5
$\begingroup$

I am very confused about the concept of alignment in a magnetic field. Perhaps I am also confusing two different phenomena and that may be the issue.

My classical understanding of a dipole says that when placed in a magnetic field, the moment's precession about the magnetic field axis corresponds to the direction of the moment when the field is turned on.

Then, the component of the dipole in the direction of the magnetic field should oscillate between antiparallel and parallel.

But why then do I keep reading about parallel or antiparallel alignment in a magnetic field? What does this alignment mean?

When I read that a nucleus having spin 1/2 is like a bar magnet, I get more confused because I don't understand why a bar magnet would align parallel OR antiparallel to a field. It should just align parallel, shouldn't it? Is spin angular momentum the determining factor in the observation?

Clarification on this topic would be much appreciated.

$\endgroup$

4 Answers 4

2
$\begingroup$

This is probably a burning question which many students of physics skip over. My own struggle to find a satisfactory answer led me to what follows: The fundamental magnetic dipole moment is closely related to the "spin" angular momentum; related by the gyromagnetic ratio. Both vectors are collinear.

The torque, experienced by this moment when placed in a uniform magnetic field, is orthogonal to the angular momentum causing it to precess.
Note that there is no kinetic energy associated with this angular momentum and with the precession. This is a key difference from other mechanical analogues like a spinning top in gravity. In the case of a spinning top under gravity, the fact that the spinning and the precession possess kinetic energy leads to an extra component in the motion called nutation.
Note that there is no torque causing any alignment.

Now consider a bar magnet which is spatially extended and comprised of many such spins that are co-aligned so, that there is a magnetic polarization along the length of the magnet. When placed in a uniform magnetic field, all the spins would just precess, causing the polarization within the body of the magnet to precess as well. No bodily movement of the magnet is expected.

When placed in an inhomogeneous magnetic field however, each of the spins will feel a force which will depend upon the relative orientation of the spin and the local field gradient. Overall, this will cause the body of the magnet to feel a torque that will cause it to align with its length along the field gradient.

So in essence, the alignment that we see in the case of bar magnets is due to a differential force along its length and will present itself only when the external magnetic field is non-uniform.

$\endgroup$
0
$\begingroup$

I think that you are confusing more classical arguments about magnetism. Maybe you already know what I am going to write, but I think this is the point.

First, precession is typical of the classical model of diamagnetism (here and here you may find something relevant). Consider an hydrogen atom in the Bohr model; you can demonstrate that the orbiting electron generates an orbital angular momentum. When we apply a magnetic field, on that momentum will act a couple that can be written as: $$\vec{τ} = \vec{μ} \times \vec{B} $$ with $B$ magnetic field and $μ$ magnetic moment of the system (in this case, due to the orbiting). If you call $L$ the angular orbital momentum of the electron, you can write the Euler's equation for the dynamic of a rigid body: $$ \vec{τ} =\frac{d\vec{L} }{dt}. $$ By reasoning on the derivative of this vector and on the direction of the torque, you can see that this will cause a precession of $L$ around $B$. The projection of $L$ in the direction of $B$ is constant.

The second part, about alignment, make me think about the classical model of paramagnetism (I can't actually find a really satisfying link to it, but I think it will be explained in all the books about magnetism from a classical point of view). In this theory, you assume that your atom has a certain magnetic dipole moment (cause by the angular orbital momentum or by the spin, with a semiclassical approach) that in a magnetic field will have an energy: $$ U = - \vec{μ} \cdot \vec{B} = - μB \cos{θ} $$ where $θ$ is the angle between the two vectors. Using Boltzmann's statistics, you can evaluate the mean value of $\cos{θ} $ and then explain the properties of paramagnetic materials. As you can see, the energy of this interaction between the magnetic dipole $μ$ and the magnetic field $B$ will the smallest possible if the two vectors are aligned.

In these classical or semiclassical models, the spin is usually added "artificially" as another magnetic dipole that can precede or can interact with the magnetic field. Even if they lead to a correct result, these classical models aren't right; other models, based on quantum mechanics, have been developed to discuss magnetic properties of materials.

I would like to add some references or sources, but I don't have any book written in English, only my teacher's notes. I will add them in the future, eventually.

$\endgroup$
0
$\begingroup$

The potential energy of the magnet (magnetic dipole) in a B field is given by $$E = - \vec \mu \cdot \vec B $$ If this energy scale is large compared to all others (e.g. interaction between dipoles, thermal energy, ...) the dipole "favors" the parallel alignment with respect to the B field, because this is the state of lowest energy. However, once the projection of the magnetic moment with respect to the B field is fixed, the atom/electron/nucleus must conserve "angular momentum". Hence, it will not flip to a state with a different $m_s, m_j, m_f$.

You might want to read about the Zeeman effect.

$\endgroup$
-1
$\begingroup$

I believe there will be no alignment without some type of friction. Imagine an inverted top (say, a top with an iron tip hanging from a magnet). I can give the top some rotation so that it will precess. In a perfectly "lubricated" environment, that is, no air and no friction at the tip, the top will rotate and precess forever, and will not align with the vertical. The final alignment, that minimizes the energy of the top, comes from friction.

A similar situation happens to a compass: if it is not initially pointing north, the magnetic field of the earth will put it to oscillate. Only the friction in the pivot gets it to stop, pointing north.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.