# Tag Info

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Have you heard of the Ampère's circuital law? It states: $$\oint_C \vec B\cdot d\vec{l} = \mu_0I_{enc}$$

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From your link: Perpetual motion describes motion that continues indefinitely without any external source of energy. This is impossible in practice because of friction and other sources of energy loss. Furthermore, the term is often used in a stronger sense to describe a perpetual motion machine of the first kind, a "hypothetical machine which, once ...

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Looking at it from a very different perspective than my other answer: Advanced technology or storing energy does not help with perpetual motion just as they do not help with making time go backwards. It is not in the scope of technology to begin with. The difference is: it looks like it is in the scope of technology - even if you look closely. That ...

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For creating perpetual motion, storing energy is not important. It does not help that we can store energy in magnets. We can store lots of energy in batteries too - nothing new here. So maybe you think about the force a magnet can have in attracting or repelling something for infinite time without getting weaker? But if you just sit on a chair - there is ...

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To answer this question, I will first give my own definition of the transverse current, and then I will show how both of your formulas agree with this new definition of transverse current. My definition For any current $\vec{j}(\vec{r})$, I define the transverse current $\vec{j}_t (r)$, which is a functional of the current, $\vec{j}_t=\vec{j}_t[\vec{j}]$, ...

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What the question refers to as "band intensity" is also referred to a "line strength" $S$. To calculate an absorption coefficient $k$ from $S$, a line shape function $f(\nu - \nu_0)$, where $\nu_0$ is the center of the line. $$k = Sf(\nu - \nu_0)$$ Then "optical depth" = $ku$, where $u$ is called "path length" but is really a measure of the absorbing ...

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Probably not, but it depends on the geometry of your coil. For a couple of dollars at the hardware store you can get a big stack of those coin magnets. If the answer to your question were yes in general, it'd be harder to break apart the big stack of magnets that to separate two of them. That's not consistent with my experience. In general for a dipole ...

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My question is this: is this energy flux that is described by the Poynting vector a form of electromagnetic radiation? Poynting vector quantifies energy that is being lost from a region of space. Integral of this vector over its boundary gives total energy region loses per unit time. EM radiation is a different concept: when you are far from the ...

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First, and most important of all: as said in the comments, you are not using the Einstein summation convention! This way you will have mistakes at your computation! Solve it. Solving that problem you may find an expression like $$E = \int d^3 x \left[ F^{\sigma 0}\partial^0 A_\sigma + \frac{1}{4} F^{\mu \nu} F_{\mu \nu} \right]$$ As you can see, it's very ...

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Also, it is obvious that the work done is due to force of "magnetic field" on "moving electrons". This part is problematic, as you probably already know (you've put the quotes). There is macroscopic magnetic force on wire 1 due to wire 2 given by $\int \mathbf j_1 \times \mathbf B_2 \,d^3\mathbf x$. Often this force is present but balanced by other ...

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Maxwell's equations are based on COE of electromagnetism. All electromagnetic phenomena are covered under maxwell's equation. The following link covers the relationship between maxwell's equations and COE http://farside.ph.utexas.edu/teaching/em/lectures/node89.html

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Your interpretation is partially correct. Complex numbers are superimposition of two independent variables. Coming to electromagnetic(EM) waves we consider two components of EM wave which are perpendicular to each other. This two components do not interact with each other but they are both part of a single wave. So the vector some of these components is the ...

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The work done will be the change in electric potential . If we assume current density to be constant and electric current to be constant(DC current) then we can equate this wire to a linearly charged wires(Static field). The amount of charge per unit length is always constant since outgoing electrons are replaced by incoming electrons. When two charges are ...

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Here is one way to think about it: When a charged particle travels in a magnetic field, it experiences a force. If the particle is stationary but the field is moving, then in the frame of reference of the field the particle should see the same force. Now let's take a conductor wound into a coil. In order to increase the magnetic field inside, I could take ...

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By assumption, there is electric current (with density $\mathbf j$) flowing in the wire (think of a circuit or a rod on rails). When placed in the magnetic field, there will be magnetic force $\int \frac{\mathbf j}{c} \times\mathbf B \,dV$ acting on the wire (on the nuclei it is composed of) so some of its parts will begin to move (with velocity $\mathbf ... 3 A very tempting mental model of an atom, reinforced by many illustrations in books, has protons and neutrons as "large" spheres in the nucleus and electrons as "small" spheres somewhere near the nucleus. If you assume that all of these particles are made of some "stuff" that has roughly the same density (which is the case for everyday solid and liquid ... 2 His demonstration was not capable to show how the electric current depends on the radius of the circuit; the magnetic field change was an abrupt and short pulse made by hand, which makes the pointer in the ammeter to oscillate too quickly to do any reliable measurement. He probably wanted to stress that the induced emf does not depend on the size of the wire ... 0 The force is perpendicular to the motion of the charge carriers. But the resulting motion of the wire is in the direction of the induced force, so work is done on the wire. 1 Yes, Coulomb’s law is only for point charges separated by a distance$r$. The inverse square law is there because of the diverging or converging nature of electric field from a point charge which is the crucial point. Here is how I explain: Imagine spherical surfaces with increasing radius around a point charge. As we increase the radius, the intensity of ... 1 Gauss theorem is of great importance. Those situations, in which the calculation of electric field by applying Coulomb's law or the principle of superposition of electric fields becomes very difficult, the results can be obtained by applying Gauss's theorem with great ease. You can notice that electric field due to a point charge decreases inversely as ... 0 It's a "tunneling" behavior. In effect, all the light is "pulled back" into the medium unless there's another body of high-index (well higher than the$n_1 = 1$) material within the distance covered by the evanescent wave. If that material is close enough, then that part of the evanescent wave, which you can view as a probability wave, is in a region ... 2 Coulomb's law is indeed a special case between two point charges. to find the force between a point charge and a plate, you would have to integrate the equation over the plate surface to calculate the contributions from all infinitesimal charge elements. It's more practical to figure out what the electric field is, and then use$\vec{F}_e = q\vec{E}$to ... 1 The term "Rosenberg-Coleman effect" originates from the article Heliographic latitude dependence of the dominant polarity of the interplanetary magnetic field. It is also referred to as the "dominant polarity effect". As the Earth orbits the Sun, the Earth travels above and below the equator of the Sun. According to Rosenberg and Coleman, the polarity ... 2 Yes you are correct the answers should be the same in either frame and in fact the answers you give are. You just have to convert back to the inertial frame Remember that i and j in the rotating frame are fixed in the rotating frame and so rotate in the inertial frame. Using i' and j' as the axis of the reference frame and i and j for the inertial frame. ... 5 Covariant notation is a simple way to say how something transforms under Lorentz. An object with an index, e.g.$z^\mu=(z^0,\vec z$), transforms under Lorentz as, $${z^{\prime}}^{\nu} = {\Lambda^\nu}_\mu z^\mu$$ where$\Lambda^\mu_\nu$is the matrix you have written down in your question.$z^\mu$is called a four-vector. This transformation property is ... 0 According to classical electromagnetic theory... I think this is the key assumption you are building your question on, and for atoms/nucleons/electrons/everything smaller this assumption just doesn't hold true. All these objects have to be described with quantum mechanics, so there is no trajectory of a localized charge or something alike - all you're ... 0 To address the follow up question, which requires more characters than a comment allows: In a simple idealized view, the Fermi level is the top energy level in the solid occupied by electrons. In silicon with no doping it sits at mid-gap: the valance band is full, the conduction band empty. In a thought experiment, if you had two separate chunks of ... 0 It all depends on what you call a 'strongly coupled' force. If you simply use a larger value for the coupling constant, then there is not much reason to believe a lot will change qualitatively, except that perturbative methods will get worse. The wikipedia article on this topic may give you a good point to start looking for a more specific question or to ... 2 The rate of formation is much higher in the presence of dust. There needs to be a mechanism for the energy of formation of the hydrogen molecule to be dissipated. Dissipating energy via a photon involves a forbidden transition. Instead, the energy can be transferred to the vibrational lattice of a dust particle. See The Interstellar Abundance of the ... 3 Just to add to Danu's Answer, which I believe to be right. The relative scalings of the "electro" and "magnetism" parts of the unified electromagnetism whole are somewhat arbitrary; we're only required to ensure that$c=\frac{1}{\sqrt{\mu_0\,\epsilon_0}}$to achieve a valid set of Maxwell equations. As we change these relative scalings, we change the ... 8 Although you might not like to hear it, the answer really DOES lie in the definition of$\mu_0$(and$c$).$\mu_0$is defined to be exactly$4\pi *10^{-7}\ \text{H m}^{-1}$. Similarly,$c$is defined as exactly$299792458\ \text{ms}^{-1}$. It immediately follows from the relation $$\epsilon_0=\frac{1}{\mu_0 c^2}$$ that$\epsilon_0$also has no uncertainty. ... 1 The scalar potential and magnetic vector potential are combined into a four-vector,$A_{\mu}=(\phi,\vec{A})$which is a gauge field, and in the language of differential geometry, a 1-form. The Lagrangian of the field theory (i.e. Maxwell theory) is, $$\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ where$F_{\mu\nu} = \partial_{[\mu}A_{\nu]}$is the ... 2 You didn't really simplify the expressions in the same manner, so they're hard to compare. Even if they're not the same, it's often useful to have a clear idea of what differs in the final result. Simplifying the first expression to share the denominator$(b^2-c^2)^2: \begin{align} W&=\frac{1}{2}\int\rho V\,d\tau\\ &=\frac{\pi ... 2 If you want to include "all real world effects" in your analysis, you need to make sure you include all effects. At the very least, include parasitics. And include the fact that your "real world voltage source" has finite impedance, output capacitance, inductance in the leads, ... So when you state Say it starts of at a voltage V when you connect it to ... 1 But why when I connect a volt-meter across the whole diode will I not see this built in potential? Because at equilibrium (V=0) the Fermi-level throughout the whole device is flat. A voltage will only exist when there is a gradient in Fermi-level over the component (i.e. a voltage drop). Edit Maybe a more visual explanation would help? Think of the ... 1 One has to define in what framework one is talking of electric and magnetic fields. In the classical framework the field is defined, for simplicity lets take a point charge, as proportional to 1/r^2 and exists up to infinity. Thus classically there is no transmission for a static charge, it just is. When a charge is moving, i.e. changing its (x,y,z,t) ... 6 Are there experiments that could show that light waves resemble more say square waves than sine waves? Are there experiments that could show that light waves resemble more say square waves than sine waves? Temporarily looking at your example of a square wave, a square wave of spatial wavenumber k can be represented in a Fourier expansion as a ... 1 So, my best understanding: The basic solution to the wave equation is\Psi(x,t)=Ae^{ikx-i\omega t}$$Where the signs are arbitrary. If you combine this with the good old Euler Formula this expands to$$\Psi(x,t)=A\cos(kx-\omega t)+B\sin(kx-\omega t)$$Where the imaginary part is absorbed into that B 1 I have always thought that electrons can and do spend some (tiny) fraction of their time in the nucleus, depending on the orbital they occupy - the same quantum mechanics that says "they must remain in orbit" does in fact allow for orbitals that, while strictly speaking having zero probability at r=0, have a very small probability at a radius comparable to ... 1 So firstly, it is not the strong nuclear force that keeps electrons in fixed orbits around the nucleus. The strong nuclear force, that is, the Quantum Chromodyanmic interactions, hold the protons and neutrons together in the nucleus, as well as holding the quarks and/ or antiquarks together in other hadrons and mesons. These interactions do not come into ... 3 It's because magnetic and electric fields transform into one another depending on your intertial frame, and therefore field lines aren't an invariant intrinsic property of space. In one frame, you could have a region where there's only a uniform magnetic field \vec B, so that any stationary charge remains classically at rest. Yet viewed from a frame ... 2 The metal is detuning both the tag's antenna and depending on how close the phone is, the phone's RFID antenna too. When a piece of metal is placed in the near field area of an antenna it becomes coupled to the antenna and it's resonance frequency drops, the impedance decreases (causing a large signal loss) and the bandwidth widens (Q decreases). In an ... 1 As you have written,$$ \epsilon = {\Delta \phi \over \Delta t}$$This equation says that \epsilon is greatest when the change in flux with respect to time is greatest, not when the flux itself is greatest. In order to find when the change in flux is the greatest, you need to come up with an equation for the flux, then take the derivative with respect to ... 1 Maybe you have seen the experiment where a permanent magnet is dropped through a pipe made of conducting metal. The magnetic flux through a cross-section of the pipe will be changing, so a current is induced. The induced magnetic field is such as to oppose the change in magnetic flux, so it will slow down the falling magnet. Eventually an equilibrium is ... 2 Simply consider Maxwell equation :$$\vec{\nabla}\wedge\vec{E}=-\frac{\partial\vec{B}}{\partial t}$$If you interger this on a given closed surface \Sigma, it follows :$$\oint_\Sigma \left(\vec{\nabla}\wedge\vec{E}\right) \cdot d\vec{S} =-\frac{\partial}{\partial t}\oint_\Sigma \vec{B}\cdot d\vec{S}$$where d\vec{S}=dS\,.\vec{n} with dS the ... 0 It is immediate from the definition of the cross product. Write$$\mathbf{B} = B_x \mathbf{e_x} + B_y \mathbf{e_y} + B_z \mathbf{e_z}$$and use that$$\mathbf{e_x} \times \mathbf{e_y} = \mathbf{e_z}$$and$$\mathbf{e_x} \times \mathbf{e_z} = -\mathbf{e_y}$$If you can't get an identity by applying the physics it's sometimes useful to make sure you ... 6 Divergence means the field is either converging to a point/source or diverging from it. Divergence of magnetic field is zero everywhere because if it is not it would mean that a monopole is there since field can converge to or diverge from monopole. But magnetic monopole doesn't exist in space. So its divergence is zero everywhere. Mathematically, we get ... 13 Your intuition about the meaning of the divergence operator is wrong. In physics it's easiest to think intuitively about divergence by using the divergence theorem which states$$\int_V dV \ \nabla \cdot \mathbf{B} = \int_{\partial V} \mathbf{B} \cdot d\mathbf{S}$where$\partial V$is the surface area surrounding the volume$V\$. The magnetic field has ...

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If you look at a light wave as a rotating x and y axis which propagates forward in the z direction, the equation which might result takes the appearance of a screw [or helix]. The equation of the wave is not only a function of time, but also in z. y = A e^ (i( B*z + w*t )) , i is the square root of -1 Note an equation of a helix ...

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Electrochemical cells (batteries) are not passive components, instead they're active charge-pumps having internal feedback effects which produces a relatively constant voltage at the output terminals. If an external field impinges on a battery's terminals, this will produce a temporary small change in potential on the terminals. But the battery then ...

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