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The neutron is not an elementary particle. Its overall charge is zero, but depending on the energy available the fact that it is composed out of charged quarks becomes important. Thus both the strong force and the electromagnetic will play a role depending on the energies of the scatters and the targets.
For low energies , like the ones coming from fission ...
10
Steve's intuition for the measurement operator is in the right direction, but it's not quite the right interaction potential. Getting that right is simple, but it requires the complex task of taking the naive answer to its ultimate consequences and then understanding what it's really saying. The hamiltonian is the same as always: a direct coupling between ...
10
Has anyone tried to incorporate the electrons magnetic dipole moment into the atomic orbital theory?
To be crystal clear:
Has anyone tried to incorporate the electrons magnetic dipole moment into the atomic orbital theory?
YES. They've tried and they've succeeded. The electron's spin magnetic dipole is a standard part of atomic physics and quantum chemistry. Anybody attempting to claim that this isn't the case is simply describing their own ignorance ...
9
Your question is best taken in two parts: (1) Are magnetic (or mostly-magnetic) waves possible, and (2) If they do exist, would it be possible to use such waves to implement the science fiction concept of "tractor beams" -- that is, as a way to exert a spatially focused, selective pull (or push) on a relatively distant object? So, here goes:
(1) Are mostly-...
9
An electron is not a spinning ball of charge and the intrinsic spin of particles cannot be understood in such terms. Not only is it difficult to make sense of what it means for a pointlike particle to spin, but also when treating the electron as a spinning ball of charge one finds a value of the ratio between the magnetic moment and the angular momentum that ...
9
It is an experimental fact that neutrons have no net charge. They have a magnetic dipole moment, which points to having charged constituents (quarks), the charges of which algebraically sum to $0$. (It is not the only evidence for constituents in the neutron: scattering experiments also indicate the presence of constituents, and exclude the possibility of ...
9
Ampere's law says that
$$\nabla \times \boldsymbol{B} = \mu_0 \boldsymbol{J} + \epsilon_0\mu_0 \frac{\partial}{\partial t} \boldsymbol{E}$$
so "far away from sources" means that the current density $\boldsymbol{J}$ can be taken to be zero, and that there are no time-varying electric fields. The latter is actually a general approximation that can often be ...
8
The current loop creates a magnetic field. This field looks just like that of a magnet, which has two poles (a dipole).
7
Elementary particles have quantum mechanical spin. This induces a spin magnetic moment, independent of the presence (or, indeed, absence) of a (net) electric charge. This is how the neutron attains its magnetic moment (as you already mentioned).
The case of the neutrino magnetic moment is slightly confusing, as they are not completely understood yet. ...
7
The easiest way to see the equality is to use a more general formula for the magnetic dipole moment of a particle. For a flat planar loop of current, it's true that $\mu = IA$, with the direction of the dipole normal to the loop. However, the more general case is that of a a volume current $\vec{J}$ in some finite region of space. In this case, the the ...
7
There are two important interactions in magnetic materials: the dipolar interaction and the exchange interaction.
The dipolar interaction corresponds to the magnetic fields from the Maxwell equations. This field is dominant on the macroscopic scale. Therefore, if we look to the macroscopic bar magnets, they nicely follow the behavior of the magnetic field ...
6
It is zero. If the magnetic dipole moment $\boldsymbol \mu$ were different from zero, the particle would have a preferred direction in space, which is not possible for a scalar.
Recall that, in general,
$$
\boldsymbol \mu\propto\boldsymbol S
$$
which, for a scalar, implies $\boldsymbol\mu\equiv\boldsymbol 0$.
answered Jan 2 '17 at 18:18
AccidentalFourierTransform
35.1k14 gold badges87 silver badges161 bronze badges
6
As far as we know, electrons are point particles, so you don't want to try to calculate anything when they are at zero separation. You'd just get infinite forces.
Classically, the magnetostatic dipole-dipole interaction dominates over the electrostatic interaction when two electrons are separated by less than about one Compton wavelength. This is because ...
6
It is not necessary for them to move. Even an electron at rest has a magnetic field. This is related to its intrinsic angular momentum or “spin”. It is not actually spinning like a little ball, and this magnetic field cannot be understood as arising from a current.
5
Let's first look at the classical situation. A charged particle moving round a circular loop had an angular momentum and because it is also a current, it produces a magnetic moment. Therefore, it can be considered to be a magnetic dipole with a moment $\vec{m}$. The magnetic moment and the angular momentum are proportional to each other with the constant of ...
5
The Earth's magnetic field is caused by eddy currents in the liquid parts of the planet's interior. We believe the field is not due to a permanent magnet because: (1) Its direction and strength change over time, and (2) the planet's interior is hotter than the Curie temperature of its elements, and so a permanent magnet would not retain its magnetism.
...
5
$\mathbf{S}$ is the spin operator. It is a vector operator that acts on spinors. It will have three components $(S_x, S_y, S_z)$ and for example if you take the $z$ axis as your spin measurement axis, you define spin up and down as the two eigenstates of $S_z$.
It can be shown that in matrix form $S_i$ is proportional to the Pauli matrix $\sigma_i$.
...
5
An electron acts like a very small magnet. We say "magnetic moment" rather than "magnetic field" because a field is a local property (move further from the electron and the field gets weaker), while the magnetic moment is a property of the electron from which you can deduce the field at any point, and from which you can calculate the energy required to align ...
5
Since gravity uses energy to push us down to earth
This is incorrect. Gravity does not use energy to pull in us.
If we started falling, then yes, gravity used energy do make us move. But that is only in the special case where gravity makes us move. In general, gravity spends no energy pulling in us.
In general, a force spends no energy. An apple lying on ...
5
The electron can be seen as a small magnet with the field of a magnetic dipole of strength $2\mu_B \vec S$. So my answer is yes in both cases. I interpret as at rest a state in which the electron has zero momentum expectation value. Note that there is no accepted model of spin as a rotation, although spin is an angular moment to every effect.
5
It turns out that if you work out the physics behind the magnetism of a material, the magnetisation $M$ obeys a transcendental equation of the form
$$M=\tanh\left(\frac{\alpha M+\gamma H}{T}\right)$$
where $\alpha$ and $\gamma$ are some constants, $H$ is the magnitude of the external magnetic field and $T$ is the temperature. What this says is that at a ...
5
The origin of magnetism is an complex problem in solid physics, (may be the the most lasting discussion in condensed matter physics). My point is:
if caring about interaction of spin in system, we need to be very careful and dipole-dipole interactions sometimes are not the leading term. I will list four kinds magnetic origin which may help you.
dipole-...
5
The radiation for a neutron in a magnetic field has bravely been calculated.
They conclude:
The calculations in this paper are mainly of theoretical interest, as good pedagogical examples in the classical and quantum theories of radiation. Physically the process is not observable,because the radiation rate of the neutron is very small,
They give an ...
4
This is actually the least hypothesis for this configuration.
Four nucleons comprising all four allowed $\text{spin} \times \text{isospin}$ states can all be expected to be in the s1 state so that they have no orbital contribution and for each pair the intrinsic magnetic moments cancel as the spins are opposed. Boom. Zero and done.
answered Jan 25 '13 at 19:55
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Less than g.
Since it generates energy in the conducting ring when passing through the ring, by conservation of energy, the magnet loses some kinetic energy (there are forces acting on the magnet.
See Lenz's law.
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There doesn't have to be eddy currents to obtain diamagnetic. In fact you can have eddy currents in iron plate, which is ferromagnetic. Diamagnetics can be insulators, so there don't have to be any currents at all. In some cases it's energetically beneficial to align so that magnetic field is lowered. There are always diamagnetic and paramagnetic tendencies ...
4
There is no classical analogue to visualise what spin is. We found out from experiments that particles have an intrinsic property which we named spin which produces a magnetic moment. You cannot visualise it since fundamental particles are zero-dimensional points in space so the term "spinning around its axis" makes no physical sense.
Its a strictly ...
4
No. The Lorentz force
$$
\vec F = q\left( \vec E + \vec v \times \vec B \right)
$$
describes the interaction of a charge with electric and magnetic fields.
A magnetic dipole moment is a convenient way to describe the source term for a very common field distribution — there are tons of problems where you compute the magnetic field for a rotating charged ...
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