Studiot said:
Why do responders keep claiming either
'the only force acting is electrostatic'
'no other forces are acting except electrostatic ones'
BUT
Introduce other forces, eg the nuclear force, to complete their argument.
When did I introduce a nuclear force? Also, it is more than a slight misnomer to say that the only forces are electro
static in nature, they aren't. For starters, the electric fields involved are usually time-varying, and not produced by static charge distributions, but rather by time-varying magnetic fields via Faraday's law. They are electric forces, not electro
static ones. More importantly, magnetic forces certainly are involved,
but they don't do any work...they simply redirect the work that is being done by the electric forces, from horizontal to vertical (loosely speaking).
I requested a derivation of a practical formula, such as the one I presented, from this much vaunted Lorenz vector formula, if I actually needed to calculate the force between the car and the crane I could do.
I asked for an opinion of what would happen I interposed an electrostatic screen to prevent the transmission of any electrostatic force between the crane and the car.
None of these questions have been answered.
Oye!
The derivation of the formula you gave from first principles (i.e. the Lorentz force law) can be quite involved and there are many different ways to go about it. The formula is often constructed in bits and pieces over several chapters of an EM text and to fully understand what is going on, you really need to go through the rest of the material. But, since you seem to have an aversion to electrodynamic textbooks (Have you even looked at the textbook example born2bwire and I referenced earlier?), and I have some spare time, I'll present one short proof, direct from the Lorentz force law.
(1) Begin by considering a bulk material, carrying a current density \textbf{J}(\textbf{x}) in an external magnetic field, \textbf{B}(\textbf{x}). The Lorentz force law tells you that the force on said material is
\textbf{F}=\int \left[\textbf{v}(\textbf{x})\times\textbf{B}(\textbf{x})\right]dq
Where \textbf{v}(\textbf{x}) is the velocity of an infinitesimal charge carrier dq in the current, located at position \textbf{x}. Using the fact that current density \textbf{J}(\textbf{x}) is defined as the amount of charge passing through the plane perpendicular to \textbf{J}(\textbf{x}), per unit area, per unit time, we see that \textbf{v}(\textbf{x})dq=\textbf{J}(\textbf{x})d^3x (where d^3x is an infinitesimal volume element of the material) and hence
\textbf{F}=\int_{\mathcal{V}} \left[\textbf{J}(\textbf{x})\times\textbf{B}(\textbf{x})\right]d^3x
Similarly, the force on a surface current \textbf{K}(\textbf{x}) in an external magnetic field is given by
\textbf{F}=\int_{\mathcal{S}} \left[\textbf{K}(\textbf{x})\times\textbf{B}(\textbf{x})\right]da
(where da is an infinitesimal area element of the material's surface)
(2) We then consider a magnetized slab of material, with magnetization \textbf{M}. We model this magnetization as coming from a bound surface current density \textbf{K}(\textbf{x})=\textbf{M}(\textbf{x})\times\textbf{n} and a bound volume current density \textbf{J}=\mathbf{\nabla}\times\textbf{M}(\textbf{x}) (See section 6.2.2 of Griffiths).The force exerted on it by an external magnetic field is then given by
\textbf{F} = \int_{\mathcal{V}} \left[\left(\mathbf{\nabla}\times\textbf{M}(\textbf{x})\right)\times\textbf{B}(\textbf{x})\right]d^3x+\int_{\mathcal{S}} \left[\left(\textbf{M}(\textbf{x})\times\textbf{n}\right)\times\textbf{B}(\textbf{x})\right]da
We then use the identities
\mathbf{\nabla}(\textbf{a}\cdot\textbf{b})=\textbf{a}\times(\mathbf{\nabla}\times\textbf{b})+\textbf{b}\times(\mathbf{\nabla}\times\textbf{a})+(\textbf{a}\cdot\mathbf{\nabla})\textbf{b}+(\textbf{b}\cdot\mathbf{\nabla})\textbf{a}
and \textbf{a}\times(\textbf{b}\times\textbf{c})=\textbf{b}(\textbf{a}\cdot\textbf{c})-\textbf{b}(\textbf{a}\cdot\textbf{b}) along with the fact that \mathbf{\nabla}\cdot\textbf{B}=0 (one of Maxwell's equations) and \mathbf{\nabla}\times\textbf{B}=0 inside the material as long as it contains none of the source current that were responsible for the external field (Ampere's law), together with the divergence theorem and integration by parts to obtain
\textbf{F} = -\int_{\mathcal{V}} \left(\mathbf{\nabla}\cdot\textbf{M}(\textbf{x})\right)\textbf{B}(\textbf{x})\right]d^3x+\int_{\mathcal{S}} \left[\left(\textbf{M}(\textbf{x})\cdot\textbf{n}\right)\textbf{B}(\textbf{x})\right]da
(3) We now apply this to the problem at hand by approximating the car by a chunk of linearly magnetizable material
\textbf{M}=\frac{1}{\mu_0}\left(\frac{\chi_m}{1+\chi_m}\right)\textbf{B}
And consider the crane's magnetic field to be approximately uniform when its magnet is close to the car. Hence, the force on the car is
\begin{aligned}\textbf{F} &= -\frac{1}{\mu_0}\left(\frac{\chi_m}{1+\chi_m}\right)\int_{\mathcal{V}} \left(\mathbf{\nabla}\cdot\textbf{B}\right)\textbf{B}(\textbf{x})\right]d^3x+\frac{1}{\mu_0}\left(\frac{\chi_m}{1+\chi_m}\right)\int_{\mathcal{S}} \left[\left(\textbf{B}\cdot\textbf{n}\right)\textbf{B}(\textbf{x})\right]da \\ &= \frac{\left(\textbf{B}\cdot\textbf{n}\right)\textbf{B}A}{\mu_0}\left(\frac{\chi_m}{1+\chi_m}\right)\end{aligned}
where A is the surface area of the car. If we assume \chi_m=1 (a very large susceptibility) for the car, and the field is perpendicular to the top of the car, we recover your formula exactly.
Since this is derived directly from the Lorentz force law, and said law tells us magnetic fields do no work, we must conclude that the role of the magnetic field in this case was to redirect work being done by electric forces at play inside the crane's magnet and the car at the microscopic level from horizontal to vertical, hence lifting the car.
As for your second question, if you can come up with a material that screens only electric fields and not magnetic ones and place it between the crane and the car, there will be absolutely no change. The crane will still lift the car. The electric forces that are directly responsible for the work are internal to the crane's magnet, and the car and will not be screened by placing your screen between the two objects.
