A moving charge in an uniform magnetic field An electron is in a region where there is an uniform magnetic field created by a large disc-shaped magnet. As is well known, if the electron is moving with velocity $\vec v$ parallel to the flat surface of the magnet (ie, perpendicular to the magnetic field) experiment the Lorentz force, $\vec F=q(\vec v\times\vec B)$. But what would happen if the magnet is moving with velocity $-\vec v$  while the electron is at rest? How we can explain the origin of the force? (Just make a qualitative analysis.) Do not use the Lorentz transformation, we want find the physic origin of the force.
 A: The explanation is that the movement of the magnet causes the magnetic field to change, and a changing magnetic field induces an electric field. The relevant equation is $\vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$. This equation says that a changing magnetic field sources an electric field the same way a current sources a magnetic field (up to a sign). This explains the origin of the force. In principle, it should be possible to derive that the electric field you get in the uniform region is $\vec{v} \times \vec{B}$.
In fact, here is the proof: Before the transformation, in the frame where the magnet is not moving, we have a constant magnetic field and therefore a constant vector potential. When we switch frames, we the vector potential is not constant, and therefore it generates an electric field given by $\vec{E} = -\partial_t \vec{A}$. In the quasi-static approximation, this $\vec{A}$ is just the galilean transformed $\vec{A}$ from the frame of constant magnetic field. Now $\partial_t \vec{A} = -\vec{v} \cdot \vec{\nabla} \vec{A} = -\vec{v} \times (\vec{\nabla} \times \vec{A}) - \vec{\nabla} (\vec{v} \cdot \vec{A}) = -\vec{v} \times \vec{B} -\vec{\nabla} (\vec{v} \cdot \vec{A}) $, where $\vec{v}$ is the velocity of the charge in the frame where it is moving, and therefore minus the velocity of the magnet in th frame where the charge is stationary.
Now we have $\vec{E} = \vec{v} \times \vec{B} +\vec{\nabla} (\vec{v} \cdot \vec{A}).$ Thus we see we have the $\vec{v} \times \vec{B}$ term as expected, but we also have a term that looks like there is a potential of $-\vec{v} \cdot \vec{A}$. This second term is zero in this case because $\vec{A}$ points in the same direction as $\vec{J}$. This term is due to relativistic effects. To see exactly what this term is, recall that $\vec{A}(\vec{r}) = \frac{\mu_0}{4 \pi} \int \frac{\vec{J}(\vec{r}_0)}{|\vec{r}-\vec{r}_0|} d \vec{r}_0$ so that $\vec{v} \cdot \vec{A} = \frac{\mu_0}{4 \pi} \int \frac{\vec{v} \cdot \vec{J}(\vec{r}_0)}{|\vec{r}-\vec{r}_0|} d \vec{r}_0=\frac{\mu_0}{4 \pi} 4 \pi \epsilon_0 \int \frac{\vec{v} \cdot \vec{J}(\vec{r}_0)}{4 \pi \epsilon_0 |\vec{r}-\vec{r}_0|} d \vec{r}_0 =\frac{1}{c^2} \int \frac{\vec{v} \cdot \vec{J}(\vec{r}_0)}{4 \pi \epsilon_0 |\vec{r}-\vec{r}_0|} d \vec{r}_0  $, so the apparent potential $-\vec{v} \cdot \vec{A}$ appears to be coming from a charge distribution $-\frac{1}{c^2} \vec{v} \cdot {\vec{J}}$. This charge distribution comes about because if you lorentz boost a stream of moving charges, the charges will contract more if you speed them up and hence the density will increase, or they  will "uncontract" if you are slowing them down, and the density will decrease. Thus we can already see there is a problem with non-relativistic electrodynamics. A relativistic treatment would give you the right answer.
