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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 ...

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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 ...

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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 ...

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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 ...

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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. ...

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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 ...

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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) ... 3 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)$$Where the imaginary part is absorbed into that B 0 I have always thought that electrons can and do spend some (tiny) fraction of their time in the nucleus, depending on the orbital they occupy - the same quantum mechanics that says "they must remain in orbit" does in fact allow for orbitals that, while strictly speaking having zero probability at r=0, have a very small probability at a radius comparable to ... 1 So firstly, it is not the strong nuclear force that keeps electrons in fixed orbits around the nucleus. The strong nuclear force, that is, the Quantum Chromodyanmic interactions, hold the protons and neutrons together in the nucleus, as well as holding the quarks and/ or antiquarks together in other hadrons and mesons. These interactions do not come into ... 3 It's because magnetic and electric fields transform into one another depending on your intertial frame, and therefore field lines aren't an invariant intrinsic property of space. In one frame, you could have a region where there's only a uniform magnetic field \vec B, so that any stationary charge remains classically at rest. Yet viewed from a frame ... 2 The metal is detuning both the tag's antenna and depending on how close the phone is, the phone's RFID antenna too. When a piece of metal is placed in the near field area of an antenna it becomes coupled to the antenna and it's resonance frequency drops, the impedance decreases (causing a large signal loss) and the bandwidth widens (Q decreases). In an ... 1 As you have written,$$ \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}$$where \partial V is the surface area surrounding the volume V. The magnetic field has ... 0 If you look at a light wave as a rotating x and y axis which propagates forward in the z direction, the equation which might result takes the appearance of a screw [or helix]. The equation of the wave is not only a function of time, but also in z. y = A e^ (i( B*z + w*t )) , i is the square root of -1 Note an equation of a helix ... 1 Electrochemical cells (batteries) are not passive components, instead they're active charge-pumps having internal feedback effects which produces a relatively constant voltage at the output terminals. If an external field impinges on a battery's terminals, this will produce a temporary small change in potential on the terminals. But the battery then ... 1 F_{\mu\nu} is a Lorentz tensor, easy to see by \partial_\mu A_\nu - \partial_\nu A_\mu, which is a 2-form. Contractions of Lorentz tensors are Lorentz tensors. \tilde{F} = \star F is the Hodge dual of F, which is also a 2-form, hence a Lorentz tensor, therefore the same applies about its contractions. By these definitions, they are also tensors in ... 6 The first equation is only valid for massive particles. If you see the formula of the Lorentz factor:$$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$If v=c (the case of massless particles), it is undefined. You can also see as v \rightarrow c, \gamma \rightarrow \infty, which "compensates" for m_0=0. The second doesn't imply zero energy, because ... 0 Well, as the people said in the comments, the Theorems of Green, Stokes and Gauss will do the job, and are about as mathematically rigorous as you could hope for here! The two different sets of formula follow directly. I don't want to write all four of them out, you should be able to do them yourself, but for example, let's consider the Gauss Law. ... 0 According to Ampere's Circuital Law, ∫B.dl=μi. [i.e. circular integral of product of magnetic field along imaginary circulating path and path length is equal to permeability of medium multiplied with current flowing through the imaginary cross section made by circulating path] So, ∑i must be zero, hence net current comes zero, i.e.Jb+Kb=0. This implies that ... 0 (Still can't comment!) How are you doing a course on CED without knowing (Classical) field theory?! Do you mean to say, you are doing a course on Electromagnetism? Assuming that part of the aim of writing the essay is to learn things along the way, and given that you don't yet know Classical field theory, this would be exactly a good thing to write about! ... 1 The conceptual problem here is that of EMF, \mathcal{E} vs Electric Potential, V. They aren't really the same thing despite being measured in the same units. For instance the EMF is caused by an external agent that isn't the conservative electrostatic field, like say a chemical reaction in a battery or a solar cell. Work is done to cause a charge ... 0 Stick a piece of paper to the magnet over the approximate pole (found from sticking the magnet to a fridge door etc), then stick the magnet back and spin it around on the pole, it will mark the paper with a dot where it was touching the metal. 0 A more comprehensive and deeper explanation of the Lorentz-force is based on relativistic electrodynamics as given in: http://chip-architect.com/physics/Magnetism_from_ElectroStatics_and_SR.pdf by Hans De Vries. 0 Let's focus on radiating EM waves first and forget about energy. When you jump into accelerating train and see charge accelerate away from you, this is all in a non-inertial frame. In this frame, electromagnetic theory has to be formulated with modified equations and new appropriate boundary conditions. That being said, nothing forbids static field in ... 0 I've seen this question before. :) You can model a DC motor as an ideal resistor, an ideal inductor and a back-emf in series with the voltage source. When you are talking about steady state operation like your question is, you can treat the inductor as a short, leaving you with just the resistance and the back-emf. From this circuit, you can see that ... 1 The whole (pedagogical) point of the slide wire generator is to illustrate that not only do changes in the magnetic field generate current in the loop, changes in the area of the loop - in a constant magnetic field - also generate a current. It's the change in magnetic flux that matters. As long as the wire is moving with some velocity, the magnetic flux ... 1 No, your reasoning is incorrect, because there's no reason for the forces to cancel in general. Actually, the charge in general will be attracted by the field of the induced opposite charge on the inside surface of the conductor. This is easy to see by use of the fact that \nabla^2 V=0 in the region devoid of charges implies that V is a harmonic ... 2 Argue by dimensional analysis. The force on a charged particle is usually taken to be$$\mathbf F = q(\mathbf E + \mathbf v \times \mathbf B)$$and this defines the \mathbf E and \mathbf B fields. With this definition [\mathbf E] = [c][\mathbf B]. However you could take as definition$$\mathbf F = q(\mathbf E + \frac{\mathbf v}{c} \times \mathbf B)$$... 0 In my physics textbook, it says that a qualitative way to envision pressure from EM waves is as follows: the electric field drives charges in the x direction, and the magnetic field then exerts on them a force qv⃗ ×B⃗ in the z direction. The net force on all the charges in the surface produces the pressure. This is a bit inaccurate, the pressure is ... 5 The difference has to do with the units in which \vec{B} is measured in. In SI units Faraday's Law reads as,$$ \nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t} $$In Gaussian and Heaviside-Lorentz Units it reads as,$$ \nabla \times \vec{E} = - \frac{1}{c}\frac{\partial \vec{B}}{\partial t} $$Basically this amounts to redifining ... 0 The magnetic field polarizes orthogonal to the electric field in free space. We generally only talk about the electric field because Maxwell's equations define a one to one relationship between the two. It would make just as much sense to only talk about the magnetic field. We choose the electric field because, in general, when light interacts with matter ... 0 The magnetic field does not vanish when light is polarized. A changing electric field induces a magnetic field, and a changing magnetic field induces an electric field. This is why, in the propagation of an electromagnetic wave, there is always an oscillating electric field coupled with a magnetic field oscillating perpendicular to this electric field. You ... 0 Yes this is common practice for stepper motors only that the strategy is a little bit different. Instead of the voltage value one controls the time average of the voltage through a pulse-width modulation. This works because the coil works like an integrator:$$ i_L(t) = i_0+\frac1{L}\int_{0}^t v_L\left(\bar t\right) d\bar t $$v_L would be$$ v_L(t) = ...

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It is true. A straight line segment is not a closed loop. But, it is like an important mathematical building block if the closed integration path $c$ can be decomposed into straight line segments $c_k$ $k=1,\ldots,n$. In this case you compute $$\textbf{A} = \frac{\mu_0 I}{4\pi}\oint_c \frac{dl}{R} = \sum_{k=1}^n\frac{\mu_0 I}{4\pi}\int_{c_k} \frac{dl}{R}. ... 0 Before answering your question i will state two maxwell laws These laws are intricately linked. A changing magnetic field instantaneously produces a changing electric field but with displacement current (J). Once electric field and displacement current are produced the cycle ends owing to conservation of energy. The same happens with changing electric ... 0 Light of any wavelength can cause heating. It depends on the material which absorbs it. Few materials absorb light present in microwave range while others simply reflect them . Consider a material which absorbs light at microwave frequencies. If the intensity of the light wave at microwave frequencies is increased then heating effect will also be observed. ... 0 The force on the charge q is given by the electric field \mathbf E_m = -\nabla \phi_m of charges of the metal shell surrounding it. This field does not vanish inside the metal, because total field \mathbf E_m + \mathbf E_q does. It follows that the potential of metal charges \phi_m is not constant throughout the metal and its inner surface is not ... 2 Can we conclude that B=0? For a general field it is wrong because every constant vector will satisfy those conditions. But for the magnetic field is it enough? It depends on what facts about magnetic field you want to admit into your hypothetical situation. If you assume the Maxwell equations with vanishing sources and the condition \nabla \times ... 1 The answer to your question is "no". Put the point charge close to the wall at some spot of the wall. There will a surface charge be collected at this spot of the wall that attracts the point charge. In the following I give an example for which one could even calculate the attracting force analytically. Nevertheless, I keep a bit informal here since the ... 2 No, that is not enough to say that B=0. You must also consider that$$\nabla\times E=-\frac{\partial B}{\partial t} which means that for a magnetic field that is constant spatially but not in time, your conditions would be true but your $B$ field would not be $0$ If, however, we had a case where $\nabla\times E=0$ as well, then (aside from being a very ...

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Your problem is clearly and comprehensively treated by Hans De Vries in: http://chip-architect.com/physics/Magnetism_from_ElectroStatics_and_SR.pdf The quintessence is that a current carrying wire appears electrostatically charged to an observer in relative motion to that wire, even when the same current carrying wire appears uncharged to an observer at ...

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