Calculating Mutual Inductance in Parallel Wires
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.