New answers tagged maxwell-equations
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Electromagnetic field has energy density, and has momentum. According to relativity, rest inertial mass has a strict relationship with energy and momentum magnitudes as a conic-shell invariant
$$ m^2 c^4 = E^2 - c^2 P^2$$
So this allows the possibility of defining inertia for an electromagnetic field. Look at eq.(5) in this paper: ...
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The correlation is the Special Relativity itself. It connects some different physical values to each other and elucidates them as being spatial and temporal counterparts of one unified value, which "lives" in the space-time being indifferent to the particular space and time aspects. From this point of view, we have better correct our unit measures, so they ...
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Let's look at Ampere's law, $\nabla\times\vec{H} = \vec{J}_f + \partial\vec{D}/\partial t$, where $\vec{J}_f$ is the free current density. The partial derivative of electric displacement vector $\vec{D}$ is the displacement current density. It arises because of a non-steady, that is time-dependent, free charge density $\rho_f$.
One example of a non-steady ...
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Contra Ben Crowell's answer, I'll argue that "displacement current" is a fine name, because Maxwell's equations treat the rate of change of electric flux density (the "displacement current") exactly the same as a charge current. When you hear the term, you know exactly what's being discussed, and even how it fits into the equations.
Without the ...
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Maxwell's equations in a vacuum have induction terms. (1) There is a term saying that a time-varying magnetic field produces an electric field. (2) There is a term saying that a time-varying electric field produces a magnetic field.
Among people who insist on giving hard-to-remember names to all the terms in Maxwell's equations, #2 is called the ...
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The displacement current is the 'phantom' current that passes through a capacitor in a circuit, since no real current runs between two plates of a capacitor. This is given by finding the rate of change of the electric flux with respect to time, and multiplied by epsilon nought. A great video on this can be found here: ...
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A long time before Maxwell wrote down a unified theory, Oersted discovered a connection between electricity and magnetism. In the development of electromagnetism, there were many bits and pieces of partial knowledge that were discovered and formulated by many different scientists. The popular ones (after whom we've named partial "laws") that immediately come ...
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Author gives a clue on the transition:
Let us assume that $\delta\vec{A}$ vanishes at infinity and integrate (formula (1)) by parts...
This is the usual step in the Lagrangian theory of field (actually, of anything). At first, we have the variation of action written in an awkward form:
$$\delta S=\int_{\substack{\text{domain of least}\\\text{action ...
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I intuitively feel that there is some relation between self-inductance and inertia. Actually, self-inductance is basically an electromagnetic inertia. The only difference is that moving mass is closed, concealed inside the massive body, electric inertia has its guts outside -- we can see its lines of force. If flux shows nothing more than the current, then ...
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One can see the consistency with the Heisenberg Uncertainty Principle by the definition of wavelength and frequency of the electromagnetic wave:
lamda*nu/c=1 where c is the velocity of light
Multiplying both sides by h and considering lamda as delta(x) and p=h*nu/c for a photon,
lamda*h*nu/c~h
delta(x)*delta(p)~h
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I see two questions here. The first is why self-inductance is not considered when solving Faraday's law problems, and the second is why an EMF can ever produce a current in a circuit with non-zero self-inductance. I will answer both of these in turn.
1. Why self-inductance is not considered when solving Faraday's law problems
Self inductance should be ...
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Yes you're correct . We willingly don't consider it , it is perfectly correct . However , like in a DC circuit , we think current is established instantaneously . It isn't , first there is a very Induced EMF , and no current flows in circuit . and then there's falling of the EMF with time and it is only at$\infty$ time , that we get the so called current . ...
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I simply misfactorised the quadratic - I knew it was a stupid mistake. I'm amazed that I didn't see it, but even more amazed that nobody else did! Here is the correct solution.
$$\phi_{01}(t,x,y,z) = \frac{1}{2\pi i}\oint d\xi \frac{\xi}{(x^{01'})^2(\xi -\xi_1)^2(\xi - \xi_2)^2}$$
The residue at $\xi_1$ is
\begin{align*}
r_1 &= ...
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Your first mistake is
$$ E_x = F_{01} = \phi_{01}$$
You have apparently confused spinor and vector indices here. The identity $E_x=F_{01}$ holds if $0,1$ are interpreted as the vector indices with four possible values corresponding to $0123=txyz$. But then you can't write that it's equal to $\phi_{01}$ because the latter has the spinor indices with two ...
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