Larmor Precession and a bar magnet

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As stated in the link below, when a magnet bar (or a current loop) is placed in an external magnetic field it (its magnetic moment) becomes aligned with the magnetic field while an orbiting electron would have precession with Larmor frequency. What is the reason for these different behaviors?
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/larmor.html
 
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Charles Link
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The equilibrium state when a static magnetic field is applied to a ferromagnetic material is to have the magnetic moments aligned with the static field because energy ## U=-\vec{\mu} \cdot \vec{B} ##. ## \\ ## Ferromagnetic resonance at the Larmor frequency has been observed experimentally in ferromagnetic materials when an r-f field is applied in addition to the static field. See https://en.wikipedia.org/wiki/Ferromagnetic_resonance ## \\ ## I also recommend the textbook by C.P. Slichter, Nuclear Magnetic Resonance, to see how the magnetic resonance results. The principles are the same, whether it is electron spin or nuclear spin, and Slichter's book treats the subject very well in the first couple of chapters. Basically the resonance occurs because torque ## \tau=\vec{\mu} \times B=\frac{d \vec{L}}{dt}= \frac{d (\gamma \vec{\mu)}}{dt} ##. The solution to this last equation is a precessing ## \vec{\mu} ##. Slichter's book discusses this in detail=much more so than what I have shown.
 
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The equilibrium state when a static magnetic field is applied to a ferromagnetic material is to have the magnetic moments aligned with the static field because energy ## U=-\vec{\mu} \cdot \vec{B} ##. ## \\ ## Ferromagnetic resonance at the Larmor frequency has been observed experimentally in ferromagnetic materials when an r-f field is applied in addition to the static field. See https://en.wikipedia.org/wiki/Ferromagnetic_resonance ## \\ ## I also recommend the textbook by C.P. Slichter, Nuclear Magnetic Resonance, to see how the magnetic resonance results. The principles are the same, whether it is electron spin or nuclear spin, and Slichter's book treats the subject very well in the first couple of chapters. Basically the resonance occurs because torque ## \tau=\vec{\mu} \times B=\frac{d \vec{L}}{dt}= \frac{d (\gamma \vec{\mu)}}{dt} ##. The solution to this last equation is a precessing ## \vec{\mu} ##. Slichter's book discusses this in detail=much more so than what I have shown.
Thanks. I can understand the precession of the magnetic moment by the applied torque. However, it remains two questions:

1) We know that a magnet bar itself can be considered as a magnetic moment. Why the magnet bar becomes aligned with the external magnetic field while the current loop (as another magnetic moment) has precession and doesn't become aligned with the external magnetic field?

2) Why, as well as doing precession, the current loop doesn't become aligned with the field?
 
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Charles Link
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This question came up previously=I'm going to try to find the thread=edit: Yes, here it is: https://www.physicsforums.com/threads/a-magnetic-dipole-in-a-magnetic-field.864607/#post-5427813 ## \\ ## I believe the answer to (1) is that, because of the heavy mass of the iron magnet, the angular momentum of that system would be far greater than that which would result from the applied torques. ## \\ ## For (2), I believe it can, and damping forces may be one reason why the ferromagnetic resonance is not observed in many cases.
 
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Charles Link
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Additional comment: You might also consider the analogous spinning bicycle wheel that is spinning on an axle with a pivot point on one side of the axle. If you spin it fast enough, the bicycle wheel will basically hold itself up, and precess around. Simple mechanical equations explain the precession=the torque from gravity is in the same direction as the change in angular momentum if the wheel is spinning fast enough so that the angular momentum of the wheel is that from its spinning rather than falling over sideways, which happens if you don't spin the wheel. Let me see if I can find a video of this demonstration: Yes=See ## \\ ## In the case of the bar magnet, there is insufficient angular momentum from the spins to make the bar magnet precess like a bicycle wheel does. Note that the torque from the magnetic field on the bar magnetic is analogous to the gravitational torque on the bicycle wheel. The insufficient angular momentum from the electron spins could be equated to a slowly spinning bicycle wheel, and the system simply aligns itself to the magnetic field with little precession. ## \\ ## You could even set up a bar magnet demo, where you have both gravitational torque and magnetic torque acting in the same direction with a magnetic field pointing upwards, and fasten one end of the bar magnet to a rope. There simply isn't enough angular momentum from the electron spins in the bar magnet for it to precess like a bicycle wheel. e.g. you could momentarily hold the bar magnet horizontal (like the fellow does in the video to the bicycle wheel). The bar magnet will not stay level and precess=it will simply rotate to align itself with the vertical magnetic field.
 
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vanhees71
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For the "precession" effect with macroscopic bodies look at the Einstein-de Haas experiment, which is also a drastic lesson for the experimental physicist not to listen to theoreticians too didgently, even if it's a genius as Einstein; de Haas missed a great discovery, namely that there's more to magnetism than the classical "Amperian atomic currents", which inevitably leads to a gyro-factor of 1. In fact in ferromagnets most of the magnetic moment is due to electron spins with a gyro factor (very close to) 2.
 
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Charles Link
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I can add something to the discussion that the OP @hokhani might find useful: ## \\ ## There are, in general, two types of ferromagnetism: ## \\ ## 1) The permanent magnet, where a large static magnetization ## M ## occurs in the absence of an applied field.## \\ ## 2) A magnetization ## M ## that results from the presence of an applied field ## H ## (e.g. A uniform ## H ## in the z-direction can be created by the current of a solenoid, and ## M ## is the magnetization of the sample inside the solenoid). Because this response is normally frequency dependent, it is often written as ## \tilde{M}(\omega)=\tilde{\chi}(\omega) \tilde{H}(\omega) ##, where the ## \tilde{M } ## and ## \tilde{H} ## are Fourier components of the respective quantities,## \\ ## [it assumes an applied signal ## H(t) ## that is sinusoidal with time,(oftentimes the current in the solenoid is made to be sinusoidal), with frequency ## \omega =2 \pi f ##, i.e. beginning with an arbitrary ## H(t) ## and ## M(t) ##, you then take Fourier transforms of ## H(t) ## and ## M(t ) ## to get their amplitudes, ## \tilde{H}(\omega) ## and ## \tilde{M}(\omega) ##, as a function of frequency], ## \\ ## and ## \tilde{\chi} ## is a frequency dependent susceptibility. This equation assumes a linear response=otherwise the ## \tilde{\chi} ## can even be amplitude-dependent in the non-linear case. ## \\ ## In a complex type Fourier transform, ## \tilde{\chi} ## can even be complex, and contain phase information for the sinusoidal response=i.e. it may contain a factor ## e^{-i \phi}=e^{-i \omega t_o} ## for a phase/time delay between ## M ## and ## H ##. ## \\ ## It may also be of interest that in his Nuclear Magnetic Resonance book mentioned in post 2 above, C.P. Slichter presents ## \tilde{\chi}(\omega) ## as the susceptibility, giving the response of the sample ## \tilde{M}(\omega) ## to an applied r-f field ## \tilde{H}(\omega) ##, but he doesn't give the reader a whole lot of introductory material. The introductory material that is supplied above might be useful in attempting to read through the first couple of chapters of Slichter's book.
 
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