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In my lecture notes for magnetostatics, my professor has this explanation of why H is not necessarily $0$ that I dont understand.

$$\nabla \times \bf{H} = \bf{J} \\ \bf{J}=0 \Rightarrow \nabla \times \bf{H} = 0 \not\Rightarrow H = 0 \\ \nabla \cdot H \neq 0 $$

"H is only fully defined by its curl and divergence."

I thought that the divergence of B is always zero and since B and H are related by only a constant divergence of H should also be $0$ so I am not sure why the last expression is there. Please help me understand his explanation.

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The field $\vec{H}$ in a medium is defined as $$\vec{H}(r,t) = \frac{1}{\mu_0}\vec{B}(r,t)- \vec{M}(r,t)$$ where $M(r,t)$ is the magnetization field which depends on the medium properties. Maxwell's law states that $\vec{\nabla}.\vec{B}=0$, i.e., magnetic monopole does not exist. But that does not imply that there is not magnetic polarization inside the medium. The presence of the $\vec{M}$ field may give rise to the nonzero divergence of $\vec{H}$.

However, if we are talking about the vacuum or a medium that follows $\vec{H}(r,t) = \frac{1}{\mu}\vec{B}(r,t)$ where $\mu$ is just a constant independent of spatial coordinates (usually linear media), then from Maxwell's law it follows that $\vec{\nabla}.\vec{H}=0$.

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    $\begingroup$ I cant fathom how adding curl to curl results in divergence. $\endgroup$
    – R. Emery
    Commented Jan 9, 2021 at 5:08
  • $\begingroup$ I am not sure how is that relevant to this question. Anyway, you can find the derivation of the expression of $\nabla\times\nabla\times \vec{A}$ in any standard vector analysis book. $\endgroup$
    – abhijit975
    Commented Jan 9, 2021 at 18:45

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