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1

Yes and no. Charges and currents curve the $U(1)$ gauge connection. We experience this curvature every day so we even have a special name for it: an electromagnetic field. Just like spacetime curvature is called gravity. However, the choice of the word 'curvature' is somewhat unintuitive here due to the fact that it is not our spacetime that gets curved. ...


1

When a wave travels through a rope, the rope goes up and down, the position of all the 'rope-particles' changes, they oscillate and this makes up the wave. With light, it is indeed the electromagnetic field oscillating, but you shouldn't think of the arrows that represent that field in your first picture of light as 'extending into the rest of the space'. ...


0

The answer appears to be neither a) nor b) According to this discussion, the radiation pressure varies with the phase velocity of the light, not the group velocity. This has been experimentally demonstrated and described in this paper, in which it was found that the radiation force felt on a mirror will vary in direct proportion to the index of refraction ...


1

According to Maxwell's equations $$ \textrm{curl}\,\textbf{E} = -\frac{\partial}{\partial t}\textbf{B} $$ therefore a variation of the magnetic field in time generates a non-zero curl for the electric field, whose solution, together with the other set $$ \textrm{div}\,\textbf{E} = \frac{\rho}{\epsilon_0} $$ describes the electric fields at any point ...


3

I know of another geometric interpretation of the electromagnetic field that is invariant under Lorentz transformations and more general transformations of coordinates. It is not well known and requires you to think of two-dimensional surfaces in four-dimensional spacetime. But it can be visualized (as easily as any visualization of spacetime), is coordinate ...


6

This sketch might help ;) It is an interactive picture of a charged particle in a plane, with the electric field lines and projections of the magnetic field lines also drawn. Here is what I thought about while making this visualization. The original question had depictions of an electric field due to a static charge and a magnetic field due to a current. My ...


-1

I would go for: $$F=\frac{2P}{c}$$ This is my reasoning: the force exerced by the light beam is due to momentum conservation. Therefore, we just have to know what the momentum of the light beam is, multiply this by 2 (because of the perfect reflection) and voilà, we have the answer. Now, as in General Relativity, we can define the energy-momentum tensor (or ...


1

The wavelength is not defined as the length after which the waves repeats itself: that is only a pictorial representation that works in one dimension for simple one component waves but it is not valid in general. Instead, given any solution of a wave equation represented as Fourier transform $$ \psi(\textbf{x},t)=\int ...


1

Suppose you shine a linearly polarized laser at the wall. Let's call the direction of laser propogation $\hat{z}$ and the direction of the electric field polarization $\hat{x}$. Then if you plot the $x$-component of the electric field vs. $z$, you will get a sine wave. The wavelength of the light is the wave length of the sine wave. So if one peak was at ...


2

You can certainly draw pictures which contain both the $\vec E$ and $\vec B$ field lines in different colors, and together they specify the electromagnetic field. The only problem is that, in this form, the vector fields themselves are not Lorentz-covariant so if you, say, assume that the electron's stationary E-field and zero B-field are the same in an ...


3

I personally struggle with visualizing the electron's electromagnetic field $F_{\mu\nu}$ at a stroke because there are too many variables involved: six interrelated quantities at each point in space. However, as Bosoneando points out, those six quantities are trivially related to the usual electric and magnetic fields by \begin{align} E_i &= -F_{0i} ...


2

You have two questions, and they have different answers. First of all, let's be clear about what Gauss's law is in integral form: $$ \int \vec{E} \cdot \mathrm{d} \vec{A} = \frac{Q_\mathrm{encl}}{\epsilon_0} $$ In words: the total flux integrated over a closed surface is equal to the charge enclosed in that surface, divided by the permittivity of free ...


-1

Your number of electrons is per unit volume. That affects your dimensional analysis. ne has units of meter^(-3)


2

As far as I understood from my so far cursory look into a living review article by Poisson, Pound and Vega on The Motion of Point Particles in Curved Spacetime, it's a bit messy. But I think if you manage to go through GR, this should be manageable, as well. It will probably help if you've dealt with Green's functions before and even better if you've seen ...


1

The rotation of the Earth's dipolar magnetic field produces an electric field in space. Because the electric field is zero in the rotating frame, it is equal to $$ \mathbf E=-(\omega\times \mathbf r)\times \mathbf B $$ in a fixed frame, where $\omega$ is the angular velocity of the Earth, $\mathbf r$ the radial distance and $\mathbf B$ the magnetic field. ...


0

The electric field around high voltage transmission lines (or "high tension" lines) is extremely high, and can be close to the breakdown threshold of air. That's why the highest voltage lines use multiple (often three) parallel conductors, to increase the effective conductor radius and reduce the peak electric field. Now, introduce a human into that field, ...


18

I don't think that nobody seriously thinks that unification of electricity and magnetism never happened or that it is unimportant. You just have to look at the standard model, and you'll find a term in the lagrangian (in units with $\hbar = c = \varepsilon_0 = 1 $): $$\mathcal{L}_{EM} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$$See? No electric or magnetic fields, ...


1

When a motor moves it also acts as a generator and the current trough the windings is given by the difference of the external voltage and the induced voltage. When the motor stands still, though, the generated voltage is zero and the windings will draw the max. current they can based on their DC resistance. In other words, the faster the motor runs, the ...


1

The result given to you by your professor is OK. Special relativity allows you to solve the problem in any reference system, and then going back to the original reference system. So the easiest way to solve your problem is going to the reference frame where the two electrons are at rest. There you have only a electric field $$\vec{E}'(\vec{r}) = ...


0

My question is why does light move in such a way rather than vibrating or rotating, or any other actions, in-situ? Is it due to absence of Higgs boson or perhaps I should ask why does not all elementary particles move in a straight line? You have managed to mix two different frames of reference in the above paragraph. Light belongs to the classical ...


1

Constantine, take a look at what Minkowski said in Space and Time: "In the description of the field caused by the electron itself, then it will appear that the division of the field into electric and magnetic forces is a relative one with respect to the time-axis assumed; the two forces considered together can most vividly be described by a certain analogy ...


1

Reducing eddy current does not change property of conductor or circuit Eddy currents (also called Foucault currents) are circular electric currents induced within conductors by a changing magnetic field in the conductor, due to Faraday's law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field. so ...


0

The second formula (where an overall minus sign is missing) is obtained using the product rule and discarding total derivatives. On the level of the action, it's just integration by parts assuming that surface/boundary terms vanish with fields decaying sufficiently fast at infinity. Thus: $$L = -\frac{1}{2} (A^\mu g_{\mu\nu} \partial^2 A^\nu - A^\mu ...


1

As stated in the comments, any loss in the system will contribute to the Johnson noise, so you are right about the skin effect and the Eddy current. I want to add that, interestingly enough, this apply not only to electric circuits, but to other linear dissipative systems. A very interesting paper from 1951, Irreversibility and Generalized Noise, proves it ...


0

There is only one connection to the wrapper, at the power supply end. One can read the problem to imply that the resistor at the other end is connected to the wrapper, but you shouldn't. If there were a connection there, the resistor would be shorted out. As the wrapper has only one connection, at long $t$ it will have come to the proper equilibrium ...


0

Fringing in cases of electric fields and magnetic fields are mandated by the laws of electromagnetism. For electric current, there is no such law which would be violated in case fringing does not happen. It must be made clear that current in itself is not a 'field'. Therefore, even if fringing does happen, it is not 'similar' to the case of the capacitor or ...


3

Ionizing radiation loses energy in matter by creating electron-ion pairs. Suppose you have an 1 MeV charged particle stopping in a silicon crystal. The first ionization energy for free silicon atoms is about 8 eV. The ionization energy will be a little different for silicon atoms on the lattice, but not grossly so: your 1 MeV charged particle is going to ...


3

According to the Review of Particle Physics (Section 33.7.4 of the 2014 edition) there are two main causes of radiation damage for electronic devices: Bulk damage due to displacement of atoms from their lattice sites. This leads to increased leakage current, carrier trapping, and build-up of space charge that changes the required operating voltage. ...


0

If the charges are moving parallel to each other, in the reference frame of of the one charge, the charges aren't moving. So the magnetic field is zero and the force is $\bar F=q \bar E $. For the charge to be still, the force must be zero, and thus the electric field also. But if the electric field can't be zero, then there is another force that holds the ...


-1

If you adopt a photonic model where the photons are seen as little particles like gas particle then the following simple demonstration indicates : The fundamental principle of mechanics gives : $N\frac{\triangle p}{\triangle t}=pS_{yz}$ Where N is the number of photons inside the cavity, p the radiation pressure and ${\triangle p}$ the change of ...


2

You are correct that the protons' linear charge density doesn't change because any Lorentz contraction happens perpendicular to the wire. You are also right that the relativistic velocity addition formula predicts that the total speed of the electrons isn't simply $\sqrt{v^2+v_0^2}$. It is actually $\sqrt{v_{(\parallel)}^2+\frac{v_{0,(\perp)}^2}{\gamma^2}}$, ...


0

If you want to use a single coil of wire, then you can begin by wrapping the wire one way and halfway through, U-bend the wire and begin wrapping the other way. As noted elsewhere in answers and comments the two halves will repel each other just like and two magnets in that situation, so don't amp up the power too much.


0

Well a router antenna is simply a dipole. It will have maximum radiation in its broad side direction and it's radiation pattern looks like a donut. Check the link below for the illustration of the dipoles far-field https://www.cst.com/Academia/Examples/Wire-Dipole-Antenna Of course the router itself and any metallic objects nearby will influence the ...


2

$\partial \vec{B}/\partial t \neq 0$ on the surface, but you don't need it to be zero. By your logic, you have $$ (E_{1t} - E_{2t}) l = \iint_S [\nabla \times \vec{E}] . \vec{n}\, d S = - \iint_S \left[ \frac{\partial \vec{B}}{\partial t} \right] . \vec{n}\, d S = - \frac{d}{dt} \left[ \iint_S \vec{B} . \vec{n}\, d S \right] = - \frac{d\Phi}{dt}. $$ ...


1

The conductivity of a material of length $l$, cross section $A$ and resistance $R$ is found by $\sigma = \frac{l}{R A}$. See Wikipedia. The capacitance $C$ of two parallel plates is $C = \frac{\epsilon_0 A}{l}$. You can find it on this page. Substitution leads you to $\sigma R = \frac{\epsilon_0}{C}$.


2

Let's maybe clear up a (possible) misunderstanding first. Mathematically speaking, to say that $V$ has no dependence on $z$ implies the following: $V(x_1,y_1,z_1) = V(x_1,y_1,z_2)$. The simple and intuitive way to see this is to observe the symmetry along $z$: There is no distinguishable difference between the points ${x_1,y_1,z_1}$ and ${x_1,y_1,z_2}$; ...


0

The configuration doesn't depend of z axis because there is no limit its axial. Otherwise those variable $y$ and $x$ does. Pain attention about this figure: In $ z = 0 $ is just a $x-y$ plane and Its too complicated to say what the potential all this plane. So because of this the Laplace equation must be solve with boundary conditions.


1

One reason for the box is the Fourier expansion of field in stable macroscopic condition (thermal radiation, cavity oscillations) works well only for finite volume. For infinite volume, the Fourier integral of such stationary field is problematic, because the field function is not L2 integrable.


1

Quantizing in a finite volume is not specific to the electromagnetic field, and it is not a necessity, neither for the electromagnetic field nor for any other. It is generally more well-behaved to quantize in a finite volume because no infrared-like divergences appear from allowing arbitrarily low momenta (since no arbitrarily long wavelengths fit into the ...


8

I spent a long time researching this question for Carver Mead (mentioned by Art Brown) in 2008, because we were both curious what Feynman meant. Carver thought Feynman's "better way of presenting electrodynamics" would be something along the lines of his own "Collective Electrodynamics," but that turned out to be only partly true, as I discovered in four ...


2

Measuring T-line power flow from the ground without contact is tricky, especially near the tower. The tower is a solidly grounded structure and tends to sink the EM fields to ground, and the conductors are far from your position (high in the air). Better to move to mid-span where there is no tower and the conductors are closer to the ground. Also better to ...


1

On point (1) I can see no reason why it should be impossible, but nobody has done it to the best of my knowledge. On point (2), there is a paper claiming that the AB experiment is entirely a result of local interactions between fields, and that it does not occur if the field interactions are totally shielded. The author claims there was a flaw in the ...


1

Give it a go and you may find a surprising result. In order for Faraday's laws to give you the exact result you would need to include the near-field terms in the E-field that you have implicitly ignored when you said $kr \gg 1$. Applying Faraday's law to the E-field as written will give you multiple terms that will approximate the right result for $kr \gg ...


1

It's important to separate the component of the field we're talking about from the spatial dependence. Ok, so you have the whole electric field, which is some vector quantity: $$ \vec{E} = E_x \hat{x} + E_y \hat{y} + E_z \hat{z} $$ But of course the electric field can be a function of position $\vec{E}(x,y,z)$, which means that so are all of its components. ...


1

Your sign is wrong when computing $$\frac{\partial{(B^2)}}{\partial{(\partial_{y} A_x)}}.$$ The only term in $B^2$ that contains $\partial_{y} A_x $ is $(B_z)^2 = (\partial_{x}A_y - \partial_{y} A_x)^2 ,$ and clearly by the chain rule, $$ \frac{\partial{(B_z)^2}}{\partial(\partial_{y}A_x)} = -2B_z$$ which disagrees with what you have by a sign. Fixing ...


3

From a practical point of view, the development of Iron Nitride magnets which are potentially more powerful than NeFeB, and of course do not need rare earthe elements


2

You should be able to show by direct substitution that your proposed wave function $\tilde{\psi}(\mathbf{x},t)$ solves the Schroedinger equation, where the vector potential is included via the minimal coupling prescription $$ \mathbf{p} \to \mathbf{p} - \mathrm{i} q \mathbf{A}(\mathbf{x})$$ (up to a sign convention for $\mathbf{A}$ and some dimensionful ...


0

Where do Maxwell's equations come from? Oliver Heaviside. They aren't Maxwell's equations. What Anna referred to about Coulomb, Ørsted, Gauss, Biot, Savart, Ampère, and Faraday is all well and good, but see where it says this in the Wikipedia article: "The powerful and most widely familiar form of Maxwell's equations, whose formulation is due to Oliver ...


3

The answer lies in another wiki article, In electromagnetism, one of the fundamental fields of physics, the introduction of Maxwell's equations (mainly in "A Dynamical Theory of the Electromagnetic Field") was one of the most important aggregations of empirical facts in the history of physics. It took place in the nineteenth century, starting from basic ...


0

The way you pose the question it seems you have in mind a solution with full translational symmetry in space, and rotational symmetry about the magnetic field direction at each point. I don't know if such a solution exists; if it does must be time-dependent. (That symmetry would imply that the spatial sections are flat, and the energy density is constant. If ...



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