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5

I actually agree with Ruslan's comment. You cannot say that the integral blows up when $\textbf{r} = \textbf{r}'$, with $\textbf{r}'$ spanning the integration domain where $\rho \neq 0$. The reason is simply that this is a triple integral and that the volume form in the integral may compensate the diverging behaviour of the Green function. To see this you ...


4

First of all, there is no real or observable lines. Even the magnetic and electric fields are nice and abstract fields which describe observable forces. The term "line" you read is an old unit of magnetic flux. One line is the flux of a uniform magnetic field of one gauss across a surface of one square centimeter perpendicular to the field, $$1\ line = 1 ...


4

The term "gradient" implies a change in some quantity versus a change in second quantity, usually over a distance. It's very much like a slope. For example, the gradient of a roof line on a house is given as rise/run like 15 cm/m or 5 inches/foot. The gradient of potential is the electric field magnitude, with SI units of volts/meter. A magnetic field is ...


3

Replace "magnetically charged" by "magically charged" - I would rather agree to that :) (and that's actually how I read it by mistake! ;)) Writing of magnetic charges (=monopoles) actually doesn't increase the plausibility. I think it's just a hoax (I mean, that the university is involved, not the whole business) - though this word would be not ...


3

I would say yes, the electric field is well defined in a volume charge distribution. You raise some interesting points, that I think ultimately comes down to: Is the electric field consistent and well defined mathematically? What is the physical interpretation of the mathematics? Is a volume charge distribution physical and realistic? To answer the first ...


3

A 'gradient' measures how quickly something changes with respect to something else. In this case, it's how much the magnetic field strength changes per unit length.


2

Field lines are a good concept for imagining things, but it does not reach too far. Imagine for example the field of two distinct sources -- the field lines would cross if you just draw them both. But this does not represent the sum field. Field lines are drawn by convention so, that their density is approximately proportional to the field strength. This is ...


2

Your reasoning goes astray when you speculate that overheating would cause an open circuit. The insulation melts before the copper, causing a short across many of the coil wraps. This reduces the strength of the electromagnet dramatically, causing the motor to not move.


2

The electric field at a point outside the volume charge distribution is well defined, certainly. What about the electric field at a point inside the charge distribution? Let $\vec{r'}$ denote a point that lies in the charge distribution and $\vec{r}$ denote a point where the electric field is to be determined. The problem is the determination of the electric ...


2

We put (four pi times) the permittivity $4\pi\varepsilon=1$ equal to one from now on (in the SI unit system) for simplicity. Let ${\bf r}_0\in \mathbb{R}^3$ be a fixed position. The electric field in the $i$'th Cartesian direction at ${\bf r}_0$ is $$\tag{1} E^i({\bf r}_0)~=~- \int_{[0,\infty[\times [0,\pi]\times[0,2\pi]} ...


2

At the start as the rings fall the magnetic flux through them increases and so an emf is induced in each of them. According to Lenz if there is an induced current it would flow in such a direction as to oppose the motion producing it. In the example given if there was an induced current it would be in a clockwise direction if one observed from the top. Such ...


1

What does the Poynting flux represent? Energy flow associated with wave motion. See this Blaze Labs picture: The Poynting vector is pointing in the direction the wave is travelling, transporting E=hf energy. Or energy-momentum if you prefer. I know that the Poynting flux is the cross product between $\vec E$ and $\vec B$ I'm afraid there ...


1

People are not understanding your question. I think you want someone to verify explicitly that the fields produced in your special case do (or don't) obey the generally valid wave equation. After all, the field produced does not look like a wave. A general solution of the wave equation for a disturbance traveling in the $x$ direction is ${\bf{E}}({\bf ...


1

The Poyntings vector is given by $${\textbf{S}=\frac{1}{\mu_0} (\textbf{E}\times \textbf{B})}$$ or $${\textbf{S}=\textbf{E}\times \textbf{H}}$$ It is known that electromagnetic waves carry energy with them. The purpose of the Poynting vector is well explained by the Poynting's theorem which is the work energy theorem in electrodynamics. ...


1

Removing a part of the rings prevents currents from going round in the ring. If there is no current, there is no magnetic field, and therefore there is no force to break the fall. Thus the "broken" ring and the plastic ring fall with the same velocity.


1

You are very close. First consider how the different lights colors interact with each other, please see additive color for an explanation. Then if you consider a double slit experiment as shown on wikipedia with coherent monochromatic light. Now we know that sunlight is composed by continium of wavelength of slightly different colors, but you can think of ...


1

I assume that A is neutral to begin with. Then inside of A cannot be any charge by Gauss law. This means that the inside of A must neutral. Then you can take the volume between B and A. The boundary of that are the metal spheres A and B. On the whole boundary surface, there cannot be any electric field as that surface lies inside the metal. If there was any ...


1

I agree that the volume form in spherical coordinates cancels the $1/r^2$ divergence from Coulomb's law, but I think there's a more physical way to interpret this mathematical fact: remember that strictly speaking, for a continuous charge distribution with no delta functions in the density, $\rho({\bf x})$ is not the amount of charge at point ${\bf x}$. ...


1

Wave velocity, when not otherwise specified, usually means the phase velocity: the rate at which the phase travels in space. This site animates the differences between phase and group velocities. The group velocity is the speed at which energy is transported. So if the context implies that the wave is carrying energy, the wave velocity is also the group ...


1

Of course in theory nothing stops you from doing the same thing on a bigger scale. Consider however that the effect is rather feeble and you need really strong field for heavier objects, which is expensive and quite cumbersome. The relevant equation is given here. If you are mainly looking for a fascinating project, I would consider magnetic levitation, ...


1

The power in the value of force is not right. There is a negative sign there. $F=1.602*10^{-17}$N $m=9.1083*10^{-31}$kg is not the right way to use here. The particle is moving close to the speed of light. The value of m you used here is the rest mass of electron. You need to apply the relativistic mass here given by the formula ...


1

There is a simple answer: Symmetry. Suppose the material is isotropic, and consider the initial condition of the p-polarized case, with a p-polarized light wave about to hit the surface. In this case, reflecting in the plane that contains the incident and scattered wave vectors leaves both the (vector) electric field and (pseudovector) magnetic field ...


1

The definition of s-polarised light is that the electric field is polarised so that it is perpendicular to the plane of incidence. Where there is a specular reflection, the plane of incidence contains the k-vector of the incoming wave and the reflected wave. Since the electric field of an EM wave must be perpendicular to the k-vector. This then leaves the ...


1

Here's a simple explanation based on the dipolar nature of the medium: For most of the materials, we can assume that the source of the reflected and refracted waves are the induced tiny dipoles in the dielectric medium. In an isotropic medium, polarization vector is proportional to the (total) electric field vector with a constant (as opposed to a tensor ...


1

It does't really make much sense to talk about a tree-level truncation (it helps for calculations, but that's it) or to take the first Feynman diagram as a true representation of reality. By the way, in your $e^- e^- \to e^- e^-$ example, the whole notion of spatial separation is ill-defined since this is a t-channel process. If going from virtual to real ...


1

Much of chapter I, section 9 in "Foundations of potential theory, Oliver Dimon Kellogg, Berlin: Verlag von Julius Springer, 1929", parts of which appear in the answers by Mathaholic, Procyon and Qmechanic, is devoted to answer this question. Let $v$ be a small region of arbitrary shape, containing $P$ (defined by $\vec{r}$) in its interior. We consider the ...


1

This sort of calculation, especially when the speed of the electrons from an observer's point of view is close to $c$, has to be done using special relativity, in that sense that the transformation between reference frames is determined by Lorentz rather than Galileo transformations. As you mentioned, you can put your reference frame origin at one of the ...


1

The imaginary part of the relative permittivity accounts for absorption in materials: \begin{equation} n = \sqrt{\epsilon / \epsilon_0} \end{equation} for the refractive index. Once you have this quantity, you're in a better position to start thinking about what happens when EM waves pass through a material. When loss is neglected, $n$ and $\epsilon / ...


1

X-rays are produced when the electrons hit the screen. But the glass envelope is leaded, so little if any radiation escapes.



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