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Let's say that I have an electromagnet, consisting of Helmholtz coils with an iron core.

David Griffiths, in his "Introduction to Electricity & Magnetism" text claims that $H$ is what we set, by controlling the free current in our magnet. This seems patently untrue to me.

Here is my understanding:

  1. We run current $I_f$ through our coils
  2. This induces a magnetizations $M_c$ in our iron core
  3. To determine the total field, in the region between the two cores, we have: $H=H \text{ (from current)} +H\text{(produced by Magnetized cores)}$. We don't have any good way to "set this" on our dial either, right?

Is this picture wrong? (I guess that I hardly even have an actual question here. It just seems like every treatment of magnetism that I have seen quickly glosses over the fields produced by actual magnetic materials.

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  • $\begingroup$ Do you mean "Introduction to Electrodynamics" by Griffiths? $\endgroup$
    – AV23
    Apr 28, 2015 at 15:48

3 Answers 3

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The $H$ field is, by definition, (upto the constant $\mu_0$) essentially the contribution to the magnetic field if there were no material. We then relate that to the measurable magnetic field in the material through its permeability: $$B = \mu H$$ or $$B = \mu_0H + (\mu -\mu_0)H$$ $$ = B(\text{from current}) + B(\text{due to magnetization})$$ This is the decomposition of the magnetic field in terms of field due to the current alone and the field due to the magnetization of the core, the magnetization being expressed through $\mu$.

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I'm not sure what you question is. I think that your logic is a bit faulty:

H is what we set

Is what you think is a wrong statement.

You claim that

We run current

Which means that we set the current and given that, you further state

H = H (from current) + ...

That means that H depends on the current, it's a function thereof so to speak.

In conclusion: You say that you set I. You say that H depends on I (among other things)

And that actually confirms the statement. By setting I, you are able to set H. Or in short: H is what you set.

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  • $\begingroup$ We are NOT able to "set H" though: just one (actually somewhat small) part of it. $H \text{(from magnetized cores)}$ is definitely the dominant contribution, and we don't set that. Couldn't we just as well claim that we set $B$ inside the cores? After all, $B=H+4\pi M$ so if we set $H$ we are setting $B$ by that argument. $B$ depends on $H$. $\endgroup$ Apr 28, 2015 at 14:21
  • $\begingroup$ To set H is what you just said: you can influence its value. The fact that it is influenced by other values doesn't render his statement patently untrue. And it doesn't matter if other influences are more dominant either. Yes, because H influences other values, it allows you to influence those in turn. Following your logic, it wouldn't be possible to "set a current" by turning a dial in the first place. $\endgroup$
    – Name
    Apr 28, 2015 at 14:44
  • $\begingroup$ @Name - This is a horribly twisted way of saying (your essential point, viz.) you set $H$ because you set $I$, and $H \equiv H(I)$. That one sentence is your entire answer. This circular logic will only confuse people. $\endgroup$
    – 299792458
    Apr 28, 2015 at 14:54
  • $\begingroup$ @TheDarkSide I agree and stated that in my answer already but Dan is insisting on this kind of logic, hence I have to argue in it to render the conclusion he derives from it faulty in order to answer his question. $\endgroup$
    – Name
    Apr 28, 2015 at 15:12
  • $\begingroup$ I still don't get how you're setting $H$ any more than you're setting $B$ or anything else. $B$ is also a function of current (among other things), is it not? After all, in the region of actual interest (the gap between the cores), $H=B$ (up to a factor of $\mu_0$). I get what is happening: we set the current, which determines $H$. Inside the cores, $\mu H=B$. Then, we can calculate the field outside of the cores, generated by that magnetization. So I guess you could argue that you are setting $H$ and based on that (and your core material), you have everything. Is that what you are saying? $\endgroup$ Apr 29, 2015 at 15:02
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The total internal field $\mathbf{H}$ is the sum of two component vectors $\mathbf{H=H_c+H_d}$. Here $\mathbf{H_c}$ is the result of the external conduction currents or external permanent magnets, what you may wish to call the external or directly controllable field. It is the field in the absence of the material being magnetized. The other term $\mathbf{H_d}$ is usually called the demagnetizing field. It is related to the material's magnetic properties and its shape while the material is in the prevailing field. As an example, for a sphere $\mathbf{H_d}=-N\mathbf{M(H)}$, where $N=1/3$ is the demagnetizing factor. For an ellipsoid $N$ is a tensor. For a linear material $\mathbf{M}=\chi_c\mathbf{H_c}$ where $\chi_c$ is the external susceptibility. With the total internal field this can be written as $\mathbf{M}=\chi\mathbf{H}$ where $1/\chi = 1/\chi_c -1/N$. For small $\chi_c$ we have $1/\chi \approx 1/\chi_c$. For a toroid and soft magnet $N=0$ and therefore $\mathbf{H}=\mathbf{H_c}$, and that why is the terminology you are asking about.

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