s\$ would be directly in series with \$V_n\$ - so the ground lift will result in an SNR of \$V_s/V_n\$).
Ground lifts with other connections
If a signal cable is able to reject the ground noise across the line as a common-mode signal (e.g., balanced line fed into differential receiver), or if the noise can be tolerated for some other reason, then lifting the ground of a cable shield will improve things overall, since ground currents are stopped (preventing undesirable voltage drops elsewhere), while issues from noise voltages across the disconnected shield are avoided.
Current ratings
Another issue with induced ground currents is that it prevents loop currents from circulating through ground conductors that may not be rated for those currents (I don't know in practice how often ground currents can be that large and have only heard about it being a theoretical issue from other people).
Common-impedance coupling
People sometimes use the term "ground loop" to refer to issues where common-impedance coupling through mains ground conductors causes there to be undesirable voltage drops between the mains ground of two devices due to multiple devices pulling current through a shared ground resistance. This causes the devices to couple: current sunk by one device will by seen as a voltage shift in the ground of another device. This is a different phenomenon from what was discussed in the question.
In this case, the ground shield is in parallel with the rest of the ground wiring and helps to reduce the total resistance between the two grounds, resulting in a lower voltage drop (despite a slightly higher current). In this case, opening the ground shield will make this voltage drop larger and most likely worsen overall performance (unless, as before, parasitic drops within signal circuitry were more dominant). Ground currents from another device exceeding the rating of the ground shield is a possibility, but is a separate issue to noise.
Conclusion
Ground lifts have limited effectiveness when applied to the ground of single-ended lines, and will most likely make the noise worse. Even when they can improve the noise, they are limited in how effective they are. They make much more sense for balanced lines (in which case a ground lift on the receiver side is preferable to minimize filtering due to the line capacitance - as suggested by Handbook for Sound Engineers, 4th ed).
If the noise across the line is the dominant factor (and not internal \$IR\$ drops), then a ground lift will worsen the noise, and lowering the resistance of the ground shield will improve the noise. However, this is also of limited effectiveness in practice since it would be hard to get a cable with a shield resistance much below the mains wiring resistance (limiting noise attenuation to ~6 dB according to the formulas I derived in my question). It also has the disadvantage of increasing the loop current, increasing potentially problematic voltage drops elsewhere. To conclude: