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## New answers tagged electromagnetism

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If electron radiates photons towards proton, then due to 3rd law of Newton electron will move away from proton. But since it's not happening, this means electron and proton are emitting photons in same direction and thus the whole atom moves(or revolve) as the electron spins. Thus atoms are unstable until they complete there incomplete shell/shells.

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It's actually quite simple. If the electromagnetic field (i.e. photons) were charged, that would imply non-linear self-interactions such as those occurring for the gluon octet that mediates the strong force. The gluonic Lagrangian takes the form $$\mathcal{L}_\text{SU(3)} = -\frac{1}{2} \, \mathrm{tr}(F^2) = -\frac{1}{4} \, \sum_{a = 1}^8 F_{\mu\nu}^a \, F_a^... 0 Deriving the electric field for a 2D world can be done in several ways. It would depend on what behavior of the electrostatic interaction you want to preserve in that world. I you asked Coulomb, when he published his expression for the interaction, he would probably have said that the expression should be the same (1/r^2) just that the distance r would ... 0 A 2013 paper by Shtanov and Sahni (already mentioned by Ben Crowell in the comments) says that the modes grow exponentially in conformal coordinates, and Barrow et al overlooked the fact that the conformal time changes very little during and after inflation. A 2014 preprint by Tsagas, one of the authors of the original paper, cites Shtanov and Sahni and ... 1 Suppose, as you said, we have a thin solenoid pointing upward, going through the middle of a loop one light-year wide. Then the three equations$$\nabla \times E = - \frac{\partial B}{\partial t}, \quad \mathcal{E} = -\frac{d\Phi_B}{dt}, \quad B = \mu_0 n I$$together imply violation of causality, as shown in your question. Maxwell's equations are already ... 0 I would like to add to @M.Enns that the magnetic lines of force is a brilliant discovery of the great mind like Faraday. They are just the visualization tool. You can visualize the direction and strength of the magnetic field from the direction and density (which is also virtual) of magnetic lines. Sometimes 1\times10^8 field lines are associated with 1 ... 0 The answer to the second part of your question is yes your approach is OK. I have gone through the link you have posted and want to say that the inferences drawn from formulae written above are for non collisional case i.e. dielectric constant \epsilon is real. However if \epsilon is complex you can not draw the same conclusion from above formulae. ... 2 No, magnetic lines of force don't flow. They have a direction, which shows the direction of the magnetic field but there is nothing flowing. If you were to place a small magnetic dipole at the location of the magnetic field line its north pole would feel a force in the direction of the line of force. The phrase "line of force" was introduced by Michael ... 0 Most materials used in optics experiments are not magnetic. By this we mean that the magnetic permeability is equal to that of free space. This results in a magnetic susceptibility of zero.$$\mu_r = \frac{\mu}{\mu_0}=1\chi_m=\mu_r-1=1-1=0$$Therefore the magnetic field of the optical wave generally does not interact with the things in the experiment. ... 0 the ratio of the change in an electric charge in a system to the corresponding change in its electric potential is called capacitance I.e the ability of system to store charge. You can find more info here. https://en.m.wikipedia.org/wiki/Capacitance 0 If you have a system of independently oscillating point charges as radiators and they do not have a coherence among themselves. Then, If you take single dipole it radiate in a dumbbell shape. If you orient these radiators randomly oriented in space the radiation will propagate as a spherical wave. If you let this wave pass through a slit then you will see ... 1 Nothing is problematic with it. As FraSchelle says above—and is also true in the development of many other physics theories, in that they are, over time, purified of the scaffolding that helped construct them*—the original motivation doesn't affect the content of the developed theory. *cf. the top of p. 90 (PDF p. 91) of Stefano Bordoni's When ... 0 One could say that the AC current is steady if its RMS (Root Mean Square) value is steady. So in a statistical sense it's steady. It's most likely a matter of semantics than definition. Or a different definition for the particular application. 2 Let us make things clear. Protons and electrons are quantum mechanical entities and there is little meaning to project classical electrical attractive behavior to the micro framework of quantum mechanics, nor classical electric field calculations . Classically, a negative charge attracted to a positive charge will experience acceleration, and accelerating ... 0 The general notions, to wit, recirculation and highly frequency-selective amplification of noise, carry over almost exactly from electronic oscillators to lasers. From 30 year old memory of electronic oscillators, though, I think some of the details of the dynamical equations are a little different. Also, recirculating a light beam is much more complicated ... 7 Without touching on electromagnetism, I'd like to bring up this construction from mechanics (it's in the Feynman lectures). Consider two equal particles approaching each other with equal speed. A----> <----B You can argue from first principles that if they stick together they will not be moving afterwards -- any argument you could make ... 0 We might think of the integral form of this$$ \oint\vec H\cdot d\vec l~=~\int\int\vec J\cdot d\vec s $$We have two integrals, on a line integral and the other an area integral. With each unit of line d\vec l this is projected onto the mangetic field \vec H. Each \delta l length is associated with a pie shaped region with area ds~=~\frac{1}{2}rdl =... 0 If you naively use a Bohr-like model for the hydrogen atom, then the electron in its ground state is imagined as moving in a circular orbit of radius r and moving with a speed v. In this case you could argue the electron is moving, moving charge is current, current creates a magnetic field. Following this model you might expect the magnetic field at the ... 10 As the wiki article you quote states, momentum is defined as the product of the velocity times the mass of an object. Classical mechanics developed theoretically on the lines explained by WetSavanna in the other answer, the conservation of momentum and energy being cornerstones of the theory. Classical mechanics is a very successful theory, and ... 44 Momentum / energy are the conserved Noether charges that correspond, by dint of Noether's Theorem to the invariance of the Lagrangian description of a system with respect to translation. Whenever a physical system's Lagrangian is invariant under a continuous transformation (e.g. shift of spatial / temporal origin, rotation of co-ordinates), there must be a ... 1 The magnetic field of the permanent magnet is much stronger than that of the coil, hence we expect no effects from the coil onto the magnet. The coil is a linear element, so the problem becomes that of simple superposition.. that is the sum of the case of a magnet dropping inside a copper cylinder (assuming a close winding) + effect of the field from the ... 1 I'll answer more in a clinical perspective. I don't know about extreme situations when the spins of your organism's atoms are rearranged in a lethal way, but as far as MRI magnets go, the first concern when using MRI equipment is the possible induction of electric currents inside the human body. These concerns are more prevalent for investigation MRI ... 2 If the loop of wire is square and the second wire with a current passes directly through the center of the wire, then everything is symmetric. There is some \vec{B} which you could find - it has magnitude \frac{\mu_0 I}{2\pi r} in the right-handed direction. Lets imagine you were being silly and ignored the symmetry of the problem. Instead you ... 1 Every complex equation is a set of two real equations: the one with the real parts and the other with imaginary parts: A = B \iff \begin{cases} \operatorname{Re}(A) = \operatorname{Re}(B)\\ \operatorname{Im}(A) = \operatorname{Im}(B)\\ \end{cases} As long as we add complex equations together and/or multiply both sides by a real number, the real ... 2 The phenomenon you are talking about is called dielectric absorption. The way it works is this: Let's say you've just discharged a capacitor. An ideal capacitor would remain at zero volts after this. However, in real life, the capacitor will develop a small voltage from time-delayed dipole discharging (also known as dielectric relaxation). Dielectric ... 0 The plane is an equipotential surface. Since the electric field is the (negative) gradient of the potential, the electric field can have no non-zero component parallel to the surface. 0 Assuming y is perpendicular to the plane, you are correct, by symmetry you cannot have a preferred direction along the plane, so the electric field components must be zero 1 I will briefly explain the lasers and pulsed lasers. Lasers: As you know lasers are coherent source of radiation. When light is passed from a normal medium it get absorbed due to the fact that most of the atoms are in ground state, if by some means we can place most of the atoms in excited state (known as population inversion) then the passage of light ... 3 It is not a matter of being superior. The pulsed laser and the CW laser have different applications. The reason for using pulsed lasers is that if the time interval is very small you can probe events in nature on very small time scales. You can also examine aspects of quantum mechanics, for if the time interval of a laser pulse is very small the Heisenberg ... 1 All of the AC motors will create a "back pressure" once power has been removed. I do however not believe you when you say you have a properly grounded piece of equipment. Please note...the grounding system must be complete ALL The way back to a earth ground (grounding rod ect)...grounding cables can have a tendency to create a thin film of oxidation on ... 0 It's prettily answered by user35952 in his comment above; nevertheless here comes the explanation: The relation$$\nabla\times \mathbf E= 0\qquad\qquad:\qquad\qquad \textrm{True only for statics}. $$The general relation is$$\nabla \times \mathbf E= -\partial_t\mathbf B\;.\tag 1$$The static fields were conservative and hence could be expressed solely ... 2 This shows a schematic illustration of the Pound-Rebka experiment: At the bottom we have ^{57}Fe source that emits gamma rays upwards with an energy of 14.4 keV. The frequency of the gamma ray is 3.48 \times 10^{18} Hz, but let's just call this \nu_0 to avoid messing around with figures. The gamma ray travels upwards, and as it travels it is red ... 1 An isolated charge (like, an electron) is produced by ionizing an atom, using energy to pull a single electron free of the atom and pulling that charge far from the opposite-charge ion. So, it DOES take energy to isolate the charge. The isolated charge has an E field around it, but the original uncharged atom had none. Similarly, when you apply a ... 0 In a sense, the size of the universe limits the wavelength of a photon: any photon that has larger wavelength than the size of the universe, cannot exist entirely within this universe. It is not clear that this can ever be tested, however. In high energy (short wavelength) the lack of a limit to thermal radiated light was an important reason for the ... 0 I can show this backwards and then explain the motivation. The uniqueness of the solution follows from the constraint on the transformation to be unphysical. Say, we take your equation and transform \psi:$$ i \partial_t \psi = \left( \frac{\Pi^2}{2m} + e \phi \right)\psi $$becomes$$ i \partial_t \left(e^{i\frac{e}{c} \Lambda}\psi\right) = \left( \...

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This answer is motivated by the Aharonov-Bohm effect and proves what the OP asks for, but in the special case $$\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A} =\boldsymbol{0}=\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{A}' \quad \text{that is} \quad \mathbf{B} =\boldsymbol{0} \tag{01}$$ To simplify the expressions we : set ...

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Brian Dodson has posted a brief explanation to this at Quora. Basically it is suggested that "the largest energy that is sustainable as an electromagnetic wave is approximately 1 MeV, or a wavelength of 0.01 Angstrom." https://www.quora.com/Whats-the-longest-possible-wavelength-lowest-possible-frequency-lowest-possible-energy-of-electromagnetic-radiation ...

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Hint: Look at the Poynting-vector associated with an electric charge Q and a bar magnet, and then consider $\int\int {\bf E}\times {\bf B} . d{\bf S}$ around a closed surface.

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Even electrostatic fields still contain potential energy in them. The issue here is, an electron (for example) cannot interact with itself. This means that its own field cannot give it a potential energy. So yes, in the case of a lone charge completely isolated, it will create an $\vec E$ field which contains potential energy, but it will do no work until ...

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Just like the curiousone told you, since the curl of E is not zero, you're dealing with a non-conservative field. So if you consider the potential difference to be the work done by E to get from the point P to Q it does matter the way you take. Griffiths answer is based on the trajectory to go from P to Q in a straight line.

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I did a bit of research and will give my own answer to my question for those that are interested. The return current follows the path of least impedance in the ground plane. There are two sources of impedance: the resistivity of the conductor, and the coupling between the trace and the ground plane: $$Z = R + jX\omega$$ At low frequencies the resistivity ...

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They are trying to explain the resonant cavity thrusters , controversial propelling engines. It seems to me that the paper you quote confuses photons with light waves, attributing a real space wave nature to the photon. From their fig1: Our reasoning is that when light waves combined with opposite phases, the photons do not vanish for nothing but ...

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How did this field magically appear? Induced electric field is caused by the variation of current-density in the stationary loop and not by magic. The electric and magnetic fields are correlated by $$\mathbf \nabla \times \mathbf E~=~ -\partial_t\mathbf B\;.$$ I am also aware that a current carrying wire exerts a magnetic force on a moving charge due ...

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I am not supporting or seconding any finding in the article. They appear to be just incoherent chatting. However the pressure excerted by a electromagnetic wave is $P_{rad}=\frac{I} {c}$ where I is the intensity of light and c is light velocity. In case of totally reflecting surfaces this will be doubled. Here you can see that the pressure is not of one ...

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No, many other couplings are possible. For example, in the very simple Lagrangian $$L = \frac{1}{2} mv^2 - mgh$$ we have coupled the particle to the gravitational field $\phi = gh$. This is already in relativistically invariant form, since both $m$ and $\phi$ are scalars. (Of course the real story for coupling to gravity is more complicated, but this works ...

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The field didn't appear "magically". Electric and magnetic fields are part of one and the same phenomenon: electromagnetism. They can not be separated. The only reason why we teach electric fields and magnetic fields independently, at first, is to make them easier to understand for students. When you are learning electromagnetic induction, you are being ...

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You are referring to the formula for capacitance of a uniformly charged disk. A metal disk has uniform potential, but NOT uniform charge: the charge density at the periphery of the disk is higher than at the center.

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Short (but cryptic) answer: complex numbers arise in quantum mechanics because we would like find solutions to the differential equation $$\frac{\partial}{\partial x}f(x) = cf(x)$$ which don't blow up as $x\to \pm\infty$. Long answer: Fundamentally, the shift from classical mechanics to quantum mechanics is replacing functions (observables) and numbers (...

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There reason I can think of is the strength of interaction. The force exerted by the electric field on a charged particle is $eE$ which is much stronger than the force by magnetic field $ev\times B$. Magnetic force only get comparable to electric force when velocity of particle (usually induced by electric field) approaches light velocity. Due to this ...

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I am answering the question formulated after the "edit" in a newer version of the text because that one seems well-defined. Indeed, a situation with a uniform field $\vec E$ may be said to be "uniform" or translationally invariant in space. Noether's theorem says that this "uniformity" (spatial translational invariance) implies the existence of a conserved ...

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