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77

Consider this: A charged particle at rest creates an electric field, but no magnetic field. Now if you walk past the charge, it will be in motion from your point of view, that is, in your frame of reference. So your magnetometer will detect a magnetic field. But the charge is just sitting on the table. Nothing about the charge has changed. Evidently ...


19

The arguments from special relativity given in the other answers is correct. What is charge according to one observer is current according to another observer that is in relative motion to the first. But this is, from a historical perspective, somewhat backwards. This consideration is what led Einstein to develop special relativity -- the paper is called On ...


12

The equations are entirely equivalent, as can be proven using Gauss' and Stokes' theorems. The integral forms are most useful when dealing with macroscopic problems with high degrees of symmetry (e.g. spherical or axial symmetry; or, following on from comments below, a line/surface integrals where the field is either parallel or perpendicular to the ...


8

You are right, the electric field and the magnetic field are distinct fields that have different properties. The reason why they are still classified as the cause for the "electromagnetic force" are the following: In higher theories, like the field theory, the electric and the magnetic field are caused by the same gauge principles. There is just "one" ...


6

It probably does not mean anything. That paper concerns the quantization of electromagnetic waves in less than three spatial dimensions. In fact, there are a number of decades-old results showing that it is often possible to evade the spin-statistics relationship in lower-dimensional systems. While these kinds of results (including this new one) may be ...


6

Several answers have given a physical explanation as to why electric and magnetic forces are tightly coupled, and why you can't develop independent theories of "just electric" and "just magnetic" fields. Your subquestions (especially #1) make me think you're looking for some kind of symmetry. It turns out, there's a really nice one! All the asymmetry ...


5

Your confusion lies in failing to recognize that they are exactly the same equations. Take for example Gauss's law $$ \vec \nabla \cdot \vec E = \dfrac{\rho}{\epsilon_0}$$ You can see that there $\rho$ is the charge distribution, and in general can be a funcion of the position. Now consider a volume $V$, you can just integrate the density to obtain the ...


5

A typical voltmeter contains an internal Ohmic resistor with known and very high resistance $R$ (called the "input resistance" or "input impedance"), and an extremely sensitive ammeter that measures the current through that resistor. When the voltmeter is connected in parallel across some circuit elements, then ideally the internal resistor has resistance ...


5

First, the gauge invariance means that the solutions $A_\mu(x^\alpha)$ are not unique. For every solution, the gauge transformations of it are solutions, too. That may be a problem because sometimes we want to have specific values of $A_\mu(x^\alpha)$ that answer a physical question. Second, we sometimes gauge fix because the equations simplify. For ...


4

Regarding 1) observe that there is a pattern in common - namely that there is some region (volume for Gauss and a surface for Ampere) and integral of the source on this region is equal to the integral of the field on the boundary. This is a striking similarity. 2) currents are nothing else than moving charges. So both fields are generated by charges. These ...


4

The problem is that they have too many solutions if the gauge is not fixed. Imagine you have some initial values, and want to solve it on the computer. Then you have to solve the equations for the next time step given the values for the previous one. But you have to compute the values for, say, four variables but have only three equations. You somehow ...


4

Dynamo Effect : The dynamo effect is a geophysical theory that explains the origin of the Earth's main magnetic field in terms of a self-exciting (or self-sustaining) dynamo. In this dynamo mechanism, fluid motion in the Earth's outer core moves conducting material (liquid iron) across an already existing, weak magnetic field and generates an ...


3

All this tells you is that the fields satisfies both the inategral and the differential equations. The two are related by the mathematical identities called the divergence theorem and Stokes' theorem. So which do you apply? Well, which ever one you want! If you run into an integral, you use the integral form, and if you're ever asked for the divergence or ...


3

As is written here the two remaining equations follow from the Bianchi identity which says that the anti-symmetrized derivative is zero, ie. $$ \partial_{[a} F_{bc]} = \partial_{a} F_{bc}+\partial_{b} F_{ca}+\partial_{c} F_{ab} = 0 $$ (remember the $F_{\mu\nu}$ is antisymmetric itself!)


3

The mass of the electron is thousands of times less than that of the ions - about 1,800 times lighter than a proton. The motions move the entire ion core, so inertia tends to resist the change of motion much more than is possible for an electron. For example, see Improved Two-Temperature Model and Its Application in Ultrashort Laser Heating of Metal Films. ...


3

The classical electromagnetic effect is perfectly consistent with the lone electrostatic effect but with special relativity taken into consideration. The simplest hypothetical experiment would be two identical parallel infinite lines of charge (with charge per unit length of $ \lambda \ $ and some non-zero mass per unit length of $\rho \ $ separated by ...


2

This answer has been hinted at in the others, but it's worth stating their collective knowledge as a succinct one liner that every physicist should know: Electric and Magnetic force only make sense in the light of special relativity if they are unified because if they were thought of as separate entities, then relatively moving observers would reach ...


2

If you define $C:q\rightarrow -q$, $P:(x,y,z)\rightarrow (-x,-y,-z)$ and $T:t\rightarrow -t$, then all the Maxwell eaquations are invariant under $C$, $P$, $T$ or any combination of them. To see this you just have to notice how the transformations act on sources and coordinates. The charge conjugation acts non trivially as \begin{align} C\rho&=-\rho,\\ ...


2

If you ignore the coefficients, your function will be: f(x) = x / (x2 + r2)3/2 differentiating w.r.t x will give you: simple product rule f(x) = u(x)*v(x); u(x) = x; v(x) = 1 / (x2 + r2)3/2 f'(x) = u'(x)*v(x) + u(x)*v'(x) f'(x) = [1 / (x2 + r2)3/2] + [(-3/2)2x2 / (x2 + r2)5/2] to maximize/minimize you substitute f(x) = 0: 1 / (x2 + r2)3/2 = 3x2 / (x2 ...


2

Too long for a comment: Consider the Maxwell equations: $$\nabla \cdot {\bf E}=\rho/\epsilon_0 \qquad \nabla \cdot {\bf B}=0$$ $$\nabla\times {\bf E}=-\dfrac{\partial {\bf B}}{\partial t} \qquad\nabla\times {\bf B}=\mu_0 {\bf J}+\dfrac{1}{c^2}\dfrac{\partial {\bf E}}{\partial t}$$ According to Heras (when commenting on a paper by Griffiths and Heald) the ...


1

Vector plots like this should not be shown to students. Some falsely begin to believe, that the blue line actually represents something in real space. That is not true. So what is shown here? This is, in fact, a snapshot of a single moment in time. I have arbitrarily chosen a convention (let me stress arbitrarily again), where I represent the electric ...


1

I have found your question and the diagram a little difficult to interpret. I have redrawn you diagram to show a charge of $+Q$ on the outer shell and a charge of $-Q$ at the centre together with two conducting shells shaded grey. What else the electric field inside the conductors is zero. If there was an electric field then the mobile charge carrier ...


1

In general your relation is $$ \vec{B}(\omega) = (1 + \chi_m(\omega))\vec{B}_0(\omega) $$ or in the time domain $$ \vec{B}(t) =\vec{B}_0(t) + \int\limits_{-\infty}^\infty \chi_m(t,t') \vec{B}_0(t') \;\rm{d}t' $$ Only in the case of instantanous material response, i.e. $\chi_m(t,t') = \chi_{m,0} \cdot \delta(t-t') $, your equation is correct. This already ...


1

International Handbook of Research in History, Philosophy and Science Teaching quotes the English translation of Guisasola et al. (2008), which discusses some of the early history of the EMF. The man who coined the term "electromotive force" was Alessandro Volta, who stated that there was a force separating the charges in current flowing in a closed circuit. ...


1

Hint : A key to the solution is what is meant by the complex wave 3-vector $\:\mathbf{k}\:$. This vector is not any complex 3-vector in $\: \mathbb{C}^{3}\:$ $$ \mathbf{k} \ne \left(k_{1}, k_{2}, k_{3} \right) \in \mathbb{C}^{3}, \:\:\text{that is with} \:\: k_{\rho} \in \mathbb{C} \tag{a-01} $$ but $$ \mathbf{k}=\left(k_{1}, k_{2}, k_{3} \right) ...


1

Traversable - Overlapping (actually intersecting) region would not be Traversable even if the gravity at some parts of the region may be zero. For exampple, between earth and moon, gravity will be zero at some point. That does not mean something in that region can go out of earth/moon system. As soon as an observer leaves that region, it either falls towards ...


1

The magma has temperature between 700 and 1300 Celsius degrees. The Curie temperature of iron is at 770 degrees Celsius. Above that temperature, iron loses magnetism. Note that right above 770 °C, iron is still solid because the melting point is around 1500 °C. So magma almost never can be magnetic because it's just too hot for that. Incidentally, if it ...


1

Firstly, Bluetooth devices do in fact interfere with one another, just that the interference is not significant, and the communication protocols of the devices make it seem like they don't (i.e. my moving of my BT mouse does not move the cursor on your screen which is paired to your BT mouse) You can isolate the system by putting the two electromagnets ...


1

The answer is that it depends on the field on the outside (boundary conditions) and the dielectric constant of the insulator. For a imaginary insulating sphere of vacuum in a vacuum, it should be obvious that the sphere does not affect the electric field at all. Inside a dielectric, the field will be weaker than on the outside. For a dielectric sphere in a ...



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