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How to calculate the magnetic dipole moment of a bar magnet?

A bar magnet is not a pure magnetic dipole, but we can calculate the dipole moment by comparing the magnetic fields of a bar magnet and a magnetic dipole at large distances where contributions from ...
David Bailey's user avatar
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$xy$-symmetry of the electric potential: Charged sphere at an external uniform electric field

The conducting sphere has an induced charge distribution with no net charge, and the external electric field is uniform and in the $\hat z$ direction. So the resultant field is such that it looks like ...
Albertus Magnus's user avatar
1 vote

Time for ferromagnet to align with magnetic field

For large polycrystaline ferromagnets, the magnetization response to a high strength external magnetic field is usually limited by eddy currents. For magnets that are very small or have very low ...
David Bailey's user avatar
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14 votes

Could relativity be consistent if there are multiple light-like fields with different invariant speeds?

No, there cannot be two different invariant speeds, because an invariant speed is a feature of mechanics, not just of the electromagnetic (or other) field. For example, given an invariant speed $c$ ...
Eric Smith's user avatar
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Writing the most general form resistivity tensor

You need to write the vector equation $$\vec{E}=\frac{1}{ne}(\vec{j}\times\vec{B})+\frac{m}{ne^2\tau}\vec{j}$$ in components. This gives $$\begin{pmatrix}E_x\\E_y\\E_z\end{pmatrix} =\frac{1}{ne}\...
Thomas Fritsch's user avatar
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Imaginary part of permittivity tensor

We can split the permittivity $\epsilon$ into to part 1) refractive part $\epsilon_r$ and 2) absorptive part $\epsilon_a$ as follows: $$\epsilon = \epsilon_r + i\epsilon_a (1)$$ permittivity ...
Sancol.'s user avatar
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1 vote

Calculating mechanical power of an electrical motor

Your mistake: You calculated your resistor with U/I. So you assume there is an resistor that eats all the current of the drill. Certainly then its power is U*I and then clearly your calculation ends ...
mond's user avatar
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2 votes
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Vector Magnetic Potential and pointwise current density source

The notes I'm following proceed to define the current density as: $$\mathbf{J}=\delta(x)\delta(y)\delta(z)\hat{\mathbf{a}}$$ which I totally agree it only exists at the origin, but my question is: ...
Thomas Fritsch's user avatar
1 vote

Calculating mechanical power of an electrical motor

This is an ill-posed question. First, some useful background: A drill motor's mechanical power rating is its torque times its RPM at its design point under load and is expressed in horsepower where ...
niels nielsen's user avatar
1 vote

Charge moving in an electric field and magnetic field?

You appear to be assuming that $\vec{v} = \int \frac{q}{m} \vec{E} \, dt$ and then plugging this into the Lorentz force law. But this isn't valid unless $\vec{F} = q \vec{E}$, i.e., there is no ...
Michael Seifert's user avatar
3 votes
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How to calculate the time derivative of electromagnetic field?

As the problem is posed, this is impossible. just knowing the spacial variation of an electric field at a specific moment in time does not mean you have any knowledge of its time variance. I assume ...
jensen paull's user avatar
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Dipole Moment between two charges of different magnitudes but opposite signs

It should be -2q*(d/2) + q*(-d/2) = -3q/2d Assuming the origin as the center of the line joining -2q and q and positive r vector to the right of origin. Also i have assumed the +q charge to be left to ...
Abhishek's user avatar
1 vote

Charge Distribution and Stability in a Conductive Solid Sphere

In a static state, there can be no charge inside a solid conductor. This means the charge at the center will be in unstable equilibrium. An infinitesimal displacement of the charge will produce a ...
Jerrold Franklin's user avatar
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Electromagnetic Dipole Radiation Derivation

It's more complicated than you think. You have to use ${\bf E}=-\nabla\phi-\frac{1}{c}\partial_t{\bf A}$ BEFORE any approximation. With everything at the retarded time, this is a difficult derivation. ...
Jerrold Franklin's user avatar
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How can I estimate the force of an electromagnet?

The wikipedia article on Electromagnets might help you: https://en.wikipedia.org/wiki/Electromagnet The forces are highly dependent on the geometry. But if you have core then this will mostly ...
mond's user avatar
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1 vote

How does capacitance of a capacitor not depend on the distance of the capacitor to a battery?

You are correct: the capacitance does depend on the geometry of the wires connecting it. We commonly account for that extra capacitance as a separate "stray capacitance" in parallel with the ...
John Doty's user avatar
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1 vote
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Conserved current from a symmetry

Let me do the calculation just for the vectorial current to show you the derivation. The fields transformations reads: \begin{eqnarray} \psi &\rightarrow& e^{i\alpha} \psi, \\ \bar{\psi} &...
T. ssP's user avatar
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How does capacitance of a capacitor not depend on the distance of the capacitor to a battery?

For a plate configuration of a certain capacitance, C, the charges on the plates are determined just by the value of $C$ and the applied potential difference, $\Delta V$, between the plates. $Q=±C\...
Philip Wood's user avatar
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1 vote

Could you calculate the force between two NON-PARALLEL, straight current carrying wires?

Assuming we already know how to calculate the $B$-field of one infinite straight wire, we only need the integral of its force over the second wire. Without loss of generality we can assume that the ...
Jos Bergervoet's user avatar
-1 votes

Charge Distribution and Stability in a Conductive Solid Sphere

Adding that there is originally a charge Q2 on the sphere, per the OP's edit: If you put a charge Q1 in the center of the conductor, either perfectly centered or just hold it in place so it can't ...
BaddDadd's user avatar
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Current density of moving charge distribution - mobile charge density vs. "ordinary" charge density of the distribution?

Perhaps considering an example will help. You apply a voltage across a metal wire. The wire itself remains neutral, so the charge density is zero inside (or very nearly so). Yet, the current density ...
Puk's user avatar
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1 vote

Current density of moving charge distribution - mobile charge density vs. "ordinary" charge density of the distribution?

It's less complicated than you think. "Mobile charge density" refers to the portion of the charge density that is due to charge that is actually moving. If the total charge density $\rho$ at ...
Brian Bi's user avatar
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1 vote

Linear momentum of an EM wave

For an entity that travels at the speed of light, the rest mass is defined to be zero. However the quantity $m$ in the above derivation is not the rest mass, but the "relativistic mass" ...
Brian Bi's user avatar
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-1 votes

Relativistic mass and relativistic charge

Why won't he perceive any alterations in the particle's charge? This is assumed because electric charge is operationally defined (practically determined) by the electric field it creates, e.g. via ...
Ján Lalinský's user avatar
3 votes
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Doubt about the derivation of Liénard-Wiechert Potentials?

You're right that some problems could arise for a kind of circular dependence of the variables involved. Let's get some partial results first and then put everything together: gradient of the ...
basics's user avatar
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2 votes

Current density of moving charge distribution - mobile charge density vs. "ordinary" charge density of the distribution?

Is easier if you write the equations in integral form: $$ I = \iint_{\Omega} \mathbf J \cdot d \mathbf A $$ $$ I = \int_{C} \mathbf J_s \cdot d \mathbf \Gamma $$ As you might know, from calculus, the ...
Álvaro Rodrigo's user avatar
5 votes
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Why does $\oint_C \vec{E}\cdot d\vec{\ell}=0$ imply $\nabla\times \vec{E}=\vec{0}$?

The first equation is only for all closed loops, not for all contours. That’s why you can’t conclude $\vec{E}=0$. The only thing you can conclude is that $\vec{E}=\text{grad}(V)$ for some function $V$....
peek-a-boo's user avatar
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2 votes

Why does $\oint_C \vec{E}\cdot d\vec{\ell}=0$ imply $\nabla\times \vec{E}=\vec{0}$?

For the very reason you say at the end of your question: while the second relation holds for all surfaces, the first one only holds for all closed loops, and not any arbitrary line. Let's do some ...
basics's user avatar
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0 votes

Differences in the direction of energy and wave propagation in anisotropic media

I recently asked myself the same question. The direction of the wave vector $\vec{k}$ does not coincide with the direction of the Poynting vector $\vec{S}=\vec{E}\times\vec{H} $ in an anisotropic ...
MauvaiseFoi's user avatar
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How exactly classical electrodynamics fails to explain the Compton effect?

In classical EM theory, the scattered radiation has the same frequency as the primary wave when the electron oscillates around a central point in space that is at rest. Frequency of the scattered ...
Ján Lalinský's user avatar
1 vote

Repulsive force between electrons at relativistic speeds

The Lorentz force law is already a fully relativistic equation and Lorentz covariant, which is not too surprising when you consider the Lorentz transformations of special relativity are named after ...
KDP's user avatar
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12 votes
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Relativistic mass and relativistic charge

The way we deal with electric charge in relativity is to introduce it as part of a complete set of ideas about electromagnetic fields and charged bodies. You are quite right that in this formulation ...
Andrew Steane's user avatar
6 votes

Is it plausible that Earth's magnetism can induce heart attacks by affecting red blood cells?

Is it plausible that Earth's geomagnetism causes the rate of heart attacks to double? No, it is not even remotely plausible. First, a doubling in the rate of heart attacks is huge. You would expect ...
Dale's user avatar
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1 vote

Is it plausible that Earth's magnetism can induce heart attacks by affecting red blood cells?

I am going to say no that it is not plausible or likely simply because every single cell in the human body has a cellular wall around it that acts as an insulator for the cell itself and even though a ...
Richard Morgan's user avatar
1 vote
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Understanding radiation mechanism of inset fed microstrip antenna

There are two apertures (you called them slots) radiating, these are fed approximately $\lambda_g/2$ apart therefore the charges on the metal that induce the electric fields parallel with the aperture ...
hyportnex's user avatar
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3 votes

Time for ferromagnet to align with magnetic field

Well, a bunch of factors come into play. First, you've got the material itself. Different ferromagnetic materials have different properties that affect how quickly their magnetic moments can rearrange ...
Testina's user avatar
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3 votes
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Is it possible to build electrical space engine based on Electromagnetic induction

Assuming radiation losses are negligible this will not provide any net propulsion. The momentum gained by pushing the magnet will be lost by stopping the magnet. And the distance moved will be ...
Dale's user avatar
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0 votes

Variations in Refractive Index of Materials

Refractive index defines the change in speed for propagating the light through the material relative to propagating the light in vacuum. $$ n \equiv \frac{c}{v}$$ Higher values indicate slower ...
Jeffrey J Weimer's user avatar
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Variations in Refractive Index of Materials

TL;DR Refractive index is determined by the way electrons interact with incoming electromagnetic waves. Impurity atoms in glass change the way they interact. Pure materials have principly known, ...
Jono94's user avatar
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24 votes

On a tinted (reflective) window, why do I need to look from up close to see inside?

Things are called a one-way mirror when the amount of reflected light overwhelms the amount of transmitted light. When you put your head up to the glass (and maybe put your hands up surrounding your ...
BowlOfRed's user avatar
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2 votes
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Reason Why 1 Coulomb was redefined in SI unit system?

Why definition of SI unit of charge was moved from that one related to current strength ? Because the new standard is more accurate and stable than the previous standard. Can it be because time is ...
Dale's user avatar
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0 votes

Interaction of two orthogonal magnetic fields

I assume that you are trying to understand the quote that you wrote. Yes, it is true that for a magnetically staurated material in the $x$-direction will still affect the $y$-direction. This is ...
Oscarcillo's user avatar
2 votes
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Energy of Monochromatic Beam of Light

You've confused several things: Photons. 𝐸beam=𝑁ℏ𝜔, N is an average number. If you measure the number of photons within a given time interval using a photo-multiplier tube, T, then you'll find ...
JQK's user avatar
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-1 votes

How to distinguish Coulomb and radiative parts of the electromagnetic field?

If a distinction ought to be made between "bound" and "free" fields, it is at a purely technical level (e.g. when applying asymptotic analysis to radiation by accelerating point ...
TLDR's user avatar
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0 votes

Graphical interpretation of complex electric fields

I will not return to the well-known interest of complex representation for scalar quantities and I will insist in a little more detail on the representation of vector quantities. I start from the ...
Vincent Fraticelli's user avatar
-1 votes

Energy of evanescent wave in total internal reflection

Those are two different approaches to look into the problem. We know that the reflective coefficient in total internal reflection is 1, which means that all energy is reflected. This is true in the ...
ondas's user avatar
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2 votes
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For these gauge transformations in electromagnetism, $\phi\to \phi-\partial_t \lambda$ and $\vec A\to \vec A+\nabla\lambda$, why do the signs differ?

We already know the 4-position is $x^\mu = (t, \vec{x})$ and the 4-gradient is $$\partial_\mu = \frac{\partial}{\partial x^\mu} = \left(\frac{\partial}{\partial t},\vec{\nabla} \right) \tag{1a}$$ or ...
Thomas Fritsch's user avatar
1 vote

For these gauge transformations in electromagnetism, $\phi\to \phi-\partial_t \lambda$ and $\vec A\to \vec A+\nabla\lambda$, why do the signs differ?

The perhaps underwhelming answer is, it doesn't matter, but you need to stick with the convention that you choose (when raising and lowering indices.) To see why, consider the following thought ...
TLDR's user avatar
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2 votes

For these gauge transformations in electromagnetism, $\phi\to \phi-\partial_t \lambda$ and $\vec A\to \vec A+\nabla\lambda$, why do the signs differ?

This sign comes from the fact that ($g^{00}=+1$ assumed) $$\partial^\mu = \left ( \frac{d~}{dt}, -{\vec \nabla} \right ) \,.$$ You can avoid such issues by using covariant notation throughout.
my2cts's user avatar
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2 votes

For these gauge transformations in electromagnetism, $\phi\to \phi-\partial_t \lambda$ and $\vec A\to \vec A+\nabla\lambda$, why do the signs differ?

Your result is almost correct, you just have two sign errors. Lets start with your transformation (B) and be careful with the signs. The four derivative is defined in such a way, that they have a ...
Eru's user avatar
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