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Why is the induced M field proportional to H, instead of the B field?

  1. Dec 9, 2013 #1
    The M field is the density of induced or permanent magnetic dipole moments. It is analogous to the P field in electrostatics. In electrostatics, the induced P field in a dielectric is proportional to the applied electric field. This is intuitive to me. Why though, in magnetostatics, is the M field proportional to the Demagnetizing field(the H field).

    M=X[itex]_{m}[/itex]H=X[itex]
    _{m}[/itex]([itex]\frac{B}{u_{0}}[/itex]+M)

    I can't seem to wrap my head around this. My text book declares it without any justification. Can anyone provide me some insight into this?

    Thank you.
     
  2. jcsd
  3. Dec 10, 2013 #2

    Meir Achuz

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    It is the B field in the material that produces M. When magnetization is produced by placing iron in a long solenoid or a torus, the H field in the iron is directly given by nI because H_t is continuous, so engineers like to consider M a function of H. But the iron knows that is is the B field that is producing M. When removed from the long solenoid or torus, end effects become dominant. That is why B and H are in opposite directions near the end of a magnet. At this point mu and chi become irrelevant. The H field is not 'demagnetrizing', although calling it so is a fairly common error of UG texts. It is B, not H, that affects M
    You are right, but don't tell your professor, because he/she probably believes the textbook.
     
  4. Dec 10, 2013 #3

    Jano L.

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    The reason behind the use of ##\mathbf H## in the definition of magnetic susceptibility is the fact that in common situations, ##\mathbf H## field is easily calculated from the electric currents in the wiring, which are easily measured by Am-meter. The field ##\mathbf B## is hard to calculate from the controlled variables such as the applied current; it has to be inferred based on the intensity of induced currents in secondary wiring.

    This is similar to definition of electrical susceptibility in electrostatics: there it is defined by

    $$
    \mathbf P = \epsilon_0 \chi_e \mathbf E,
    $$

    because the ##\mathbf E## field is easily calculable from the voltage applied to capacitor, while the ##\mathbf D## field is not.

    In macroscopic EM theory, there is not much reason for such assertion. Both fields are only macroscopic fields, describing the physical state in a different way.

    Only in microscopic theory we could attempt to identify the basic field that acts on particles and makes them magnetized. The common view is that the macroscopic field ##\mathbf B## is a kind of probabilistic description of the much more complicated microscopic magnetic field ##\mathbf b## actually exerting force on the individual charged particles. So iron responds to the presence of microscopic fields ##\mathbf e, \mathbf b## and magnetizes; but it would be a stretch to say iron feels ##\mathbf B##, because ##\mathbf B## is just a simplified concept invented by humans, incapable to account for all the myriads of nuclei and electrons in the material.
     
  5. Dec 10, 2013 #4
    Thanks for the responses guys. After reading them, my understanding is that the M field is defined this way conventionally because it is more convenient. This is allowed because, by rearranging the equation M=X[itex]_{m}[/itex]H in order to solve for M, one gets:

    M=[itex]\frac{X_{m}}{X_{m}+1}[/itex] [itex]\frac{B}{u_{0}}[/itex]

    Thus, defining M proportional to H is equivalent to defining M proportional to B, within an arbitrary constant of proportionality.
     
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