New answers tagged

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You are simply forgetting that the first $\vec{r}$ and putting $\hat{r}$ in it's place. Remenber: $\vec{r}=r\hat{r}$ and $\hat{r}=\dfrac{\vec{r}}{r}$. A simple dimensional analysis shows that it`s $\dfrac{1}{r^4}$ the correct proportionality. Just look the first line. Since $\hat{r}$ is dimensionless and $[\vec{r}]=[r]=$meter ([X]= dimension of X) we have ...


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There is an $\vec r$ factor in the front of the first big expression and $\vec r=\color{red}{r}\hat r$.


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It is easy to show, using the Lorentz force, that the work done per unit of volume and time on free charges by EM plane waves is $W = \mathbf {E.J}$, where $\mathbf E$ is the electric field of the wave and $\mathbf J$ is the density of current. The magnetic field contribution for the Lorentz force cancels out, and doesn't contribute for the work. Every ...


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This is my second answer. Don't worry, it starts diverging from my first answer at some point. Let's first consider two astronauts hovering near the event horizon of a black hole. When they communicate, using laser beams, they notice that they must aim the beams almost straight up, and that there is almost no delay, the beam is very fast. (In their ...


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If I said you had a cannon with a fixed charge in it and you wanted to know the angle required to hit a specific target, would you be surprised to find it's a quadratic with two solutions? Near 45 degrees, the ammo overshoots the target. Near 90 and 0 degrees, the ammo does not reach it. But at two points between, the target is reached. A similar process ...


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Maxwell's theory has been validated by the entire body of physics since publication.


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The Ampere's law is valid only for steady currents and so when you use the Ampere's law in between the loop and piece of wire, you won't get the correct result because the current in the piece of wire isn't steady. You see charge keeps accumulating on one side of the wire and hence the current isn't steady.


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We can't apply ampere circuital law in case of finite length wire . Ampere circuital law is applied in case of a wire with infinite legth because infinite wire is considered to be a closed circuit . Finite wire is not a closed circuit and the is "Ampere Circuital law" so it can be applied in a closed circuit only .


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The entire theory (quantum electrodynamics) used by Uehling to derive this potential, is based on the assumption that the electron is a point particle. So the mainstream interpretation of the extra charge density “outside” the electron is that it is polarization of the vacuum by the point electron. Any other conclusion contradicts the premises of the theory.


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A linear dipole antenna (thickness << length AND thickness << wavelength) is sensitive to the electric field of the EM wave, a circular loop antenna (thickness << diameter AND thickness << wavelength) is sensitive to the magnetic field of the EM wave.


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Let's consider an equivalent example: The mass density of water is $\rho_m \approx 1kg/liter = 1000 kg/m^3$. We are able to calculate the mass of the volume $V$ of water, but in order to do so, we have to know the volume, $$ M = \int \rho_m dV = \rho_m V $$ The second equation is true if the mass density is constant over the volume. Just transferring mass ...


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Though user BowlOfRed has provided an excellent answer clarifying the flaw in your argument, here's my take at trying to provide a visual reasoning as to why this idea might not work. First of all, the direction of the forces in your illustration which comprise the couple aren't in the correct direction. Here I'm assuming your "very good" disk has ...


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So it is said that, as the areas $\delta A_1$ and $\delta A_2$ shrink together, the total charge remains finite. But why would the two areas not shrinking together cause the total charge to become infinite? This isn't what they're saying. The misreading you've done here is analogous to the following: "As $x \to 0$ in $(\cos x)/x$, the numerator remains ...


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This idea has popped up ever since magnets were first discovered and reappears at regular intervals since then. It is the so-called magnetic motor and you'll find lots of mostly worthless writings about it on the web. The proof I recall about why it is in principle impossible to make one of these relies on the fact that the divergence of the magnetic field ...


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Let's first consider two astronauts hovering near the event horizon of a black hole. When they communicate, using laser beams, they notice that they must aim the beams almost straight up, and that there is almost no delay, the beam is very fast. (In their accelerating frame the speed of light somewhere above them is billion cees) An outside observer would ...


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Thus, EM waves would not spread spherically anymore, but only towards the singularity This is not true because inside the horizon "radially inward" is the only possible radial direction but the metric is still spherically symmetric, so the restriction is only abut the radial coordinate.


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Does the radio (between two co-moving astronauts) stop working when crossing the event horizon? Assuming that the black hole is massive enough that there are negligible tidal effects at the horizon then their radios would continue to work and their conversation would carry on without a pause. Now there are others, who suggest that inside the horizon, ...


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You are incorrect that there is a torque on the wheel. While the field from the permanent magnet may pull a bit on some sections of the wheel, the total pull averages to exactly zero. Yes, some of the bits of the wheel want to move clockwise to be closer to the magnet, but some other bits of the wheel want to move counterclockwise to be closer. We don't ...


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"H is "opposed" to M, hence the "demagnetizing field"." H is NOT a demagnetizing field. Only B acts on the atoms to polarize them. H is only used to find B. Any 'demagnetization' that occurs is because B is half as large at the end as at the middle. H being opposite to B and M is an end effect. Because B gets weaker toward ...


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If you are asking for the name of the squared ring of iron that connects the two coils in a core-type transformer, it is called the “yoke.” The iron yokes together the two coils, like a wagon yoke held together the two oxen pulling a wagon.


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The infalling observer who is free falling with negative escape velocity v=-c√(rs/r) will receive redshifted signals from the far away observer all the way down to the singularity (if he falls in with less than the escape velocity the signal he receives might as well be blueshifted). The far away observer will receive redshifted signals from the infalling ...


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This mathematical shape is a particularly simple example of a toroidal polyhedron.


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That can be called a core-type transformer (with a single window) in contrast to the shell-type transformer. Note also that those types of cores are also constructed with laminated materials to reduce eddy current losses. The linked definitions from the Electrotechnical Vocabulary of the International Electrotechnical Commission (IEC) give the above terms as ...


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Although sound waves and electromagnetic waves are both waves these phenomena are nothing alike. The major difference is that sound is a mechanical wave, while an electromagnetic wave (=light) is non-mechanical. Think of a mechanical wave as "many" harmonic oscillators (=springs) which are coupled. The sound disturbs the equilibrium position of ...


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The speed of a wave is determined by the medium in which it propagates. There are actually two types of 'motion' here. In the presence of a wave, the medium 'oscillates'. The pressure or Electro-Magnetic (EM) field go up and down for sound and light respectively. The speed of these oscillations (more precisely their period) is determined by the wave ...


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Sound waves are not in the electromagnetic spectrum. You should hence be careful about comparing the two. Sound waves always need a medium to propagate in, and hence their speed always depends on the medium. An EM wave (light) can also propagate in vacuo, where its speed is a constant (and it’s invariant). This is the fundamental postulate of special ...


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Yes, the magnetic force will be zero, but an electric field will be present in the moving frame to make sure the same force is present in both frames.


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Maxwell's equations place a constraint on the current, namely that it be conserved. To see this, take the divergence of Ampere's law for $$0 = \mu_0 \nabla \cdot \mathbf{J} + \mu_0 \epsilon_0 \nabla \cdot \frac{\partial \mathbf{E}}{\partial t}$$ which is equivalent to $$\nabla \cdot \mathbf{J} = - \epsilon_0 \frac{\partial}{\partial t} (\nabla \cdot \mathbf{...


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Does this imply that any arbitrary time evolution of current density can be defined beforehand, and the corresponding fields always found that satisfy Maxwell's equations? Yes, given a charge density $\rho(\mathbf r,t)$ and a current density $\mathbf J(\mathbf r,t)$, you can find fields $\mathbf E(\mathbf r,t)$ and $\mathbf B(\mathbf r,t)$ satisfying ...


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Nuclei, like molecules, have vibrational states, but the vibrational character of these states is not as pure for nuclei. You can model them as shape oscillations (usually ellipsoidal deformations about a spherical or ellipsoidal equilibrium shape), but in reality they're usually more like a slightly coherent superposition of a few different particle-hole ...


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There isn't a derivation; it is a definition of the B-field.


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Maxwell's equations describe the electromagnetic fields, but they make no statement about force. Although you can find crackpot papers claiming to "derive" the Lorentz force law, it is actually necessary to have a separate equation to relate the fields to the force.


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The claim that it could rip apart the sun is totally false. This is equivalent to a pretty standard textbook problem: what happens to a conducting sphere when placed in an electric field? Sure, at first the protons and electrons move relative to each other, but that builds up a field that counteracts the applied field extremely quickly. In practice, the ...


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You want to compare the strength of the two fields. If you place the galvanometer so that it's field is at right angles to that of the earth, then the ratio of the two gives the tangent of the angle measured by the compass. (Note that the compass needle is balanced and mounted so that it only responds to the horizontal component of the earth's field.)


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The answer is 'yes, sort of.' The approach isn't necessarily to get two identical wavelength photons to be absorbed simultaneously, but rather to have one photon "bump up" an electron to a higher orbital, then the second photon kicks the electron free before it can reradiate back to ground state. In fact, 2-photon absorption is an effect used for ...


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Where did the negative sign go? It's simply a matter of notation: your textbook starts talking about electron because indeed they are the particles that move in a wire; but as you probably already know the convention is to think about currents of positive charges and not negative ones. So when the book starts talking about currents it readily flips the sign, ...


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Every magnetic object must have some magnetic potential for my magnet That scientifically accurate but this type of force is so rare and significant compared to the other forces like gravity, electrostatic forces that its effect will be negligible. A good way to think about this, did you ever feel or see Alpha Centauri's gravitational pull or even the Sun'...


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I am no expert in how computers exactly work, but I have read that computers release some radiation. It is nothing to worry about, it won't do any harm, but I think that may be causing some static charge in your clothes, especially since you sit in front of the computer for long times. That is why they make the sounds.


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Because while the magnet drops, it induces currents in the copper pipe, and in turn produces a magnetic field that exerts magnetic force opposing the movement of the magnet. Like air resistance, this magnetic force will eventually be as strong as gravity and thus net force on magnet becomes zero so the magnet reaches a terminal velocity check out this video ...


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If i get the picture correctly, then the problem is that the metamaterial will also respond to changes in the EM field. That's in fact what makes it slow light down. So it doesnt matter how fast you change the field, your system will excert an opposite force not allowing it to move forward.


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Let me start by saying it's a very difficult mathematical job to calculate the force between macroscopic magnets. See this Wikipedia article. But I'll try to explain in words why conservation of energy isn't violated as you ask in the question body. Say we start with one magnet. Now say we put a lot of magnets (or substances that react to it by getting ...


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if all electrodynamics experiments are done in earth frame how do we know that the speed of light 𝑐 we calculated from it is not 𝑐 with respect to the earth? First, not all electrodynamics experiments are done at rest with respect to the earth frame, nor necessarily close to the earth. However, for the sake of discussion let’s limit ourselves to ...


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well my friend you seem to be committing a very trivial mistake in perceiving the $'r'$ in the charge density and the $'r'$ in the denominator of the potential as the same! so to avoid the confusion use different notation for the problem as described in the figure above !


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Yes, the voltage a distance $r$ from a point charge $q$ is $$V=\frac{1}{4\pi\epsilon_0}\cdot\frac{q}{r}$$ where we have set $V=0$ at $r\to\infty$. Therefore, if you want to determine the potential at some location $\mathbf r$ of a charge distribution with volume charge density $\rho=\text dq/\text d\tau'$, we treat each little charge element $\text dq=\rho\,\...


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Even in the case of total internal reflection there is evanescent leakage of the optical field into the less rare medium. The further above the critical angle you are the smaller this effect. The light penetrating into the cladding in this case is due to this “normal” evanescent leakage and not due to something like over bending the fiber. This leakage would ...


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One minor thing that will help is that no object has potential energy on its own. Potential energy comes from the structure of how objects are found in space. We do indeed sometimes say "this object has gravitational potential energy," but what we really mean is that there is potential energy in the system of the Earth and the object. It's not ...


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You can ask the same question about gravity or electric charge. Where does the asteroid get its energy from when it is captured by the earth's gravitational field and crashes into the earth? The asteroid gains its kinetic energy from its gravitational potential energy and similarly the magnet gains its energy from its magnetic potential energy. How did this ...


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All three of the expressions are "derived" from experimental facts. You can in principle try to derive them from Lagrangian dynamics, but that would require you to assume a Lagrangian, which in the first place was written down so that it reproduces the experimentally observed gravitational and Lorentz forces. Another way in which one could arrive ...


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The OP is not asking this question, but the title is misleading, so I am answering the question of the title. The use of complex scalars in any scientific measurement is a convenience based on the fact that the complex numbers are the only two-dimensional scalar field. In the case of electromagnetic energy, one can calculate the phase angle and the ...


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