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I am studying some magnetism, and i am curious about bar magnets.

Lets say you have an uniform prism magnet of Iron for example. How would you calculate the demagnetization field H which the bar magnet produces? As I understand, first you need the magnetization M which is the magnetic moments per volumen. But then what ? Or would you apply and external field to the bar magnet and see how it reacts, if this is the case, how does this work ?

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If you know M then H is just $\frac{M}{\chi_m}$ where $\chi_m$ is the magnetic susceptibility. This only holds for paramagnets and diamagnets, however. In the case of a ferromagnetic material like iron, you would have to refer to the hysteresis curve. If you know the net (= applied+induced) magnetic field B in the material, then you can use any of these relations

$$ \begin{align*} H=& \frac{B_{net}}{\mu}\\ H=& \frac{B_{applied}}{\mu_0}\\ H=& \frac{B_{net}}{\mu_0}-M\\ H=& \frac{B_{applied}+\mu_0M}{\mu}\qquad{(\because B_{induced}=\mu_0M)} \end{align*} $$where $\mu=$permeability.

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  • $\begingroup$ Thank you, but how come do you need an external field to determine the H/B magnetic field? $\endgroup$ – Elias S. Mar 13 '17 at 17:53
  • $\begingroup$ In the case of para- and diamagnets, one needs an external field to magnetize the matter: only then will the matter create its own field by aligning its domains. Once the external field is removed, the matter loses its magnetization and goes back to the random alignment of magnetic moments. However, in the case of a ferromagnet, the induced field is retained even when the external field is removed--this is called magnetic hysteresis, so in this case you wouldn't need an external field to determine H. $\endgroup$ – GeeJay Mar 14 '17 at 6:45

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