# Magnetic Hysterisis and Irreversible Losses

1. May 17, 2005

### Modey3

Hello. I'm still confused about the thermodynamics involved in magnetic domain wall motion. I understand the different pinning mechanisms, but I fail to see how this phenomena leads to Irreversible losses during magnetic hysterisis. For instance, how can precipitate which can put a section of a domain wall in an "energy well" lead to irreversible losses. I'm sure the answer is really simple, but at this moment I'm not seeing it. Is there a mechanical anaology? Thanks

Modey3

2. May 18, 2005

### Gokul43201

Staff Emeritus
All that follows is what I've just thought through, so take it as nothing more than my opinion (though, one that I feel reasonably confident in).

The key is in the difference between the driving forces for domain wall motion during magnetization and demagnetization. During magnetization, the driving force is the applied field, H. As the domain wall passes a "particle" it falls into a potential well. Increasing H eventually provides sufficient energy to raise it out of the well and continue moving it. During demagnetization, the driving force is actually thermal excitations at ambient temperature (and spin-spin interactions, which are small enough to neglect, for this discussion).

Consider a magnetized sample with one part of a domain wall pinned at a particle in some applied field +H. Increasing the field will move the wall beyond this particle, but let's stop at this particular value of the field, and instead of increasing it, simply turn off the field. Now the probability that the wall gets unpinned depends on the the temperature. At low temperatures, thermal excitations may not be enough to unpin the wall, but at high temperatures they may be. So, if thermal excitation is insufficient tp unpin the wall, it will take an extra reverse field (the coercive field, Hc) of sufficient strength to achieve this unpinning. So, extra work needs to be cone to bring the domain wall back to its original position (or the sample back to zero magnetization).

This extra energy is the hysteretic loss per cycle.

Note : It is a natural conclusion from this, that if at ambient temperature, kT >> U(particle), there will be essentially no hysteresis. On the other extreme, at 0K, hysteresis should be "perfect" and the M-H curve must look like a rectangular box. I'm a little rusty on this but I believe these are consistent with experiment, where Hc decreases with increasing T (< Tc).

PS : Forgive the sloppy figure.

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3. May 19, 2005

### marlon

Just to add to what Gokul has written down. When a DC magnetic field is applied onto some ferromagnetic sample, the domains will all align in the direction of the applied field. There are two processes at hand here:

1) for weak DC B-fields : de volume of the domain that has the same direction of the B field will increase.

2) For strong DC B-fields : the domains are forced to rotate their magnetization towards the direction of the field.

Actually, both processes take place ;only when the external DC B-field is weak, the first process will dominate and when the B-field is strong, the second process will dominate over the first.

Magnetorestriction is the phenomenon where a material changes its dimension when an external DC B-field is applied.

Also, magnetoresistance is very intruiging : conduction electrons are forced to move in helical trajectories when a strong enough extern magnetic field is applied. The field needs to be strong enough so that the electron trajectory is curved within a distance that is equal to its mean free path (average distance between two collisions)

Further info at "introduction to nanotechnology" by Charles P. Poole and Frank J. Owens
Look at pages 181 and 334

regards

marlon

4. May 19, 2005

### Modey3

Thanks Gokul for thaty insightful explanation. This gives me a more intuitive "feeling" for the process.

Modey3

5. May 19, 2005

### Gokul43201

Staff Emeritus
In most experiments that I've done or heard of, the B-field is usually ramped slowly from 0 to B(max). So, the above two mechanisms almost always occur sequentially.

Initially, there is domain growth, where favorably aligned domains "eat up" unfavorable ones, till at some field, there's only a single domain. Further increasing the field beyond this point rotates the spins in this domain along the field direction. So usually, the mechanism involves growth followed by rotation (very roughly speaking).

An alternative mechanism to domain growth (in fact, this often happens in parallel) is nucleation (which is really just rotation in the vicinity of an impurity), but that's a different story.

6. May 19, 2005

### Modey3

"Initially, there is domain growth, where favorably aligned domains "eat up" unfavorable ones, till at some field, there's only a single domain. Further increasing the field beyond this point rotates the spins in this domain along the field direction. So usually, the mechanism involves growth followed by rotation (very roughly speaking)."
I thought that if the field isn't applied along the "easy" direction, the domains will be aligned at a intemediate point between the applied H field and the intial domain orientation.
"An alternative mechanism to domain growth (in fact, this often happens in parallel) is nucleation (which is really just rotation in the vicinity of an impurity), but that's a different story."
When I was taking Phase Transitions in grad school, I was taught nucleation always happens before growth. So at a domain boundary junction (or even with a domain itself), an appropriate overpotial will nucleate a domain towards the applied field. After this the domain grows by Bloch Wall motion.

7. May 19, 2005

### Gokul43201

Staff Emeritus
A more accurate way to say that would be : "nucleation is always followed by growth". To have growth, you do not need nucleation...like in this case, where the domains already exist at zero field.

True.

Another important underlying condition that rarely gets mentioned is the fact that any magnetization measurement is a non-equilibrium measurement. If the applied field were changed infinitely slowly, the material will have sufficient time to thermalize and hence will show no hysteresis.

The rate constant for domain equilibration goes something like
$$e^{-<U>/K_BT}$$, where <U> is the mean depth of the impurity potential.

8. May 19, 2005

### Modey3

"A more accurate way to say that would be : "nucleation is always followed by growth". To have growth, you do not need nucleation...like in this case, where the domains already exist at zero field. "
Yeah I forgot about 2nd order phase transistions (non-nucleation phase tranformations) like a order-disorder transition. I think you would only get this type of tranformation in magnetics if materials was heated/cooled above/below the Curie Point, Tc, in an applied field. The domains would cease to exists above Tc and the material would be paramagentic.

Modey3