# Tag Info

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This is a really interesting question. It turns out that your body is reasonably conductive (think salt water, more on that in the answer to this question), and that it can couple to RF sources capacitively. Referring to the Wikipedia article on keyless entry systems; they typically operate at an RF frequency of $315\text{ MHz}$, the wavelength of which is ...

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The Apollo 11 flag was included almost as an afterthought. It was just a month or so before liftoff, and someone at NASA slapped themselves on the head and said, "we need an American flag to plant at the landing site!" Someone rushed out to a local store (Sears?) and bought a standard nylon flag, which went to the Moon. Besides being bleached out by solar ...

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Photons or cosmic rays don't (normally) emit gravity waves. Consider the comparison with radio waves. A moving electron doesn't emit radio waves. It has to be accelerating to emit EM radiation. Specifically radio waves are only emitted when there is a changing dipole moment. So you wouldn't expect a particle moving at constant velocity (photon or ...

5

Remote "key fob" designers intentionally limit size so they conveniently fit in your pocket. However, the convenience comes at a big price - the tiny loop antenna inside is extremely inefficient, transmitting less than 10% of the energy pumped into it, while the rest is simply converted into heat. When holding your remote to your head, your arm, shoulder ...

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The way it works has nothing to do with your body. Remotes have their antenna as a more or less circular trace on the board (a loop antenna). The strongest signal is when the top or base of the remote is pointed at the receiver. The weakest signal is when the fob is pointed 90 degrees away, such as when pointing it like a TV remote. Guess which way most ...

5

$400$ - $700\text{ nm}$ corresponds to about $430$ - $750\text{ THz}$ ($10^{12}\text{ Hz}$), not $\text{MHz}$ ($10^6\text{ Hz}$). To convert from wavelength to frequency, use $$f = \dfrac{c}{\lambda},$$ where $\lambda$ is the wavelength, $f$ is the frequency and $c$ is the speed of light. So, for $400\text{ nm}$, this is: f = ... 4 Internet propagates with radio waves. Radio waves take advantage of a wave guide generated by the charged ionosphere and the ground for long distance propagation. Storm fronts with lightning and charged clouds do interfere with the propagation of a signal. Sudden changes in the atmosphere's vertical moisture content and temperature profiles can on ... 4 I would make a flag from iron oxide (red), platinum (white), and lazurite (blue). It won't wave in the wind, but it will retain the color. The base would be a platinum plate, of course. I would made a really large one, so that people wouldn't complain that it was too cheap. 3 Here's the logic (well a particular rendition): Recall that n is defined as the ratio of the speed of light c in vacuum to the speed of light v in the given medium; \begin{align} n = \frac{c}{v} \end{align} Note that in a linear medium, Maxwell's equations are exactly the same as in vacuum, except \mu_0 and \epsilon_0 are replaced by \mu and ... 3 In light of this, why do photons traveling from the most distant reaches of the observable universe not lose energy due to the gravitational radiation they must emit? There is a misconception here in "gravitational radiation they must emit" . There does not yet exist a unified theory of elementary particles and the three interactions well described by ... 3 Kyothe was on the right track, but in fact we do radiate in the visible, just in such small amounts that it's not detectable for all practical purposes. If you look at the referenced Planck (black body) curves for objects around human body temperature, the short-wave tail is nonzero in the visible range, but it's there. 3 If you read the wikipedia article on orbital angular momentum of light you will see that in the first place it is a classical electromagnetic concept, where the light has a vorticity, i.e. a helical motion around the axis of the vortex. When one goes to the quantum detail of photons one can define an OAM against this classical axis for each photon in this ... 2 The acoustic wave in the material causes a variation in refractive index and the amount of light scattered at a particular angle (its similar to Bragg scattering) depends on the intensity of the modulation. The scattered light does carry both frequencies in the form of blue or red shifting. It shifts by mf where f is the driving frequency and m is the order. ... 2 The Wikpedia article on solar cell efficiency gives a number of reasons that solar cells are less than 100% efficient. One of the large ones is the thermodynamic limit-a photon of less energy (longer wavelength) than the silicon band gap cannot produce an electron and one with higher energy can only produce as much voltage as the band gap. Even if you ... 2 When you say "without altering the actual momentum of it" is that really true? E^2 = p^2c^2 + m^2c^4 $$so for a photon E = pc, since rest mass is zero. Now according to your first "traditional" calculation of m, we would have E = pc = m_1c^2, and therefore p=m_1c, where m_1 is mass according to the first "traditional" calculation. For your ... 2 The field itself has to be real, of course. The imaginary part is added as a computational aid. Dealing with an exponential, whose derivative is itself, is much easier than dealing with a cosine. So we add the imaginary part, do our computation, then dump the imaginary part of the result. It's all simple when the computations are all linear. Sometimes, ... 2 Alfred Centauri has almost answered the question for you (actually he has), but he's using knowledge about and properties of Maxwell's equations that it sounds as though you haven't yet met. Maxwell's equations are covariant with respect to Lorentz transformations. That's a fancy way of saying that they keep their exact same form, and must foretell the same ... 2 The Earth and the Sun has magnetic fields which shields us from cosmic rays, as a charged cosmic ray particle will loose kinetic energy when its direction is perpendicular to the magnetic field. So what happens to the kinetic energy of the cosmic ray particle? According to the first law of thermodynamics it can't just disappear. It goes to the magnetic ... 2 Yes, alternating current will radiate electromagnetic waves. For example, in telecommunication, the transmitter itself generates a radio frequency alternating current, which is applied to the antenna. When excited by this alternating current, the antenna radiates radio waves. 1 "What does the peak stand for?": If you consider infinitesimally small ranges of wavelength values, the energy density (intensity) will be maximal at the peak. "And what does the graph tell us?": Considering a place of uniform temperature, with radiation in equilibrium with the surroundings, such as in a uniform temperature box, the graph tells us how ... 1 The term "Tokamak" refers to a design, not a size. The planed ITER reactor has the goal of 500 megawatts output. So it would take approximately 300,000,000 such reactors to produce the same power as the solar energy reaching the Earth. https://www.iter.org/factsfigures 1 Usually the Newtonian limit is described as taking v << c but a much better way to express it is saying that the kinetic energy is much less than the rest energy$$ \frac{1}{2}m v^2 << m c^2 $$of course this runs into trouble when we talk about photons since we don't have a well defined concept of velocity, in the Newtonian sense. This is ... 1 For a particle of fixed mass m moving in a fixed gravitational potential \phi(\vec{r}) the motion is independent of the mass of the particle. The equations are$$ \vec{F}=-m\nabla\phi $$and$$ \vec{F} = \frac{d\vec{p}}{dt} = m \frac{d\vec{v}}{dt}  It's clear that the $m$'s cancel when combining these equations. So from this point of view it doesn't ...

1

Firstly, if your waveguide is a hollow conductor, it cannot support TEM modes. There must be at least two separate (electrically insulated from one another) conductors in the waveguide's cross section for TEM modes to propagate. The reason is that the transverse field dependence of a TEM mode is the same as that of a static field, as I explain in detail in ...

1

I am not sure if my answer is correct, from what I understood: (i) At the relativistic limit, $m<<E$, so the second and third terms in (6.15) will be negligible, just as you said. (ii) P&S is aiming at $\hat{k}$ parallel to $\mathbf{v}$ or $\mathbf{v}'$ and integrating around $\theta=0$, since (6.15) peaks there (also ref my comment below ...

1

No energy conversion process completes at 100% efficiency. Using the sun, gasoline or nuclear, none of these will generate power at 100% of total potential power available. That figure of 1KW/m^2 is for total available light energy correct? Due to how solar panels work (I don't think I have enough knowledge to speak on this part directly) solar panels are ...

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With circularly polarised waves, the effect can only be observed interferometrically, because the Faraday rotation simply becomes the imposition of different phase delays on the two circularly polarised components. To understand this, witness that the Faraday rotation on a general polarisation state expressed with linear polarisation state basis is: ...

1

Although I have worked fairly near to these topics (albeit a long time ago), I found good information googling "wavelength band fiber". The encyclopoedic site at http://www.rp-photonics.com of the engineer Dr. Rüdiger Paschotta is an excellent reference (I have no links to this guy; I simply go to his site as a reference when I need reminding of this stuff)> ...

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I think your question is either: what FREQUENCY of alternating current, as in the case of a local open core transformer, or else what frequency of RF energy (as in microwave, radio wave, etc.) could be used to power or provide charge to laptops batteries 'wirelessly'? Not just any frequency you want, that's for sure. In the United States, the FCC takes a ...

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