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This is kind of my own interpretation that is based on De Broglie's wavelength, but I think it is worth mentioning, because I don't think anyone really tried to explain why. This is a simplification, of course, but De Broglie's wavelength comes from Einstein's famous equation $E=mc^2$. If we replace $E$ with the energy of a photon $E=\frac{hc}{\lambda}$, we ...


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All that upon are in my point of view (Information Science) no definitive answers. Only some mathematical relations can't answer this question: Are electrons and photons excites of the same electric field? Photons are said to be that particles which interact in the electric field. Therefore they are also the massless kernel elements of electrons, but in ...


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In my understanding the major problem with solar flares are the (near) DC currents that get induced in power lines (phone lines have DC blocks, so that's not a problem). Since both ends of a power transmission line are terminated by a transformer, the DC current will induce a DC magnetization in the transformer cores. Since the transformers are run close to ...


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In general, the higher the frequency, the greater is the range. RF radiation has more energy at the higher frequencies That's not correct as stated. Yes, higher frequency photons have more energy than low frequency photons. But your transmitter isn't counting photons. You can have a $1W$ transmitter at low or high frequencies. The high frequency ...


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If the flare drops energy into the atmosphere, it sets up a changing electric field. The strength of the electric field is measured in $V/m$. This means that the longer the distance, the greater the potential difference. If you have a short wire, the potential between one end and another is not very large. But a long wire can connect points with much ...


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A very understandable confusion. When a charged particle travels in a magnetic field, it experiences a force perpendicular to its velocity. This force will change the direction but not the magnitude of the velocity - in fact an electron will travel in a circle in a uniform magnetic field (ignoring energy it loses from radiation). However, if you have a ...


1

since there is no resistance. That's not quite correct; in fact, there is infinite parallel resistance or, better, zero parallel conductance. Recall that, for a parallel RLC circuit, the circuit elements are parallel connected. If the parallel resistance were zero, the Q would be zero since the resistance is effectively an ideal wire shunt across the ...


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One way of thinking about a $Q$ factor is as a measure of how many periods of oscillation it takes for the amplitude to dissipate (for concreteness, lets say the number of periods required for the amplitude to decrease by a factor of 2). Then an LC circuit (that is, a series RLC circuit with $R=0$) has no dissipation and therefore will undergo infinitely ...


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In my opinion, Purcell is actually a mathematical step down from Griffiths, and certainly covers fewer topics than it. It's great for intuition-building (and every serious physicist should own it), but if rigor is what you're after, it's not the best choice. Jackson and Landau & Lifshitz are going to be the standard answers here. Another option, and ...


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Neither. "In the XY plane" means "perpendicular to the Z axis". It doesn't mean they are either parallel to X or Y - it could be at 45° to either of these axes. The equation of a plane is sometimes written as $$(\vec{x} - \vec{x_0})\cdot \vec{n} = 0$$ Where $\vec{x}$ is any point on the plane, $\vec{x_0}$ is a known point on the plane, and $\vec{n}$ is ...


1

The Biot-Savart Law does not obey Newton's Third Law for two open current segments. It only works if at least one of the loops is closed. Proof: The force between two wire segments can be written as a double path integral using the Biot-Savart Law and the Lorentz force Law. It works out to be: $$ \vec{F}_{21} = \frac{\mu_0 I_1 I_2}{4 \pi} \int_1 \int_2 ...


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So basically, I missed out an important symmetry in the question which is that the $B \propto \frac{1}{r}$ meaning that the flux integrated around the loop circumference $ = 2\pi r$ results in the $r$-dependence cancelling out simplifying the integral greatly. The result is that: $\vec{N} = \frac{ \vec{E} \times \vec{B} } {\mu_{0}} = \frac{10 I }{\pi r} ...


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For an electric field usually you have three components like $(E_x, E_y, E_z)$ in Cartesian coordinate system. Now you want to rewrite the same vector in a spherical coordination, what you should do is as follows: first you write the vector like the electric field as $\mathbf{E}=|E|\mathbf{e_r}$ where $|E|$ is given by the $|E|=\sqrt{E_x^2+E_y^2+E_z^2}$ and ...


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You can't take the root mean square of the spherical coordinate parameters because they aren't all the same units (one is a length measurement while the other two are angle values). Well you can, but the output is meaningless. To convert spherical parameters to Cartesian coordinates, you use simple trig: \begin{align}E_z & = r \cos(\theta) \\ E_x & ...


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The constant of magnetic permeability = $1.2566370614… \times 10^{−6} H \times m^{−1}$ The electric constant ≈ $8.854187817620... \times 10^{−12}$ These constants have to do with the resistance of an electric or magnetic field permeating through a classical vacuum.


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This isn't really a "boundary conditon" problem in the sense that we aren't trying to take knowledge of $\vec{A}$ on some surface and extend it to a solution of Laplaces equation in some region, which has the surface as a boundary. The reason you can't do this is that you have no apriori reason to know how $\vec{A}$ should look for a given $\vec{J}$. Rather ...


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$$ \mathbf{\bigtriangledown \times B = \mu J} $$ and $$ \mathbf{B = \bigtriangledown \times A} $$ so $$ \mathbf{\bigtriangledown \times \bigtriangledown \times A = \mu J} $$ From the definition of the vector Laplacian we have $$ \mathbf{\bigtriangledown \times \bigtriangledown \times A} = \mathbf{\bigtriangledown}^{2}\mathbf{A} - \mathbf{ ...


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Why isn't any current induced in a rectangular loop when it is moved in a uniform magnetic field. Because, there is no change to the flux passing through the loop. Suppose we're looking at a diagram of a 3cm tall rectangle on a page and the uniform field is into the page (it won't matter how wide the rectangle is for this explanation). Now, if you slide ...


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The reason for the sparking is the photo-electric effect. The microwaves energize outer orbital electrons in metals and causes emission. Due to the conductive nature of the Al and insulting nature of the air (it's a dielectric), the electron charge can build. However, recombination of electrons into the orbitals is possible, but very slow compared to the ...


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I have read no less than six different explanations of why aluminum foil sometimes sparks in a microwave. This is my best guess: Microwave ovens operate at about 2.45 Ghz. Water absorbs the non-ionizing radiation and becomes more active, creating heat. But aluminum foil, like all shiny metal, does not absorb microwaves - it reflects them. Because the ...


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I can't answer your last two questions, but I'll have a stab at the first. Imagine if I had a device which rapidly changed charge from positive charge to negative charge. What would happen if I brought a charged particle (of either sign) near this device? Suppose my charged particle is positive. While my device had a negative charge, the charged particle ...


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Gauss's Law states that $$ \oint \mathbf{E} \cdot \mathbf{dA} = \frac{Q_{\text{enclosed}}}{\epsilon_0} $$ As you mentioned, since $Q_{\text{enclosed}} = 0$ , $\mathbf{E} \cdot \mathbf{R}$ cannot be everywhere positive or negative since that would make the integral $\oint \mathbf{E} \cdot \mathbf{dA}$ nonzero. Your intuition is correct. In addition, even ...


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Thermo-electric coolers use electricity to cool things. Not sure if that's what you're looking for, as it requires physical contact. But in general, it's a lot easier to make heat than to remove it.


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As far as I know, there used to be a Kilpatrick limit on sparking at a given RF frequency. At the typical frequencies around 2 or 3 GHz one can get up to 50 MV/m, which is the max field I've heard of in the context of the ILC and other linear colliders. According to "New Techniques for Future Accelerators II: RF and Microwave Systems" edited by Mario ...


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When the pseudoscalar invariant $\vec E \cdot \vec B$ is zero, we have three cases. If $E^2<c^2B^2$ then you can switch to a frame moving with speed $E/B$ in a direction mutually orthogonal to $\vec E$ and $\vec B$ where there is no electric field in the new frame. Solve in the new frame. Then bring it back to the original frame. If $E^2>c^2B^2$ then ...


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The field along the axis of a finite solenoid can be found by integrating the formula you found (for a current loop) from $z_1$ to $z_2$. The current in a "slice" of width $dz$ is just $n I \, dz$, so we have $$ d B_z = \frac{\mu_0 n I}{2} \frac{R^2}{(z^2 + R^2)^{3/2}} dz \Rightarrow B_z = \frac{\mu_0 n I R^2}{2} \int_{z_1}^{z_2} \frac{dz}{(z^2 + ...


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I'm not sure if this was clear in the previous answers, but you cannot use ${\bf B}=\mu{\bf H}$ in ferromagnetic materials.


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Gravity can change frequency. A light beam going towards a massive body is blue-shifted by the gravitational field. If it is escaping a gravitational body, then it is red-shifted.


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The photon is an elementary particle. There are two ways do measure the frequency and therefore the energy of the photon since its energy E=h*nu . 1) using a diffraction grating which analyses the wavelengths in a beam of light , as below: This is the spectrum of iron. Each line is composed of zillions of photons with that frequency. If one sent one ...


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In regards to your question: Can the frequency of light change in propagation from one media to another, the answer is no. I found a previous response to a similar question that might help you: Think of it like this: At the boundary/interface of the medium, the number of waves you send is the number of waves you receive, at the other side, almost ...


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There are several ways to define the direction of a surface $A$. If the surface is closed the vector $d \vec A$ by convention points out of the surface. If the surface is open then the boundary of the surface is some curve, let us call it $C$ the direction of $d \vec A$ is then given to point in the direction of the right hand rule. If you have a negative ...


3

We do not "swallow" the index. You must distignuish between the geometrical object and its components, and that is not unique to forms, but occurs for all vectors: If you have a vector $v\in V$, where $V$ is some vector space, it has no indices. It's just an element. Now, if you choose a basis $e_1,\dots,e_n$ of the space, you write $$ v = v^\mu e_\mu$$ and ...


0

Why is it important to incorporate both electric and magnetic forces into one single expression? Because the Lorentz force is "due to electromagnetic fields". I know people talk about electric fields and magnetic fields, but see Wikipedia: "Over time, it was realized that the electric and magnetic fields are better thought of as two parts of a greater whole ...


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As Danu said, indices are not "natural" part of tensor fields, they are just a pretty outdated formalism of dealing with them. Unfortunately or fortunately, it is also a pretty well-working and efficient formalism, at least in general relativity. Electrodynamics and classical mechanics would be better off using differential form notation imo... Anyways, the ...


2

In the context of the mathematics of differential geometry, the concept of differential forms is made more precise. It is an invariant object, and does not transform under coordinate changes. In particular, they are not objects that 'inherently' come with indices, so a 2-form like the electromagnetic field tensor would be expressed by mathematicians as ...


3

Well, you could treat them separatedly, via two equations, say $$\mathbf{F}_\text{elec}=q\mathbf{E}$$ $$\mathbf{F}_\text{mag}=q\mathbf{v}\times\mathbf{B}$$ but since Newton's second law holds, in presence of an electric field and a magnetic field, the total force will be the sum of both, that is, the Lorentz force. I would say is just as simply as that.


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A constant charge density does not imply a zero magnetic field. Even considering a set of isolated charges, suppose they were (mechanically) moved along a circular path. The charge density could remain the same but there would be a current flow. The curl of the magnetic field produced would be $\mu_0 \vec{J}$, where $\vec{J}$ is the current density. If the ...


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First, note that the way you have written the electrical wave-field isn't anything more than exactly a way to write a wave function in general. This is because the term $e^{i(kx-ωt)} $ can be written as: $$e^{i(kx-ωt)} = cos(kx-ωt) +i sin(kx-ωt) $$, and from here you can keep in general the real or the imaginary part as you wish. As for $κ$, the imaginary ...


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I'll answer to 1 Maxwell equations are already relativistic, but - in a flat spacetime. You can write Maxwell equations for a general metric $ g_{\mu \nu} $ (The original Maxwell equations are formulated for a flat spacetime - $ g_{\mu \nu} = \eta _{\mu \nu}$ ). One way this can be done is by the following algorithm: Transform coordinates to a local ...


0

I want to give an answer for the case, that the mentioned particles are electrons. Let us consider that the magnetic dipole moments of this two electrons are aligned in a straight line through the points (0,a,0) and (a,a,0). Since both electrons are moving their magnetic dipole moments begin to turn when the electrons leave the mentioned points. Perhaps it ...


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I don't have time to do calculations, but maybe the mechanism is as follows (if this video is not a fake): the electrode (or the leads) cause a crown discharge, charges are captured by fog droplets, and then the charged fog droplets move in the electric field. EDIT: See also ...


0

We can use the following the equation to solve for $B$. $$E = cB$$


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You are right in stating that potential and hence potential differences are dependent on field. The relation in fact is $\mathbf{E} = -\nabla V$ Hence, as we can see, if $E$ = 0, then $\nabla V$ is in fact constant, not $V$. Now, to compute the potential, we can rely on coloumb's formula, taking $V$ at infinity t be zero, for a differential ...


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Definition of potential difference is the amount of work per unit charge to move a charged particle from one place to the another place. The potential difference between point $a$ and point $b$ is as below, $$ V_a - V_b = - \int_{\mathbf{r}_b}^{\mathbf{r}_a} \mathbf{E}\cdot \mathrm{d}\mathbf{r}.$$ What we call as potential with $V=\frac{kQ}{r}$ is the amount ...


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(i) Roger Shawyer Shawyer's output seems to be mostly available on emdrive.com. Among the theoretical explanations he provides there are A Note on the Principles of EmDrive force measurement Principle of Operation Theory paper None of these appear to be peer-reviewed. (ii) NWPU group Applying Method of Reference 2 to Effectively Calculating ...


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Because observations made by physicists have found that this is what nature does.


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There are many different levels of explanation for this question. Strangely enough most of them will dive into quantum electrodynamics, Feynman diagrams and exchange of virtual photons... I will try a simpler path that still carries some explanation. When you put two charges at a distance, they deform the -- otherwise flat -- electromagnetic (EM) potential ...


1

you can draw feynman digrams and then calculate scattering amplitudes and it is in the non relativistic limit is proportinal to potential.so if the potential is positive it means they repel. this sort of claculation is done in peskin book and A.Zee book.in peskin book page no 125. this is the most rigorous work to prove gravity is always attractive. by ...


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Firstly, I would like to say that there is no particular terminal separation between negative charges and positive charges. Actually you will understand it better if I would clarify in this way that scientists first saw that having even follow the same statistical distribution i.e. Fermi Dirac distribution some of them actually repel others and some do ...


0

A proof, or atleast an intuitive understanding can be achieved, I think, from Stokes theorem. For the rigorous proof, you need to know a bit of vector calculus but for an inuitive understanding, pure logic will suffice. The curl of a vector function $\overrightarrow{F}$, shown as $\overrightarrow{\nabla{}}\times\overrightarrow{F}$ represents the amount by ...



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