# Tag Info

## New answers tagged electromagnetism

1

In general, it is not. Assume a constant current flowing through a cylindrical conductor. Applying Ampere's circuital law for a surface inside the cylinder: $$\oint {\vec Bd\vec l = {\mu _0}\int\!\!\!\int {\vec J} d\vec s}$$ $$B2\pi r = {\mu _0}J{{\pi {r^2}} \over {\pi {a^2}}}$$ $$B = {{{\mu _0}} \over {2\pi {a^2}}}Jr$$

0

According to Wikipedia In perfect conductors, the interior magnetic field must remain fixed but can have a zero or nonzero value. Require a constant magnetic flux - the magnetic flux within the perfect conductor must be constant with time. Any external field applied to a perfect conductor will have no effect on its internal field configuration. ...

0

Yes, it is zero. In short Ideal conductor, means no change in electric potential along it, which means no electric field inside. In details 1) Assume there is a non-ideal conductor of length L. 2) Assume there is some non-zero current flowing through it causing potential drop across the ends of the conductor. 3) Electric field inside a conductor is uniform ...

1

If the charges are moving at (near) c relative to a given reference frame (there is no mention of a reference frame in the question, but there must be one, otherwise we wouldn't know there is any - inertial - movement whatsoever), but they are at rest to each other, then according to SR postulates we may as well assume that they are simply not moving at all. ...

2

Any electric charge would experience a force due to an electric field. Therefore, the electric field in electromagnetic waves produces currents in antennas. It happens all the time in wireless communication.

0

The magnetic field is not 0, but the integral around the contour is. The reason you cannot apply Ampere's law in this case is that the magnetic field is not homogeneous along the circle

3

The problem lies in what we learn about good old constrained dynamics from traditional Dirac approach is not complete and is somehow inconsistent, and the above is one example of this. This was the message of Pitts' paper mentioned in the question above, who reviewed a bunch of previous work on this very matter. I will mention couple of references from that ...

0

A Birkeland current is a set of currents that flow along geomagnetic field lines connecting the Earthâ€™s magnetosphere to the Earth's high latitude ionosphere. A Birkeland current are also refers to the electric currents in a planet's ionosphere that follows magnetic field lines (i.e. field-aligned currents), and sometimes used to described any field-aligned ...

1

According to this website, A Birkeland current is a set of currents that flow along geomagnetic field lines connecting the Earthâ€™s magnetosphere to the Earth's high latitude ionosphere. These then are driven by solar wind in the Earth's magnetosphere. Birkeland currents are caused by the movement of plasma perpendicular to the magnetic field. They ...

16

First, the strong force acts on scales where our classical idea of forces as something that obeys Newton's laws breaks down anyway. The proper description of the strong force is as a quantum field theory. On the level of quarks, this is a theory of gluons, but on scales of the nucleus, only a "residual strong force", the nuclear force remains, which can be ...

2

1) It is necessary for a plane EM wave. If one assumes solutions to the Maxwell's equation to be plane waves, it is not hard to show that $\vec B \cdot \vec E = 0$. Namely, take the third Maxwell's equation and dot both sides with $\vec E$. $$\nabla \times \vec E = - {{\partial \vec B} \over {\partial t}}$$ $$i\vec k \times \vec E = i\omega \vec B{\rm{... -1 From the current knowledge of physics, they should not get the same force. Think about it this way: Let's fix the frame of reference to that where all three electrons move at near speed of light to the right. For the boson photon to transmit the force from B to A it can only travel at speed 0.00000000001c relative to A which means photon seems to travel at ... 1 In the book "Plasmonics and Plasmonic Metamaterials: Analysis and Applications" edited by G. Shvets, Igor Tsukerman, we read in section 2.1: In other words - they clearly state that the enhanced reflectivity is a result of the presence of a inverted dye - that is, a dye with a population inversion, meaning that it can be subject to stimulated emission. ... 0 I have seen the answers and your edit of question. It is not the serial or parallel connection of the wire that matters, both your coils will generate same magnetic field if the number of turns are same and current flown (in each turn) is same. Suppose you have 10 turns and the current flown in each coil is 1 ampere than in second case the current flown in ... 0 CuriousOne's comment has it exactly. Maybe some elaboration is in order, though. I'm only going to cover the DC case -- as CuriousOne points out, when you move to AC inductive effects enter the picture. Thinking of the coil as a fused cylinder of fixed shape is an excellent mental move for understanding the problem. No matter how you slice up the ... 1 You understand that mechanical devices such as levers, gears, springs and pulleys all conserve energy. Do you think that some elaborate combination of such devices can violate conservation of energy? The same applies if magnets are included - we know that interactions between magnets conserve energy, so any combination of mechanical devices and magnets also ... 1 Once in it won't come out, it just becomes part of the BH, adding mass and charge (plus or minus), and probably angularity momentum, to the BH. When something gets inside the horizon, it can never escape, out of any side of the BH. There is another process where one can use charged particles to extract energy from the electric energy of the BH. This is ... 1 No. The reason: Below the event horizon, there is no outgoing timelike direction. It means, it doesn't matter how do you accelerate a particle, it will move inwardly: The cones are the light cones of the object, what means if it would send out a radio signal in every direction, it would go in these cones. The best reachable orbit (-> the most delayed ... 2 If you have a single tube, the current will flow on it directly without making the N loops. It will result a different direction, i.e. different magnetic field, its magnetic field will be much weaker. Having the loops, the magnetic fields created by the induvidual loops is added. Actually, you have "the same current" using N times to produce the ... 3 All objects and fields that have a nonzero mass, energy, or momentum interact gravitationally, and so do neutrinos â€“ although they're very light and hard to produce so the gravitational force from any neutrinos we know is undetectable at this time. Neutrinos also have negligible but nonzero interactions with the electromagnetic field. They're uncharged and ... 1 A neutrino is thought to interact only through the weak force and gravity. They interact primarily, though, through the weak force (perhaps explaining the Martin/Shaw comment). Interestingly, since the neutrino has a minuscule mass (as opposed to none at all), it could have tiny neutrino magnetic movements, therefore allowing the possibility that it could ... 2 Even a "perfect" Faraday cage does not block EM radiation. If a EM radiation hits a cage the incoming photons or get dissipated in the mesh and re-transmitted with longer wavelength to both sides of the mesh (in principle a sort of black body radiation), heating the inside and the surrounding of the cage, or some amount of photons are going through the mesh ... 1 Google mathematical methods in the physical sciences pdf and you will be able to download an ebook by Mary Boas, which was written for people like yourself. As Jacob says above, calculus is a must learn, and lots of websites give you examples of different levels of calculus problems. Conceptually, a good textbook is Halliday and Resnicks Physics, which sets ... 12 An "RF cage" is commonly used to keep signals IN as well as OUT (see for example the mesh door of a microwave oven.) The short answer is - reciprocity says "if it works in one direction, it works in the opposite direction". The induced charges on the sphere (in the case of a charge inside it) are just enough to cancel the field outside exactly - because ... 2 In stimulated emission in a laser the emitted radiation has the same phase (and hence direction) as the incident radiation. The mirrors select some of those for regeneration back through the amplifier, where the process continues, and intense radiation builds up between the mirrors. Some of the atoms will decay by spontaneous emission, and that radiation ... 1 The statement "On the cylindrical surface \mathbf{J}\cdot\hat{n}=0..." refers to just inside the wire so \sigma\neq0 and \mathbf{E} cannot be whatever it wants. 2 So the total current supplied to your two solenoids will be 4 times larger. And that means that you are heating the supply lines, or fry your circuit breakers 7 The radiation pattern of any dipole antenna looks similar to what you are showing in the 2D plots - but in your interpretation of the 3D pattern you have the axes wrong. A dipole antenna with the main axis vertical will transmit power in the horizontal plane, with less and less power as you go further away (inverse square law). If you measure the power as a ... 0 This was one of the questions raised during the early stages of QM. If this were to happen atom would not exists as the electron would spiral towards the nucleus. Interference effects at the atomic level are usually assumed to give rise to the stationary states as we know them. The picture of electron actually going round the nucleus has been replaced by the ... 1 The problem with this question is all of the assumptions that go into it. When we're taught physics, we are given analogies that help our understanding, but mislead us when we try to dig deeper. Firstly, charged particles like electrons are always surrounded by an electromagnetic field. Changes in that field propagate through space at the speed of light and ... 6 This is a relatively tricky one, because it involves the differences between the \mathbf B field and the \mathbf H field in the SI and CGS systems, and those relationships change in the different systems. In short: Oersteds are used to measure the \mathbf H field in CGS units. Teslas are used to measure the \mathbf B field in SI units. In the SI ... 2 https://en.wikipedia.org/wiki/Magnetic_monopole You are correct in your intuition that there would be a "Magnetic Current" if monopoles existed. Indeed it is fun to imagine a beautiful symmetry between the Electric Force and Magnetic Force, but alas, nature only gives us half of it. No it isnt possible to simplify this relationship mathmatically. It takes ... 9 They are technically units for incommensurate quantities, but in practice this is often just a technicality. The magnetic field that makes sense (B) is measured in teslas (SI) or gauss (CGS), and the magnetic field that people spoke about 100 years ago (H) is measured in amps per meter (SI, also equivalent to a number of other things) or oersteds (CGS). ... 4 From a quick google search, it seems that Oersteds are used for defining magnetic field strength and Teslas are used for defining magnetic field strength in terms of flux density. They seem to not really be meant to be converted between, though you technically can (as evidenced by the other answers here). This website and this website might be helpful to ... 0 Implicitly, what you are doing in this problem is taking the limit \epsilon_0\rightarrow 0 in an electrodynamics problem. Let q_+(t) and q_-(t) be the two charges of the spheres at time t: q_+(0)=Q, q_-(0)=-Q. Since the velocity of light is (large but) finite, it will take some finite time for charge to move through the wire, so |q_-(t)|-|q_+(... 0 Any plane slices space in two half spaces. For some electromagnetism problems is a convenient simplification to assume that a half space is made of a (usually homogeneous and isotropic) conductive material. For example, for some geophysical problems, it's useful to model the Earth as a conductive half space. And another example (with a diagram): in this ... 0 Yes, extremely low-energy "soft photons" exist and can have important effects even when their energy is too low to be directly detected. 1 Eddy currents. Conductors resist changes in magnetic field. This is the basis of the magnet-in-copper-pipe demo. When the conductive metal encroaches on the plasma ball it encounters the ball's (spheromak's) magnetic field. The eddy current effect pushes back on the metal and pushes inward on the plasma, compressing and adiabatically heating it. The plasma'... 1 I would like to add that if we do not consider the elementary particles but think of those charged spheres made of metal, they can actually break. If you keep on removing electrons from a material block and protect the discharge from the neighboring atmosphere, after a stage the repulsion among the like charges become stronger than their cohesive force of ... 1 @Holger Fiedler Hi, Thanks for your comment on my answer here. Here I would like to say that I am an experimental physicist and working on the ultrashort coherent XUV radiation. I have first hand experience with the interference of XUV radiation using double slits. Your question is very genuine. Long ago I have thought over this question for quite a some ... 0 Quoting from the Wikipedia page on the CGS system: The e.s.u of charge, also called the franklin or statcoulomb, is the charge such that two equal q=1\:\mathrm{statC} charges at a distance of 1\:\mathrm{cm} from each other exert an electrostatic force of 1\:\mathrm{dyn} on each other. The e.m.u. of current, also called the biot or abampere, is the ... 3 Lets treat this classically, for simplicity. Neutron has no charge. However, it has a intrinsic magnetic dipole moment \mathbf\mu. The force of the magnetic field applied to this dipole:$$ \mathbf F = \nabla(\mathbf\mu\cdot\mathbf B)  Since you are assuming the magnetic field does not depend on space (its homogenous), then its "gradient" will be zero. ...

1

The situation described here is not that simple. First of all we see a simple situation. You have a solenoid and generate an appreciable magnetic field inside it. Now you put a charged particle (electron) into it (not the wire with current), It is also assumed that the mean free path of this charged particle is quite large such that it do not collide with ...

1

I think the resolution to the op is quiet simple and the entire paradox is because the solenoid in the third figure, should be moving to the left and not to the right! (If you are in a car moving to the right, then according to you the trees are moving in the opposite direction, ie to the left!) In other words, the system: in the reference frame of the ...

0

Yes lens law is used to determine direction of current whether it is clockwis or anticlockwise direction with respect to observer

2

Related question Is an alpha particle's curvature in a magnetic field visible with a homemade cloud chamber? You give no link for your photo and no details about the magnetic field imposed on the cloud chamber. Here is the case of alpha particles in a strong magnetic field: An alpha particle is not readily deflected by a magnetic field. For this ...

1

You should probably read up on the Stueckelberg action and the Affine Higgs mechanism it sends you to. Your boldface supposition "I'm not supposing that my matter has any global symmetry here, that I might be able to gauge" is unwarranted for the specific model you propose, $J_\mu=\partial_\mu\phi$. There is a global symmetry, $\phi \to \phi+\alpha$, whose ...

0

You're right that that Lagrangian isn't in general gauge invariant. In addition to making the $A^\mu$ terms gauge invariant, the $\mathcal{L}(J)$ term must also be gauge-invariant. And not just the equations of motion for $J_\mu$, either - the specific algebraic expression for $J_\mu$ in terms of fundamental fields must be such that if you literally plug ...

1

You are confusing the drift velocity of the electrons (which is < 1 mm/sec) with their Fermi velocity (which is $1.57\cdot 10^6 ~\rm{m/s}$ for copper) - source. If any "bunching up" of electrons were to happen, it would very quickly resolve itself. As was pointed out in the comments, the "signal" that travels in an electrical wire is essentially carried ...

2

As you say, it is because of the nuclear force. The charges do repel each other, but not strongly enough to overcome this other, stronger force. However, this is why in nature the neutrons in a substance tend to outnumber the protons--the neutrons can provide additional nuclear force without adding to the electrical repulsion. Having only protons isn't ...

Top 50 recent answers are included