Well, your lecturer certainly shouldn't have put it like this, however it's true that you have got a lot wrong here. It's stuff you definitely will need to understand better if you're studying power engineering.

First, you seem to think that electrons are attracted by magnetic north poles. They aren't; in fact stationary charges and magnetic fields aren't concerned with each other in any way at all^{1}!

Next, you're talking about electrons in circular orbits about the nucleus. That's roughly the Bohr model, which kind-of-sort-of-works, but not really. You want to familiarise yourself to the orbital model, which describes very well how bound electrons actually behave.

Even in an orbital, you might be inclined to talk about "the nucleus is off the center by a distance proportional to the voltage". That's again kind-of-sort-of-right since the nucleus lies in a locally-harmonic potential which can be read as "pertubation by an electric field (which in a fixed capacitor is proportional to the voltage) will cause a proportional displacement of the nucleus", but the way you phrase it it's still nonsense. Voltage "is" not a distance, it's a potential (i.e. energy).

Anyway, this isn't actually relevant to understanding rotating-magnet phenomena, i.e. inductance in coils. These are concerned only with *conduction electrons*, which aren't bound to any particular atom at all but "move" through the entire conductor, which is why there can be currents. It is these moving electrons that experience a significant force in the presence of a magnetic field. What current actually is is the number and "speed" with which these electrons move through the conductor, while even a strong displacement of the bound (valence) electrons would not consolidate a current^{2}.

Now, all of this seems to say there isn't any such thing as inductance. *Sure there is*! Only, it's rather more complicated: electrons at rest aren't affected by stationary magnetic fields, but in the same way that moving electrons are affected by such fields, *moving magnetic fields* (or, more generally, time-varying magnetic fields) also cause a Lorentz force upon resting electrons. So, effectively, what you're saying about electrons being moved around by moving magnetic fields isn't all that wrong again, it only works quite a bit differently. A moving magnetic field will in fact "push resting conductance electrons" through a wire a bit, i.e. induce a voltage. But that voltage really can't be read as anything displacement-like, it's a fundamental electrodynamic phenomenon. In fact, the voltage in its pure, exact value can only be measured if you *prevent* the conductance electrons from moving, as otherwise they would themselves cause a magnetic field cancelling the inductance etc. pp..

As you see, the whole subject is quite a bit more complicated than you thought. I'm sure you *are* capable of understanding it, but probably not in a few minutes, which is why your lecturer can't really be blamed for not trying to explain it right away.

^{1}Actually, electrons are also small magnets themselves (they have an instrisic quantum-mechanical *spin*) and therefore *are* attracted to inhomogenic magnetic fields, but that's quite another issue.

^{2}Actually, it would... but that's mostly relevant in the high-frequency-regime, i.e. bound electrons that jiggle back and forth very quickly.

"why can we still consider the ball as a ball with uniform charge density ρ ?"

In reality, we cannot. The question doesn't ask about a realistic model. The electron cloud is considered to remain spherical with constant density simply by hypothesis. This is what I infer from "Assuming that the electron cloud moves, but not distorted".

"what does it mean that the atom is in equilibrium at each step of the process ? if this is the case the nucleus shouldn't be moving at all."

Yes, a true equilibrium process is not possible for this reason. But imagine the process takes 1 microsecond. Then imagine it takes one second. Then imagine it takes one year. As we allow the process to take longer and longer times, it can get closer and closer to being in equilibrium at all points. The work done on the atom will always be slightly higher than that calculated in the problem, but it can be made arbitrarily close by taking long enough times. The question is asking about the limit.

"Also, if I change the field gradually, why does the atom is in equilibrium state ? I thought that for small values of E⃗ total electrical force is −x^, since the force of the attraction is higher than the force from the external field, and so the nucleus would go back to the center and then pushed to the right and goes back to the center again, until |E⃗ | reaches some minimal value to overcome the attraction."

I find your wording very confusing, but I think you are asking why there are no oscillations. That is again by assumption - it is assumed the atom is always in equilibrium. If it were oscillating, it would have had to have been out of equilibrium at some point. Again, this is only an idealized model.

## Best Answer

The electron bound to a proton should not be thought of as a point particle. Its position should be thought of as a superposition of all possible electron positions, weighted by their probability. The object which describes the likelihood of observing the electron at any one position or momentum is the wavefunction $\Psi$, which is a function of the coordinates of space, $r,\theta,\phi$.

The electron bound inside a hydrogen atom has a wavefunction that obeys the following Hamiltonian $$\left[-\frac{1}{2\mu}\frac{\partial^2}{\partial r^2} - \frac{2}{r}\frac{\partial}{\partial r} + \frac{\ell(\ell + 1)}{2\mu r^2} - \frac{e^2}{r}\right]\psi(r) = E \psi(r)\,.$$ Here, we've separated the electron wavefunction into a radial part $\psi(r)$ and an angular part $Y_{\ell m}(\theta,\phi)$, so the total wavefunction is written $\psi(r)Y_{\ell m}(\theta,\phi)$. $\ell$ is a quantum number that represents the amount of orbital angular momentum carried by the electron. The coordinate $r$ represents the separation of the proton and the electron, and $\mu$ is the reduced mass of the electron and proton $$\mu = \frac{m_e m_p}{m_e + m_p}\approx m_e + \mathscr{O}(m_e^2/m_p)\,.$$ The Hamiltonian consists of four terms, although the first two should be thought of as a single term representing the radial kinetic energy of the electron-proton system. The third term represents the fact that two bodies with mutual angular momentum will not be able to collide head-on. The fourth term is the attractive electric potential between the proton and electron.

In nature, the electron-proton system will tend to occupy the lowest energy level accessible to it. In this case, the $\ell = 0$ state, corresponding to a spherically symmetric wavefunction, is the lowest energy configuration of the system. Up to a normalization factor, the ground state is $$\Psi(r,\theta,\phi) \propto e^{-r/2r_b}\,.$$ Here, $r_b$ is the bohr radius of the atom. We are now in a position to address your question.

In order to ask "where is the electron?" we need to be a bit more precise in what we mean. The electron can in principle be anywhere, since the electron wavefunction is non-zero everywhere. Therefore, the correct question to ask is "where am I most likely to find the electron?" The job of answering this question is given to a set of objects called "expectation values," typically written as $\langle x \rangle$ where $x$ is the quantity you are measuring and $\langle\cdot\rangle$ represents an ensemble average. In quantum mechanics, the expectation value is computed as follows $$\langle x\rangle = \frac{\int{\rm d} V \Psi^* x \Psi}{\int{\rm d} V \Psi^*\Psi}$$ In our case, we're interested in measuring the expectation value of the position operator $\vec x$. Therefore, we must compute (up to a normalization) $$\langle \vec x\rangle = \int{\rm d} V \Psi^* \vec x \Psi = \int{\rm d} V \left[\begin{array}{c}r \sin\theta\cos\phi\\r \sin\theta\sin\phi\\ r\cos\theta\end{array}\right] e^{-r/r_b}\,.$$ Performing the angular integrals yields the result $$\langle \vec x\rangle = \vec 0\,.$$ Recalling the physical meaning of the $r$ coordinate is the separation of the proton and electron, we find that the most probable position of the electron is precisely on top of the proton. In other words, the center of mass of the electron is aligned with the center of mass of the proton.

Another question you might ask is "how far away from the proton should the electron be on average?" Now we're interested in the expectation value of the operator $r$, instead of $\vec x$. $$\langle r \rangle =\frac{ \int{\rm d} V \Psi^* r \Psi}{ \int{\rm d} V \Psi^*\Psi} = \frac{\int{\rm d} V r e^{-r/r_b}}{\int{\rm d} V e^{-r/r_b}} = r_b\,.$$ Thus, we expect to find the electron a Bohr radius away from the proton!

To summarize: the electron will most likely be found at the origin, but its typical separation from the proton is a Bohr radius.