Tag Info

New answers tagged

1

OP asks (v1): How one can know the gauge field emerging from the local gauge invariance is actually the EM field? Assuming that OP is pondering about gauging theoretical models (rather than concerned with our actual world and phenomenological inputs) then the answer is: One cannot know. For starters, the gauge group $G$ could be different than $U(1)$. ...


0

Propagation of electromagnetic waves on Earth is highly wavelength dependent, which is also true for the question whether the atmosphere has anything to do with it. What you are looking at is actually a formula for the geometric attenuation (aka free space or path loss), which has absolutely nothing to do with the atmosphere. As ACuriousMind points out, your ...


0

Virtual particles never been considered "real" in any serious interpretation of QFT for two reasons: 1) what we call virtual particles are just terms in a series at some order of a coupling constant, 2) there's no way to interact with them. An interaction between two physical objects is a black box in QFT. The way people usually use to "see" inside this ...


0

Although you can (as you obviously know) think of electromagnetic radiation as either a particle or a wave, it's easier in this case to think of it as a wave. As a thought experiment, if you wave a magnet near a piece of wire, an electric potential will be induced in the wire. Likewise, if you pass current through a wire, a magnetic field will be produced ...


0

It seems to me that a plane wave description of the field in the dielectric medium can be valid only at distances(-y direction) large compared to the wavelength of the wave. Here I am assuming that the xz plane itself is the upper boundary of the dielectric. The modified variation of the field below the dielectric boundary must be used to evaluate the fields ...


0

Your question is pretty vague, but does this help-- The E-M wave amplitude is a complex oscillating function, as you can see at the wikipedia page . Now, the power in any wave is the square of the amplitude, or more precisely, the product of the amplitude and its complex conjugate. It turns out that E-M energy is quantized, so we can assign a specific ...


1

It is the electric field that does the work, not the magnetic field! When one has current in the loop, it can undergo a voltage drop or rise according to the inductance of the coil. Inductance relates to the electric field and its work. See: http://en.wikipedia.org/wiki/Faraday%27s_law_of_induction "The induced electromotive force in any closed circuit ...


2

The only naturally occurring symmetry breaking radiation of this kind is the CMB. Unless you are talking about charged particles of more than approx. 1e19eV energy (in the CMB rest system), the effects are negligible, as far as I know. For those ultrahigh energy particles, however, this so called Greisen–Zatsepin–Kuzmin limit (GZK limit) forms a cosmic fog ...


0

Depends on the angular velocity of its rotation. Electromagnetic interactions propagates at light speed. This means, if the coil is rotating in a frequency low enough, and if the induction circuit is close enough, the magnetic fields will arrive in the other circuit, almost "statically". Therefore, as good approximation in this case, you can consider the ...


5

White dwarfs with strong magnetic fields ($>$1MG) make up only about 10 per cent of the white dwarf population. A further few per cent have fields in the 10-1000 kG range (e.g.Liebert et al. 2003). So it is not clear that the Sun will end up as a "magnetic white dwarf" at all. The production of magnetic white dwarfs is thought to arise via at least two ...


1

The idea is that since the steel beam has conduction electrons that are free to move, the movement of the charges in a magnetic field causes a magnetic force to act. The magnetic force causes the electrons to accumulate at one part of the curved surface of the rod, thereby creating a potential difference. The charges keep accumulating till the potential ...


0

I think what I do is similar to what lionelbrits has posted. I calculate the voltage around a square loop anenna two ways: first, as the rate of change of magnetic flux from the peak of the magnetic wave passing through; and second, as the maximum difference of the voltages along the two vertical legs as calculated by integrating the e-vector. If you assume ...


3

The hand-wavy way to do it is to consider a wave solution like the one below, and apply Faraday's law to loop 1, and Ampere's law to loop 2: If you make the loops narrow enough, i.e., their widths are $dx$, then $$\oint_1\!\vec{E}\cdot \vec{ds} = -\frac{d\Phi_B}{dt} \to \frac{\partial E_y}{\partial x} = -\frac{\partial B_z}{dt}$$ $$\oint_2\!\vec{B}\cdot ...


2

The differential and integral forms of Maxwell's equations are truly equivalent; they are essentially the same set of equations. One can convert between the two using two mathematical theorems: Divergence Theorem (Wikipeda - Divergence Theorem) Stokes' Theorem (Wikipedia - Stokes Theorem) The divergence theorem states that the flux over a closed surface ...


0

In the book i have says that the non-electrostatic force of the source moves charge form the negativive terminal to the posotive terminal. This is true for an open circuit or a close one?


2

Expanding on Jan Dvorak's comment: When you change the magnetic field inside a loop, an emf (electromotive force) will be generated. Now if you have two loops, each of these will experience the same e.m.f. When you put them in series, you have a coil with two loops, or two coils with one loop. No matter which way you look at it the voltage across them ...


1

The wavelength does of course remain the same. Think about two circles (representing wavefronts) evolving over time. Their respective radii increase at the same rate, such that the distance between them (the wavelength) always stays the same. Now think about drawing a box of fixed size and placing it over the wavefronts. This represents your detector (or ...


-1

You'll need to first calculate the Poynting vector or the momentum density of the fields. The angular momentum density is L= r x p. Once you have that, you can find torque easily.


0

A accelerated electron emit photons which travel through space without any changes until they will be received by another particle. Radio waves are made from a lot of accelerated electrons in an antenna rod. The radio wave one detect is modulated from antenna generator and contains the photons from the periodically accelerated electrons. The radio wave get ...


2

Since the emitter is a point, the wave will never get "flatter". It will always look like a dipole pattern (for the non-relativistic case, for the relativistic case the forward and backward lobes become asymmetric, I believe, with a strong amplification of the emission into the forward cones, which should also become narrower). I think you are mistaking your ...


0

Yes there is an electric field outside of a current carrying wire, in a direction along the wire axis (i.e. parallel to the wire). This is true in both the AC and DC case. There is also of course a magnetic field in the azimuthal direction. For a resistive wire oriented along the z-axis, the electric field inside the wire is given by Ohm's Law $E_y=\eta ...


0

To start with, the microwave oven is a Faraday cage. Only objects inside get the full power of the magnetron. As far as I know, the microwaves penetrate all the way through objects in the oven, although the microwaves are not evenly distributed (which is why there is the rotating plate). The WIFI is meant to broadcast the signal in all directions. Although, ...


0

The trick is to (a) simplify the geometry and (b) simplify the effect of the "battery". So I chose Cartesian geometry with an infinite slab of electrons in y and z. There is variation in x only. At some value of x the electron density rises from zero up to some arbitrary value representing the density of electrons in the wire, then it drops back down again ...


1

I guess the question is whether the LED will also act as a photodiode, i.e. whether incident light will excite electrons and therefore generate a current. If so, then reflect the diodes light back onto it will indeed reduce the current flowing in the diode by generating an opposing EMF. This turns out to be surprisingly hard to Google, but Wikipedia ...


0

Actually Alesssandro Power's answer is the answer you want since you can express electric and magnetic field in terms of vector and scalar potentials. But if you want full terms explicitly written then you can look at Jefimenko's equations. Somehow you need to have a distribution of charges and their velocities in order to get the fields. If you don't ...


0

Yes..there will be induced current since when you rotate the coil the electrons in the coil also rotate. and by Lorentz force there will be an induced current. May be small but there will be an current.Faraday's paradox


0

In general, a inductive 'loop' for picking up a magnetic field (as shown connected to the galvanometer in you diagram) is 'symmetric' with respect to rotation about the axis parallel to the magnetic field from the coil connected to the battery. If the 'loop' is square (as shown), the symmetry is perfect for 90 degree rotations. There may be some small ...


0

It is extremely useful to use the vector and scalar potentials when solving time-dependent problems in electromagnetism. The vector potential $\mathbf{A}$ by definition satisfies $$ \mathbf{B} = \nabla \times \mathbf{A} $$ while the scalar potential $\phi$ is defined such that $$ \mathbf{E} = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t}. $$ The ...


0

For a wire carrying a steady current, all the time derivatives are zero, and $\rho$ is zero. Therefore the electric field is zero. This leaves Ampere's Law, and the solution of that satisfies $\nabla\cdot\vec{B}=0$. That uses all of the equations. But I wonder, like @JavierBadia, whether you intend a solution that ramps the current up from zero.


0

Act on the first equation with $\gamma^\nu D_\nu + m$. We find \begin{align} (\gamma^\mu D_\mu + m )( \gamma^\nu D_\nu - m ) \psi &= (\gamma^\mu \gamma^\nu D_\mu D_\nu - m^2 ) \psi \\ &= (\frac{1}{2} \left\{ \gamma^\mu ,\gamma^\nu \right\} D_\mu D_\nu + \frac{1}{4} [ \gamma^\mu , \gamma^\nu ] [ D_\mu , D_\nu] - m^2 ) \psi \\ &= (D_\mu D^\mu - ...


0

The answers posted by CIA and dmckee are great, and they correctly point out that textbooks usually handwave about why electrons don't easily leave the surface of conductors, but I would add that the electrons actually can move through the air, even if the electric field is not strong enough to ionize the air and form a plasma. Anyone who has tried doing ...


1

One way to think of a vector field like $\mathbf{E}$ is to separate it into a divergent part and a curling part. Roughly, at a given location in space, the divergent part spreads out from that location and the curling part curls in a closed contour around that location. The divergent part (also called "irrotational") has $\mathbf{\nabla}\times\mathbf{E} = ...


2

The freespace dispersion equation is $\omega^2 = k^2\,c^2$ and this cannot change: this simply follows from considering plane wave components of propagating fields, which all fulfil the Helmholtz equation $$\nabla^2 A_j + \frac{\omega^2}{c^2} A_j = 0\tag{1}$$ which is fulfilled by all Cartesian components of the moncrhomatic EM field vectors and, for a ...


0

Wavenumber k is the number of waves per metre. Frequency w is number of waves per second. The number w/k is the speed of the wave.


0

The first equation is the definition of the polarisation field. The second equation is not a definition and is only true in limited circumstances - those dielectric materials where the polarisation is indeed linearly dependent and in the same direction as the applied electric field. The connection between the two in those materials where this is true, is ...


3

Technically, $\omega^2/1^2-k^2/c^2=0$ is a degenerate hyperbola if that counts. But I don't think you can derive an equation of the form $\omega^2/a^2 - k^2/b^2 = 1$ for waves propagating in free space. You may however find something of the kind if you consider materials with fancier dispersion relations than $\omega = ck$, like e.g. plasmas.


2

A propagating EM wave is a field that needs no charges. Likewise, a propagating gravitational wave is a field that needs no masses. However in both cases there is no way (classically) to create the wave without a charge/mass. Considering EM, the divergence of the field is zero unless there is a charge, or put another way since field lines can only begin and ...


1

I don't know cases in which the e.m. field can be produced without charges, however I can tell you how to produce an e.m. field without electric charges. Take a neutral particle, meet it with its antiparticle, have them clash, and you'll get gamma rays. The latter are electromagnetic. Also, gamma emission from excited nuclei, has not much to do with the ...


0

Knowing it comes from Feynman, I would think he meant that all contact forces here on earth are electrical (between the electron shells of the atoms). That includes the drive of a screwdriver, and the friction (and pleasure) generated when having sex.


2

I assume it refers to Fleming's Left Hand Rule for motors


0

Electric displacement D is not the same as E by definition. In vacuum both quantities are equal but in matter D differs from E due to the presence of polarization P. We know that in electrostatics curl E is zero everywhere but curl D may not be since it depends on curl P. Curl D depends on distribution of polarization in matter. Curl P would be zero in a ...


1

Perpendicular and parallel polarisation usually refer to linear polarisation of the E-field direction with respect to an interface between two media. Specifically, parallel refers to parallel to the plane of incidence - that plane containing a normal to the interface and the k-vector of the wave. Perpendicular polarisation means polarisation perpendicular ...


0

The electric field must be polarised so that it is perpendicular to the wave propagation. I think this is what option 3 is trying to say, though it is poorly and imprecisely phrased. This could be true whatever the polarisation state of the electric field if the plane referred to is perpendicular to the wave motion. But I think what is meant is a plane that ...


-1

What type of polarized light is he talking about. If it is linearly polarized then 3 is correct. If it is elliptically polarized light then 2 is correct. For elliptically polarized light if you are at one location then as time progress the electric vector would rotate thus its plane of vibration would keep on changing.


1

You seem to be asking about mode transformers; these are extensively used in antennas as they connect to waveguide feeds. The waveguides usually employ TE10 (rectangular) or TE11 (circular) modes, but if you want to feed a horn, say, then you have to shape the field properly to avoid reflection, reduce sidelobes, and reduce cross-polarization coupling, etc. ...


1

Refractive index, as a number, actually varies with the frequency of the EM wave, so when someone quotes it as just some number (like: water's refractive index is n=1.33) usually it means in the optical frequency range ("for optical signals"). I suspect that is what's going on here. As to VF being greater than one or negative, I don't think I can explain ...


0

A magnetic field can't do any work. Instead of an electromagnet a permanent magnet has the same effect and this magnet will not get weaker in time when charges moves. What happens is that the resistance of the wire gets higher, so the electric source is doing the work. But why the resistance increases? This is because the free moveable charges (the ...


6

You say: The Coulomb potential comes from classical electrodynamics but actually the Coulomb potential is predicted by quantum electrodynamics as a low energy limit. Quantum field theory describes the interactions between charged particles as the exchange of virtual particles, and it's not immediately obvious that it would lead to an inverse square ...


6

Is the Coulomb potential also used to solve the hydrogen atom in relativistic quantum mechanics? Yes, the Coulomb potential is there in the solution of the hydrogen atom with the Dirac equation, which is formulated in the relativistic framework. Now it is time to specialize to the hydrogen atom for which $$\frac{V}{\hbar c}=-\frac{Z\alpha}{r}$$ ...


1

What I assume the book is trying to say is that, as the electrons move downward (because they are part of a current), the magnetic field bends their path toward the left. This is the horizontal motion that the book mentioned. But of course the electrons can't run off the edge of the bar, so they pile up at the left side, leaving unmatched positive charges ...



Top 50 recent answers are included