1
$\begingroup$

I understand that coercivity is the field/force required to demagnetize/magnetize a ferromagnetic material. What if we had two opposite magnetic fields of different strengths values H acting on the same ferromagnet what would happen?

H1 = 50 Oe H2 = 10 Oe

The coercivity of the material is 0.5 Oe, what would happen? Will it stay magnetized to the stronger field(H1)? Or magnetization = 0?

$\endgroup$
0
$\begingroup$

Magnetic field is additive, so if I understand you correctly your magnet would be in 40 Oe field in direction of H1. As this is more than coercivity magnet will be magnetized by that effective field.

As a sidenote: in ferromagnetic materials it's not that easy to obtain zero magnetization. In zero field you would have spontaneous magnetization. But this can be (almost) done by applying alternating field with amplitude lowering over time. See degaussing on Wikipedia for example (section about CRTs to be precise). Magnetization is not entirely removed, but randomized, and it's amplitude vastly diminished.

EDIT: If you have no magnetization and you don't exceed coercive field you will have magnetization which is not permanent, and will vanish soon after you take your field away. If you exceed coercivity you will get permanent magnetization which will decrease slightly if you take field away. Now if you change direction of field to opposite you will get decreasing magnetization with growing field. At field strength equal to coercivity you will get zero magnetization, and then direction of your magnetization will switch if you keep increasing field any further. Look up histeresis curve. Here you have it really nicely detailed. In your example all of this is unimportant as field strength is much higher than coercivity so sample will magnetize in direction of stronger field no matter what.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So the ferromagnet's magnetization is "dropped". However, we are ignoring the coercivity of the material being 0.5 Oe. It does not matter how much field strength is present only, but also the coercivity of the material itself how strong of a field it "needs" to become magnetized and demagnetized. $\endgroup$ – AxtII Nov 15 '13 at 11:32
  • $\begingroup$ H is a vector field, a measure of strength. It's not considered a kind of force. So by introducing the ferromagnetic material to an external force(Mechanical Stresses) what would be the result? Since my previous argument was two H vectors, and the ferromagnet will react to H = 40 Oe, they are fields opposing each other. Now if a force is applied to cause a disruption to the domain's alignment what would be the result? Assuming the field still exists. $\endgroup$ – AxtII Nov 22 '13 at 6:29
  • $\begingroup$ But H field is a bit like force. Certainly so if we consider magnetic system. Mechanical stress doesn't have to influence magnetization much. There are few effects that can couple stress to magnetization. Most important are magnetostriction and piezomagnetism (which of these two depends on material). There could be other indirect couplings, but they are much weaker. So to answer question what would happen - not much, probably just minor change of magnetization, probably hard to detect. $\endgroup$ – Jarosław Komar Nov 22 '13 at 10:31
  • $\begingroup$ Mechanical stress would not influence magnetization, however, could influence demagnetization if it was constant. Imagine causing constant mechanical stresses to such a ferromagnet, it's domains will disrupt. Even if H 40 Oe is present. My theory is that here we are applying to forces. Even if we did not apply an equivalent amount of force in mechanical stress it still weakens the magnetization by a huge factor. I'm I correct here? $\endgroup$ – AxtII Nov 23 '13 at 20:43
  • $\begingroup$ For very high stresses (or pressures) there could be phase change into non-ferromagnetic one. But this is usually huge (tens of GPa) pressure. Stress doesn't influence magnetization that much, because it's related to spin. And a slight bit to other magnetic moments, like nuclei moments, and orbital moment of electrons. You would have to change (or at least deform) crystal structure to change spin system. It's very weakly coupled to anything. First order approximation would be to treat it as something completely separate, dependent only on magnetic fields. $\endgroup$ – Jarosław Komar Nov 25 '13 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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