# Tag Info

2

You're perfectly correct. Referring to Classical Electrodynamics by Jackson, we see that the index of refraction $n$ is given by: $$n=\sqrt{\frac{\mu}{\mu_{0}}\frac{\epsilon}{\epsilon_{0}}} = \sqrt{\mu_{r}\epsilon_{r}}.$$ But Jackson notes that for most optical frequencies (and non-meta-material media), $\frac{\mu}{\mu_{0}}=\mu_{r}\approx 1$, which is why ...

1

I did the 1st example in Schneider's online book both ways and got the same graphs, so I'm convinced the answer is that they are equivalent but that Yee's algorithm is faster by a factor of 4 (it uses 1/2 as many points and 1/2 as many time steps).

0

Now per QED, electrical charges interactions are effected by photons. Suppose you are one of the two charges. How do you know to attract or repel the other charge? You want something that does not exist - intuitive picture of physical process within a theory which is a demonstration of how far can one go with mathematisation of experience and ignoring ...

0

The attraction of unlike charges and the repulsion of like charges is an experimental observation that has to be included in any model of electromagnetic reactions When talking of photons one is in the quantum mechanical regime. As in classical electrodynamics the sign of the charge defines the potential, attractive or repulsive, between the two charges, ...

3

That is the point of a Faraday cage. Although some of the EM field can penetrate inside the shell, under various circumstances, there is an effect where a conductive shell forms a void in a pervasive EM field. To be clear, though, EM fields do not just 'stop' - they are formed of photons, which are reflected at such an interface. Consider the microwave ...

7

Your misconception is the application of Gauss's law. There are solutions to Maxwell's equations (including Gauss's law), that permit an electric field inside the shell. If there are no charges inside the shell, all Gauss's law tells you is that the number of electric field lines entering the shell must equal the number coming out. An example of a field ...

9

Very simply: just because something is true in the static case doesn't automatically make it true in the dynamic case. To reject a static electrical field from inside a shell, it is sufficient for the charges on the surface on the shell to move however slowly they want in order to arrange themselves so as to cancel the field. This is Gauss's Law. When an ...

0

The thesis you refer to puts values of 0.02 S on the conductivity of freshwater and 4 S for seawater. I'm assuming that the S here stands for the SI unit of Siemens per metre, the value of $\sim 5$ Siemens per metre is one I have used for seawater in the past. Submarines communicate with frequencies as low as 100 Hz.To test for a "good conductor" we compare ...

0

What is required for sparks across a gap is a high voltage, while what determines whether something will do you harm is the amount of current. You can theoretically survive almost any voltage so long as the current is sufficiently low. It's just that most sources of electricity that have high voltages tend also to have high current.

2

Please read my answer here Collapse of the wave function and Heisenberg uncertainty where I address this part of the question. 1) there existed the spectra from hydrogen and other atoms that could be explained by quantized orbits 2) the hydrogen atom spectra were fitted with a series and the Bohr model explained the series 3) Postulates are as axioms for ...

1

The problem here is in the way you take the volume to be small. You assert that $$\int \mathbf E \times \mathbf B \mathrm dV \to 0$$ while $$\int_S \mathbf T \cdot \mathrm d\mathbf S \neq 0.$$ But this limit is not well-defined. $\mathbf S$ is the surface of the volume over which you integrate and will vanish as well in your limit $V \to 0$. If you ...

0

This looks an incredibly difficult way of solving what is quite an easy problem. If you use Faraday's law in integral form, constructing a small, rectangular loop that goes into and out of the interface, it is easy to show that the component of the E-field that is perpendicular to the normal surface vector (i.e. the E-field parallel to the interface plane) ...

1

An electromagnetic wave consists of an electric and magnetic field that moves through space. If the space that the wave moves through is filled with electric charges, such as the ions in salt water, then the electric field will start to push these charges around. This pushing requires energy, and this energy is drained from the electric field. This is the ...

0

This looks pretty much correct to me, except that it is not true that any object that approaches the poles will be attracted because they are much more permeable than air Any object that is ferromagnetic will be very strongly attracted. Paramagnetic materials may also be attracted, but less strongly. diamagnetic matrials would be repelled. An ...

4

The electromagnetic field itself contains energy distinct from the energy of charged bodies, the energy in a given volume of empty space can be found by integrating the energy densities $\frac{1}{2}\epsilon E^2$ and $\frac{1}{2} \frac{B^2}{\mu}$ over the region. When the EM fields increase the kinetic energy of charged particles, there is a corresponding ...

1

"Vacuum" actually means the ground state of space. When there are real particles in space, "it" is in an excited state, and no longer the vacuum. But we usually think of a small region of empty space as approximating a vacuum, for reasons of locality, etc. If you have two real particles in space, and talk about their mutual interaction, then you necessarily ...

0

But in case of a stationary conductor with v=0 how can variable magnetic field apply force on stationary charges inside a stationary conductor? $$\nabla \vec E = -\frac{\partial \vec B}{\partial t}$$ $$\vec F = q(\vec E + \vec v \times \vec B)$$ There is a non conservative electric field associated with a time changing magnetic field and thus, a ...

0

Let me explain the difference using a simple model of an atom - as consisting of electrons revolving in orbits around a heavy nucleus. When a material is exposed to a magnetic field, the electrons move in such a manner as to oppose it. The material mildly repels the magnetic field. However, if a material has unpaired electrons, it has a net molecular ...

0

Magnetic forces are not easy to apprehend. Personally, I dislike magnets, so in a first step I will use coils. Consider two coils $S_1$ and $S_2$ along the $\vec{z}$ axis at a distance $d$ one from each other. They are fed by a current $\vec{i}=I\vec{e_\theta}$. As you know, the magnetic field induced by each coil is like : The magnetic field from $S_1$ on ...

1

Your calculation is right: it is telling you "what goes in, comes out again"! The plane wave does indeed bear energy. The two nonzero parts of your calculation: $$P_1=\int_{s_1} Re \,\, \bar S \,\,\hat i_n dS = - \frac {| \bar E|^2} {2 \zeta} A$$ $$P_2=\int_{s_2} Re \,\, \bar S \,\,\hat i_n dS = \frac {| \bar E|^2} {2 \zeta} A$$ are opposite in sign. ...

0

As it looks like another question I've supplied an answer to might be duplicated here (and hence closed), I am going to provide a similar but not identical answer here. In words - divergence is the flux of something into or out of a closed volume, per unit volume. The best visual picture I have of this is a fluid flow. Imagine water spewing out of a tap - ...

2


0

Can someone help provide me with an argument why the electric field must be zero in a perfect conductor? It's not clear exactly what you're looking for. In a sense, any argument attempting to prove that the electric field must be zero in a perfect conductor will beg the question. For example, here's an excerpt from "Electromagnetics for High-Speed ...

1

The electrical field $\mathbf{E}$ is an external field, which "drags" the conductor electrons through the conductor "lattice". The conductivity $\sigma$ describes the resistance of the "lattice". When the resistance is zero, a non zero current can exist in the conductor without necessity to support it with an external field. If $\mathbf{E}$ is non zero, the ...

0

What makes you think they don't? When I was a wee lad, my physics teacher showed a video to me of exactly that. A man was driving down a road listening to a radio station (BBC if memory serves). At regular intervals, the music would get quieter and cut out. The issue was that the road in question was nearly along the line between two radio transmitters, ...

1

Or rather, if the magnitude of the EM wave is not of concern, then what exactly is the information which is sent and what information is received and decoded by the receiving antenna? Let me give you some perspective on this, as it's a point that is often overlooked. The whole reason you can receive a (weak, as you say) signal in the distance, is ...

1

Good question. If you look at a list of radio stations in the UK you will see that most of them (and all of them that cover the whole country) use several different frequencies to broadcast precisely because of this problem. (look at the AM list which is clearest on this point - AM = amplitude modulation). Radio stations that cover a large area use several ...

1

So, in the case of waves traveling through a pond, if you imagine the transmitter as the rock hitting the pond and the receiver as, say, a frog sitting in the water, then you can see that the frog feels the waves that bounce back or interfere with each other, but the first wave that hits him will be the strongest. While a wave is traveling, a significant ...

0

Your argument against is based on the fact that in one frame there is no magnetic field in one frame, but there is a magnetic field in a different frame. So there must be magnetic virtual particles in some frames but not in others. Hence magnetic virtual particles can't exist. However there aren't separate magnetic and electric virtual particles. There are ...

1

To start with the electric field of the electron is as far as possible to measure symmetric, the electric dipole moment is very small. In this article The electron's EDM must be collinear with the direction of the electron's magnetic moment (spin). Within the standard model of elementary particle physics, such a dipole is predicted to be non-zero but ...

2

The Lorentz force experienced by a charge $q_1$ is: $$\mathbf F = q_1(\mathbf E + \mathbf v \times \mathbf B)$$ where x means vector-product. The electric field $\mathbf E$ between charges does not come from magnetic properties of the charges. With the magnetic field the things change. A moving charge $q_2$ produces a current, and a current produces ...

0

There is a 90 degree phase difference between the input current and the induced voltage in the ring because of Lenz's law. Then, there is a 90 degree phase difference between this induced voltage and the induced current because of the self-inductance of the ring. This amounts to a 180 degree phase difference between the input current and the induced current ...

-2

I think the reason why the magnetic force is perpendicular to the magnetic field is, when the magnetic field exerts a force on the current carrying wire, the huge force at the centre moves either upward or downward depending on the direction of the current.

3

If the cross product is zero, the two vectors are parallel (here $\hat n$ and $\vec E_1 - \vec E_2$). If the difference of the two fields has only a normal component the tangential component is zero or: $\vec E_{1,t}-\vec E_{2,t} = 0$ or $\vec E_1 = \vec E_2$. A more detailed way to look at it, is by splitting the vectors into normal and tangential parts: ...

3

If the cross product of two vectors is zero, that means that the two vectors are parallel. That is, $\hat{n}$ is parallel to $\vec{E_1} - \vec{E_2}.$ Since $\hat{n}$ is normal to the surface, that means that $\vec{E_1} - \vec{E_2}$ is normal to the surface. That is, the only difference between the two fields lies normal to the surface. Therefore there is no ...

0

The impact on the Earth's magnetosphere causes it to fluctuate and that induces high currents in things like power lines. These are potentially so great they can cause surge arresters to trip and close down parts of the electricity grid. A Carrington Event would be so great that it could melt grid power transformers and cause massive damage that might take ...

6

The Lagrangian provided is Maxwell's Lagrangian, supplemented by a gauge fixing term: $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - \frac{1}{2}(\partial_\mu A^\mu)^2$$ The equations of motion are, $$\partial_\mu F^{\mu\nu} + \partial^\nu (\partial_\mu A^\mu) = \partial_\mu \partial^\mu A^\nu = 0$$ Instead of making a gauge fixing procedure a ...

4

The extra term, in general $$\mathcal{L}=-\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}-\frac{1}{2 \xi}(\partial_{\rho}A^{\rho})^2$$ is called gauge fixing term. This term is needed in order to be able to quantize the field $A_\mu$. Without this extra term the photon propagator is ill defined D^{\mu\nu}={-i\over k^2+i0}\left(g^{\mu\nu}\,+\,(\xi-1){k^\mu ...

1

The result that is most relevant here for understanding this is Gauss' Law. Gauss' Law says in particular that the electric field 'through' a surface (i.e. the flux) depends only upon (and in fact is proportional to) the charge contained inside the surface. Here this implies that the force on a test charge inside the outer shell does not depend at all on ...

0

Electrical sources, like current sources are conventional electrical elements that behaves according to a definition. A current source is said to impose a current. For example, when studying the circuit, you do assume that the current in the wire is chosen by the current source only. Thus, when there are two different current sources in series, you have two ...

2

No. The difference in behavior between radio waves and light is entirely due to the difference in frequency (and hence, wavelength). The photoelectric effect would work equally well with radio waves if the energy needed to liberate an electron were very small. The difference in that case is with the material not the radiation. Light and radio both ...

1

In the context of the homopolar motor you can find lots of resources on the Internet: e.g. http://blog.first4magnets.com/what-is-a-homopolar-motor-and-how-does-one-work/ In electromagnetism an EMF can be produced by changing the magnetic flux through a conducting circuit. However you can also produce a "motional EMF" by moving charged particles in a ...

0

If I understand correctly, the power that power sources output can be constant, but the current and the voltage is purely dependent on what loads you have in the circuit. Say my power source is a coal burner, that burns 5 coals a minute to produce 10 watts of power. Using P=IV, the current and the voltage can both be seemingly a wide variety of variables. ...

1

Electrons such as all protons, neutrons, positrons, ... have magnetic moment(s). If a electron lay in a magnetic field this field will align this moment, such like every magnetic dipole will be aligned. Now if the electron is moving into a magnetic field his magnetic dipole will be aligned too. But this alignment is accompanied by a gyroscopic effect. ...

0

No the textbook is correct. Are you using Flemming's right hand rule? Because according to the right hand rule if B (the magnetic field) is into the page and the motion of the wire is towards the left then the current will be pointing downwards. Relate that to part of the circuit parallel to the moving "arm PQ" and the circuit will be counter clockwise. :) ...

1

Cores of the transformers a made of ferromagnetic material, which can change the shape due to magnetostriction phenomena. When transformer works with typical grid frequency of 50Hz..60Hz, it can be heard. As it was answered above, generated noise depends on mechanical construction. New power supplies use frequency converters, which drive the transformers ...

6

Transformers generate oscillating magnetic fields at the mains frequency and the fields produce an oscillating force on: anything nearby that's ferromagnetic (like the core) anything nearby that is carrying a current (like the windings) The sound you hear is because various bits of the transformers are moving in response to the oscillating fields and ...

1

90% of Astrophysics is to do with electromagnetic phenomena. Bar neutrinos or directly grabbing stuff in our own solar system, there's not much else you can do but observe the electromagnetic radiation coming from out there. Your question is therefore massively broad. But here are some examples you could research. Rayleigh scattering observed in the ...

2

It might not actually answer your question, but to throw it into the bowl: There are some advances in MRI using permanent magnets and even conventional electromagnets with static magnetic fields of about 0.5 Tesla. As far as I know one can do imaging with a reasonable resolution with these devices without the need for extensive cooling. They are used for ...

1

I think your argument is completely flawed. Consider a uniform H-field. The closed line integral of this field around any loop is zero - and there must be no free current through the loop. Hence there is no free current, yet the H-field is non-zero. You might be better off thinking about this in terms of free current density. In this case we can say ...

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