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One form of this interaction which is real and has consequences is called radiated electromagnetic interference. This happens when the circuits are carrying high frequency signals, under which conditions a simple piece of wire in the circuit or a trace on a PC board becomes an antenna, broadcasting EM waves. When those waves strike other circuit elements, ...

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The figure I was looking for is about 0.9997c according to https://www.tau.ac.il/~tsirel/dump/Static/knowino.org/wiki/Electromagnetic_wave.html *

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...if the right part removed, would electromagnetic waves still be produced near the inductor and what the changing electromanetic field are in this situation? Every accelerated electron emits EM radiation. If you operate an LC circuit with its oscillating current, the electrons are accelerated back and forth and emit photons. In addition, the charges in ...

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There is no simple answer because it will depend on the humidity of the air and the frequency. From the humidity you get the fractional water vapor content and average permittivity; having the RF frequency and the dispersion relationship for water $\varepsilon_w=\varepsilon_w(\omega)$ weighted by the relative vapor content will give you $\langle\varepsilon\... 1 The relative permittivity is defined as the ratio $$\epsilon_r = \frac{\epsilon}{\epsilon_0}$$ 0 I would say, yes, but in a different way. Radio communications are usually frequency modulated (FM) or amplitude modulated (AM). The fundamental frequency is the "carrier" wave. Frequency modulation induces harmonics, which is this case are a lot more complex than those than a musical instrument. You can produce frequency modulation by taking a signal which ... 0 To address the core misconception: Note that any plane wave in the$x$direction can be written as$f(k(x - ct))$. This does not imply that it's monochromatic and does not in any way define$k$. With$f$being an unspecified function,$k$is a completely arbitrary parameter that corresponds to rescaling the argument of$f$. That is, if the plane wave was ... 3 There are cases where tunneling is suppressed. For example, if you have a square potential barrier and send a particle towards it, the transmission (measuring the tunneling rate) coefficient oscillates as a function of energy. That is, for some energies the tunneling rate is lower than one might expect. For low energies the particle bounces back with high ... 0 The above explanation by LLlAMnYP holds fully in the limit case$T=0$, where the$R$would indeed be exactly$1$for$\hbar \omega<2\Delta(0)$. For$T>0$, unpaired thermally excited quasiparticles, whose density goes to zero as$T$approaches$0$, absorb and dissipate energy even at frequencies$\hbar \omega<2\Delta(T)$. A more complete and ... 5 The electromagnetic field in absence of charges and current is described by a wave equation. The solutions of the wave equations in a general setting (without symmetry contestants and without boundary conditions) are plane waves with a fixed frequency and momentum. These are purely sinusoidal$\propto \sin(k x-\omega t)$. Similarly, sound waves are the ... 15 If the wave is truly monochromatic then it will be sinusoidal. If it has a different profile then Fourier's theorem tells us that it can be built from an infinite series of (co)sine waves with increasing integer harmonics of the principle frequency (i.e. not monochromatic). 0 The force from reflection is only exactly twice the force from absorption if the light rays retrace their original paths after reflection. In this situation, this isn't true - the direction of outgoing rays is different from the direction of incoming rays, and the force on any given point is going to be perpendicular to the surface at that point and related ... 5 A plane wave is a single frequency , a sinusoidal variation in space and time by mathematical construction. Mathematically : the traveling wave solution to the wave equation ... is valid for any values of the wave parameters, and since any superposition of solutions is also a solution, then one can construct a wave packet solution as a sum of ... 1 The division of the EM spectrum into spectral regions is independent of the medium - but when the medium has a dispersion relation that differs from vacuum significantly enough to make a difference, the classification is done in terms of frequency (which is the same in every medium) instead of wavelength (which, as you noted, doesn't). 1 Cut-off voltage in the photoelectric effect is a characteristic of the material and is in no way dependent on the intensity or frequency of light shown on the material. Why? Because cutoff voltage represents the amount of energy needed to knock-off the electrons from an atom and why would that change if you change the frequency or intensity of the incident ... 0 X rays penetrate matter because their energy is much higher that of any matter excitations. The electrons in matter are to slow and too heavy to react and compensate the field, as they do for optical frequencies. Fort radio wave the opposite applies. They reflect off matter, especially off metals, unless you apply very special coatings. By reflection and ... 1 The interaction of photons with matter is complicated. The electromagnetic spectrum covers many orders of magnitude in frequency and photon energy, and there are qualitatively different processes that occur in different regimes. The results depend on the electrical properties of the material, such as conductivity and permittivity. We have materials like ... 1 To understand what radio waves are, please read about What are photons, electromagnetic radiation and what are radio waves. The thickness of a wall You should have no problems with X-rays. The moment of such photons is simply strong enough to penetrate materials. Without interacting with the wall, they exit the wall with the same wavelength (I prefer to ... 0 In the Lorenz gauge the Maxwell equations reduce to $$(\partial_t^2 - \nabla )A^\mu = -j^\mu /\epsilon_0 ~.$$ For the time independent case this reduces to Poisson's equations for the scalar and the components of the vector potential. It follows that in the static case the Coulomb law holds for all four components of the potential. It is the concept of gauge ... 0 "Components in fixed direction" just means that the coordinate system basis vectors don't vary in space. For example$\mathbf{\hat{x}}$is the same for all$x,y,z$while$\mathbf{\hat{r}}$depends on the angular coordinates$\theta$and$\phi$. The magnetic vector potential in the Lorenz gauge is specifically constructed so that we can use the scalar Green ... 0 What the Planck-Einstein law implies is that if take any frequency$\nu$, irrespective of it being a natural number, rational number etc., then that light would come in photons, particles which can only have specific amounts of energy given by$E = nh\nu$. So basically,$n$can only be a natural number. The energy of the photon is discrete. Not the frequency.... 0 Frequency isn't even a property of photons, it's an artifact of the frame in which they are measured, so in a world with "only integer frequencies", a simple boost would ruin it. (Planck units aside). The point of: $$E = h\nu$$ is that it is the minimum energy observed in electromagnetic wave with frequency$\nu$, and that the only possible energies are:... 1 NASA Armstrong avionics tech here - RF specialist. The old medical studies on RF state a safe voltage level would be below 5,000 volts for the transmitter. Above that, the possibility of parasitic X-rays increases with voltage, especially in radar equipment. From what I understand so far, microwave oven magnetrons typically need about 4,000 volts from ... 0 The spatial resolution,$x_{min} = R\sin\theta_{min}$is the radius of the Airy disk projected on the photo sensor (or film) of a camera. If the size of a pixel is bigger than$x_{min}$, it will be the resolution bottleneck (rather than the fundamental limit). Note that there are also other factors that may reduce the resolution in respect to the ... 2 In physics, we use the term "acceleration" to refer to more than just an increase in speed. The word "acceleration" refers to all situations where the velocity vector changes. This could mean that the speed increases, or the speed decreases, or it could mean that the speed stays the same and the direction of motion changes. So, when we say "an accelerating ... 1 A classical electron is a point charge, or at best a sphere the size of the classical electron radius. A point charge (or small sphere) moving in a circle does not constitute a steady current. For a point charge moving in a circle of radius$R$centered at the origin in the$xy$-plane, the current is given by: $$I(x,y,z,t)=\delta(R\sin(\omega t),R\cos(\... 1 Normally the carrier frequency is modulated (amplitude or frequency) by a much lower frequency carrying the data. If you decrease the ratio of data frequency relative to carrier frequency, you can reduce data error/incorrect signals. For low fundamental frequency carriers signals, this significantly reduces the data rate below what many find useful. ... 2 The displacements field equations can be derived for a cubic small element \Delta x \Delta y \Delta z of a body, by taking the components of the net force on it:$$F_{x+\Delta x} - F_x = ma_xF_{y+\Delta y} - F_y = ma_yF_{z+\Delta z} - F_z = ma_z(\sigma_{xx(x + \Delta x)} - \sigma_{xx(x)})\Delta y \Delta z + (\sigma_{yx(y + \Delta y)} - \... 2 I am reaching way back in my memory, but I recall that in seismology there are at least two distinct types of mechanical wave propagation. Both are wave phenomena (and are thus modeled as solutions of wave equations), but one is more similar to electromagnetic waves than the other is. Shear waves or S-waves are transverse waves, in that the displacement ... 0 In the figures showing the intensity due to two dipoles, the direction of dipole moment oscillation (and thus acceleration of the charged particles involved) is not in the plane of the paper(screen), but perpendicular to it. So the far fields from both fields add up constructively at points of b) denoted by "4". These fields are almost zero at any point ... 1 It comes from the freespace impedance, convert it to same scale (20*log10(377Ω)) = 51.5 In dB multiplication becomes adition, realizing this, you have something that looks like ohms law. 0 Amplitude is just how strong the light is. If you take a red LED and you hold it close to a wall, the amplitude is "high". If you have it far away, the amplitude is low, and equivalently, it looks less bright. (see also the comments below). The color does not change. They look the same. (Simple answer above, the harder answer has to do on the fact that ... 0 You're correct they won't be closed loops. Instead they'd look like diagrams of electric fields. Note the equations would be very similar then: you can get Gauss's law for the electric field by replacing$\vec{B}$with$\vec{E}$, and$\rho_m$with$\rho_e$, and shifting some constants. Intuition for electric fields carries over to magnetic fields, in that ... 2 the real question should be why the CHARGED electrodes are necessary Because electrons are charged particles. The electrons in the negatively charged cathode feel attraction to the positively charged anode, but the pull is not quite strong enough to free them unless they happen to be hit by a photon of light. Then, after they fly through the vacuum to the ... 1 Equations of the form below are solutions of the wave equation:$\mathbf E = \mathbf E(u)$, and$\mathbf B = \mathbf B(u)$, where$u = \mathbf {k.x} – \omega t + \theta$,$\omega = |\mathbf k|c$, where$\mathbf k$is a constant vector and$\theta$is a constant. The expression for$u$means:$u = k_xx + k_yy + k_zz - ωt + θ$and$|\mathbf k| = (k_x^2 + k_y^...

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For a traveling plane wave E and B are in phase. For a standing plane wave they are 90 degrees out of phase. This follows from the Maxwell equations.

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after some research it seems difficult for me to give a comprehensive answer. But here are the key notes: The displayed pictures, where $E$ and $B$ are in phase are correct for the $\textbf{far field}$ --> So look up near and far field The $\textbf{curl}$ of $E$ and $B$ are related to the time derivatives of $B$ and $E$, respectively. So that involves ...

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Maxwell's equations relate spacial derivatives with time derivatives, e.g. $$\nabla\times\mathbf{B}=\mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}$$ So your assertion However, it is just as frequently stated that the amplitude of a magnetic field is related to the derivative of the electric field at any given point in space and/or time is ...

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$\mathbf{J}$ is non-zero on the wire but zero everywhere else. So you have your wave equation. To compute the fields, you can get them from retarded potentials: $$\mathbf{A}(\mathbf{r},t)=\frac{\mu_0}{4\pi}\int \frac{\mathbf{J}(\mathbf{r'},t_r)}{|\mathbf{r}-\mathbf{r'}|}d\mathbf{r'}$$ $$V(\mathbf{r},t)=\frac{1}{4\pi\epsilon_0}\int \frac{\rho(\mathbf{r'},t_r)... 0 If a charge is moving at constant velocity for inertial frame A, it is always possible to make a change of coordinates, applying Lorentz transformation, to the inertial frame of the particle, say B. For frame B, the field is static all over the space. And frame A is moving around this static field. So, there is no signal or information to be ... 0 In-depth research shows that clearly: Magnetic field and electric field are unified, magnetic field is just the one appearance of electric field, there is not independent magnetic field in the world; To say it simply, magnetic field interaction is the electric field interaction that be effected by motion. Motion can take weak effect to the electric field ... 1 Árpád, sometimes it is helpful to use analogies to better understand what is happening. The next one I am talking about is only for the better imagination and is not the only correct explanation. The excited state in an atom is given to an electron by a pulse. It can come from another electron that wiggles around, or from an incident photon. The wiggling ... 0 An axially homogeneous waveguide has a special discrete set of modes that can "propagate" (including evanescent modes, too) along its axis without dispersion. A suitable linear combinations of these modes can form almost any kind of possible solutions in the waveguide. (If the waveguide has a homogeneous cross section then it can be shown that the modes form ... 10 This statement: On the otherhand RF environments are incredibly chaotic, it's essentially impossible to design an antenna with the same limited receptive FoV as a photo-receptor or accurately get information from a point target in a cluttered environment using RF signals. isn't actually correct. It's fairly straightforward to design an antenna that ... 4 Lightwaves are just as "poorly" behaved, but on a different scale. All of the effects you are interested in are proportional to wavelength. A 400nm blue light source on our scale has the same behaviors as a 400kHz RF source has on a scale 1 million times larger! If you had an antenna the size of the empire state building, you could resolve RF the same way ... 5 The full formula for the sum of two waves of different frequencies f_1 and f_2 is:$$\cos(2\pi f_1t)+\cos(2\pi f_2t)=2\cos\left(2\pi\frac{f_1+f_2}{2}t\right)\cos\left(2\pi\frac{f_1-f_2}{2}t\right)$$In other words, the combined wave can be thought of as oscillating with a frequency that is the average of the two frequencies that produced it, while its ... 1 The frequency,9,192,631,770 Hz. does not refer at all to the number of photons. The number of photons is given by the number of atoms that undergo that transition. You need to determine the intensity of emitted light to determine the number of photons. So your question about the number of photons is not answerable. The hyperfine transition most likely emits ... 2 Perhaps it may help to note that if the transition frequency is f, and the total energy which has been emitted is E, then the number of photons which have been emitted is E/(h f) where h is Planck's constant. A single excited atom will emit just one photon on its journey to the lower-energy state of any given transition. (There do exist much rarer ... 0 You did not quite explain how you failed to obtain the target result. I would not like to spoil the fun of catching your factors and signs involved, so I will strictly deal with significant proportionalities.$$ \sigma^{\mu\nu} F_{\mu \nu}= \sigma^{0i} F_{0 i}+\sigma^{i0} F_{i 0}+ \sigma^{ij} F_{ij}=2\sigma^{0i} F_{0 i} + \sigma^{ij} F_{ij} . $$Now,$$ \...

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Now this could mean that with every single transition, a single photon is emitted, meaning in one second, exactly 9,192,631,770 number of photons need to be produced You're confusing the frequency of light and frequency of transition events. Frequency $\nu$ of light, which is $9\,192\,631\,770\;\mathrm{Hz}$ here, is the number of periods of EM field at a ...

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