How do you model the magnetic moment that's induced by an external magnetic field? So basically if you have an external B field, and we put a ferromagnetic rod in that field, what would be the rod's magnetic moment?
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$\begingroup$ Are you referring to "soft" magnetization which disappears when the external field is removed, or "hard" magnetization where the magnetization remains after the external field is removed? The answers given are sort of a mix of both. $\endgroup$– Mariano GCommented Aug 20 at 19:37
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3 Answers
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In Gaussian units, ${\bf B=H}+4\pi{\bf M}$. If it is long and thin, $\bf H$ is negligible, and ${\bf M =B}/4\pi$ for most of the length. The magnetic moment is $\bf M$ times the volume. When $\bf B$ is removed, $\bf M$ will reduce to smaller value, depending on the material.
answered Jul 28, 2020 at 19:48
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So basically if you have an external B field, and we put a ferromagnetic rod in that field, what would be the rod's magnetic moment?
First, you would have to know if the ferromagnetic material is above or below its Curie temperature.
If it is above its Curie temperature, then it is in a paramagnetic state and the equation for the magnetization, M, is: $$M=\frac{B_a C}{(T-T_c)^{1.33}}$$ where, $B_a$ is the applied magnetic field, $T$ is the temperature, $T_c$ is the ferromagnetic transition temperature, $C$ is the Curie constant, and the 1.33 is approximate.
If the ferromagnetic material is below its Curie temperature, then, in the absence of an external magnetic field, real materials will have magnetic domains set up internally to minimize the magnetic energy. These domains will have their magnetization vectors pointing in different directions so that the overall magnetization will be small.
Application of an external magnetic field, $B_a$, will favor an increase in the overall magnetization in the direction of the field. There are two processes:
- In weak applied fields, the volume of domains aligned with the field grow at the expense of those not aligned, and
- In strong fields, the magnetization vector rotates toward the field direction.
Kittel, at least my 5th edition, covers this very well.
The full change in magnetization is obviously dependent on how long the magnetic field is applied, how strong the field is, and likely other parameters. There is no simple equation here that I am aware of.
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First you need to find the magnetization $M$, this is equal to $\chi_mH$ where $H=\frac{B}{\mu}$. Then the magnetic moment is just $\iiint{MdV}$, over the volume of the rod.
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$\begingroup$ The problem is that $ M = \chi_mH $ applies only in diamagnets and paramagnets. But my rod is made of iron, a ferromagnetic material. Do you have any idea of the link between $M$ and the H field with ferromagnetic materials? $\endgroup$ Commented Mar 30, 2017 at 19:25
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$\begingroup$ You'll need to refer to the hysteresis curve of that particular ferromagnetic material. $\endgroup$– GeeJayCommented Mar 31, 2017 at 12:04