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The FM radios operate at frequencies close to 100 MHz, $f$. If you divide the speed of light, 300,000,000 meters per second, by $f$, you get the wavelength which is about $\lambda=3$ meters. Ideal dipole antennas have length $\lambda/2$ or, if they are monopole antennas emulating one-half of the half-wavelength antenna, $\lambda/4$. That's 150 or 75 ...


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TL,DR: Magnetic coupling results in lower transmission of sound energy than physical contact Controlling what surfaces vibrate gives more control over sound generation The same benefit could be achieved with other forms of isolation (e.g. foam) but it wouldn't look as cool. It is bunk, mostly. A magnetically levitating speaker maintains a ...


2

This advertising strategy is basically using pseudoscience to get naive people to buy a product. The efficiency problem in speaker design has nothing to do with momentum transfer from the speaker to the air. That's trivial, since the mass of air a speaker moves is typically orders of magnitude less than the mass of the speaker itself. Instead, the (low ...


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It's not a fundamental feature of electrical potential, but: If you have a polycrystalline metal and you cut and polish a smooth surface, the differently-oriented regions will present a different lattice plane to the outside. Crystals cut along different planes may have slightly different work functions, and so the electric potential very close to such a ...


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A steady current cannot flow in a perfect conductor. In deriving Ohm's law to describe currents in conductors, we must assume that the conductor is "good" but has some resistivity. The electric field inside the conductor is not zero, but the force it produces on the flowing electrons is counteracted by the resistive drag force from the conductor. When a ...


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Imagine you standing some distance from me, and you move a charge back and forth along the line joining us. Waldir, you are quite right that the electric field I observe will fluctuate, and that these fluctuations will not reach me instantly - they will travel at the speed of light. However, this is not electromagnetic radiation. Why?- The electric field ...


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First consider Faraday's law, which states that $$ \nabla \times \textbf{E} = -\frac{\partial \textbf{B}}{\partial t}. $$ We can interpret this as follows: whenever we are generating a magnetic field that changes with time, there is an associated electric field, and vice-versa. An equivalent interpretation is that a changing magnetic field causes a ...


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Electrical amplification is about using an input signal to modulate a larger amount of power that comes from a separate power supply of some sort. And yes, there is such a thing as a magnetic amplifier that works on a very similar principle (even though the inputs and outputs are usually electrical). But you can't get an output value that's greater than the ...


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For static charges, the relationship is V (voltage) = Q (charge) / C (capacitance). Capacitance is a function of the shape, size and distance between objects, which are all continuous values. (Well, I suppose you could argue that shape and size are quantized to the atomic spacing of the object's material, but you can't say the same thing for distance.) So ...


2

The coating has nothing to do with it - it's the metal of the pan that matters. As was mentioned in the comments, many Teflon coated pans are made of aluminum (or aluminium, depending on where you live...). The issue is skin depth: for a non-ferromagnetic material, at the frequencies of the induction heater a large fraction of the volume of the pan becomes a ...


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Does magnetical levitation like this really provide any conceptual benefit to the isolation of the speaker from its support? I.e. does actual "contact" make any fundamental difference to sound propagation? Conceptually, yes if you are worried about noise induced by the speaker. Speakers have a tendency to shake things they are attached to, so if you ...


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Voltage is a continuous function. If you are a certain distance from a (point) charge $q$, the potential is $$V=\frac{q}{4\pi\epsilon_0 r}$$ By adjusting the value of $r$ to anything you want (not quantized), you can get any potential you want. And so yes, when you do any analog-to-digital conversion, you will "destroy" a certain amount of information. ...


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Voltage doesn't come directly from the charge of the electron. It's the energy per charge. The charge carriers may be discrete, but the energy is not. We can easily generate a potential by moving a wire through a magnetic field. The potential is proportional to the speed of the wire, which is a continuous value. $$V = vBL\sin{\theta}$$


0

Two things: First of all, Newton's laws of motion still hold, so all the energy applied to the voice coil is going to move something. Since $ m_1 v_1 + m_2 v_2 = 0 $ (momentum conservation),and a speaker bolted to a wall has a seriously large mass, it's more likely that it'll send a higher proportion of its energy into the air than a "free-floating" ...


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I see it as follows. Let us imagine standing in the pathe of an oncoming gush of a fluid. We feel the oomph of momentum. A particle of electric resistor in the path of electric current (perhaps) feels same oomph when electric current flows. The higher the current density the higher the oopmh the particle of electrical resistance feels.


1

A simplified picture for DC circuits is as follows: A charge develops at the surface of the wire, having a gradient along the length of the wire. This surface charge distribution causes an electric field within the wire, pointing along the wire, and having a uniform cross section. This field accelerates the charge carriers in the bulk of the wire. The ...


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No, current is supposed to flow through all parts of the conductor, although not necessarily with equal current density. If the conductor is perfect, the electric field vanishes but there may be non-zero current density inside. Surface component of current can be probably neglected for situations with stationary current.


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There are several ways to amplify the magnetic field, though the mechanism is not same as for electrical signal amplification, but still they are fruitful. compression:- since a magnetic flux through a surface remains conserved, if we compress the field lines or stretch (or fold) the field line then we can increase the energy by working against the field ...


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I'm leaving an answer because the following intuition needs some proper debunking: Since the E field inside a "perfect" conductor is zero, do the electrons(the current) flow only on the outer surface? The logic is perfectly clear, and applied totally incorrectly in this case. I think most people would start with this approach, but more careful ...


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firtree is correct - I will just try to flesh out his answer a bit. (1) Your last question first - charge (or current) at a point is like mass at a point. For finite masses, if you want to see how much is contained in an infinitely small volume (i.e., at a point), the answer is zero. So instead, people consider the mass density which can have non-zero ...


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Note that in a perfect conductor (zero resistance) nowhere can a static electric field be sustained. For real conductors, DC current flows evenly over the cross-section. But AC current flows mainly near the surface (the so-called skin effect). AC current can exist only on the surface of a perfect conductor.


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This article explains how electrons flows in a wire. Basically, the penetration depth of the electrons in a conductor depends on the frequency of the varying electric field to which they are subjected. For DC, electrons flow into the wire; as the frequency increases, the flow starts to move toward the surface. Since the E⃗ field inside a "perfect" ...


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The key to understanding EMP is that this is an induced effect, so you need rapid changes in magnetic field, $\frac{dB}{dt}$. In order to generate a rapidly changing field, you have to have a rapidly changing current in an inductor - as you may recall, $$V = -L \frac{dI}{dt}$$ For this rapid change in current you not only need a high voltage - you need a low ...


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There appears to be some confusion. The ratio is the number of windings of the primary coil divided by that of the secondary coil. Thus 1:2 and 50:100 is the same thing. However, the number of windings itself is far from a trivial matter. To learn more have a look at: http://en.wikipedia.org/wiki/Transformer


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The value of the line integral $\oint_C \vec B \cdot d\vec l$ really does only depend upon the current bounded by the closed path $C$. That's a consequence of Ampere's law. However, the value of the magnetic field $\vec B$ at any point along the path $C$ depends on every current, even those outside. Knowing the value of the line integral is not the same ...


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But magnetic field at that point can have only on "CONSTANT VALUE". Not true. If a particle is acted on by some combination of electrical and magnetic forces in one frame of reference, then in another frame of reference, it will be a different combination of electrical and magnetic forces. It's possible to have a force that's purely electrical in one ...


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So there is not an easy way - I had to grind out all the algebra. Starting with $$\nabla^2 E +k^2E = 0$$ The Electric field forms are $$E_r = P_n^0(cos\theta)[\frac{A_n j_n(k r)}{k r} + \frac {B_n y_n(k r)}{k r}]$$ $$E_\phi = P_n^1(cos\theta)[C_n j_n(k r) + D_n y_n(k r)]$$ Using $\nabla \cdot E = 0$ and the symmetry over $\phi$ I find $$E_\theta = -\frac ...


0

No problem - I think this one is clear, rj. The answer is No: Amperes law is correct and exact irrespective of whatever currents are flowing outside the loop without crossing it. Of course these currents still affect the total field at any give point - it's just that these contributions sum to zero. A good example is the field near a long straight wire. ...


0

It would be nice to provide a picture or diagram to make the question a little more clear. The answer is that the curve of integration (to find the associated magnetic field) should cover the area or loop into which the current flows, as such there is no current outside which is not taken into account. Of course one can divide the calculation in areas and ...


1

Well, the details are important. From the abstract in your link of the paper we see that: 1) it is a publication from 1978 2) it calculates rates for positronium annihilation in the very high magnetic fields found in astrophysical situations, 10^12 Gauss It explicitly states that the momentum contribution comes from the magnetic field. In the relevant ...


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How does one show the momentum imparted to a perfect conducting resonance cavity (boundary) of any shape by a classical standing electromagnetic wave inside is zero? For a standing EM wave in a cavity, the Poynting energy of the EM field inside is constant. This implies no energy is being transferred to the matter of the cavity from inside so the ...


3

Uniformity of $\mathbf B$ in space and the equation $$ \nabla \times \mathbf E = -\frac{\partial\mathbf B}{\partial t} $$ implies uniformity of $\nabla \times \mathbf E$, but not uniformity of $\mathbf E$, because these conditions do not determine electric field uniquely - there is infinity of solutions. It is an artificial situation which lacks further ...


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You missed the negative sign in your equation. I misread your question at first. You are asking about a solenoid. Because of the cylindrical symmetry of the problem, we know the electric field has to have cylindrical symmetry with respect to the central axis. Integrate both sides of the equation over the cross sectional area and use Stokes theorem. The ...


0

I looked briefly at the Blandford and Thorne notes. The analogy between $\omega$ and B appears to be mainly illustrative and not to be extended too far. (I don't see the reference to the analogy between E and $\omega \times \bf{u}$ there at first glance, but it does not seem to be apt at all.) It seems to be intended to draw upon any previous intuition ...


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I think you can calculate the electric field it generates using the Maxwell equations and it should be a constant field. Then with the field you can calculate the torque exerted on the disc and thus you can get the $\omega(t)$.


1

It seems that the dimension is wrong. $[\hbar] = J\cdot s$, $[e] = A\cdot s$, where $J=$joule, $s=$second and $A=$ampere. Then, $[\hbar/e]=Wb=T\cdot m^2$, with $T=$tesla and $m=$meter. The light speed $c$ is erroneous [see this wikipedia article http://en.wikipedia.org/wiki/Magnetic_flux_quantum]. Now you plug $\hbar\approx 10^{-34}J\cdot s$ and $e\approx ...


2

I think that the problem is that you are considering an electric current at distance $r=0$ with the Biot-Savart formula. It's like when you have a wire with a current and you want to find magnetic field on the wire, or electric field on a point-like charge. In your problem current $J$ is a linear function of distance, but in Biot-Savart you have something ...


1

There is an electric field due to the changing magnetic field, which has direction left -> right. Due to this field, there is an immediate charge seperation caused in the metal rod (-ve near M and +ve near N) which creates another electric field of equal magnitude and opposite direction to perfectly cancel the external electric field. Thus the net electric ...


0

My answer is more of comment on other correct answers: you cannot build a delta-function for the photon in 3D becase the longitudinal component of a massless vector field is missing. But that does not mean there is no useful and meaningful concept of a wave function in the single-photon sector. This is just a peculiar fact about free electromagnetic field, ...


0

There are several different waves associated with a photon. In QED the photon is associated with a classical solution of the (4-)vector potential. The vector potential contains features that are not physical, as a change of gauge is not reflected in any change of physical properties. Thus its role as a wave function might be somewhat questionable.But still, ...


1

The confusion you face is a historical one. Originally the interactions of different bodies was thought to happen at a distance more or less instantly, such as the case in the time of Newton and his gravitational theory. But when we discovered electromagnetism, and in particular, when Maxwell completed his formulation of Electromagnetism as contained in ...


3

A wavefront (your signal) has a fixed amount of energy given to it by the transmitter. Whatever happens to the wave once it leaves the transmitter is independent of the transmitter, thus receiving a signal does not drain any additional energy from the transmitter (though it can drain energy from the wavefront itself). EDIT: As pointed out by @Alfred ...


0

I hope the cylinder is, although long, not infinitely long? I mean the field must extend beyond the cylinder’s ends. The problem is not well-formulated because the resulting state in the second case depends on how exactly the transition to superconductivity is effected. If you cool the cylinder from inside, it will expunge the magnetic field outwards and ...


1

You make many questions in one, all of them have their own answer. Just to clarify, nuclear decay and nuclear reaction are two totally separated and different things. Radioactivity occurs naturally, spontaneously. You have to sit and wait for the nucleus to decay. A nuclear reaction is forced, is something you obtain by, for example, shooting a particle ...


0

There is no such thing as “conduction of electric wave in conductor” (and I am unsure about where “electric waves” can be observed). There is a conduction of electric current in a conductor. One can say that electric potential in a piece of conductor is always the same (so the electric field is zero inside it), although it is not always so due to resistance, ...


-2

They mean magnetization of ferromagnetic core: http://hyperphysics.phy-astr.gsu.edu/hbase/solids/magpr.html https://en.wikipedia.org/wiki/Magnetization And yet one (possibly) useful link: https://en.wikipedia.org/wiki/Magnetic_core


0

Here is an excellent tool for "visualising" the electric and magnetic fields of waves being produced by a variety of sources (including antennae). This applet allows you to slow down time or even freeze it at a particular instant. You can show the fields, but also the rate of change of the fields and the currents. What is doesn't do is show this in full 3-D, ...


1

Two coils (around the same core) with the same number of turns, connected in parallel, will work and have less resistance. But one can reach the same objective using a thicker wire. Two coils (around the same core) with different numbers of turns, connected in parallel, are effectively a short circuit for a coil with a magnetic core, and anyway not a good ...


0

First of all, I would suggest you to read the comments I have made in the Danu's answer to check whether I have understood your question or not. See, $\oint B\cdot d\ell~=~ \mu_0I$ has been derived only on the basis of $\vec{\nabla}\times \vec{B}=\mu_0 \vec{J}$. But actually the Maxwell equation is $\vec{\nabla}\times \vec{B}=\mu_0 (\vec{J}+\epsilon_0 ...


2

After a lot more searching, I have found the answer to my question! :D Below is a summary of the information I found. There is no specific webpage I can link to because I relied on sources who quoted other sources which no longer exist, but maybe this information can be useful to someone else someday. Most of what I learned comes from Professor Lou ...



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