What is the physical explanation of why a stationary fermion is experiencing zero force from an external static magnetic field? Since fermions like free electrons have an intrinsic spin magnetic dipole moment shouldn't they when subjected to an external static magnetic field experience a magnetic force even assuming they are stationary (stationary in a vacuum for example)?
I understand that because their angular momentum they may not be moved by an external magnet and resist any linear motion but saying that they will experience zero magnetic force $F_{(M)}=q(\mathbf{v} \times \mathbf{B})$ when they are stationary $v=0$, I find peculiar and had like to know the physical explanation for this if any?
Are there any cases, experiments or theory that prove otherwise?
Question Update 17 Feb 2022:
According to an answer here by @Hoody and other comments, a free unbound hypothetical stationary electron under a static homogeneous magnetic field would not experience a force that would force it to translational motion but nevertheless would experience a torque the would change its initial magnetic moment $\vec{m}$ orientation in space:
$\begin{array}{l}
\vec{F}=q \vec{V} \times \vec{B}=0 \\
\vec{\tau}=\vec{m} \times \vec{B} \neq 0
\end{array}$
However, this still does not answer the core of my question here which is, even so with this applied torque mentioned, why is then a free electron as suggested by known physics,  under this torque not forced to any translational motion (i.e. orbital or linear)?
Is this above  paragraph actually true or just never experimentally or analytically verified up to now?
1) What happens when I approach the pole of a permanent magnet (non-homogeneous field) close to a copper rod? Will the free electrons inside the copper rod initiate an orbital motion besides there known gyromagnetic motion under the influence of this torque applied by the external magnetic field, opposite to the orbital direction of the unpaired electrons inside the permanent magnet and therefore generate also the diamagnetic property of the copper material?
2) What about a free stationary electron inside a vacuum environment? Will it start to move towards a magnet even if it was initially stationary under the influence of a non-homogeneous magnetic field? Will the same behavior occur with a homogeneous magnetic field and if not why?
If this is the case for the free electrons in the copper rod or the vacuum is this then not an actual prove that electrons DO move translational (free electrons forced to orbital motion besides their gyromagnetic motion) under a static magnetic force even if they were initially stationary?
Is it possible and is this ever experimentally verified, given a very strong external magnetic field so that the arc trajectory of this orbital motion under the applied torque the magnetic moments of the free electrons are subjected with, to be so large that it will force a free electron practically in a linear motion?
Finally, does not the above mentioned, leads in the inescapable conclusion that magnetism is not an exclusive emergent (side effect) phenomenon of electron charge translational motion but actually originates from the intrinsic property of spin magnetic dipole moment of the electron similar as electric phenomenon originates from the intrinsic property of the charge of the electron?
Therefore the origin of the electromagnetism phenomenon in general  must exclusively be attributed to the whatever unknown intrinsic mechanics of the electron itself?
As an implication of the above mentioned and conclusions reached is that the classical electromagnetism, magnetic force equation of the Lorentz force:
$$F_{(M)}=q(\mathbf{v} \times \mathbf{B})$$ for initial $v=0$ should be amended with a torque factor.
 A: Most generally, the force $\vec F_m$ and torque $\vec \tau$ due to the magnetic field $\vec B$ acting on a particle with a charge $q$ and magnetic moment $\vec\mu$, moving with velocity $\vec v$ can be written as
\begin{align}
\vec F_m &= q \vec v \times \vec B + \vec\nabla (\vec B\cdot\vec\mu),
  &&&&(1)\\
\vec\tau &= \vec r \times \vec F_m + \vec\mu \times \vec B. &&&&(2)
\end{align}
The first term in (1) is the standard Lorenz term, while the latter is usually not mentioned as a fundamental law since classical electrodynamics did not assume point-like particles with spin and magnetic moment but rather modeled magnetic dipoles by an electric current loop, behaviour of which in $\vec B$ can be derived just from the first term (focres act on parts of a the electric circuit because there are moving charges inside).
In Eq. (2), the first term $\vec r \times \vec F$ is the usual textbook definition of a torque in the case of a force acting on a point particle - it is just a tool to understand how will the tracejtory of the particle be curved due to $\vec F_m$ (i.e., it tells us something about orbital motion).
Apparently, it depends on the choice of your coordinates.
In contrary (but analogously to the previous paragraph), the second term in Eq. (2), $\vec \mu \times \vec B$, is an entirely new piece of physical law for a point-like particle. It is not present usual chapters on classical mechanics since they do not assume that that particles have spin (it is hard to imagine a rotating point-like particle). Again, if you modeled magnetic dipoles by an electric-current loop, you can derive the second term in (2) from $\vec F_m = q \vec v \times \vec B$.

However, this still does not answer the core of my question here which is, even so with this applied torque mentioned, why is then a free electron as suggested by known physics, under this torque not forced to any translational motion (i.e. orbital or linear)?

Now, let us have a look at what forces and torques actually do: Changes in motion of the center-of-mass of a body are always determined solely by the force, according to the famous Newton's law $\vec F = m \vec a$. Denoting the body's momentum by $\vec p=m\vec v$, this law can be also reformulated as $$\vec F = \frac{\mathrm d\vec p }{\mathrm dt}. \qquad (3)$$
In contrary, torque causes changes in angular momentum $\vec J$ (i.e., in "rotation"):
$$ \vec \tau = \frac{\mathrm d \vec J}{\mathrm d t}. \qquad (4)$$
Generally, angular momentum of a moving particle with spin is a sum of two terms, $$\vec J = \vec L + \vec S$$ where $\vec L = \vec r \times \vec p = m\vec r \times \vec v$ is the usual orbital angular momentum while $\vec S$ is the particle's internal angular momentum, i.e., spin.
Now, if $\vec F_m = 0$, Eq. (3) tells us that there will be no acceleration. This is perfectly consistent with Eq. (4) which yield
$$
 \vec \tau = \frac{\mathrm d\vec J}{\mathrm dt}
= \frac{\mathrm d\vec L}{\mathrm dt} + \frac{\mathrm d\vec S}{\mathrm dt} 
= \frac{\mathrm d(\vec r \times \vec p)}{\mathrm dt} + \frac{\mathrm d\vec S}{\mathrm dt} 
=\vec v \times (m\vec v) + \vec r \times \vec F + \frac{\mathrm d\vec S}{\mathrm dt} 
=0+0+\frac{\mathrm d\vec S}{\mathrm dt}.
$$
I.e., in the absence of a magnetic force, the magnetic torque acts solely on the spin of the particle. Using Eq. (2):
$$\vec \mu \times \vec B =\vec \tau =  \frac{\mathrm d\vec S}{\mathrm dt}. $$
Btw., notice that since the particle's magnetic moment  $\vec \mu$  is proportional to its spin $\vec S$, the torque is always perpendicular to $\vec S$ and hence may cause a change in the orientation of the spin (i.e. precession) but not of its size (i.e. electrons always have spin of 1/2 $\hbar$ - but let's not dig into the quantum world in this discussion).

Is this above paragraph actually true or just never experimentally or analytically verified up to now?

It is true as far as classical (non-quantum) electrodynamics is true. As noted by others in the comments, a theory cannot be verified but only disproved. Electrodynamics is the most precisely measured part of physics whatsoever.


*

*What happens when I approach the pole of a permanent magnet (non-homogeneous field) close to a copper rod? Will the free electrons inside the copper rod initiate an orbital motion besides there known gyromagnetic motion under the influence of this torque applied by the external magnetic field, opposite to the orbital direction of the unpaired electrons inside the permanent magnet and therefore generate also the diamagnetic property of the copper material?


Yes, diamagnetism is a property of all materials. But notice that the origin of the torque causing the orbital motion is not the magnetic moment of the electron but the first term, $\vec r \times \vec F_m.$



*What about a free stationary electron inside a vacuum environment? Will it start to move towards a magnet even if it was initially stationary under the influence of a non-homogeneous magnetic field?


Yes.

Will the same behavior occur with a homogeneous magnetic field and if not why?

No. For a non-moving electron in a homogeneous magnetic field, both terms in Eq. (1) are zero. The electron might only undergo spin precession.

If this is the case for the free electrons in the copper rod or the vacuum is this then not an actual prove that electrons DO move translational (free electrons forced to orbital motion besides their gyromagnetic motion) under a static magnetic force even if they were initially stationary?

As explained above, this does not happen in vaccum. In the material, we must be careful about what we mean by stationary. In the quantum language, an electron in its atomic orbital is stationary; however, its classical analogue is that it is somehow circling around the nucleus and hence it undergoes an orbital motion eve initially before you apply the external magnetic field. Hence, the answer is NO.

Is it possible and is this ever experimentally verified, given a very strong external magnetic field so that the arc trajectory of this orbital motion under the applied torque the magnetic moments of the free electrons are subjected with, to be so large that it will force a free electron practically in a linear motion?

Are you asking wheather we can compensate the curving effect of the magnetic force $q\vec v \times \vec B$ by the interaction of the electron's magnetic dipole $\mu$? I think the answer is NO, since both effects are directly proportional to the same field $\vec B$.

Finally, does not the above mentioned, leads in the inescapable conclusion that magnetism is not an exclusive emergent (side effect) phenomenon of electron charge translational motion but actually originates from the intrinsic property of spin magnetic dipole moment of the electron similar as electric phenomenon originates from the intrinsic property of the charge of the electron?

Indeed. If you accept the semi-classical picture of a point-like electron with non-zero spin and magnetic moment, then there are two sources of magnetism (moving charges and particle's spin), which also both respond to an external magnetic field (first and second terms in Eqs. (1) and (2).)
Nevertheless, note that in relativistic quantum theories, both "sources of magnetism" naturally emerge from a single physical law (i.e. Lagrangian of quantum electrodynamics).

Therefore the origin of the electromagnetism phenomenon in general must exclusively be attributed to the whatever unknown intrinsic mechanics of the electron itself?

No, not all magnetic phenomena are caused by the magnetic dipoles of elementary particles - see the first terms in Eqs. (1) and (2). Note that pions are charged particles without spin or magnetic moment but their trajectories can still be bent by magnetic fields (and it is an experimental routine).
A: I think the confusion here is that you intuitively know that a small rare-earth magnet, released between the poles of a large horseshoe magnet, will be pulled linearly (and will snap into contact with the horseshoe magnet).
But which pole of the magnet will it be pulled to? The answer is the closest one. Regardless of which way the magnet was initially oriented.
What if you released the rare-earth magnet exactly half way between the poles of the large horseshoe magnet? This is difficult to perform because this position is unstable but the answer is that it would not be pulled linearly (not until some tiny perturbation knocks it slightly closer to one pole than the other).
The confusion is that in everyday experiences of a magnetic dipole in a magnetic field, that magnetic field also has a strong gradient. It is stronger near the poles. The only place where the field (of the horseshoe magnet) is approximately homogeneous is exactly half way between the two poles.
Classically, if you place a magnetic dipole in a uniform/homogeneous external magnetic field, you can think of the force upon the north end of the dipole being exactly balanced by the force upon the south end of the dipole, so that the net force is zero. (There may be some torque, to cause the dipole to align with the external field, but no linear net force results.) Whereas if the external magnetic field has a gradient, then the spatial extent of the dipole means that each of its poles will experience a different external field strength, and hence the forces will be slightly out of balance, resulting in a nonzero net force and a linear acceleration. The logic is similar in the quantum case, although without attributing a spatial extent to the dipole. The reason that the spin ½ charged fermion is not accelerated in an external static homogeneous magnetic field is because you constructed the field to be homogeneous, not any reason peculiar to fermions or quantum mechanics.
A: *

*If something makes you accelerate, it is a force.

*If it makes you rotate, it is a moment of force, i.e., a torgue.

A spin-1/2 particle in a homogeneous (!!!) magnetic field will experience a torgue but not a force.
A: I don't understand the question or the comments and answers.
The magnetic force on a stationary electron is $F=\nabla(\mu\cdot{\bf B})$.
