36

Consider the two paths $ABCDA$ and $EFGHE$. Path $AB$ contributes a positive value to the $\vec B\cdot d\vec l$ integral but the other parts of the loop contribute nothing, so overall there is a finite value for the $\vec B\cdot d\vec l$ integral but no enclosed current which violates Ampere's law. Again path $EF$ contributes a positive value to ...


14

Yes, it is possible to guide magnetic field lines using a shaped magnetic material. Just as field lines concentrate when entering the south pole of a magnet from a large area, an external magnetic field can be "gathered" using, for example, a cone-shaped piece of iron. The cone can be positioned such that the static field spread over a large area enters the ...


10

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

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

The electric field is nonzero. For a cylinder of finite length, it's nonvanishing everywhere. In the limiting case of an infinitely long cylinder, the field is only nonvanishing inside the cylinder. The easiest way to solve this is to use the fact that the electric and magnetic polarizations $(-\textbf{P},\textbf{M})$ transform in exactly the same way as ...


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. ...


6

This is basically what a solenoid does. You have multiple current rings, and "within" the solenoid the magnetic field loops are concentrated whereas outside they are very weak and actually divergent in the limiting case. A much more interesting questions is if one could design a solenoid or solenoid like structure which "minimizes" the magnetic fields and ...


6

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 ...


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 ...


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

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 $\boldsymbol\mu\equiv\boldsymbol 0$.


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.


4

From the magnetic moment $$\mathbf{\mu}=\mathbf{\mu_L}+ \mathbf{\mu_S}=(g_l \mathbf{L}+g_s\mathbf{S})\frac{\mu_N}{\hbar}\tag1$$ take the scalar product with $\bf{J}$ $$\mathbf{\mu_J} ·\mathbf{J}=(1/2(g_l +g_s)\mathbf J^2+1/2(g_l -g_s)(\mathbf L^2-\mathbf S^2))\frac{\mu_N}{\hbar} \tag2$$ so with commutator relation $$\mu=(1/2(g_l +g_s)j+1/2(g_l -g_s)\frac{(l-...


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.


4

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.


4

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 ...


4

Almost. Whereas I guess the E-de-H effect does require the magnetic moment to be along the spin direction, what the effect is really used to show is that an electron spin flip imparts $\hbar$ of macroscopic angular momentum to the cylinder. Just because we know that the electron is a two state object transforming under rotation by the s=1/2 representation (...


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