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0

The only thing I can see them going for is the fact that only two of $\epsilon_0$, $\mu_0$ and $c$ are independent, and typically, a modern view will fold $\epsilon_{0}$ into the definition of charge, and declare $c$ to be the fundamental constant used to transform space into time in special relativity, making $\mu_{0}$ a prediction of the theory. I ...


1

Hint: The canonical stress-energy tensor from Noether's theorem is not necessarily symmetric, and often needs to be improved with appropriate improvements terms. This is e.g. the case for EM. See also e.g. this Phys.SE post and links therein. References: Landau and Lifshitz, Vol.2, The Classical Theory of Fields, $\S$33.


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I'll take a slightly different take on the question than the other answers already posted, because I want to address a different side of the question. Just because a theory makes incorrect predictions doesn't make it useless. Take Newtonian gravity, for instance. It doesn't correctly predict the bending of light or the precession of Mercury, both of which ...


0

The magnitude of the forces $q_1$ and $q_2$ exert on each other is equal. According to Coulomb's law, the magnitude of the force that a charge $q_1$ will experience due to a charge $q_2$ is $$|\mathbf F_{12}|=k_e{|q_1q_2|\over r^2}\ ,$$ where $k_e$ is Coulomb's constant and $r$ is the distance between the charges. But that equation is symmetrical in $q_1$ ...


0

He may have been thinking about teaching physics top-down, rather than bottom-up. There is nothing wrong with that. That's exactly what Landau/Lifshitz do in Volume 1 of their "Course of theoretical physics", by introducing a least action principle and deriving much of Newtonian mechanics from it. One could do the same thing for electrodynamics, but the ...


-3

Not exactly. I suggest you read up on the writings of Oliver Heaviside and the teachings of Professor Eric P. Dollard who are not bound by any threats to teach the absolute truth in reality. You will be surprised at what you don't know. Such as the speed of longitudinal magnetic dielectric transmissions through the earth which is 291,000 miles per second. ...


0

Typically, if you force two magnets together with the same poles facing each other they will repel. The energy stored by forcing them together is released by releasing one or both magnets then it (they) will convert that energy from potential (stored) into kinetic (speed).


2

The energy you supply by the force is transformed into changing the flux tube configuration. Microscopically this means that certain parts of the superconductor go from superconducting to "normal" in a small areas (flux tubes) to allow the magnetic flux through and the energy supplied is used to rearrange these regions. In superconductors that stay fixed ...


2

In term of Maxwell's equations what you need to do is take the two curl equations, and then isolate and substitute: $$ \nabla\times E +\frac{\partial B}{\partial t}=0\\ \nabla\times B =\frac{1}{c^2}\frac{\partial E}{\partial t} $$ So you take a time derivative of, say, the first, exchange nabla and differentiation by time (linear operators that are ...


1

For a metal, the permittivity can is typically described by the Drude model with a permittivity given by, \begin{equation} \epsilon = \epsilon' - i\epsilon'' = \epsilon_\infty - \frac{\omega_p^2}{\omega(\omega - i\gamma)} = \epsilon_\infty - \frac{\omega_p^2}{\omega^2 + \gamma^2} + i\gamma\omega\frac{\omega_p^2}{\omega^2 + \gamma^2} \end{equation} where ...


-1

I found the following quote on the American Institute of Physics website. It is a continuation of Feynman's quote above. I believe it answers your question about his new approach. "When I planned it, I was expected to teach electrodynamics, and then to teach a subject which would really be all the different branches of physics, using the same equation — ...


0

There are three separate possibilities: A theory correctly predicts an experiment result A theory predicts something but an experiment contradicts it. A theory makes no prediction whatsoever regarding some experimental result. You're treating the second and third possibility as if they were the same thing, but they're entirely different. For example, ...


0

The electric and magnetic forces are defined via the Lorentz force on a charged particle $$\vec F = q(\vec E + \vec v \times \vec B)$$ The magnetic force comes from the second term which defines it to be perpendicular to the velocity and therefore displacement $\vec {dr}$; so it doesn't do any work.


3

To add to Lordrain's answer, I'll give you two scenarios where this equation would be useful. First, notice that this is Maxwell's extension of Ampere's law, meaning you can apply it to situations where a magnetic field exists but there is no current (e.g inside a capacitor with an alternating current). Let's first consider the situation where you'd like to ...


2

You are right. These two equations cannot be valid at the same time. In particular, AL is only valid for steady current situation. Indeed, current induces a circulating magnetic field, as indicated by AL. However, current is not the only thing which can produce circulating current. The time-varying electric field can also produce magnetic field. This is what ...


4

I understand your confusion here, and I had the same concern when I learned it at the first time. Since the left-hand side is an integral of B, at best we can obtain the flux through some loop, not the value of B at some certain point. Is this your question? Note that the equation do contain some more information. It holds for ANY loop. Intuitively ...


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I believe that Ampere's Law is wrong in some situations. When Maxwell looked into it, he discovered that Ampere's Law is not always true, so he modified it to get the Ampere-Maxwell Law. According to my understanding, this can be shown if you have a wire with a current flow that is charging a capacitor. If you calculate the magnetic field between the ...


5

First let me make two comments before answering the question. The difference between metal and insulator rest in the existence of the itinerant electron Fermi surface or not. Ising (or Heisenberg) model is just an effective theory of local moments (localized electrons in the atoms), which contains no information of the itinerant electron, so there is no ...


3

The main justification for considering the Ising model is that it's exactly solvable in one & two dimensions (and that it shows critical behavior which is universal in some sense). It is not particularly meaningful as an approximation to a real physical system. The Heisenberg model does a much better job, but it is also a lattice model. If you really ...


2

Short answer: even in non-uniform fields, the speed won't change, but the guiding center can drift with some velocity. In a magnetic field (uniform or otherwise), the force on a charged particle is: $$ F = q\; \vec{v} \times \vec{B(x)} $$ The direction of this force will be tangential to the velocity vector because of the cross product (the result of the ...


1

All matter is made out of elementary particles , these combine to form protons and neutrons, among other composite particles, and protons and neutrons form the nucleus of what we see as an atom: a nucleus surrounded by electrons. The electron is the elementary particle with which matter we encounter every day, the very keyboard I am using , we have most ...


0

Can't really answer, but I think you have posed an interesting one. It goes to the physical meaning of the vector potential, which is a sort of momentum. Of course, if one can reflect $\vec{E}$ and $\vec{B}$ then the $\vec{A}$ which generates them should reflect as well. This, of course, assumes that $\vec{E}$ and $\vec{B}$ are causally generated by ...


1

If there is a magnetic field present, it will dictate the direction of the magnetization, as you anticipated and as @user3683367 said. This is then not referred to as spontaneous symmetry breaking but the external magnetic field breaks the symmetry explicitly. In the absence of an external magnetic field, the alignment is indeed random. You intuition that ...


0

As far as I understand it is determined by an external field. The ferromagnetic system has two ground states for its magnetization both with equal energy (degenerate ground states). If you distort the system with an external magnetic field you make one of those states the preferred one so that your system magnetizes in this direction.


1

There is a confusion between the terminology "perpetual" , which means "continuously", devices that almost move forever, and a machine that can produce energy. As the other answers point out energy is conserved and if it looks as if energy is provided from nothing a closer analysis shows the mistake, as in the drinking bird perpetual setup. In the case of ...


3

You might be able to get it to work for quite a while, depending on your skill as an engineer. But there is a critical difference between a well-engineered machine that runs for a while, and a perpetual motion machine that runs forever without input. The latter is impossible.


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No, there are several mistakes in your derivation, although you miraculously end up with the right expression. The term $\sin \theta$ comes from taking the horizontal (X) component of the electric field - not from the expression you used for $dq$. The diagram I envisage for your problem is this: So I would say that $$dq = \lambda dy$$ and then $$dE = ...


4

The key here is the antimagnetic strip, quite aside from whether or not such a device can be built. When you insert the anti-magnetic strip, you must change the shape of the magnetic field. You must force the magnetic field to "leave" the high permeability ball. The same magnetic induction $|\vec{B}|$ in a high permeability $\mu$ material represents a lower ...


4

If the magnet is strong enough to pull the ball up the bottom slope, it will be too strong to let the ball fall. Even worse, as you have drawn the diagram the magnet is pulling the ball down the lower slope when it is toward the left end. Anywhere to the left of where the perpendicular from the magnet to the ramp, the magnet is pulling more right than up. ...


-1

Too many abrupt changes in direction. Also, is this spring powered? If so, I would think it to be simpler by using an electromagnet to apply and cut voltage. More moving parts mean more opportunities of failure.


3

Perhaps a slightly different way of thinking about this would be in terms of the Poynting vector. The force exerted is given by $$F = \frac{1}{c} \int {\bf S} \cdot d{\bf A}, $$ where ${\bf S}$ is the Poynting vector and in your terms $S = c E$. In this case I assume everything is at normal angles, so no need to worry about that. The Poynting vector is ...


0

These are two separate questions. 1) Does light move at c in non-inertial frames? First to get a bit pedantic, I assume you mean the coordinate velocity. As light does not have a well defined proper velocity, this seems to be the most reasonable way to interpret your question here. While rewriting Maxwell's equations in a non-inertial frame, and then ...


1

The first layer are Maxwell's equations. They look like this: $$\nabla \cdot \vec{E} = \rho, \; \nabla \times \vec {E} = -\frac{\partial \vec{B}}{\partial t}$$ $$\nabla \cdot \vec{B}=0,\; \nabla \times \vec{B} = \mu_0( \vec{j} + \epsilon_0 \frac{\partial \vec{E}}{\partial t})$$ Where $\vec{B}$ is the magnetic field, $\vec{E}$ is the electric field, $\rho$ ...


-1

Yes gravity affects voltages and forces in circuits but has very small effect suppose you have vertical wire on globe directly which is 1 meter ,electron mass is 10 power -31 kg,and you know globe mass and radius , by simple calculations depending newton's law of gravity and work's law you get: the voltage gravity applied on wire is 10 power -26 and it ...


2

Let charge A be at the origin, moving to the right (along the positive x axis). Let charge B be at coordinates (1,0), moving in the positive y direction. A's magnetic force on B vanishes, since by symmetry the magnetic field due to A is zero at B's position. B's magnetic force on A doesn't vanish. Does the law already fails in magnetostatics? I guess ...


-1

GOOD question and thinking But the thickness still effective whatever the value is If thickness value is big and battery has limited charges it will be discharged for example in one second If thickness is very very big for example 1000 times than before the battery will be discharged in one millisecond that's it!


0

According to Stoke's theorem, $$\oint_C\vec{E}\cdot d\vec{l}=-\int_S\partial_t\vec{B}\cdot d\vec{S}=-\frac{d\Phi}{dt}.$$ If a circuit is of laboratory size and $\partial_t \vec{B}$ not too large, then the integral of $\vec{E}$ around the circuit $C$ is approximately zero. This is Kirchhoff's circuit law (KVL). Kirchhoff's current law (KCL) is simpler: it is ...


2

It is an old conundrum how and why Newton's 3rd law fails for the differential form of Biot-Savart. To quote Bleaney & Bleaney:"Page and Adams {1945) have shown that there is no real violation, since the electromagnetic field of the current elements possesses momentum which is changing at a rate just equal to the difference of the two forces." Leigh Page ...


10

Magnetic effect on gravity: In technical terms: yes. For practical purposes: no. As Danu states, a magnetic field is a form of energy, and Einstein showed that energy can be equivalent to mass. A sufficiently large amount of energy (of any kind) collected together will produce measurable gravity (aka bending of space-time). This source puts the total ...


3

Typically, with inductors, we use the complex impedance, $Z=i\omega L$ with current frequency $\omega$ and inductance $L$, for the voltage: $$ V_{ind}=IZ,\quad V_{rms}=I_{rms}|Z| $$ where the left equation is the voltage of the inductor and the right equation the root-mean-square. Surely, however, what goes on inside the inductor doesn't matter, it only is ...


0

KVL is essentially a statement of energy conservation, while KCL follows from conservation of charge. KVL, as the name implies, depends on a definition of voltage (scalar potential). This definition is only possible when the electric fields are conservative such that $\nabla \times E = 0$, so that we can define $E = -\nabla V$. In an inductor, this is ...


16

The electromagnetic field tensor $F_{\mu\nu}$ which encodes all the information about the electric and magnetic field, certainly contributes to the energy-stress tensor $T_{\mu\nu}$, which appears in the Einstein Field Equations: $$G_{\mu\nu}= 8\pi G T_{\mu\nu}$$ The left hand side of this equation encodes the geometry of spacetime, while the right hand side ...


0

The force on a magnetic moment (magnet) in a magnetic field depends on the magnetic field gradient. (How fast the field is changing in space.) So you don't really want a big field, but a big field gradient. The force from a single coil has a rather complicated dependence on position. Maybe I can sell you one of these? ...


1

The attenuation constant is specifically the imaginary part of the wave number (ki), while the wave number in dissipative media is the real part of the wave number (kr). Let me know what you think, because I just started studying this.


6

I am hoping someone could explain, rather in-rigorously, the use of the word inconsistent. Essentially, in this context, inconsistent means the two theories give different, incompatible answers to the same question. In this specific case, the question is: If light (an electromagnetic wave) is measured to propagate at speed $c$ in an inertial frame ...


12

Inconsistency between two theories just means that there are statements that one theory says are true, and the other says are false. An easier example than the one you're asking about is the inconsistency between Newtonian mechanics and special relativity. Newtonian mechanics says that if you keep applying a force to a material object, it will eventually go ...


1

This has a three-part answer. The first part concerns large-scale, quasi-static electric fields. The nice thing about electric fields is that one can always do a simple galilean transformation to a frame where this quasi-static electric field does not exist. So that is the first part (not very satisfactory, but true and practical). The second part ...


2

Experimental data is given in http://journals.aps.org/pr/abstract/10.1103/PhysRev.98.889 - unfortunately I only have access to the abstract. It may be worth taking a look. The shape of the cathode does not matter. The material does. Key to solving this problem is knowing the work function of the material - that is the minimum energy that an electron needs ...


0

$j=\sigma E$-This defines conductivity $\sigma$ for a material-it is the starting point.I think you are confused with signs.here $\sigma$ is conductivity and you are mixing it up with Surface charge density of electrostatics.So is the case for $\rho$.You are mixing it with volume charge density but it is actually resistivity here. Note:This obviously is a ...


2

The equation you give is for the static situation in which the iron is already fixed near the solenoid (cf this site). In this set-up, see the image below, we have a coil with surface area $A=\pi d^2/4$ with diameter $d$ placed a distance $g$ below a metal plate (it also seems that the plate must have a surface area > $A$). The volume of the field, $V=gA$, ...



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