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1

I think I have a proof for you (though you may find it unsatisfying). For the most part, I'm following the proof on the wikipedia page you link to. I do avoid the dirac delta function however. Starting from the Biot-Savart law: $$\mathbf{B}(\mathbf{r})=\frac{\mu_0}{4\pi}\int_V ...


1

$\newcommand{\dv}[2]{\frac{\mathrm{d} #1}{\mathrm{d}#2}}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\div}[1]{\nabla \cdot #1} \newcommand{\pdvt}[2]{\frac{\partial^2 #1}{\partial #2^2}} \newcommand{\curl}[1]{\nabla \times \vec{ #1}}$I assume you know how to solve the Maxwell's equations in vacum. In the end you get an expression of the ...


2

Below I'll use Planck units, for which, in particular, $c = \epsilon_{0} = =1$. In fact, the full system of Maxwell's equations provides the statement that the only two vector components of the EM field $\mathbf E, \mathbf B$ are independent (in general, due to a deep symmetry reason, namely that a massless particle has only two polarizations). Next, if we ...


0

This question is natural, and I needed to solve it for myself many years ago. I don't remember the derivation and don't have time to reproduce it, but at least in the case of $J=\rho=\sigma=0$ and fixed $\omega\neq 0$ the boundary conditions are not independent, and it is enough to require that tangential components of $E$ and $H$ are continuous at the ...


3

For clarity I think is best to start with Minkowski spacetime. The equation we are trying to solve to understand the radiation of a point particle is: $$\square A^{b}=j^{b}$$ with the gauge $\nabla_{a}A^{a}=0$ and $j^{b}$ is the current density. The potential \begin{eqnarray} A^{b}(t,x)&=&\int G^{b}_{a}(t,x,t'x')j^{a}(t',x')dx'^{3}dt\\ ...


1

To be clear, you can use Ampere's Law: $$\oint \vec B\cdot \mathrm{d}\vec\ell=\mu_0I_\text{enc}.$$ Specifically, it is the form without the displacement current, and it works because you are in magnetostatics. And Rob Jeffries' answer is totally satisfactory. But to specifically address your concern with the charge build-up lets look at an example of a ...


0

The fact that Marconi's machine worked! It relies on the electric field component to affect the electrons in a long wire whuch today we call the antenna. On an optical scale we have modern devices: nonlinear materials in optic fibers, wakefield particle accelerators, and metamaterials bending light as-designed. The E-field and the B-field are very real, ...


-1

This example can be done using the simple form of Ampere's law. Rather than get an accumulation of charge at the end extend the conductor out at right angles and off to infinity. You then have an infinite L-shaped conductor.


1

Well think that in vacuum, the electric field vector $\bf E$ is a good measure of the strength of the electrostatic field. But if you are measuring inside a medium, say there are some charges around where you are measuring, now the presence of those charges affects the value of $\bf E$ from the external electric field. These charges will arrange themselves ...


0

The only explanation I found to this is the following: From Maxwell's Equations in Lorentz-Heaviside units: $\nabla\times E = (i\omega\mu/c)H$ $\nabla\times H = (-i\omega\varepsilon/c)E$ where now $\mu$ and $\varepsilon$ refer to the common relative permeability and permitivity respectively. Defining a sort of new $H' = \sqrt{\mu/\varepsilon} H$ then, ...


0

You can take an infinite cylinder, the field due to it will be due to Ampere's law, equal to for a point at a distance of 'a' from the centre for $a>r$ equal to $\mu_o/2\pi a$ ($J$x $\pi r^2$). Now this can be written as written. Now the thing is you have to make the the volume go to zero, and to let the current remain the same you have to make $r ...


0

Magnetic reconnection comes from a cartoon picture of what magnetic field line motion may portray. This is not based on any physical law. Field lines are not real entity - just a means to display the lines of force when magnetic field is present in space. Field line motion is non-unique also, which is a fundamental flaw for people relying on field line ...


1

How can one meaningfully say that one field generates the other in an EM-wave? You can't because they don't. The electromagnetic wave is an electromagnetic field variation. See Wikipedia where you can read this: "Over time, it was realized that the electric and magnetic fields are better thought of as two parts of a greater whole — the electromagnetic ...


1

A changing electric field does generate a magnetic field, ( hence electromagnets) and a changing magnetic field generates an electric field. The solution for radiated energy is given in terms of electric and magnetic fields in a given reference frame. This animation is instructive for radiation: Electromagnetic waves can be imagined as a ...



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