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0

Maybe you should think about superconductor materials!! possible get upto 8 T and is very usual to use electromagnets (very big) for different applications, what can give you up to 2 T. But you didn't specified what kind of field you are thinking, dipole, quadrupoles?


-1

The "articles" a and and an has its correct usage, dear! "A" usually follows a word that starts with a consonant while "An" follows a word starting with a vowel. There are some exceptions to this, though, as its usage may also depend on the sound of the word as it is pronounced.


0

Just blindly multiplying the overall answer by some factor isn't the way to go about it. I have an alternative proposal which may work well as a zeroth-order approximation at least. You already have the expression for the 2-layer case, and if I observe correctly, you are only concerned with the reflected part, not the transmitted part. So, a smaller ...


3

Yes, it is already a reality. Permanent magnets from rare earth alloys https://en.wikipedia.org/wiki/Rare-earth_magnet can exceed 1.4 tesla while ferrite and ceramic ones only have 0.5-1.0 tesla. Making the material as big as a meter or anything you want is just a matter of accumulating a larger amount of the material (or adding smaller magnets). At ...


0

Hopefully this should dispel some of your confusion. In general, the fields $\textbf{B}$ and $\textbf{H}$ are related by $$ \textbf{H} \equiv \frac{1}{\mu_0}(\textbf{B} - \textbf{M}) $$ This is always true, regardless of the materials involved. We define linear media as materials whose fields satisfy $$ \textbf{B} = \frac{1}{\chi}\textbf{M}. $$ In this ...


4

From a physics perspective, the fundamental reason for this is something called the bandwidth theorem (and also the Fourier limit, bandwidth limit, and even the Heisenberg uncertainty principle). In essence, it says that the bandwidth $\Delta\omega$ of a pulse of signal and its duration $\Delta t$ are related: $$ \Delta\omega\,\Delta t\gtrsim 2\pi. $$ A ...


3

Note: Emilio Pisanty wrote an answer that is probably a better fit for the question and site, but I'm leaving this answer around because I feel it can contribute to an understanding of how this works in practice. For one thing, you'd need to be able to differentiate between the signals inside the frequency band. As an example, I'm going to use a Morse ...


0

See, you are required to find volume current density $J_\phi$. Though its name is volume current density, you know it is the current flowing per unit surface area. Now the subscript $\phi$ in $J_\phi$ denotes it is flowing in the $\hat{\phi}$ direction. Now in the spherical polar co-ordinate the infinitesimal length elements along the direction ...


2

I think a ferrite rod antenna in a radio receiver is an example where the magnetic component of an EM-field is picked up.


1

An electron's magnetic field is a dipole field - that is, the field strength is given by source: http://www2.ph.ed.ac.uk/~playfer/EMlect4.pdf In this expression, $m$ is the magnetic dipole moment. For an electron, this has the value $$m=-928.476377 × 10^{−26} J/T$$ The magnetic permeability $$\mu_0=4\pi\cdot 10^{-7}\frac{V\cdot s}{ A \cdot m}$$ Note - ...


0

We do not use just the electric component, this is not possible. An EM radiation is an oscillation in media or void of electric and magnetic fields. As described by Maxwell equations, the electric field oscillation generating the magnetic one and vice versa. So is kind of an electric wave and a magnetic one co-dependent with each other. It is true that we ...


0

Here's the Feynman response I read in his opening paragraphs in his Feynman lectures: The reason a proton and electron simply don't crash into each other is that if they did, we would exactly know their position-assuming one of them is stable, which one is (the proton). If we knew their position, we would be highly unaware of the momentum, meaning it could ...


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This phenomenon is not special to light or Maxwell's equations: it's a simple consequence of detector nonlinearity and I should think that this would have been pretty clear to any bright experimentalist who thought carefully about how his or her kit makes its measurements. In the simplest case, the detector's response $y(t)$ as a function of time $t$ is ...


2

If you note that $$\delta(\cos\theta)=\frac{\delta(\theta-\pi/2)}{\sin\theta}$$ Then you can see that the sine terms actually cancel out.


1

This is just a complement to the previous answers which give the correct response. If you want to think about it in an intuitive way, imagine that the interaction between electrons and photons becomes weaker. In the limit when it becomes nearly zero, the light will be almost not scattered at all and will continue in a straight path.


2

EDITED ANSWER: The delta distribution $\delta(x)$ is not unique. It is invariant under transformations of the form $\delta(x) \to f(x)\delta(x)$ where $f(0) = 1$. This is because it is really a distribution and not a function. It is mathematically improper to talk about $\delta(x)$ instead of $\int \delta(x)dx$. Derivations of the term you're interested in ...


2

It is momentum that defines the incoming direction and momentum transfer the outgoing one. The photons, quantum mechanically carry momentum equal to p=h*nu/c . Momentum is a vector and defines directions. An electromagnetic field is an emergent classical quantity built up by innumerable photons. There exists also a momentum defined for the classical field ...


1

Let me offer you a slightly modified version of your question to illustrate a way of re-formulating it your thought process. How does a pool ball know from which direction the cue ball hit it? The answer is the same in the sense that "the particle" does not know all by itself, "the system"1 has certain invariant quantities (like momentum and energy) ...


0

While both the answers given in some sense are correct, the true reason has to do with energetic considerations. It is a matter of what is stronger and can be phrased as the following question: Will the wavefunction alter itself to accommodate the flux, or will the flux quantize itself because the wavefunction is trying to remain single valued? As an ...


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If you're asking about how the mathematical model we use to explain electromagnetic interactions predicts EM radiation, this involves a bunch of math that you can look up on Wikipedia. It sounds like you are looking for a philosophical explanation, however. "Mechanism" may not be the right word to use to describe electromagnetic field theory. We have some ...


-1

Magnetic forces represent the relativistic correction to Electric forces (they are the same phenomenon!). The germ of relativity is that when objects move, information about their positions travel at the speed of light. If that information moved instantaneously or objects never moved, there would be no Magnetic forces. Accordingly, the Magnetic force on an ...


13

There is a sort of analog called gravitomagnetism (or gravitoelectromagnetism), but it is not discussed that often because it applies only in a special case. It is an approximation of general relativity (i.e. the Einstein Field Equations) in the case where: The weak field limit applies. The correct reference frame is chosen (it's not entirely clear to me ...


4

There is a gravitational analogue of the magnetic field. See gravitoelectromagnetism and frame dragging on Wikipedia.


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You must known the impedance of the medium. For the vacumm: $$ \eta_0 = \sqrt{\dfrac{\mu_0}{\varepsilon_0}} $$ then, the relation between the fields: $$ \vert\mathbf{H}\vert = \dfrac{\vert\mathbf{E}\vert}{\eta_0} $$ in vacumm, of course.


0

You are probably looking for Maxwell equations.


1

The thing which is "vibrating" is the electromagnetic field, namely its $\vec E$ and $\vec B$ vectors. The animations here show precisely this. Of course, it's not that some particles vibrate in this case. The electromagnetic wave can exist without any matter at all — all it needs is the field, which is present everywhere. But, if we have some charges ...


1

For low-frequency radiation, it's quite simple: there's some electronic circuit that works (simple case) analogous to a tuning fork, but instead of building up mechanical tension it charges a capacitor and instead of the inertia in the fork's arms it has a magnetic field in a solenoid. You can measure the voltage against time, count the oscillations in one ...


1

$ \mathcal{E} = -{{d\Phi_B} \over dt} = -8t$ (Faraday's law of induction) $ E = V/d = \mathcal{E}/d = \frac{-8t}{d} = \frac{-8*3}{15\times 10^{-3}} V/m = -1600 V/m$


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It's a quirk of units: notice that the conversion between them is dimensionful, and has the value $3\times 10^8\,\mathrm{m/s}$, which is the speed of light. In the CGS system both fields have the same units, and field-squared is an energy density. In SI units, the energy density for a configuration of fields is given by \begin{align} \frac{dU}{dV} &= ...


27

As you already indicated, physical units need to be considered. When working in SI units, the ratio of electric field strength over magnetic field strength in EM radiation equals 299 792 458 m/s, the speed of light $c$. However, the numerical value for $c$ depends on the units used. When working in units in which the speed of light $c=1$, one would ...


1

I think the best explanation of electromagnets/Ferro magnetism is in the Feynman lectures Vol II. (chapter 36.) He makes up his own units so that might be a bit confusing. But work through the electromagnet problem. (and even use a simple linear relationship B = uH) That helped me a lot.


3

It seems entirely symmetrical. But it isn't symmetrical. For the voltmeter on the right, the 'outer' loop that encloses the magnetic field consists of the voltmeter, leads, and a $100\Omega$ resistor For the voltmeter on the left, the outer loop that encloses the magnetic field consists of the voltmeter, leads, and a $900\Omega$ resistor. Moreover, ...


0

Your error is that you cannot understand ferromagnetism while neglecting hysteresis and nonlinearity. Consider the case of some iron cooled from above the Curie temperature in the absence of any magnetic field. As the iron crystallizes, it forms strongly magnetized domains — a classic case of spontaneously broken symmetry. Because there is no external ...


1

The definition of $\mathbf{H}$ is $$ \mathbf{H} = \frac{\mathbf{B}}{\mu_0} - \mathbf{M} $$ where $\mathbf{B}$ is the magnetic field in which the object is immersed in and $\mathbf{M}$ is the magnetisation of the object, i.e. the "field" ($\propto$ to a field) caused by the internal magnetic properties of the object. If there are no free currents, then ...


1

Interesting setup! I assume that you understand that for a short time there is a (induced) current flowing in the loop with the two resistors. You can get the same current by imagining a battery that is in series with the loop (you can put it at A, for example). With that battery in place, it is easy to see that the voltage across the 900r resistor will be ...


0

For problem solving, the method of images come in handy in the following cases: conducting sphere, conducting cylinder, conducting ellipsoid, and conducting plane. Another example is to regions of dielectrics, with different $\epsilon$ (permittivity). And that is it. This general rule should save you some time. So, the first problem you mentioned can be ...


2

I solved my problem numerically, using the diffusion equation $\frac{\partial V}{\partial t} = -k\nabla^2 V$, with the following boundary conditions: Voltage at point D is fixed at 1.0 Voltage along the vertical line halfway between points A and D is fixed at 0.5 (voltage at point A is 0.0, use symmetry so we don't have to simulate the left half of the ...


0

Because when you take the sine of 0º to 360º and plot the graph of these values, AC current behaves the same way. It can also be represented by a cosine function, but in this case we assume that the initial value of the AC current shouldn't be zero.


-1

To the question "What is the electric field outside a cylindrical solenoid when inside is turned on a magnetic field" the answer is that outside exists a electric field. That means that the fringes shift in the double slit experiment with electrons could be explained with electromagnetic fields and it is not necessary (but of course possible) to explain it ...


0

Now since the coil moves through this region of space, it should therefore possesses an induced electric field as well. I don't follow your reasoning here. The electric and magnetic fields are reference frame dependent; the fields 'mix' in a certain way via the Lorentz transformation. In the reference frame in which the magnet is at rest, the ...


0

In Faraday's experiment, the relative velocity between the coil and the magnet define whether a change of induced magnetic field flux occurs around the coil or not. So if you consider them at rest, and when you consider them moving at the same velocity relative to one another, then there's no change of magnetic field flux felt by the coil. Be careful there ...


1

To get a better understand of what is going on, take a look at the plot below, also linked here: http://en.wikipedia.org/wiki/File:Dispersion_Relationship.gif What the author meant by "letting the speed of light go to infinity" is that the we let the slope of the blue line become infinite. In that case, the solid red line would not curve as shown below, ...


0

By Cavendish torsion balance, similar to the measure of gravitational constant.http://en.wikipedia.org/wiki/Cavendish_experiment


2

Here is a compass, a small dipole magnet in the shape of a needle: as the person holding it turns around it moves pointing to the geographic north in the location, then the directions are all defined on the face of the compass. So that is the way a human can see/sense simply the magnetic field of the earth. Semantics on the compass is a bit complicated ...


0

Aluminum is not magnetic without an external magnetic field, however when an external magnetic field is applied or in presence of it Aluminum becomes "slightly" magnetic as its electron align to the magnetic field however due to thermal motion as described by Vintage the alignment of the electrons within the material (aluminium) is randomized thus its net ...


3

To good approximation, a magnetic dipole in a uniform field feels a torque, but not a force. The length scale for variations in the field produced by the dynamo in the earth's core is comparable to the size of the earth's core: many hundreds of miles. For a magnetic dipole the size of an animal (any animal) the earth's natural field is uniform, and you feel ...


3

Optical conductivity and AC electric conductivity experiments are indeed quite similar. However, they operate in different frequency regimes and measure slightly different quantities. Optical conductivity refers to an experiment using light, such as a reflectivity measurement and then using a Kramers-Kronig transform to deduce the real part of the ...


2

A strongly paramagnetic material has a magnetic susceptibility of a few hundred parts per million; that is, the field strength inside of a paramagnet is different from the field strength in vacuum starting in the fourth decimal place. This is comparable to the ratio between the earth's natural field (typically about 0.5 gauss) and the surface field of a good ...


4

Radioactivity is the result from a confluence of special relativity and quantum mechanics. Special relativity introduces the generalized energy, E=m*c^2 , which allows the energy conservation to count in the sum the rest masses of the particles which comprise a nucleus. In this relativistic energy conservation we find some nuclear isotopes which are at a ...


0

No, the electron cannot be inside the solenoid. The Aharonov-Bohm effect is intrinsically a topological effect caused by the solenoid effectively removing a line from $\mathbb{R}^3$, making it homotopic to the circle $S^1$. It has nothing to do with the electron "acting like a wave" or "acting like a particle" (which are notions one should not use anyway), ...



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