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I have been teaching myself magnetostatics for some time with some help from texts by Griffiths, Jackson, and other online sources. My approach is a little unorthodox in that I try to ignore electrostatics and free currents to the extent possible. Maybe not the best approach but one way or another, I think I have gained a pretty good understanding of the topic. However, some pockets of confusion remain.

One concept I still struggle with is knowing when $\nabla \times \boldsymbol{B} = \bf 0$ is applicable. I know it is true for a constant field and for a dipole field. But is there a general class of problems for which it is always true?

Another related question: I know that if there are no free currents, then $\nabla \times \boldsymbol{H} = \bf 0$ and we can write a Laplace equation for scalar potential $\Phi$. But is that also true for the analogous case of the vector potential $\bf A$ where $\nabla \times \boldsymbol{B} = \bf 0$? Or do we have to account for bound currents in the case of nonzero magnetization $\bf M$ ? The term "current" is not always defined clearly.

My two main applications of interest are 1) a permeable sphere or ellipse in a constant field, and 2) a permanently magnetized sphere or ellipse.

If these topics are summarized succinctly anywhere in a reference, I would appreciate knowing about it.

Thanks in advance.

Mariano G
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1 Answers1

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The vector potential exists because $\nabla \cdot \bf{B}=0$ always and everywhere. In the static case, $\partial_t \bf{D}=0,$ and then $\nabla \times \bf{H}=\bf{J}$ and wherever $\bf{J}=0$ there is a local potential field such that $\bf{H}=\nabla \psi.$ By local I mean outside of any macroscopic current in a singly connected domain. If the medium is both linear and homogeneous so that $\bf{B}=\mu \bf{H}=\mu \nabla \psi$ then there also holds $\nabla \times \bf{B}=0.$

But if $\mu$ is not a constant but varies spatially, $\nabla \mu \ne 0,$ then $$\nabla \times \bf{B}=\nabla \mu \times \bf{H}+\mu \nabla \times \bf{H}=\nabla \mu \times \bf{H}\ne0.$$ In fact, the so-called bound currents are the sources, writing $$\bf{B}=\mu_0(\bf{H}+\bf{M})$$ you get $$\nabla \times \bf{B}=\mu_0 (\nabla \times \bf{H}+\nabla \times \bf{M})=\mu_0 \nabla \times \bf{M}=\mu_0 \bf{J_b}$$ where the bound current is defined by $\bf{J_b}=\nabla \times \bf{M}.$

This post may also help The form of $H$ in a bar magnet.

hyportnex
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