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A Torque upon a rotating magnetized body

  1. Oct 20, 2016 #1
    Hey there!
    I would really appreciate if someone could help me with the following problem. I didn't find it anywhere (textbooks, web...).

    Not only currents but also magnetized bodies possess magnetic moment. I know that if the magnetized body is at rest in a constant external magnetic field its magnetic moment will align to the direction of the external field. One can understand this by thinking that this parallel configuratioin minimizes the (magnetic) energy of the system. That's fine.

    Thus, my problem is the following: what would happen if the magnetized body was rotating (aroung the magnetic moment axis) initially? Would the body's magnetic moment also align to the external magnetic field and stays there? Or would it exhibit a precession motion just like the magnetic moments of elementary particles do?

    It does seems to me that something would be different since the rotation would mean that there are charges in movement and therefore another magnetic moment would be associated. I happen to think now, but I'm not quite sure. Assuming, as I did, that the the rotation axis coincides with the magnetic moment (due to magnetization), the magnetic moment due to rotation would be parallel to the first, so in the end the rotation would only contribute to increase the magnitude of the magnetic moment. Therefore, qualitatively being the same process.

    I think this problem is important and is completely overlooked.
    The behavior of the magnetic moment of elementary particles does seem to be completely different from the behavior of macroscopic bodies.

    Thanks a lot!
    Cheers!
     
  2. jcsd
  3. Oct 20, 2016 #2

    Charles Link

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    If I am analyzing it correctly, when the magnetic moment is perpendicular to the magnetic field, two things can happen: 1) The magnetic moment experiences a torque that will cause the moment to tend to align with the field. This comes from torque ## \tau=m \times B ##, and the energy ## U=-m \cdot B ##. 2) A second possibility exists that results because ## \tau=d J/dt ## where ## J ## is the angular momentum. ## J ## is proportional to ## m ## so that if ## J ## rotates in the plane perpendicular to ## B ##, the equations of motion are satisfied. The torque on a current loop of an electric motor, if I analyzed it correctly, is largely of the first type. A permanent magnet in a magnetic field will also behaves largely according to the first method. The second possibility occurs in cases such as the nuclear spins in magnetic resonance. The reason for the difference appears to be that the mechanical motion (from the associated mass) that results in the first two cases dominates the angular momentum of the system as opposed to the angular momentum being determined by (proportional to) the magnetic moment, e.g. in the nuclear spin case.
     
  4. Oct 21, 2016 #3
    That's exactly the subject from where this came from, nuclear magnetic resonance. :)

    For elementary particles we know from experience that the total angular momentum and magnetic moment are proportional. Futhermore, we also know from experience that the proportionality constant (the gyromagnetic ratio) can be negative, something I think it is not possible classically.

    Following your setup my question would be: would a macroscopic body experience the second situation?
    This, it seems, amounts to the question: is the magnetic moment of a macroscopic body also proportional to its angular momentum?

    If I understood your reply correctly, the answers to both questions is no.

    Would you have a reference for me about your explanation of the difference in terms of the mass?

    Also, I think I don't agree with you. In principle I could think of a rotation so small that both angular momenta (motion and magnetic moment) would be similar (or a sufficiently large magnetic field).
     
  5. Oct 21, 2016 #4

    Charles Link

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    I'm going to need to compute the answer carefully, but I think I analyzed it correctly. The torque ## \tau=m \times B ## can be made much larger than any gravitational torque that might exist, (we can also work the problem in zero gravity), but I think the calculation will show that in the case of an iron magnet, the mechanical angular momentum from the magnet will be the dominant change in angular momentum with any rotation. If the permanent magnet were to rotate in the plane perpendicular to the static magnetic field, it would require a torque parallel to the static magnetic field. I'm going to need to give this some further study. .. editing... The magnetization of the permanent (cylindrical bar) magnet is along its axis and the angular momentum from this magnetization is also along this axis. This means that if it rotates in a plane perpendicular to the static magnetic field at the proper rate(constant rate), the rate of change in angular momentum ( ## \Delta J/ \Delta t ## from the magnetization where ## J=M/\gamma ##) could be equal to the torque (## \tau=m \times B ##) from the magnetization in the static field, if there is negligible mechanical mass. There doesn't seem to be any practical way to make the mechanical mass of the permanent magnet so small that the angular momentum from the magnetization is the dominating angular momentum. ## \\ ## editing some more: If you are studying magnetic resonance, one thing they seem to omit (at least in C.P. Slichter's book) is the treatment of the simplest magnetic system (e.g. a cylindrical sample in a solenoid) with an applied ac magnetic field (from an ac current in the solenoid) ## H(t)=H_o exp^{i \omega t} ## (real part). In that case, for a linear response, ## M(t)=\chi(\omega) H(t) ## where ## \chi ## is a complex frequency dependent susceptibility. ( Essentially ## \tilde{M}(\omega)=\tilde{\chi}(\omega) \tilde{H}(\omega) ## where we are writing Fourier transforms of a linear response convolution integral for ## M(t)=\int \chi(t-t') H(t') \, dt' ##. )The result is ## M(t)=M_o cos(\omega t-\phi) ## for some ## M_o ## and phase ## \phi ##. (magnetization and applied ac field are in the z-direction). They need to present the simple case before doing the more complicated cases. (Note: In this simpler case, there is no precession. The magnetization simply tends to align with the applied (sinusoidal) field. The phase lag ## \phi ## can even be very small.) ## \\ ## Upon doing the simpler case, the problem of the r-f ## H_1 cos(\omega t) ## field applied at right angles to the strong static DC ## H_o \hat{z} ## field is more readily understood. In this more complicated case, a resonant precession results from the static field interacting with the magnetic moments, so that ## \tilde{\chi}(\omega) ## for this r-f ## H_1 ## field is found to peak strongly at this resonant frequency. Alternatively, the resonant frequency is determined by ## \Delta E=\hbar \omega_o ## where ## \Delta E ## is the energy difference in the static field between the two magnetic spin states.
     
    Last edited: Oct 21, 2016
  6. Oct 22, 2016 #5
    Let's work without gravity as you said. The mass is irrelevant in that case, right? :)

    Would you confirm that your assumption of proportionality between the magnetic moment and angular momentum is valid for macroscopic bodies?
    As I said before, my question boils down to this one.

    I really really appreciate your effort! :)
    I'm ok with the NMR calculations, though.
    I would like to know how quantum is this Larmor precession, i.e., does it just occurs for quantum systems or it also occurs for macroscopic classical bodies?

    The thing is, although calculations are different, classical and quantum, the semi-classical vector model (as it's called) where one uses the classical equations of motion plus the assumption ## J=\gamma m## (I will call proportionality assumption) provide the insight to think about a spin 1/2 in the NMR setup. This coincides to the Bloch vector representation of spins, which would be the same as applying the classical equations to the average values of the spin operator (the Ehrenfest's theorem).
    So far so good. I just wanted to give a step back. How quantum is this proportionality assumption? Macroscopic bodies would never experience this behavior?
    That certainly would justify the name 'Semi-classical'.

    My view is that this assumption is quantum, in the sense that we know it by experience with quantum systems. That's why I begin this post. So that someone can convince otherwise.
     
  7. Oct 22, 2016 #6

    Charles Link

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    For a permanent magnet, the total angular momentum is proportional to the magnetization. The DeHaas-Einstein experiment (with a macroscopic body) is the confirming experiment for this result. The magnetization ## M ## of the body is reversed, but since there are no torques on the body (The applied field is along (opposite) the direction of magnetization), the body(held by a wire) starts spinning mechanically in the direction so that no change in total angular momentum occurs... The magnetization per unit volume ## M ## is used to compute the angular momentum from the magnetization. The relationship ## J=\int M d \tau/\gamma ## holds for a macroscopic body, just as ## J=\mu/\gamma ## works at the atomic level. (C.P. Slichter uses ## \mu=\gamma J ##). ## \\ ## Additional note: The mechanical mass is still relevant even without any gravity present. ## \\ ## Editing some more: Additional comment is the precession that occurs at the atomic level has both a quantum and semi-classical explanation. It could also occur at the macroscopic level except that the conducting wire (copper) of a macroscopic current loop is much more massive than the rotating electrons and the copper atoms do not have their angular momentum on the same axis as the rotating electrons if precession occurs. Similarly for a permanent cylindrical magnet where the collection of magnetic spins can be analyzed as a surface current on the outside of the cylinder. Here again, the iron atoms will have an angular momentum that is not along the axis of the magnetization if precession were to occur. Instead the torque of the magnetic field on the magnetization will cause the entire permanent magnet to be aligned with the magnetic field.
     
    Last edited: Oct 22, 2016
  8. Oct 22, 2016 #7

    Charles Link

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    A follow-on: It is also useful to compare to the mechanical (bicycle wheel) gyroscope=wheel mounted on a short pivoting arm. There is a torque from gravity, and this equals the change in angular momentum as the spinning wheel precesses. For the analogous macroscopic magnetic problem, we have the required torque. From the onset, starting with a permanent magnet or moment from a current carrying loop perpendicular to the field, there will be too much angular momentum in the z-direction if the precession begins. In the case of the spinning bicycle wheel, this change in z-angular momentum simply causes a slight wobble in the precession that subsides over time as the z-angular momentum stays nearly constant with a constant precession rate. In the case of the permanent iron magnet in the magnetic field, the initial change in (mechanical) z-axis angular momentum that occurs when a very initial precession occurs will be quite dominant and can't be treated as a minor perturbation. ## \\ ## It may also be worthwhile to do a quantitative calculation using 560 grams= .56 kg of iron =10 moles=6.02E+24 atoms and giving one electron per atom each ## \hbar ## units of angular momentum. ## \ ## ## \hbar=(6.626/(2 \pi)) E-34 \, joule \, sec ##. Multiplying this, we get spin angular momentum ## J_{spin \, total}=6.0 E-10 \, joule\, sec ## from what is qualitatively a typical amount of magnetization for an iron magnet. If the .56 kg of iron moves(rotates) at ## v=.1 m/sec ## with ## r=.1 m ##, (just estimates), ## J_{mechanical}=mvr=6.0 E-3 \, joule \, sec ##. The spin angular momentum is dwarfed by the mechanical angular momentum.
     
    Last edited: Oct 22, 2016
  9. Oct 24, 2016 #8

    Charles Link

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    @pac134 Please read the latest posts (#6 and#7). I would enjoy your feedback from them.
     
  10. Oct 29, 2016 #9
    Sorry for the absence. I was out of town due to work. I'm gonna have look and provide you the feedback. :)
     
  11. Oct 29, 2016 #10
    Very nice! Didn't know about this effect. Our discussion already elucidated some issues to me but I'm still a bit confused. I am partially convinced that we should see Larmor precession in macroscopic bodies and I see your argument about the relevance of the size of the macroscopic system.
    It would be nice to see this in a mesoscopic system if possible, though.

    Answer to me as best as possible please. Imagine I'm a student and I ask:

    If the magnetic moment is proportional to the angular momentum for both macroscopic and microscopic systems why don't we see the Larmor precession in macroscopic bodies? I thought the precession was a quantum phenomenon.
     
  12. Oct 29, 2016 #11

    Charles Link

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    The amount of angular momentum that occurs as a result of even a strongly magnetized object is very small compared to any amount we can introduce by rotating the object manually. If the amount of angular momentum from magnetism were about one million times as strong, we would feel the same gyroscopic type effects from trying to rotate a magnet (in the absence of an external magnetic field) that we experience when we attempt to turn a bicycle wheel that is spinning rapidly. A cylindrical magnet has angular momentum along its axis. The calculations that we performed show this angular momentum is rather miniscule and is unable to appreciably affect the motion of massive objects. For the purpose of most practical calculations involving the motion of (ferromagnetic) objects in which this angular momentum is contained, the angular momentum from the magnetization can be ignored. ## \\ ## For analyzing gyroscopic motion, you can take the angular momentum vectors ( angular momentum ## L ## points along the axle of the spinning wheel), at two successive directions (e.g. as a spinning wheel precesses), and compute the change in these vectors, by finding the vector that points from one head to the other. If the angular momentum is a significant amount, a torque must be supplied in this direction i.e. ## \tau=r \times F=\Delta L/\Delta t ##. For a spinning wheel attached to an arm in a gravitational field, the force of gravity along the arm supplies this torque. The force is downward, but the spinning wheel stays upright and precesses to have ## \Delta L/\Delta t ## equal the gravitational torque. If the wheel wasn't spinning, the gravitational force would simply pull it downward. The wheel would turn downwards (with the appropriate angular acceleration) because it has a gravitational torque on it. When the wheel is spinning rapidly, this does not occur, but instead, the wheel stays upright and precesses. Magnetization could act like the spinning wheel, but the magnitude of the angular momentum for magnetization is about one millionth of what it needs to be for this to occur.
     
    Last edited: Oct 29, 2016
  13. Oct 30, 2016 #12
    I'm convinced. Thank you very much! :)
     
  14. Oct 30, 2016 #13

    Charles Link

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    Very good. It's actually an interesting item that you get basically as much angular momentum from nuclear spins that are aligned in an object (e.g. in an MRI machhine) as you do get from ferromagnetic material with aligned electron spins, but the magnetization and magnetic moments and magnetic surface currents are about 1000 times stronger for a ferromagnet than for a sample with aligned nuclear spins. A single unit of spin gives angular momentum ## \hbar ## (or ## \hbar/2 ##), regardless of whether it is an electron spin or nuclear spin. In the case of the nuclear spin, the magnetic moment is much much smaller than for the electron spin.
     
  15. Oct 31, 2016 #14

    Charles Link

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    @pac134 Please read my post #13.
     
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