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3

Ok I'll give this one a go - the reason is that the effect is just not strong enough. Andre Geim showed that in a strong enough magnetic field you can levitate a frog. Frogs are mostly water and water is diamagnetic. You can also buy a kit from your local electronics store that has a piece of (strongly diamagnetic) pyrolytic carbon and a neodymium magnet ...


0

I'm confused what you are trying to reconcile. The magnetic field is given by the Biot-Savart Law: $$ d\vec{B} \sim \frac{Id\vec{s}\times \vec{r}}{r^3} $$ where $$ I = \int \vec{J}\cdot d\vec{A} = \int \rho \;\vec{v}_{\text{drift}}\cdot d\vec{A}$$ So taking the Galilean transformation $\vec{v}_{\text{frame}} = \vec{v}_{\text{drift}}$ leads to: $$ ...


0

I think you have the right approach for the non-grounded. For the grounded case, we can use the uniqueness theorem, which says that given a charge distribution and the voltage on the boundaries, there is only one solution for the voltage. The grounded sphere has a $V = 0$ surface at $R + \delta R$ and at infinity, and no charge outside. I can solve this by ...


1

We have, for a point charge $q$ at position $\vec r(t)$: $$\rho(\vec x, t) = q\delta^3(\vec x - \vec r(t))$$ $$\vec J(\vec x, t) = q \frac{d\vec r}{dt}\delta^3(\vec x - \vec r(t))$$ Let us for now work without worrying about what the derivative (more precisely, gradient) of the delta function actually is. We will also enforce the convention that $\vec\nabla$ ...


1

The Gravitoelectromagnetic equations are exactly the same as Maxwell's equations with $\epsilon_0$ replaced by $(-4\,\pi\,G)^{-1}$, so, to the extent that the GEM equations approximate the Einstein field equations, the behavior of mass is very much like that of electric charge. Here are the differences: The minus sign in the "gravitoelectric constant" ...


1

Maxwell's equations of electromagnetism don't allow an electromagnetic wave to just sit there without moving in empty space (in the absence of electric charges). An electric field is created by a magnetic field changing in time, and vice versa, so the coupled fields have to be changing in time (i.e. propagating) in order to exist and sustain themselves.


1

$$\frac{\nabla'\times \vec{J}}{R}=\nabla'\times(\frac{\vec{J}}{R})-\frac{\vec{J}\times(\nabla'R)}{R^2}$$ Thanks to Prof. Y. F. Chen I was able to figure it out. While in the integral the first term on the RHS can be converted into a surface integral as below: $$\int\nabla'\times(\frac{\vec{J}}{R})d^3x^{'}=\oint(\vec{n}\times\frac{\vec{J}}{R})d^2x^{'}$$ ...


1

For the moving wire situation in which $d\Phi/dt=Blv$, the wire is typically part of a larger closed loop of conducting material (a circuit). The change in flux is due to the changing area of this loop, which is immersed in a uniform magnetic field. Here's a plagiarized image illustrating this:


0

The spin of a particle flips under time reversal. It has nothing to do with the phase of the particle! You may go through time reversal symmetry for more information about spin flipping.


1

In my opinion, your solution is correct. The book's solution is correct as well, but you would need to draw a different diagram for it to make sense. Their assumption is to take by default, r as lying on the positive x axis. In this way, the distance between the $Q_2$ and the required point is the mod of $r-2$. Since this expression is being squared, it ...


1

Pay close attention to how you defined $x$: it's not the coordinate of the desired point on the $x$ axis, it's how far left of the positive charge that point is. So when your equation tells you $x = 4.83$, that means the desired point is $4.83$ units left of the positive charge. But, presumably the question you've been asked wants the coordinate of the ...


0

"The Barnett effect is the magnetization of an uncharged body when spun on its axis. It was discovered by American physicist Samuel Barnett in 1915... The magnetization occurs parallel to the axis of spin. Barnett was motivated by a prediction by Owen Richardson in 1908, later named the Einstein–de Haas effect, that magnetizing a ferromagnet can induce a ...


0

hahaha Actually, EMF is always applicable, even in a blackhole. All stars have perfect symmetry, to a degree, gravity is always trying to crush the star and thru fusion SNF / the EMF electrons repel this force outward causing this balance. Now you can have GRB's and the formation of blackholes if the star is large enough or if not Neutron stars or even ...


1

I'm not sure what you mean by macroscopic? The experiment was first done with silver atoms which, due to having 1 unpaired electron of spin 1/2, split into 2 distinct beams. This was a quantum mechanical event that was visible macroscopically. If by macroscopic you mean on a large object instead of a beam of atoms then I don't think it would be feasible or ...


0

Well, you use the amplitude $|\mathbf E|$ of the electric field to find the total amount of energy carried by the electromagnetic wave; the root-mean-square magnitude of the Poynting vector for an EM plane wave is $$ \left< S \right> = \frac{\epsilon_0 c}2 |\mathbf E|^2 . $$ You seem to be after a way to specify the degree of elliptical polarization, ...


3

What do you consider to be Ampere's law? I think the RHS is not the current that cuts some plane, it is actually a surface integral of the current density, over any surface that is bounded by the loop. In other words it does not matter what surface you define ; so long as it is bounded by the loop, Ampere's law works just fine. So usually, you choose the ...


1

In the case of a planar loop, the surface that it bounds does not have to be a plane. It could bubble out. Ampere's Law is still valid (assuming the other conditions for validity are met). Ampere's Law applies to any loop, and any surface bounded by the loop. I guess I'd better add that I don't know what happens in pathological cases where, for example, ...


0

The difficulty of the prior experiments consisted in isolating the relativately weak gyro magnetic effect against the background of the purely magnetic forces acting on the studied rod... In order to avoid this difficulties, in the variant of the experiment proposed by Einstein, the magnetic field of the coil acts on the iron rod ... for a very short ...


2

Yes ... but let's be careful to understand that the sensation of touch is a psychophysical phenomenon. The electrons at the surface of an object "push" against the electrons at the surface of your fingers. The electrons never touch each other. Your skin deforms a bit, and the nerves in your fingers detect this deformation, and send a signal to your brain. ...


0

Both voltmeters are indeed connected to the same endpoints, but if the system contains changing magnetic fields then it is no longer true that this implies that they must read the same voltage. Instead, their difference $V_2-V_1$ measures the line integral of the electric field along the outermost wire loop (i.e. the one including both voltmeters and no ...


2

Using a process called interference, we can find wavelength, because the way that waves interfere is reliant of wavelength. Interference is based off of two key principles of waves: they are made up of peaks and troughs. When troughs overlap, they go lower. When peaks overlap, thy go higher. When a peak meets a trough, they cancel. Of course, the positions ...


0

Yes light can change propagation direction when scattering off of a grating into pre-defined directions that depend on the angle of incidence and the grating period (as well as the refractive medium the grating resides in). This is just Bragg's Law. The change in direction (or momentum) is determined by the above parameters. You can think of the change in ...


1

This answer contains some additional resources that may be useful. Please note that answers which simply list resources but provide no details are strongly discouraged by the site's policy on resource recommendation questions. This answer is left here to contain additional links that provide little or no commentary. Staelin's Electromagnetic Waves. This is ...


1

While your assumption of $J$s cancelling out in the intersection is valid, your conclusion that $B_1$ and $B_2$ are along the same line only holds along the line joining the centers of the circles. More generally, that is not the case; we can get a more complete solution (which also correctly takes care of the directions) by using the vector quantities ...


0

In the dynamic case (current flowing), the redistribution of the electrons takes work to overcome resistance. As you know, $V = I \cdot R$ and $E = \frac{dV}{dx}$. There needs to be an electric field in order for current to flow - and the current flows because of the potential difference. You are right that the electrons will try to flow to cancel the ...


2

It's cause $\mu_0$ is not part of the displacement current density.


2

Actually you can't talk about poles here. The pole model assumes two hypothetical opposite poles as north(+) and south (-) analogous to electric charges just for an analogy to the Coulomb's law for the H-field of magnets. This model doesn't explain magnetism produced by electric currents. Ampere hypothesized that all magnetic fields are due to electric ...


0

To all of your question: Yes you are right. Lorentz force of moving charge in magnetic field is based on the electron’s magnetic dipol moment. The magnetic field align the electron's magnetic dipol moment in the direction to this field. The motion of the electron undergoes a - predictable and perpendicular to the two vectors of the velocity and the magnetic ...


1

Some distance away! Honestly. See Rod Nave's hyperphysics, and take a look at this picture: This is like a very short solenoid, the North pole is on the left, the South pole is on the right. Your straight current-carrying conductor is merely one section of the loop. You might think you don't have a loop, but you do. It might not be a neat and tidy ...


2

A laser is just a thin slice from the spectrum of light. Is it more efficient compared to the visible spectrum of light? It depends on the frequency of the laser and how efficiently the solar panel can turn light of that frequency into electrical energy. If a solar panel would operate better/best with light of a certain frequency, using a laser with that ...


0

They can be made of any shapes. Shapes usually we see are common because of the worldwide standard that is useful in installation and efficiency. :)


0

It can. Just search online and you can buy magnets as tubes, balls, cubes etc.


0

Okay, I think I'll answer your question with a paper, basically I'm doing a dissertation on Gamma Ray Bursts and Gravitational Waves. Look for a paper called 'Gravitational Wave Memory from Gamma Ray Bursts’ Jets' It's not mine, I'm referencing it. Basically I think the answer is yes. You will get GWs in your direction if a jet from a gamma ray burst is ...


0

When comparing light waves and sound waves in this fashion, we need to consider what is waving. In a sound wave, the position of air molecules are waving. In a light wave, the strength and direction of the electromagnetic field is waving. This does not exert any force on air molecules (actually it does, but that force is so small, and the frequencies are ...


1

Look at the original, Heinrich Hertz, 1889! If E=B=0 on a plane, S=0 => no energy Transport through that plane. I agree with Annix: "The pictures may be depicting a static wave, but certainly, not a propagating wave."


0

Why doesn't current induced by changes in flux affect the flux, while current induced by a battery does? Because electromagnetism features a "screw" mechanism. Take a look at Minkowski’s Space and Time: "In the description of the field caused by the electron itself, then it will appear that the division of the field into electric and magnetic forces is a ...


1

If we put a magnetic dipole, $\mu$, in a uniform electric field of strength $B$ then the torque on the dipole is given by: $$ T = \mu \times B \tag{1} $$ where the $\times$ symbol indicates the cross product. If the angle between the dipole and the field is $\theta$ then equation (1) can be written as: $$ T = \mu B \sin\theta \tag{2} $$ For small angles ...


0

I think you may have a slight misconception regarding the actual meaning of an electromagnetic wave. Electro-magnetic waves are thought to be functions of probability of finding a photon at a certain point in space. Read some information on the photoelectric effect this may provide some more clarification.


0

Faraday's law states that $V = - \frac{dF}{dt}$ where $F$ is the overall magnetic flux due to all sources through the area bounded by the loop. As you point out, it is precisely because of this that inductance etc. make sense. You may have likely encountered examples in your textbook where the rate of change of magnetic flux due to an external source is ...


0

Light does become polarized in a magnetic field. The magnetic field of a black hole was detected due to the polarization of light. Check this article: http://www.iflscience.com/space/black-holes-powerful-magnetic-field-observed-first-time


1

The square bracket transformation This is just the application of chain rule. The LHS means a derivative over the primed spacial coordinates while keeping unprimed spacial and time coordinates fixed. $$\nabla'[ \rho(\mathbf{x'},t')]_{ret} = \left(\sum_i \frac{\partial }{\partial x_i'} \hat{i}\right)[\rho(x_i',x_j',x_k',t')]_{ret}\\$$ But the $\rho$ is a ...


1

What is potential energy truly? It depends on the circumstances. When you compress a spring it's stress in the bonds or electromagnetic field between the atoms. IMHO at the fundamental level it's essentially spatial stress. That might sound unfamiliar, but it shouldn't, because the stress-energy-momentum tensor "describes the density and flux of energy and ...


2

You assume that the screen is perfectly absorbing when you use this field. By uses this field you are assuming that the electromagnetic field pass through the material without reflection. Furthermore, you are assuming that you don't have field in the another side of the surface. $$ \textbf{E}=E_0e^{i(kx-\omega t+\delta)} \hat{\textbf{y}}, \ \ x<0 $$ $$ ...


3

This is called Helmholtz theorem, which states that for any vector field $\vec{F}$ that is twice continuously differentiable in a bounded domain, we can perform the decomposition $$ \vec{F} = - \vec{\nabla} \Phi + \vec{\nabla}\times\vec{A} $$ See http://en.wikipedia.org/wiki/Helmholtz_decomposition for a derivation


1

Consider a system of stationary point charges $q_i$, with position vectors $\mathbf{r}_{i}$ in an external electric field, assuming a value $\mathbf{E}_{i}$ at the position of the $i$th charge. The net torque on this system, about the origin of coordinates is: $$\mathbf{\tau} = \sum_{i} \mathbf{r}_{i} \times (q_i \mathbf{E}_{i})$$ or $$\mathbf{\tau} = ...


1

Let's say the wire is in the $\overrightarrow x$ direction, while the applied magnetic field is in the $\overrightarrow z$ direction. The Lorentz force law $\overrightarrow F=I\overrightarrow v\times \overrightarrow B $ tells us that the the force will be in the $\overrightarrow y$ direction. The electrons in the wire will incur a net force in the ...


0

Ah, yes, I think I got an answer. With moving particle energy also moves in the same direction. It was strange for me that we have $r$-component of $\vec{S}$ but it seems to be normal.


0

You should understand how potentials work and how they relate to forces. It is not as simple as saying that information travels at the speed of light. As I will show below, potential can change instantaneously everywhere (but that doesn't mean that information is traveling faster than light. Lets begin with the Maxwell's equations: $$ \nabla \cdot ...


-4

Sonic booms happen when an object crosses the sound barrier. Light leaves its source at the speed of light, so it never creates the effect, even in a circumstance where the light is strongly interacting with the air, like when there is fog.


21

There are many differences between light and sound waves noted in other answers, such as the impossibility of any object with nonzero rest mass reaching lightspeed. However, there is one likeness that I don't think has been noticed yet and that is the following: a sound wave travelling at the speed of sound does not make a sonic boom! This is because the ...



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