# Tag Info

## New answers tagged classical-electrodynamics

0

The energy density in an electromagnetic field is given by $$u = \frac{1}{2}\left( \epsilon E^2 + \frac{B^2}{\mu} \right)$$ The momentum carried by the electromagnetic fields is derived from the Poynting vector, which is the power per unit area. $$\vec{N} = \frac{1}{\mu} ( \vec{E} \times \vec{B} )$$ Momentum per unit area, per unit time is just the ...

1

What is potential energy truly? It depends on the circumstances. When you compress a spring it's stress in the bonds or electromagnetic field between the atoms. IMHO at the fundamental level it's essentially spatial stress. That might sound unfamiliar, but it shouldn't, because the stress-energy-momentum tensor "describes the density and flux of energy and ...

0

You should understand how potentials work and how they relate to forces. It is not as simple as saying that information travels at the speed of light. As I will show below, potential can change instantaneously everywhere (but that doesn't mean that information is traveling faster than light. Lets begin with the Maxwell's equations: $$\nabla \cdot ... 1 The main problem is radiation reaction. It's a fact that if I take a charge and jiggle it about, it will somehow "inject" energy into the electromagnetic field around it; we know this because we can detect this energy in the form of electromagnetic radiation. This means that when I grabbed the charge and jiggled it, I must have performed extra work on it ... 0 It does interact with its own field! This is the only way to get an accelerating charge to radiate away energy -- it has to lose energy itself, so it has to be repulsed by its own field. 0 I don't really know any high energy QFT, so I don't think I can explain a QFT concept in terms of classical physics and relate the two, but if I've understood the question I can give a pretty classical guess. Coulomb interaction is mediated by the electromagnetic field, and this interaction travels at the speed of light in vaccuum. So the electron ... 1 It should be seen more like: A stationary charge generates an electromagnetic field A moving charge generates an electromagnetic field An accelerating charge generates an electromagnetic field So the hierarchy/pattern you mentioned isn't really much of a hierarchy/pattern after all. But actually, you can show that only the second derivative enters the ... 3 If somehow a magnetic field around a wire can be made to exist identical to the magnetic field produced when a current passes through the wire, will current be produced in the wire? If the magnetic field is identical, there is a current through the wire. Put another way, if there is no current through the wire, the magnetic field is not (nor can be ... 0 The answer to the thought experiment is No. Magnetic fields are produced only by charges in motion and electric current in a wire is an example of charges on electrons in motion. Current in a conductor can be produced by a magnetic field in motion and, as mentioned, a generator uses this principle. These two relations between magnetic and electric field both ... 2 If there is a current density in the wire, there needs to be an electric field. An electric field can be generated by a changing magnetic flux. This is Faraday's law. So, in your thought experiment, while you are assembling your magnetic field there will be a transient current induced that produces an opposing magnetic field. Once you have established the ... 0 Yes ofcourse, it is a very famous phenomena named "Electromagnetic induction". and there is a famous law there named "Faraday's law of Electromagnetic induction". and it doesn't work like the way that you explained. it's like this: "if you have a time-varying magnetic field around the wire, it will produce a current inside the wire" and it can be ... 3 As you know the answer should be a hyper trigonometric function instead of a trigonometric one. Your mistake is with lowering/raising of vector components$$ p^\alpha = m_0 \left( u^0, u^1 \right) = m_0 \left( \eta^{00}u_0, \eta^{11}u_1 \right) = \pm \left( - u_0, u_1\right) $$Where the \pm comes from your metric convention. This will lead to$$ r^2 + ...

2

a discrete charge distribution is approximated by a continious charge distribution if the ratio $$\frac{\bar{d}}{D} << 1$$ for $D$ the smallest diameter of the distribution and $\bar{d}$ the mean distance of discrete charges in any subvolume of the distribution (e.g this ratio has to be small in any subvolume of the distribution).

5

Even though a continuous charge distribution does not exist in nature, it is a very useful concept in situations where there are so many very small charge carriers that you don't see them individually. Think of it like water. If you are not using sophisticated tools, you can't see the molecules. It looks continuous. Hydrodynamics is the study of the water ...

0

You are roughly correct. However, you must be careful because the surface you would choose for finding the magnetic field from $\mathbf{J} \cdot \mathrm{d}\mathbf{S}$ is NOT the same surface you would use to find the electric field. The concept you want to solve this problem is self-inductance. Defining the magnetic flux $\Phi=\int\mathbf{B} \cdot ... 1 Monopoles: Either north or south pole alone. Dipoles: Both north and south pole in each other's influence The Magnetic field of lines originate from North Pole and end at south pole. Gauss's law of Magnetostatics states that total magnetic flux from a closed surface is zero. That is number of incoming field line equals the number of outgoing lines. I.E. ... 0 Take the simplest case of a uniform magnetic field and a circular wire perpendicular to the magnetic field. You know that a changing magnetic field induces an electric field. Because the magnetic field is uniform and the wire perpendicular, the electric field has the same magnitude on all points of the wire. You can find the total emf from the change in ... 0 The key to this question is the position of the voltmeter and the flux it contains. Consider the diagram below: Let us say we are finding the voltage between A and D using the volt meter$M_1$. We can do this using one of two loops: One going straight from A to D. The one going from A to B to C then to D. The first one will give an answer of ... 1 Electric fields produces a difference of potential on two points with different distances of the field source. Magnetic fields induces current on a closed loop if the loop is not on parallel in relation of the lines of field and the magnitude of the field does have to change (you have to have a flux). If you have a magnetic field interfering on your ... 0 User31782 gave the right answer, but it's quite hard to read because of formatting. Let me repeat the argument for you: The coil rotates at$\omega$, and the field is also changing at$\omega\$. At any moment in time, the area of the coil normal to the direction of the field is $$A = A_0 \cos(\omega t)$$ and the field is $$B = B_0 \cos(\omega t)$$ And ...

0

So,the integral is equal to zero because the Q that is enclosed in the Gaussian surface is zero.That does not necessarily mean that the electric field is equal to zero.The only other way for the integral to be equal to zero is if the sum of the dot products inside the integral is equal to zero.It means that you have equal negative dot products as positive ...

2

The relativistically correct electromagnetic fields of any charge are given by the Lienard-Wiechert potentials. The waves emitted during the course of acceleration do not belong to the field of the charge, since the waves are vacuum solutions of Maxwell's equations. So, the presence or absence of radiation has nothing to do with the field being "static" or ...

-1

Why can't the particle borrow some of its momentum from the field or lend some of its momentum to the field thereby increasing or reducing its velocity? Because the particle is the field. We make an electron (and a positron) out of electromagnetic waves in gamma-gamma pair production. In atomic orbitals electrons "exist as standing waves". We can diffract ...

1

It's acutally possible. The phenomenon is called Bremsstrahlung (its direct translations would be something like "stopping radiation"). It can be described purely with classical theories like special relativty and electromagnetism. If a charged particle is accelerating, it "borrows" some of his momentum and energy to the EM-field which is then radiated as ...

Top 50 recent answers are included