# Confusion about $B$-field being expressed in terms of magnetic scalar potential using auxiliary field $H$

Potentially silly question, sorry! I understand that the $$B$$-field can be expressed as the curl of $$A$$, where $$A$$ is the magnetic vector potential, and so $$B$$ has no divergence. The auxiliary field $$H$$ however, is not necessarily divergenceless, and may have curl also (if free currents are present). I am confused about the following:

If there are no free currents present, then: $$\nabla \times H = \vec{0} \rightarrow H = -\nabla \phi_m$$

where $$\phi_m$$ is a scalar magnetic potential. In linear media, $$\vec{B} = \mu \vec{H}$$. Therefore:$$\vec{B} = -\mu \nabla \phi_m$$ I am confused by this equation, because if B only has curl, how can it be expressed in terms of the grad of a scalar potential like this? Doesn't this expression imply that B has divergence, which is not possible? I am not sure where the error is coming from in this line of reasoning. Could someone please clarify?

• Why would it imply divergence of $\mathbf B$ does not vanish? This is a relation between $H$ and $B$; if anything, it means divergence of $H$ need not vanish, if $\mu$ varies with position. If $\mu$ is the same everywhere, it's like electrostatic field in vacuum and divergence of both $H,B$ vanishes. Feb 6 at 0:53

Since $$\nabla \times \nabla \phi=0$$ whenever $$\nabla \phi$$ is differentiable $$\nabla \times \mathbf B = \nabla\times (\mu \nabla \phi)\\ =\mu \nabla \times \nabla \phi+\nabla\mu \times \nabla \phi\\ =\nabla\mu \times \nabla \phi$$
$$\nabla \cdot \mathbf B = \nabla\cdot(\mu \nabla \phi)\\ =\mu \nabla^2\phi+\nabla\mu \cdot \nabla \phi\\$$
If $$\mu$$ is constant in a region then there $$\nabla \mu = 0$$ and thus $$\nabla \cdot \mathbf B =\mu \nabla^2\phi =0$$ and we get the special case of Laplace's equation for the magnetic potential $$\nabla ^2 \phi=0.$$ Otherwise, in general, the only thing we know is that $$\mu \nabla^2\phi+\nabla\mu \cdot \nabla \phi=0.$$
Even if $$\mu$$ is "piece-wise" constant, that is $$\nabla \mu = 0$$ in discrete regions, you have at the boundary of those regions jump conditions: the normal component $$B_n$$ of $$\mathbf B$$ is continuous and the tangential component $$H_t$$ of $$\mathbf H$$ is continuous. These continuity conditions can be interpreted as surface sources generating a non-zero field within the constant $$\mu$$ regions.