Here is two models for (1) magnetic coupling, and for (2) electric coupling.
(1) assume the source of the magnetic field is a long straight wire, in the plane of a rectangular loop; you will encounter lots of these topologies (a wire and a loop) in circuitry.
To compute the induced voltage, which we model as a source inserted into the loop, I use this formula (which ignores the Integral needed for accuracy when the wire is near the loop)
Vinduce = [MUo * MUr * LoopArea / (2 * PI * Distance_wire_to_Loop)] * dI/dT
For MUo = 4 * PI * 1e-7, for MUr = 1 (air, copper, Fr-4, aluminum), the formula becomes
Vinduce = [2e-7 * Area/Distance] * dI/dT
Example of use: switching regulator, I = 1 amp, switching is 100 nanosecond, Distance between Switcher wire and the circuit loop being 1cm, and the circuit has 1cm * 1cm loop.
Vinduce = [2e-7 ( 1cm * 1cm) / 1cm ] * 10million amps per second dI/dT
Vinduce = 2e-7 * 1cm * 1e+7 = 0.02 volts
Thus tis difficult to have analog circuits NEAR a switching power supply. Also the operating of a crystal oscillator near a switching power supply, or near ANY switching current (such as from MCUs) has risk of edge jitter in the clock timing.
Notice the wire is NEAR the loop, thus this equation needs the Integral (Using natural log) for complete accuracy.
(2) I will provide example of Efield interference modeling, later in the day.
simulate this circuit – Schematic created using CircuitLab
We need the capacitance between Transmitter and Receiver. To get started, pick the smaller area of the two, and use the parallel-plate capacitive model:
C = E0 * Er * Area / Distance
Suppose trace is 30mm by 2mm, and is 30mm from a 117VAC (160v peak) at 60Hz power wire, that has NO SPIKES ON THE POWER; if pure sin, we know the slew rate exactly; if power has motor spikes, assume 100X faster spikes; if next to MCU Dataline, assume 5 volts/1nanosecond dV/dt.
What is the capacitance, between Transmitter and Receiver?
C = 9e-12 farad/meter * (er=1 for air) * 30mm * 2mm/ 30mm
C = 9e-12 farad/meter * 0.002 meter
C = 9e-12 * 2e-3 = 18e-15 ~~ 20 femtofarads
That forms a voltage divider with the 20picoFarads I assumed the Receiver PCB trace to look like: metal over ground + ESD diodes of the OPAMP + FET gate capacitance of the OpAmp's diffpairs + capacitance of the 10MegOhm resistors. Thus 20 pF is a useful starting point.
Notice this is capacitive divider, just like a scope probe. We have
18 femtofarad / 20picoFarad ==== 60dB drop, or 1,000:1 smaller.
Thus the 160 volts at 60Hz will appear, on the opamp feedback path, at 0.16 volts. But there is more.
Unless the High_Pass_Filter dominates. Does it? What is the time constant of 20pF and 5MegOhms? 100 microSeconds, or 1,600Hertz HPF corner.
The ratio between 60Hz and 1,600Hz is 25:1, thus the HPF further reduces the 60Hz injection by 25:1, or 0.16/25 = 6 milliVolts.
Notice we are assuming a CLEAN PURE SIN, no spikes, at 60Hz.
If the Transmitter has 10uS spikes, the HPF would have no effect.
And an MCU dataline, with 1nanoSecond edges, would ignore the HPF.