# Does zero free current entail zero $\vec H$?

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...

• 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. – Michael Seifert Mar 11 '16 at 21:26

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.
• 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 $\vec B = \mu_o \vec M$? – quarkleptonboson Nov 15 '14 at 15:28
• @quarkleptonboson You'd have to be specific. In LIH media, what you suggest implies $\vec{H}=\vec{M}/\mu_r$. – Rob Jeffries Nov 15 '14 at 17:22