New answers tagged

1

The source of the magnetic field is the current density $\vec J$. Randomly moving charges have $\vec J =0$ and therefore produce no magnetic field.


0

What you are writting is a way to show that a wave that does not change its shape as it progresses through space and that has the general form of the one-dimensional wave function ($\psi(x, t)=f(x-v t)$) has velocity $v$ when is forward travelling and $-v$ reverse travelling. You only have two axis: $\phi$ and $x$ and you are representing $\phi(x)$ . Indeed ...


0

You can accelerate charged particles using electric fields, and you can create electric fields by various methods, including for example by a making time-varying magnetic field. There is a lot of sophisticated knowledge about this, mainly in two communities: the particle accelerator community which supports high energy physics experiments at CERN and ...


2

Yes, a changing magnetic flux will induce a current in a complete conducting loop. If there is a gap in the loop, then current cannot flow and the emf will appear as a voltage difference across the gap. Without a gap, the emf is dissipated by the current in the resistance of the loop and there will be no voltage difference between any two points in a ...


2

Of course yes. Since you know about the Faraday's law, then you may know about the Lenz's law. It says that the polarity of the induced emf is such that the current produced by it, gives rise to magnetic field which opposes the external time varying magnetic field. The emf is the one that is "induced" and it drives the current and not the other way ...


0

You are talking about a transformer (which can step AC voltage up or down), but the power out is always a little smaller than the power in (due to resistive losses). The size depends on the power being delivered. The wires need to be big enough to minimize resistance, but with enough turns to give an adequate inductive reactance on the input and the ...


1

You forgot one parameter: Length. The magnetic field B is proportional to the turn density, denoted by n. n = N/L , where N - no of turns and L - length of the solenoid. So if you wish to increase the thickness of the wire, you wouldn't be able to fit the same number of turns in the same length. The turns will be less and hence the field. So only option ...


0

The field (and flux) produced by a bar magnet is always in the same direction along the axis of the bar magnet.


0

The 2nd term above, $v\times B\cdot dl$, will be zero in a current-carrying wire since the drift velocity is parallel to the wire segment $dl$. The line integral being non-zero around a loop must be the non-conservative induced electric field, the electrostatic field due to any charges, will have a line integral equal to zero.


0

The closer the magnet gets to the coil, the bigger is the magnetic field going thrue the coil and thus the bigger the flux thrue the coil. This is due to the magnetic field being stronger closer to the magnet ($V$ is the Volume of the magnet which we integrate over) $$\vec B(\vec r)=\frac{\mu_0}{4\pi}\int_V \frac{\vec j(\vec r')\times (\vec r-\vec r')}{|\vec ...


3

[quote] But aren't the excess electrons due to the n-type impurity attracted by the positively charged holes of the p-type impurity...[/quote] Once the depletion region has formed, there exists no attraction between the excess electrons in the n-type region and holes in the p-type region. This is because of the barrier potential which is set up by the ...


3

As far as I know, Ampére tried to explain the force between conductors by the formula you wrote. The dominant notion of the time was action at a distance, that was successful for gravity and also for electrostatic. Even the magnetic forces were explained by him as resultant of forces between micro currents in the materials. The idea of a magnetic field that ...


4

Magnetostatics is, in some sense, a toy concept taught to students in preparation for the formal magneto-quasi-static (MQS) approximation. The purpose of the MQS approximation is to decouple the electrical field from the magnetic field. This is done by setting $\frac{\partial}{\partial t} \vec E \approx 0$ so that Ampere's law becomes $\nabla \times \vec H \...


0

In the case of a closed loop of a conductor without resistance (say a superconductor), the initial transient effect of the applied B-field is to induce an emf, what results in an increasing current. Once the loop is not moving in a way to change to magnetic flow, the current keeps constant, and the emf is zero. The case of a no-closed loop can be modeled as ...


0

To understand what the sign of Potential signifies it is enough to see that, work done $W=F.dr$ hence basically positive/negative work or potential simply represents the parallelness or antiparallelness between displacement and force that caused the displacement. In case of oppositely charge body, they attract by Coulombs law and so the word done by an ...


1

How is this true? Are we integrating over a loop which extends outside the conductor so that $\mathbf{E}_{s}$ contributes only in one direction? What does this result prove? The reason the loop goes to zero is that the integral: $$ \oint_{C} d\mathbf{l} \cdot \mathbf{E} \tag{0} $$ is a path integral of the electric field, $\mathbf{E}$, along segments $d\...


1

Kirchhoff's circuit laws are two equalities that deal with the current and potential difference (commonly known as voltage) in electrical circuits. They do not care whether your circuit is simple series, parallel, or some complicated circuit. You can use them wherever you want to, but of course, it's useless to use them in a place where it's simple to ...


0

The mistake you are making is similar to if someone concluded that mass (or inertia) is required for something to experience acceleration and motion. It isn't. Similarly, inductance is not required for the production of a magnetic field. It might just seem that way if you are reading about magnetic machines at a high level that does not address the lower ...


1

I would personally define a “steady current” as one that obeys $\frac\partial{\partial t} \mathbf J = 0$ everywhere. The requirement that the charge density in “magnetostatics” not change with time, $\frac\partial{\partial t} \rho = 0$, allows the student to use the tools developed during electrostatics to figure out what the electric fields do. Most ...


0

Is Dirac viewpoint still correct under the current modern physics view (in the 2021, in the 21 century)? No, it's old-fashioned. A significant percentage of physicists (I can't give you a number) have a new paradigm based on the string theory landscape and eternal inflation. In string theory there are a vast number of possible vacuum states: $10^{272,000}$, ...


0

DC coils do not have zero inductance. They have very low impedance for steady and unchanging flows of current, but significant impedance at the instant of application of the driving voltage (because at that time, the current flow is rapidly changing).


0

Yes after a short while, the battery, the resistors and the wires will establish stationary state in the circuit where electric field is constant and electrostatic everywhere. In the battery there are other forces than macroscopic EM forces (microscopic forces due to chemical reactions) that push the mobile charges against the electrostatic field inside the ...


0

Electrostatic field means that the fields are static. But that is compatible with a static flow of current. According to the Ampére-Maxwell law, that implies a static magnetic field. And a static magnetic field, according to the Maxwell-Faraday law implies that $\nabla \times \mathbf E = 0$


1

If ρ is linear and zero at the surface, then, ρ = $ ρ_o$(R – r), where, R, is the radius at the outer surface and, $ ρ_o$, is your unknown.


0

Search for the patent for an electromagnetic weapon in a 500lb dumb bomb form factor. It will give some insight but not performance parameters. The effectiveness of the weapon is not intrinsic to the weapon. It also depends on the vulnerability of the target. EU technical commissions have been working on IEC performance requirements and measurement methods ...


0

If the magnetic force is not cancelled by the electric force there is no equilibrium, so the charges will move until there is.


3

Your calculation of the gauge covariant derivative is incorrect. The $\nabla$ operator changes depending on what it acts upon, just like the covariant derivative in GR. In general a gauge covariant derivative changes based on the representation under which the object it acts upon transforms. This is actually the same as for the covariant derivative in GR as ...


3

Physics does not have an explanation why electromagnetism exists, but it does describe how it works. One could argue that parallel currents attract because of the Lorentz force and Ampère's law. Conversely, these laws are what they are because parallel currents attract. In the end the laws are as they are to explain, among others, this observation.


1

Faraday might have used an argument like this one... The magnetic fields due to two wires carrying currents in the same direction tend to cancel between the two wires, but re-inforce beyond the two wires (applying right hand grip rule). Magnetic field lines behave as if under tension but exerting lateral pressure, so if you draw the pattern you will see that ...


1

It's certainly possible for there to be both a surface charge present, and a current flowing through a conductor. The surface charge might or might not be the charge that is moving to form the current, though. For example, any time current flows in a wire that also has a DC voltage relative to another nearby conductor, this will be the case.


7

I suspect the reason why you're getting seemingly contradictory answers is that you're asking an "is it A or B?" question where the real answer is "no, it's neither of those things." On one hand, no, an electromagnetic wave is not a "pure mathematical line" — it's a fluctuation in the electromagnetic field, which permeates all ...


4

If I interpret your question correctly, you're asking if the E and B fields constitute some sort of spatial displacement (of a medium in which the wave travels, as with water waves or elastic waves in a solid), on the grounds that they are considered to point in a definite spatial direction. The answer is 'no' - they are better considered to be gradients in ...


2

Update I rephrased a few things and added some additional details after chatting with the OP. Question 1: How can I formally prove this conservative electric field to be $\mathbf{E} = -\mathbf{v} \times \mathbf{B}$ other than with dynamics arguments? The rest of this question isn't really necessary, so I only put in the first part. The electric field ...


0

While this statement is true, it should be stressed that magnetic force does not do work IN REFERENCE OF THE MAGNETIC FIELD. If viewed in other frame of reference( often when the field is moving relative to you), it may do work.


4

A conductor must consist of a union of surfaces and bulk. Therefore, a conducting sheet must have two sides separated by a neutral bulk(However thin that might be). Therefore, within a given volume of an insulating sheet and a conducting sheet, the latter has twice the amount of charge and produces twice the electric field twice.


1

What's particularly bothering me is the equation $\nabla \times \mathbf{E}=-\dfrac{\partial \mathbf{B}}{\partial t}$. If we were to put it in the integral form, (using Stokes' theorem), there seems to be no reference to $V$ or $B$, which seems to be incorrect as per definition $(1.2)$. $$ \oint \vec{E} \cdot \vec{dl} = \int_{S} -(\frac{\partial }{\partial t}...


26

In the image included in your question, one should not view the arrows as literally extending through space or having a spatial length associated to them. The length of these arrows is the magnitude of the field, and they should each be thought of as living inside of their own vector space which is "internal" or "attached" to a given ...


17

While electromagnetic waves do have lateral extension, since the electric and magnetic fields are defined on all of $\mathbb{R}^{3+1}$, most graphical representations, including the one you have included, do not show how an electromagnetic wave extends in all directions. The reason for this is quite simple: representing the entirety of an electromagnetic ...


1

Suppose our object is described by the function $g(x,y,z=0)$. We can write this function as a superposition of plane waves (Fourier decomposition), $$ g_{object}(x,y,z=0) = \iint df_x df_y \;G(f_x, f_y,z=0)e^{-2\pi i (f_x x + f_y y)} $$ where $f_x$ and $f_y$ are the spatial frequencies. Hence, the object consists of many different spatial frequencies. Now, ...


6

The plane wave depicted in the image is invariant under translations in $x$ and $y$. It fills all of space.


12

Yes. Often these can be considered plane waves: The vectors represent the electric field of the EM wave. See the description on Commons by Dave3457. Wavefronts can also be spherical, etc.


1

I think you are seeing the problem wrong, the curl of E its not zero, the text, just say that the $z$ component of $\nabla\times \vec{E}$ is zero that means that there is not magnetic field on the direction of the conductor but there exist magnetif field in other dimesions, for example the $y$ component of $\nabla\times \vec{E}$ is $\frac{\partial E_x}{\...


1

This really just comes down to the definition of A. We define A through $\boldsymbol \nabla \times \textbf{A} = \textbf{B}$. Then, using the Maxwell equation $\boldsymbol \nabla \times \textbf{E} = - \frac{\partial\textbf{B}}{\partial t}$ we obtain $$\boldsymbol \nabla \times \textbf{E} = - \frac{\partial}{\partial t} \boldsymbol \nabla \times \textbf{A} = -\...


0

The first obvious answer is yes, of course. How could it be otherwise? Momentum is conserved. But it depends on how you divide up the world. Charged particle A is getting accelerated by something. That doesn't just happen for no reason. Charged particle B interacts both with charged particle A and also with whatever is accelerating A. It's the whole system ...


0

Using the integral form of the Ampére-Maxwell law without the current, that is zero across the plates: $$\oint_{\partial \Sigma} \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \varepsilon_0 \frac{\mathrm{d}}{\mathrm{d}t} \iint_{\Sigma} \mathbf{E} \cdot \mathrm{d}\mathbf{S} $$ we can see that around the borders of the plates, the LHS is maximum, ...


0

Assuming that the approximation that the electric field is constant in space but changes with time is reasonable, then since $\vec j = 0$ between the plates, a possible solution for the $\vec \nabla \times \vec B$ equation which also satisfies $\vec \nabla \cdot \vec B = 0$ is $\vec B = \frac{1}{2c^2} \vec r \times \frac{\partial \vec E}{\partial t}$. To see ...


2

In short: There is a lower limit on the spectral width $\Delta \omega$ of unpolarized light. Therefore, strictly monochromatic unpolarized light can not exist, not even mathematically, while spectrally broad unpolarized and polarized light exists. The following holds for classical light which has a well-defined electric field at each point in spacetime. An ...


0

If we start from the charge-current continuity equation for any arbitrary charge distribution, we have: $$ \partial_{t} \rho + \nabla \cdot \mathbf{j} = 0 \tag{0} $$ where $\rho$ is the charge density, $\mathbf{j}$ is the current density (specifically the macroscopic average current density, see pages 248--258 in Jackson [1999] for definition and derivation),...


1

First of all, let me just point out that the radius of the $U(1)$ is related to the gauge coupling, so rescaling the radius is essentially the same as rescaling the gauge coupling of the theory. The only place I've seen this pointed out is in references dealing with $S$-duality in electrodynamics since part of the transformation is on the gauge coupling, and ...


-1

The answer by Wolpertinger is wrong. Reasoning does not help, so I will use the argument by authority and quote from Frank S. Crawford, Waves (Berkeley Physics Course volume 3), 1968. Unpolarized radiation defined. We are now prepared to say what is meant by "unpolarized" light. Unpolarized light is light whose two polarization components ($x$ and ...


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