It depends what is connected to the other winding, which is why "mutual inductance" is also called "coupling factor" (They are not identical, but closely related terms).
The classic way of characterising a transformer's performance (after establishing n, the turns ratio) is to first measure the inductance of the primary - with the secondary open circuit. This measurement is the "primary inductance" - effectively unaffected by the other winding since no current flows in it.
And the primary inductance is an impedance connected across the power source - effectively wasted power, and as it is a low impedance at low frequency it determines the low frequency performance of the transformer.
Then re-measure the primary, but with the secondary short circuited. This is the "leakage inductance" (technically it's the parallel combination of primary and leakage inductances, but the primary inductance is usually a large enough impedance that it can be regarded as infinite, and ignored). Anyway the "leakage inductance" is essentially the coupling factor of the transformer into a short circuit - so in a good transformer it will be a very low impedance.
(The same pair of measurements can be made on the secondary, with the primary open/short circuit. It should give you the same result, scaled by n^2).
So the leakage inductance doesn't change the winding inductances - it couples one winding to the other, allowing the load impedance (scaled by 1/n^2) to appear in parallel with the winding inductance.
And the series combination of source impedance and primary inductance determine the LF response, while the series combination of leakage inductance and (load impedance/n^2) determine the HF response.
A negative inductance would imply that the current is dependent on voltage varying with time but also negated. (i = -Ldv/dt instead of i = Ldv/dt ). No one can build a physical inductor that will automatically flip the voltage coming into it, but it makes for an easy analysis. A circuit representation is a way to model the physical world.
With models you can get results that are not physical, that model the system of interest just fine. An important part of electrical engineering is being able to model systems, but also realize the differences between the model and the real world. There are also no ideal circuit elements, there are no capacitors, inductors or resistors that have don't have parasitics.
For most things the parasitics don't matter (do you really care if your resistor has a few nanoHenries of inductance when your creating a voltage divider? No, but you will if your trying to run a GHz signal through it).
Best Answer
If you know the separation of the wires
$$ L = (\frac{\mu_o * Length}{\pi}) * \ln(\frac{separation}{radius} + SDfactor)$$
where separation is the center_to_center distance, to be greater than 2*radius
https://en.wikipedia.org/wiki/Inductance shows derivation of these formula
As the link explains, there is a fudge factor to model skin effect. For low frequencies, where current uses the entire cross-section, add 1/4 inside the "ln" arqument. At high frequencies, add nothing.
Now lets put this equation to use. Consider a MCU SPI dataline, 1mm from the sensitive signal trace, and parallel for 40mm. What is the mutual inductance? We need to know the radius, so assume its 1/2 * 0.25mm (10 mils = 0.25mm). Yes, I'm using a flat trace circuit, in an equation for "round wires".
$$L_{Ind} = \left(\frac{4*\pi*10^-7 \frac{H}{m} * 0.04m) }{ pi}\right)* \ln(\frac{1mm}{ 0.125mm})$$
$$L_{Ind} = 4*0.04 * 10^-7 * ln(8) $$ {ignoring the skin-effect factor} $$L_{Ind} = 0.16 * 10^-7 * 2.08 = 0.32 * 10*-7 = 32 * 10^-9 = 32 nH$$
Suppose the dataline has 100pF Cload, with 2.5 volts/2.5nanoSeconds slewrate; the charging current \$I = C * \frac{dV}{dT} = 0.1nF * 1\frac{V}{ns} = 0.1 A\$. We'll assume the dataline current rises to 0.1 amps in 1nanoSecond, remains there for 0.5nS, and decays back to zero in 1nS.
What voltage is induced by the mutual inductance, from the dataline into signal trace?
V = L * dI/dT = 32nH * 0.1 amps/1nanoSec [knowing the nano cancel] = 3.2 volts.
NOW.....what is the benefit of separation? If we just have wires in air (no underlying planes), we see the mutual inductance (the coupling) drops very slowly because of ln(separation/radius).
Fortunately there can be underlying planes.