# Tag Info

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$\def\vE{{\vec{E}}}$ $\def\vD{{\vec{D}}}$ $\def\vB{{\vec{B}}}$ $\def\vJ{{\vec{J}}}$ $\def\vr{{\vec{r}}}$ $\def\vA{{\vec{A}}}$ $\def\vH{{\vec{H}}}$ $\def\ddt{\frac{d}{dt}}$ $\def\rot{\operatorname{rot}}$ $\def\div{\operatorname{div}}$ $\def\grad{\operatorname{grad}}$ $\def\rmC{{\mathrm{C}}}$ $\def\rmM{{\mathrm{M}}}$ $\def\ph{{\varphi}}$ ...

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As the magnet approaches the solenoid, a current is induced. The current generates a magnetic field. The field repels the magnet, slowing it's approach. The amplitude of the oscillations diminish. If there was no resistance, this would work in reverse as the magnet receded from the solenoid. The magnetic field would accelerate the magnet. The magnet would ...

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As Mostafa says, it is macroscopically at equilibrium, not necessarily microscopically. There may be one misunderstanding you have, which is about "surface". I will talk about it later. In my opinion, equilibrium should be understood as no electron moving. It is easily to show that the electric field in conductor is zero. If the electric field is non-zero, ...

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If you have just given the voltage signal with $$\def\l{\left}\def\r{\right} v(t) = \l(2-\l|\frac t{\rm s}-2\r|\r)\rm V$$ then the current at $t=2$ is undefined. Right. But, in most cases really nobody cares. What we learn theoretically about the current from the above voltage signal definition is that  i(t) = \begin{cases} C\cdot 1\frac{\rm V}{\rm s} ...

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An atom in isolation offers a potential well, and electrons form bound states in the well. The energy of those bound states can be calculated exactly in the case of a single-electron (hydrogen-like) atoms or by variational computational methods for more complicated cases. Now when you put several atoms together in a tight and regular array, they offer a ...

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Here's the logic (well a particular rendition): Recall that $n$ is defined as the ratio of the speed of light $c$ in vacuum to the speed of light $v$ in the given medium; \begin{align} n = \frac{c}{v} \end{align} Note that in a linear medium, Maxwell's equations are exactly the same as in vacuum, except $\mu_0$ and $\epsilon_0$ are replaced by $\mu$ and ...

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You are right in that a magnetic field is build up, which generates a electric field opposing the given potential. But the consequence is not an oscillation of current, but only a damping of the increase of the current. Therefore, if you have a Heaviside step function for the voltage, it'll result in an "exponential" increase of your current ($I(t) = I_0 ... 3 Maxwell is being misunderstood. First, Maxwell makes very clear that length, time and mass are the fundamental types of units. Then he discusses a totally different convention that isn't used today, saying "in the astronomical system, the unit of mass is defined with respect to its attractive power". In other words, Maxwell is talking about a concept of ... 3 The explanation is Maxwell's text: If, as in the astronomical system, the unit of mass is defined with respect to its attractive power, the dimensions of$[M]$are$[L^3T^{-2}]$. To motivate this, it is perhaps useful to be aware of some of the different systems of units for electromagnetism used historically. One of the units of charge commonly in use ... 2 The definition of Ampere is obtained by the below equation of force between two infinitely long parallel current carrying conductors. Where$F$is force,$\triangle{L}$is small length element,$\mu_0$is absolute permeability of vaccum or free space,$I_1, I_2$are current flowing through two conductors. By calculation we can obtain that ... 2$\def\vE{{\vec{E}}}\def\vS{{\vec{S}}}\def\vA{{\vec{A}}}\def\rot{\operatorname{rot}}\def\grad{\operatorname{grad}}\def\div{\operatorname{div}}\def\ph{{\varphi}}\def\vn{{\vec{n}}}\def\vr{{\vec{r}}}$The charge floating through the wire causes a current density$\vec{S}$in the wire (not only on the surface). This causes an electrical ... 2 Emf on a conducting object induces eddy currents. These in turn decay due to the electrical resistance of the object. What you end up with is energy in the form of heat. When you compare the two objects (essentially a conductor versus a non-conductor), a portion of the potential gravitational energy goes into generating eddy currents. That means the ... 2 In a single free atom, electrons have well defined energy levels and are somewhat bound to atom. Consider the following quantum mechanical model of atom to get an idea about an isolated atom. When all this isolated atoms come together to form the crystal, the atoms do not have well defined energy levels. There will be molecular orbitals. When the atoms ... 2 Mostly, yes on both counts. It depends whether the solenoid also gets longer as it gets wider. The approximation that the field outside the solenoid vanishes is valid for points whose distance to the solenoid's centre is much smaller than the distance to both ends. This is impossible for a point outside a solenoid that's wider than it is tall. To answer ... 2 The answer can be found in the nature of gravity. It is a force that arises due to curvature of spacetime which underlies everything. A black hole is a specific configuration of spacetime where nothing can leave by definition, not even light. Since there is no concept comparable to curved spacetime underlying the other forces, we observe no such phenomena. 2 I've made a small illustration depicting the key idea. If this is in coherence with what you've asked, we could summarize some important points about the case. Total energy of the Earth and Bar Magnet system is given by the equation:$KE + PE = \frac{1}{2}mv^{2} + \frac{GMm}{R}$While PE is there for both Earth and magnet system (combined), KE is ... 2 This should go on forever, and current should keep appearing across the load resistance. This is a contradiction. Since there is current through (not across) the load resistance, there is work being done on the load:$p = i^2R$. Let's be clear on this: the coil-load system does no work on the pendulum, the pendulum does work on the coil-load ... 2 Let's analyse the case for an infinitely long cylinder first, then move to a finite one (I've just realised you mean one of finite length having written my answer). Your hypothesis is right only in the case of an infinitely long cylinder, but not for one of finite length, unless your cylinder is a perfect conductor, in which case no symmetry assumptions are ... 1 In figure (a), the voltage is continuous but the time derivative is not; the capacitor current would discontinuously change sign from positive to negative. In figure (b) however, the voltage is discontinuous. It is typically said that the voltage across an ideal capacitor is continuous since, for the current to exist, the time derivative of the voltage ... 1 Now saturation is defined as how much that core can absorb the magnetic field Not quite. The saturation point of a ferromagnetic is roughly defined as the internal$\mathbf{B}$-field strength at which ferromagnetic amplification of the external$\mathbf{H}$-field stops. It doesn't really have anything to do with the size of the material. A crude way of ... 1 Feynman explains it best in this classic video, but here are some of the essentials. Magnets attract and repel at a distance, and there is really no way of rephrasing that fact which will explain this force in terms of "winds of force" or any similar construct and which will not incur inaccuracies and inconsistencies that will render it completely useless. ... 1 The force on the pendulum only applies when the pendulum is in the vicinity of the coil. At that moment the harmonic motion of the pendulum is distorted. It 's amplitude is lessened and with it the upward motion. So the kinetic energy of the pendulum is converted into gravitational energy and electric energy. But the gravitational energy is less than without ... 1 The short answer is that there is a induced force on the magnet. This induced force will make the pendulum loose energy in the same proportions as there is electrical energy being generated. A good experiment to show this effect is by having a small bar magnet and a copper pipe Or solenoid. When you let a small bar magnet drop from a certain height it will ... 1 An inductor stores energy in a magnetic field. After current has been flowing in the inductor for a period of time, it has built up a magnetic field around the wire making up the inductor. In that state the inductor offers no opposition to current flow. If it were then disconnected from it's energy source (battery perhaps) then the the magnetic field will ... 1 The intuition is that the valence electrons are so far away from their nucleus that when they combine to form metals, they feel the attraction of all the other nuclei as strongly as from theirs. In a more rigorous description, the orbitals for the valence electrons fully overlap with their neighbouring atoms, so their "play field" extends all over the ... 1 Draw the graph of$\frac 1 x$. You can see that it is a decreasing function for positive$x$. Hence conductance decreases as resistance increases. We could have defined conductance as any other decreasing function also but$\frac 1 R$appears in many equations so we defined it that way. You might want to look at derivation of$J=\sigma E$to get better ... 1 The electrical resistance of an electrical conductor is the opposition to the passage of an electric current through that conductor; the inverse quantity is electrical conductance, the ease at which an electric current passes. Consider resistance of$0.0001$ohms, what is$\frac{1}{0.0001}ohms$? It is equal to$10000$siemens. I hope this helped you in ... 1 First, your force equation is wrong, as you're missing the electric field. Wait what electric field? That's the point! A changing magnetic field induces an electric field$\nabla\times E=-\frac{\partial B}{\partial t}\$, and this "pushes" the current. Note that the applied magnetic field is perpendicular to the circuit/wire, so that at least part of the ...

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