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

1. Dec 9, 2013

### burgjeff

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$_{m}$H=X$_{m}$($\frac{B}{u_{0}}$+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. Dec 10, 2013

### Meir Achuz

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.

3. Dec 10, 2013

### Jano L.

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.

4. Dec 10, 2013

### burgjeff

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$_{m}$H in order to solve for M, one gets:

M=$\frac{X_{m}}{X_{m}+1}$ $\frac{B}{u_{0}}$

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