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I'll take a slightly different take on the question than the other answers already posted, because I want to address a different side of the question. Just because a theory makes incorrect predictions doesn't make it useless. Take Newtonian gravity, for instance. It doesn't correctly predict the bending of light or the precession of Mercury, both of which ...


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He may have been thinking about teaching physics top-down, rather than bottom-up. There is nothing wrong with that. That's exactly what Landau/Lifshitz do in Volume 1 of their "Course of theoretical physics", by introducing a least action principle and deriving much of Newtonian mechanics from it. One could do the same thing for electrodynamics, but the ...


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I found the following quote on the American Institute of Physics website. It is a continuation of Feynman's quote above. I believe it answers your question about his new approach. "When I planned it, I was expected to teach electrodynamics, and then to teach a subject which would really be all the different branches of physics, using the same equation — ...


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There are three separate possibilities: A theory correctly predicts an experiment result A theory predicts something but an experiment contradicts it. A theory makes no prediction whatsoever regarding some experimental result. You're treating the second and third possibility as if they were the same thing, but they're entirely different. For example, ...


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To add to Lordrain's answer, I'll give you two scenarios where this equation would be useful. First, notice that this is Maxwell's extension of Ampere's law, meaning you can apply it to situations where a magnetic field exists but there is no current (e.g inside a capacitor with an alternating current). Let's first consider the situation where you'd like to ...


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You are right. These two equations cannot be valid at the same time. In particular, AL is only valid for steady current situation. Indeed, current induces a circulating magnetic field, as indicated by AL. However, current is not the only thing which can produce circulating current. The time-varying electric field can also produce magnetic field. This is what ...


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I understand your confusion here, and I had the same concern when I learned it at the first time. Since the left-hand side is an integral of B, at best we can obtain the flux through some loop, not the value of B at some certain point. Is this your question? Note that the equation do contain some more information. It holds for ANY loop. Intuitively ...


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I believe that Ampere's Law is wrong in some situations. When Maxwell looked into it, he discovered that Ampere's Law is not always true, so he modified it to get the Ampere-Maxwell Law. According to my understanding, this can be shown if you have a wire with a current flow that is charging a capacitor. If you calculate the magnetic field between the ...


1

The first layer are Maxwell's equations. They look like this: $$\nabla \cdot \vec{E} = \rho, \; \nabla \times \vec {E} = -\frac{\partial \vec{B}}{\partial t}$$ $$\nabla \cdot \vec{B}=0,\; \nabla \times \vec{B} = \mu_0( \vec{j} + \epsilon_0 \frac{\partial \vec{E}}{\partial t})$$ Where $\vec{B}$ is the magnetic field, $\vec{E}$ is the electric field, $\rho$ ...


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Are you talking about the famous derivation of the displacement current, where Ampère's law is both true and false depending on what surface you choose to integrate through, despite the same boundary, as below: (Image from WikiMedia commons http://commons.wikimedia.org/wiki/File:Displacement_current_in_capacitor.svg) The solution of this was to add a ...


3

If no charge is moving, there is no magnetic field. A point charge at rest has only an electric field, from "its" point of view. However, electric and magnetic fields are not seperate, since someone moving with respect to the resting charge would see a magnetic field due to the behaviour of the fields under Lorentz transformations. You may (for some ...


2

Hints: This is related to the Maxwell action, cf. e.g. this and this Phys.SE posts. The EL equation is Ampere's law ${\bf \nabla}\times {\bf H}_0={\bf J}_0$, where ${\bf H}:=\frac{\bf B}{\mu}$ and ${\bf B}:={\bf \nabla}\times {\bf A}$. A necessary condition for the existence of a ${\bf H}_0$ solution to Ampere's law is the (stationary) continuity equation ...



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