There are two kinds of magnetic fields (different authors give them different names), $\vec B $ and $\vec H$ which are related by the equation $$ \vec B = \mu_o (\vec H + \vec M)$$ where $\vec M$ is the magnetization.

Ampere's law for free currents states $$\oint_C \vec H \cdot d\vec l = I_{free} $$

This is my question: does zero free current entail zero $\vec H$?

My argument for that is this: since the contour integral of $\vec H$ is zero for all arbitrary curves C in a region of zero free current, $\vec H$ is necessarily zero. However this may not be a mathematically correct argument...

  • $\begingroup$ Your argument concerning the vanishing of $\vec{H}$ could be applied to any vector with zero curl. In particular, you could use the same logic to prove that the electric field vanished everywhere in space. $\endgroup$ – Michael Seifert Mar 11 '16 at 21:26

I think your argument is completely flawed.

Consider a uniform H-field. The closed line integral of this field around any loop is zero - and there must be no free current through the loop. Hence there is no free current, yet the H-field is non-zero.

You might be better off thinking about this in terms of free current density.

In this case we can say $$\nabla \times {\bf H} = {\bf J},$$ so that a zero free current density just means that the H-field has zero curl at that point, nothing more than that.

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  • $\begingroup$ oh, i didn't think of that... so given a system with only bound current (zero free current everywhere), $\vec H$ can still be nonzero? But why do some problems end up with $\vec B = \mu_o \vec M$? $\endgroup$ – quarkleptonboson Nov 15 '14 at 15:28
  • $\begingroup$ @quarkleptonboson You'd have to be specific. In LIH media, what you suggest implies $\vec{H}=\vec{M}/\mu_r$. $\endgroup$ – Rob Jeffries Nov 15 '14 at 17:22

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