# Tag Info

0

The monopole moment is the total quantity of whatever it is you're considering. For example, if you're looking at a 3-dimensional multipole expansion of the mass of the earth, the monopole moment would be the total mass. That doesn't have anything to do with treating the earth like a sphere. Instead, it's the statement that the total mass of the earth (...

0

Indeed ,E means electric field to to both free and bound charges . 1.how we are applying lorentz force - we are taking one electron (charged particle) as a system and calculating the force on it. And we are just applying the summation over to all electron .remember here individual electron is the system and not the whole body so you can apply lorentz force. ...

0

The voltage of a battery is clearly conservative. If you connect the battery to a voltmeter, it will read the same value independent of the path of the wires. If you move or prolong the wires the readout will be the same. This proves that the voltage of EMF is conservative. Now connect a coil to the voltmeter. If you now place a magnet near the coil. The ...

1

The case of two charges that magnetically interact can be discussed in a much simpler manner. We are all familiar with the concept of potential energy and know that kinetic energy is not separately conserved : only the sum of kinetic and potential energy is. In special relativity energy is the time component of four momentum. For the same reason kinetic ...

4

You cannot consistently consider a charged particle without considering its electromagnetic field as well. A system of charged particles plus their field is “closed” and its energy, momentum, and angular momentum are conserved. Without the field it is not closed and these quantities are not conserved. A single non-interacting charged particle does not ...

0

Your example with gravity is a bit misguided. The work due to gravity around any closed path is always zero. If there were a curve with gravity acting on only part of it, and being "switched off" otherwise, it would mean that gravity were nonconservative. An example of such a curve: A better gravitational analogy is a ball rolling around a closed path in a ...

0

All of the field lines don’t lie within a circle of any finite radius. Think about the field lines that start out heading almost perpendicularly away from the plate.

0

In case anyone else runs into this problem, with some outside help I managed to find a solution. If one makes the substitution $t_r = t-\frac{r}{c}$ in the last integral, the full expression reads, $$\vec{A}(\omega) \sim e^{i\omega\frac{r}{c}} \int_{-\infty}^{\infty} e^{i\omega t_r} e^{-\frac{k}{m}t_r} dt_r.$$ So now the two integrals only differ by a phase ...

3

However, if we have a battery, we know the EMF acts within that battery only. Not true. The force which moves the charges inside the battery (or whatever) does not generate a vector field in the surroundings, as Newton or Coulomb's forces. Not true. Here is a fairly realistic simulation I did of the fields in a DC circuit: As you can see, there are ...

0

Emf is defined as the work done to take a unit positive charge one complete round around the closed circuit.It can induced by changing the net magnetic flux through the loop or by connecting the circuit with a battery..In the first case it is quite obvious that the emf is non conservative as the induced electric fields due to the changing magnetic flux is ...

1

First of all, I assume that in the second term in the "right" answer 2 should be in denominator; if this is the case, it actually satisfies the equation necessary (as $\frac{\partial^2}{\partial z^2} |z| = 2 \delta(z)$ in terms of distributions) Both your answer and the other one are actually OK; different gauge transformations can lead to Couloumb gauge. ...

3

Assume your B-field points along the z-axis in cylindrical coordinates. $$\vec{B} = B_o \sin(t)\ \hat{z}$$ Ampere's law (in the absence of any currents and in a vacuum), says $$\nabla \times \vec{B} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$$ Now taking the curl of your B-field we have $$\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} = ... 0 Are you confusing black body radiation with the photoelectric effect? Classically, the black body spectrum is given by$$ f(\omega) \sim \omega^2,  which is the Rayleigh-Jeans law. It is in good agreement with experiment for small $\omega$, but it also obviously can't be true for large $\omega$, because $f(\omega)$ being a probability density has to ...

Top 50 recent answers are included