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9

They are technically units for incommensurate quantities, but in practice this is often just a technicality. The magnetic field that makes sense ($B$) is measured in teslas (SI) or gauss (CGS), and the magnetic field that people spoke about 100 years ago ($H$) is measured in amps per meter (SI, also equivalent to a number of other things) or oersteds (CGS). ...


6

This is a relatively tricky one, because it involves the differences between the $\mathbf B$ field and the $\mathbf H$ field in the SI and CGS systems, and those relationships change in the different systems. In short: Oersteds are used to measure the $\mathbf H$ field in CGS units. Teslas are used to measure the $\mathbf B$ field in SI units. In the SI ...


6

The problem is that you calculated the curl wrong; you missed a delta function arising from the discontinuity. We can write $E_z$ as $$ E_z = \ln(Cr) \Theta(R-r)$$ and, using that $\Theta'(x) = \delta(x)$, we get that $$\nabla \times \mathbf{E} = -\frac{1}{r} \Theta(R-r) \hat{\phi} - \ln(CR)\delta(r-R)\hat{\phi}$$ so when giving $\nabla \cdot \mathbf{E}$...


6

Conclusion: The particles are, in-fact, magnetic This is doubtful. There are two possibilities here: If the objects used to grind (let's call them grinders) the dust grains were made of a ferromagnetic material like iron/steel and/or nickel, then it is possible that the grinders deposited some material onto the dust. There is no reason to think that ...


4

First of all, the tangential component of ${\bf E}$ (i.e. the component parallel to the interface) is always continuous across interfaces, so $E_z$ must be continuous at $r = R$, which fixes $C = 1/R$. There is no physical configuration of sources that could produce any other value for $C$. ${\bf E}$ does not depend on time, so ${\bf \nabla \times E} = -\...


4

From a quick google search, it seems that Oersteds are used for defining magnetic field strength and Teslas are used for defining magnetic field strength in terms of flux density. They seem to not really be meant to be converted between, though you technically can (as evidenced by the other answers here). This website and this website might be helpful to ...


4

1) The case of a moving conductor and a stationary conductor is fundamentally different. When the conductor is stationary, a changing magnetic field produces an electric field everywhere in space, whose curl along any loop enclosing the varying magnetic field is non-zero, given by Curl(E) = -dB/dt .Using stokes law, we easily find the emf to be the rate of ...


3

It depends on the sense of circulation of the current in the wire of the solenoid. If you take the solenoid in your right hand so that curved fingers copy direction of current in the wires, the thumb will show direction of magnetic field inside the solenoid. It thus points to the "north pole" of the electromagnet. This picture may help to understand this ...


3

Great Question! First of all, you can solve this problem for the Magnetic Field and the Electromagnetic Field. You won't have to do any work with the Lorentz Force Equation. Instead you will use the solutions to Maxwell's Equations. $$\pmb{\mathcal{E}}=-\nabla \phi-\frac{\partial{\pmb{A}}}{\partial{t}}$$ $$\pmb{\mathcal{B}}=\nabla \times\pmb{A}$$ where $\...


3

This is a surprisingly complicated question, and I'm not sure there is a universally accepted answer. To see why this is turn off your magnetic field and give the electron enough velocity to keep it in orbit around the Earth. Now in the Earth frame the electron has a centripetal acceleration of $r\omega^2$ and therefore it should be emitting radiation. ...


3

Please note that this is not so much an answer as an extended comment. At this website (which is the Google Book website for Contemporary Newtonian Research by Z. Bechler) it says It is well known that Newton became convinced towards the end of his life that electricity played a vital role in the operations of nature. In the famous final paragraph of ...


3

For any given electric field $\mathbf{E}(\mathbf{r},t)$, and assuming no constraints on the magnetic field, there is always a set of charges and currents (both possibly time-dependent) that produce that electric field. What's more, while the charges required are uniquely determined, the currents are not. Here's the construction: Let $\rho(\mathbf{r},t) = ...


3

Maxwell's equations only give a unique electric field subject to a set of boundary conditions and an initial condition for the field.


3

In general, it is not. Assume a constant current flowing through a cylindrical conductor. Applying Ampere's circuital law for a surface inside the cylinder: $$\oint {\vec Bd\vec l = {\mu _0}\int\!\!\!\int {\vec J} d\vec s} $$ $$B2\pi r = {\mu _0}J{{\pi {r^2}} \over {\pi {a^2}}}$$ $$B = {{{\mu _0}} \over {2\pi {a^2}}}Jr$$


2

1) It is necessary for a plane EM wave. If one assumes solutions to the Maxwell's equation to be plane waves, it is not hard to show that $\vec B \cdot \vec E = 0$. Namely, take the third Maxwell's equation and dot both sides with $\vec E$. $$\nabla \times \vec E = - {{\partial \vec B} \over {\partial t}}$$ $$i\vec k \times \vec E = i\omega \vec B{\rm{...


2

The "formal law" that prevents this is called conservation of energy. The problem is that it costs energy to generate a current in the electromagnet - and more importantly, according to Lenz's law you will need to put in additional energy to maintain the flow of current when the coil moves towards the permanent magnet because the changing flux in the coil (...


2

You can't get away from special relativity, which is what "unifies" electric and magnetic phenomena. The electromagnetic field really is a single field (actually a tensor, but don't worry about that), and its components mix together in the Lorentz transform in a similar way to how the x,y,z components of a usual vector field mix together in a rotation. ...


2

If you think of a magnet as being made up of individual dipoles all aligned with one another (this is what happens in Iron or Neodymium bar magnets), then you can ask "how does each dipole transform under Lorentz transformation?" Under Lorentz transformation magnetic dipoles become electric dipoles. The electric field from the electric dipoles causes the ...


2

Switching the electromagnet on and off at appropriate times will make the arrangement into a (not particularly efficient) electric motor. In order to use it to spin up the disks -- or just overcome the friction in the bearings for an extended time -- you will find that the electromagnet requires slightly more electric energy deposited into it in order to ...


2

In answer to your 1st question, whether the magnets attract or repel depends on how their poles are oriented with respect to each other. However, as they pass the confluence point C (ie position in 2nd image) the interaction will overall have no effect on the speed of the discs. For example, N & S poles approaching each other will attract and accelerate ...


1

As far as I can understand you are proposing moving electrons mechanically rather than with some sort of electromotive force to make an electromagnet. Say we have a little electromagnet maybe 5 cm long with 100 loops of wire and we run 0.25 A of current through it. With an air core we should get a magnetic field inside or about 1 millitesla (handy ...


1

No. It is because field lines travel in closed loops i.e loops don't start or stop at any point, or even cross paths. This is why field lines are from south to north inside the the solenoid not from north to south.


1

Although the pull force (perpendicular) is magnetic, the shear force is mainly frictional. It varies depending on the type and condition of the surfaces, and is roughly 10-25% of the pulling force, unless the magnets are coated in rubber or polyurethane when it can be as much as 200%. There is no reliable way to predict the friction force, so if it is ...


1

The situation described here is not that simple. First of all we see a simple situation. You have a solenoid and generate an appreciable magnetic field inside it. Now you put a charged particle (electron) into it (not the wire with current), It is also assumed that the mean free path of this charged particle is quite large such that it do not collide with ...


1

You're hitting on some of the considerations that led Einstein to his 1905 "Elektrodynamik bewegter K├Ârper" paper. In form of the Lorentz force law you state, the force is the electron's velocity relative to your present frame. The magnetic field $B$ is also as measured in your present frame. You are right to be worried that the whole lot might not be ...


1

A magnetic field won't do anything, but if you apply an electric field you can electrolyse the iron salt and produce metallic iron. There are some other more involved ways it could be done. For example zeolites will selectively absorb some metal ions. If you choose the correct zeolite you could remove all the iron. Though arguably this is just a form of ...


1

$$m\frac{v^{2}}{r}=qvB$$ is applied when the centripetal force is provided by the magnetic field. you question says that the magnetic field is only present at the center,since the rotating particle is not moving through the magnetic field but around it ,there is no force acting on the particle due to the magnetic field(some other force must be providing ...


1

If you put your itty bitty magnet with the 1.4 tesla field at the North Magnetic Pole, and another identical one at the South Magnetic Pole, then you would have added 1.4 tesla at each pole. There would be zero effect on the Earth's magnetic field. Like all fields you're likely to encounter, magnetic field strength depends on the distance from the source. ...


1

In your title you ask about preventing a current from a magnetic field. One way to do this is to generate an equal but opposite magnetic field to the first magnetic field. If the 2 fields are time dependent then they will cancel the currents generated In the body of your question you seem to ask about nullifying a magnetic field generated by a current. ...


1

There is a name for machines which move charges around to cause the field to vary: wireless transmitters. As is well-known the disturbance propagates at $c$. No: if special relativity is correct (and it is extremely well tested) there is no way of transmitting information 'instantaneously', nor in fact is there any well-defined notion of what it would mean ...



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