\[\mathbf{\nabla} \cdot \mathbf{H} =\mathbf{0}\]
and \[\mathbf{\nabla} \times \mathbf{H} = \mathbf{J}\]
Since
\[\mathbf{\nabla} \cdot \mathbf{H} =\mathbf{0}\]
we can write
\[\mathbf{H} = \mathbf{\nabla} \times \mathbf{A} \]
for a suitable vector field \[\mathbf{A} \]
, called the magnetic vector potential.The choice of
\[\mathbf{H}\]
is not uniqu. In fact we can find \[\mathbf{A, \;A'} \]
such that \[\mathbf{H} = \mathbf{\nabla} \times \mathbf{A} = \mathbf{\nabla} \times \mathbf{A'} \]
In this case
\[\mathbf{\nabla} \times (\mathbf{A} - \mathbf{A'})= \mathbf{0}\]
.It then follows that
\[\mathbf{A}- \mathbf{A'} = \mathbf{\nabla} \phi \rightarrow \mathbf{A} = \mathbf{A'} + \mathbf{\nabla} \phi\]
Where
\[\phi\]
is some scalar function.Taking the divergence of the second equation above gives
\[\mathbf{\nabla} \cdot \mathbf{A} =\mathbf{\nabla} \cdot \mathbf{A'} + \nabla^2 \phi\]
We can choose
\[\phi\]
so that \[\nabla \cdot \mathbf{A} = 0\]
Hence, in magnetostatics, the magnetic vector potential is chosen so that
\[\mathbf{H} =\mathbf{\nabla} \times \mathbf{A}\]
and \[\mathbf{\nabla} \cdot \mathbf{A} = \mathbf{0}\]
Substitute
\[\mathbf{H} = \mathbf{\nabla} \times \mathbf{A} \]
into \[\mathbf{\nabla} \times \mathbf{H} = \mathbf{J}\]
to obtain \[\mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{A}) = \mathbf{J}\]
Now use the identity
\[\mathbf{\nabla} \times (\mathbf{\nabla} \times \mathbf{A}) = \mathbf{\nabla} (\mathbf{\nabla} \cdot \mathbf{A})- \nabla^2 \mathbf{A} \]
Since
\[ \mathbf{\nabla} (\mathbf{\nabla} \cdot \mathbf{A})=0\]
, we have\[ \nabla^2 \mathbf{A}= - \mathbf{J} \]