U-shaped magnet and the moving charge: Where is the origin of the Lorentz force from the viewpoint of the moving charge?

Question

Assume that a charge $+q$ is located at rest between the poles of an infinitely long U-shaped magnet, which is laid along the $x$-axis. If the charge slowly accelerates to a velocity $v$ along $x$ and perpendicular to the magnetic field, we know that the Lorentz force is exerted on the charge abruptly considering the instantaneous velocity of the charge. [See the Figure.] However, what happens from the viewpoint of the charge?

The charge asserts that the moving magnet produces an electric field perpendicular to the magnetic field, which exerts a force on it. However, where is the origin of this electric field in the view of classical electromagnetism? From the viewpoint of the charge, the only origin seems to be the change in the magnetic field of the moving magnet that occurs near the edges of the magnet that are located far distant from the charge. Remember that since the magnet is infinitely long, the magnetic field is uniform at any finite distances from the charge along the $x$-axis. Therefore, from the standpoint of the charge, no $\partial B/\partial t$ happens near the charge but at infinite distances away from the charge where the edges of the magnet produce a nonuniform pattern for the magnetic field.

In this case, by the motion of these edges, an electrical field of $\partial B/\partial t$ is produced, however, due to the limited speed of light, it takes time for the induced electric field to reach the charge from infinitely large distances. Therefore, the charge claims that there is a non-zero period of time during which the induced electric field has not reached the charge yet, and thus no Lorentz force is being exerted on it. This problem seems to be paradoxical since the observer at rest with respect to the magnet detects an instantaneous Lorentz force exerted on the charge, whereas the observer located on the charge does not detect any orthogonal force to its motion direction when the charge accelerates in that the induced electric field needs a long time to reach it. Where is the problem?

Either consider the acceleration and final velocity $v$ of the charge to be very small so that the charge can be considered inertial or consider the scenario after the charge reaches a constant velocity of $v$.

Answer 1

When we say that there is an uniform magnetic field, it is also necessary to say: according to a frame $X$. It is not possible to have a pure uniform magnetic field (meaning with no electrical field) for any frame.

For this frame, the complete information is $B_x = B_y = 0$, and $B_z = B$, $E_x = E_y = E_z = 0$. (Here I am neglecting the field from the test charge). That defines the electromagnetic tensor $F^{\mu \nu}$ for this frame.

From the frame $Y$, momentarily co-moving with the charge, the tensor is Lorentz transformed, and $E_y \neq 0$.

The situation didn't change, and it is reflected bt the fact that the tensor is the same. Its components however changed due to the change of frame

Answer 2

where is the origin of this electric field in the view of classical electromagnetism?

This question can be answered by considering the magnetization-polarization tensor, as described in the textbook Classical Electromagnetic Theory, by Jack Vanderlinde (2004, pp. 313–328). The magnetization-polarization tensor is given by: $${\mathcal {M}}^{\mu \nu }={\begin{pmatrix}0 & P_{x}c & P_{y}c & P_{z}c\\-P_{x}c & 0 & -M_{z} & M_{y}\\-P_{y}c & M_{z} & 0 & -M_{x}\\-P_{z}c & -M_{y} & M_{x} & 0\end{pmatrix}}$$

So, in your example in the rest-frame of the magnet (the unprimed frame) it is magnetized in the y direction, so it has a magnetization polarization tensor of $${\mathcal {M}}^{\mu \nu }={\begin{pmatrix}0 & 0 & 0 & 0\\0 & 0 & 0 & M_{y}\\0 & 0 & 0 & 0\\0 & -M_{y} & 0 & 0\end{pmatrix}}$$

Now, ${\mathcal {M}}^{\mu \nu }$ is a standard tensor, so it transforms as usual. In the moving (primed) frame we obtain $${\mathcal {M}}^{\mu' \nu' }=\left( \begin{array}{cccc} 0 & 0 & 0 & \frac{v \gamma M_y}{c^2} \\ 0 & 0 & 0 & \gamma M_y \\ 0 & 0 & 0 & 0 \\ -\frac{v \gamma M_y}{c^2} & -\gamma M_y & 0 & 0 \\ \end{array} \right)$$ Thus there is an electric polarization of $\beta \gamma M_y/c^2$ in the $z'$ direction. This polarization is the source of the E field in the $z'$ direction which exerts the force on the charge in its momentarily comoving inertial frame.

Therefore, from the standpoint of the charge, no ∂B/∂t happens near the charge but at infinite distances away from the charge where the edges of the magnet produce a nonuniform pattern for the magnetic field.

The source of the E field in the primed frame is not from any $\partial \vec B/\partial t$, near or far. It is from the polarization of the same material that is the source of the B field.

the charge claims that there is a non-zero period of time during which the induced electric field has not reached the charge yet, and thus no Lorentz force is being exerted on it. This problem seems to be paradoxical since the observer at rest with respect to the magnet detects an instantaneous Lorentz force exerted on the charge, whereas the observer located on the charge does not detect any orthogonal force to its motion direction when the charge accelerates in that the induced electric field needs a long time to reach it. Where is the problem?

The charge makes no such claim, and there is no such problem here. The E field from the polarization of the material in the primed frame is present any time the B field from the magnetization is present. There is no such period of time where the force is absent in one frame and present in another frame.