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Attempting an answer: Eddy currents are induced where there is a change in the field. The way you have drawn the situation, there is no place where the field changes while the white rectangle moves: $\frac{dB}{dt}=0$ everywhere in the conductor. So while there is a current flowing around the loop, there is no eddy current induced (that I can see). The ...

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Only in a perfect diamagnetic. In a real conductor the induced magnetic field is limited by the resistance of the material, so it will always be smaller than the inducing field.

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It's a complicated subject - but very well studied in the context of eddy current brakes, where the retarding force is used to create a braking force without mechanical friction / wear. For me, the starting point for finding out more was this post - in particular the posting by Jim Hardy contained lots of good links. It seems that some of the most ...

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You know that $$m\vec a=q\vec v \times \vec B,$$ So in particular, since $\vec B = B\hat z,$ we have $$ma_x=qv_yB,\text{ and } ma_y=-qv_xB.$$ And in our case $B=-\beta x$ so we have $$m\ddot x=-q\dot y\beta x\text{ and } m\ddot y=q\dot x\beta x.$$ You can take the time derivative of the left equation and get $$m\dddot x=-q\ddot y\beta x-q\dot y\beta ... 3 You need to watch what you mean by the ambiguous term "derive", which can mean either "was derived historically" (i.e. was motivated by or is a derivative of, in the non-mathematical sense) or "is derived logically/mathematically". Historically, I think you are correct that \boldsymbol{\nabla}\cdot ... 0 Well I dont know if we can prove it but there is a much more elegant way of formulating EM which may be helpful here. As you may know there are two potentials on EM: the scalar potential \phi and the vector potential \vec{A}, from which \vec{E}(t,x) and \vec{B}(t,x) are derived. From this two objects and following symmetry considerations you can ... 3 Maxwell derived his equations from 1) charge conservation law; 2) Coulomb's law; 3) Bio--Savart--Laplace law; 4) Faraday's law of induction. The equation \boldsymbol{\nabla}\cdot \textbf{E}(\textbf{r})=\frac{\rho(\textbf{r})}{\epsilon_0} was indeed derived from Coulomb's law and in its differential form is written using Gauss--Ostrogradskiy theorem. ... 3 No we cannot prove it; Maxwell postulated that it would hold dynamically because it made the most sense for it to do so as he pondered the displacement current problem. As you likely know, Maxwell pondered the inconsistency between Ampère's law for magnetostatics and the charge continuity equation. Ampère's law for magnetostatics reads \nabla\times ... 1 I can give a back-of-the-envelope derivation of a drag force that ignores fringe effects and other complications. Say the conductor is a plate of thickness \Delta z traveling with velocity v in the x direction. Take the magnetic field to be constant in a rectangular area, with the the \Delta y side perpendicular to the velocity much longer than \Delta ... 1 I don't have the expertise you're looking for, but here's a crack at it: This is a really hard problem whose answer depends on the material properties. In particular, the answer depends on how big the 'eddies' are: if the current moves in roughly a big circle then the effect is large, but if there are lots of little eddies the effect is small. However, ... 1 The derivation assumes the wire is a perfect conductor, and also that it is negligibly thin. If it had some resistivity, then you're right, there would be an electric field in the wire, but even in that case the electric flux \int \vec{E}\cdot\text{d}\vec{a} would be negligible, and so would its time derivative. 0 The physics creating eddy currents and EMFs in inductors is the same: Faraday's law of induction.  \oint_C {E \cdot d\ell = - \frac{d}{{dt}}} \int_S {B_n dA}  The strength of any induced current and voltage is dependent on: 1) The amount of magnetic flux (\int_S {B_n dA}) 2) The rate at which the flux is changing So for the loop in your first ... 2 There is never actually an electric field in a conductor in the electrostatic sense. An E field is always generated perpendicular to a charged surface (the wire). For any wire carrying current, the electric field tends to radiate outward from the wire. The magnetic field will be circulating around the wire such that the Poynting vector,  \vec S = \vec E ... 0 For the flat Amperian Loop, The current flowing through the wire that pierces the surface of the loop) is I. However, there is no field piercing the surface of the loop. Now, why is there is no field piercing the loop:- Of course the field between the plates of the capacitor no way pierces the surface of the loop "Isn't there a field inside the wire, ... 0 For the first figure shown, and assuming equal velocity magnitudes for both charges, I get that Newton's third law does not hold because: 1) The electric force of charge 2 on charge 1 is equal in magnitude and opposite in direction to that of charge 1 on charge. Both have x and y components. 2) The magnetic force of charge 2 on charge 1 is equal in ... 1 The eddy currents are caused when the metal plate which is subjected to a magnetic field is take out rapidly. This causes current to flow within the metal plate and thus opposing the already existing magnetic field (Lenz's law). To calculate the current I think the approach would be as follows. We have F_d, external force B, magnetic field ... 0 I like a 1941 text Stratton Electromagnetic field and Waves and also Like Born and Wolf which is elegant and clearly written. I like another old Text Rojanski. I also like the latest version of Purcell which is in MKS units and has answers to problems in the back -1 I need the Latex environment to correct the answer provided by @Urgje, so forgive me for writing a new answer. As shown by Hitoshi, the commutator between the two components of kinetic momentum is:$$[P_x,P_y]=i\frac{e\hbar}{c}B_z Let's now show that the kinetic momentum does not commute with the Hamiltonian. ...

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