Do a current carrying resistor experience a force? In a current carrying resistive conductor, an electric field within the conductor creates a force on the electrons within the conductor. In broad terms, classical theory says that the electrons are accelerated, but very soon make an inelastic collision with the fixed lattice, thus transferring their energy to the lattice and heating the material. The drift velocity is very slow. But hey, at least according to classical Newtonian mechanics, does not this mean that the force exerted on the electrons by the electric field must be counteracted by an equal and opposite force on the lattice, and thus on the conductor? The force exerted on the electrons is absolutely real - it must be counteracted by something, mustn't it?
Edit 25/8/2022. After considerable thought and many sleepless nights, I conclude that there is no net force on the resistor.
But while there is no net force on the resistor, IMO it is nonetheless true that the slowly moving electrons exert a force on the fixed lattice, and it is fun to calculate this force. For example, let the resistor be a 0.1 meter length of copper wire of diameter 0.359 mm, corresponding to a cross sectional area of 0.1 mm^2, and carrying 10 Amps of current. The resistance calculates as 0.0172 ohms, so the voltage across this resistor will be 0.172V. The dissipated power is 1.72W, and the electric field in the wire is V/L = 1.72 V/m. All typical real-world values.
OK, so what force is exerted on the resistor, by these moving electrons? Easy. What I will actually calculate is the force exerted on the moving electrons, by the electric field, and this force must be counteracted by the fixed lattice. We can use F=qE. We know E as 1.72 V/m. The number of electrons is the electron density of copper (8.5E28 e/m^3) multiplied by the volume of the resistive wire, which calculates as 8.5E20 electrons, which corresponds to 136 Coulombs.
F = qE = 136 x 1.72 = 233 Newtons  (or 23.8 kg if you prefer)
If you don’t believe that, then it can be calculated by a different method as I originally suggested, noting that for conservation of energy, VI = Fv where v is the drift velocity. So F = VI/v. If you calculate the drift velocity for this example, and knowing that VI = 1.72W, then wonder of wonders, you get exactly the same answer, a force of 233 Newtons.
OK. At first glance, that is an absurd result, a massive 24 kg force being exerted on a tiny 1.7 W resistor, as a result of a modest current through the resistor!! If this was the net force on the resistor, then people would have noticed this by now! And yet, we don’t observe resistors being ripped off circuit boards and flying uncontrollably across the room, so what is going on?
The solution to this apparent paradox is that there is an equal number of +ve charges that make up the fixed lattice, and so the electric field produces an equal and opposite force on the fixed lattice, for a net force of zero. But be that as it may, I personally find the magnitude of the equal and opposite forces (233N) to be remarkable.
But wait, because there is more to it than that. I calculated the number of electrons in the resistor, based on an “uncharged” resistor, where the number of electrons is exactly equal to the number of +ve charges in the lattice. But hey, we have a voltage across this resistor, so we have an excess of electrons at one end, and a deficit of electrons at the other end.  So different parts of the resistor will experience a force on account of this charge, acted on by the Efield. Or, if you prefer not to consider the Efield as such, then you have a Coulomb attractive force between the +ve charges at one end, and the -ve charges at the other. In fact, the resistor will experience an electrostatic force trying to compress the resistor along it’s length. Every resistor is also a capacitor, and this is analogous to the force that attracts the 2 plates of a capacitor to one another. There are two things to note about this force. First is that there is still no net force on the resistor, which is just as well as such a force would likely create havoc with the conservation of momentum. Second, the capacitance of a small resistor is very small, on order of a pF, so if you calculate the force crushing the resistor then it is found to be exceedingly small, too small to measure, and too small to ever damage the resistor.
Finally, if anyone cares, the excess charges (which are what create the Efield along the length of the resistor) reside only on the surface of the conductor, and do not contribute to the drift current. Also note that the number of excess charges is negligible compared to the number of charges in the drifting electron sea. In my example, if C=1pF, the Q=CV = 1E-12 x 0.172 = 1.72E-13 Coulomb of excess charge. Yep, that is very small compared to the 136 Coulomb of drift charge.
Any comments on any of the above?
 A: Let's not (yet) ask about the force.  Let's ask about the impulse, $\Delta\vec p = \vec F \  \Delta t$.
For simplicity, let's use a copper wire, which has roughly one conduction electron per atom.  Copper has mass density $\rho \approx 9\rm\,g/cm^3$, and atomic mass $A \approx 64\rm\,g/mol$, which gives a number density
$$n \approx 
%9/64
0.14\,\frac{\rm mol}{\rm cm^3}
% \approx 8\times 10^{22}\,atom/cm^3
\approx 0.8\times 10^{20}\,\frac{\rm atom}{\rm mm^3}
.$$
Finding the charge-carrier drift velocity is a well-known homework problem.  You know the density of the charge carriers, above; you know the current, and the rate at which charge carriers must move in order to carry this current.  For normal currents in normal conductors, this works out to a few centimeters per second.  So when you switch from "no current" to "direct current," the electrons in each cubic millimeter of copper will acquire a net momentum on the order of
$$
\begin{align}
p_\text{final} &\approx 
\left(
10^{20}m_\mathrm e
\right)
\times
\left(
\text{few}\frac{\rm cm}{\rm s}
\right)
\\&\sim
\text{few dozen}\,\frac{\rm pg\,cm}{\rm s} \quad \text{ (per cubic millimeter)}
\end{align}
$$
That mass unit is "picograms."  Now, if you're really interested in force, you can look at the resistances, capacitances, and inductances in the rest of your circuit, compute any relevant $RC$, $RL$, $LC$ time constants, and pick the slowest one as a guess about $\Delta t$ to find a force $F = \Delta p/\Delta t$.  But if you just say "fast electronics are nanoseconds," then you get a force
$$
\frac{\rm pg\,cm}{\rm s\cdot ns}
\sim 10^{-8}\,\rm N \quad\text{(per cubic millimeter)}
$$
However, this is only considering the momentum of the conduction electrons.  Each conduction electron is paired with a copper atom, whose mass is roughly $64\times1800$ times larger than the electron mass.  So if a given mass of copper suddenly gets an electron drift velocity of a few centimeters per second, the momentum of the entire wire can be held constant if the lattice drifts in the opposite direction by a few dozen nanometers per second.
If you know about electron drift velocities, you know that the electrons' overall drift is much slower than the thermal motion of individual electrons.  Current flow in a copper conductor is much more like water flow through a dammed reservoir than it is like water flow through a running river.  The same relationship holds for the lattice atoms, which are oscillating in place with typical kinetic energies $\frac12 mv^2 \approx kT$, corresponding to speeds of order $10^2\rm\,m/s$.  A net momentum of the lattice corresponding to a "lattice drift" of $\sim 10^{-8}\rm\,m/s$ is a minuscule correction to these thermal speeds, and so any overall transport of the lattice is probably modeled better by incoherent transport of phonons than by a coherent back-reaction of the entire wire.
In any real experiment, electromagnetic forces with external or internal magnetic fields would almost certainly overwhelm this tiny momentum associated with charge-carrier drift velocity.  If you turn on a big current in a low-mass wire in Earth's magnetic field, you can arrange for the wire to visibly jump: a macroscopic momentum change.
