A previous example of inverting a gradient equation was the electrostatics equation. We can do the same for the magnetostatics equation, which has the following Geometric Algebra form in linear media

\label{eqn:biotSavartGreens:20}
\spacegrad I \BB = – \mu \BJ.

The Green’s inversion of this is
\label{eqn:biotSavartGreens:40}
\begin{aligned}
I \BB(\Bx)
&= \int_V dV’ G(\Bx, \Bx’) \spacegrad’ I \BB(\Bx’) \\
&= \int_V dV’ G(\Bx, \Bx’) (-\mu \BJ(\Bx’)) \\
&= \inv{4\pi} \int_V dV’ \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } (-\mu \BJ(\Bx’)).
\end{aligned}

We expect the LHS to be a bivector, so the scalar component of this should be zero. That can be demonstrated with some of the usual trickery
\label{eqn:biotSavartGreens:60}
\begin{aligned}
-\frac{\mu}{4\pi} \int_V dV’ \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \cdot \BJ(\Bx’)
&= \frac{\mu}{4\pi} \int_V dV’ \lr{ \spacegrad \inv{ \Abs{\Bx – \Bx’} }} \cdot \BJ(\Bx’) \\
&= -\frac{\mu}{4\pi} \int_V dV’ \lr{ \spacegrad’ \inv{ \Abs{\Bx – \Bx’} }} \cdot \BJ(\Bx’) \\
&= -\frac{\mu}{4\pi} \int_V dV’ \lr{
\spacegrad’ \cdot \frac{\BJ(\Bx’)}{ \Abs{\Bx – \Bx’} }

\frac{\spacegrad’ \cdot \BJ(\Bx’)}{ \Abs{\Bx – \Bx’} }
}.
\end{aligned}

The current $$\BJ$$ is not unconstrained. This can be seen by premultiplying \ref{eqn:biotSavartGreens:20} by the gradient

\label{eqn:biotSavartGreens:80}

On the LHS we have a bivector so must have $$\spacegrad \BJ = \spacegrad \wedge \BJ$$, or $$\spacegrad \cdot \BJ = 0$$. This kills the $$\spacegrad’ \cdot \BJ(\Bx’)$$ integrand numerator in \ref{eqn:biotSavartGreens:60}, leaving

\label{eqn:biotSavartGreens:100}
\begin{aligned}
-\frac{\mu}{4\pi} \int_V dV’ \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 } \cdot \BJ(\Bx’)
&= -\frac{\mu}{4\pi} \int_V dV’ \spacegrad’ \cdot \frac{\BJ(\Bx’)}{ \Abs{\Bx – \Bx’} } \\
&= -\frac{\mu}{4\pi} \int_{\partial V} dA’ \ncap \cdot \frac{\BJ(\Bx’)}{ \Abs{\Bx – \Bx’} }.
\end{aligned}

This shows that the scalar part of the equation is zero, provided the normal component of $$\BJ/\Abs{\Bx – \Bx’}$$ vanishes on the boundary of the infinite sphere. This leaves the Biot-Savart law as a bivector equation

\label{eqn:biotSavartGreens:120}
I \BB(\Bx)
= \frac{\mu}{4\pi} \int_V dV’ \BJ(\Bx’) \wedge \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 }.

Observe that the traditional vector form of the Biot-Savart law can be obtained by premultiplying both sides with $$-I$$, leaving

\label{eqn:biotSavartGreens:140}
\BB(\Bx)
= \frac{\mu}{4\pi} \int_V dV’ \BJ(\Bx’) \cross \frac{\Bx – \Bx’}{ \Abs{\Bx – \Bx’}^3 }.

This checks against a trusted source such as [1] (eq. 5.39).

# References

[1] David Jeffrey Griffiths and Reed College. Introduction to electrodynamics. Prentice hall Upper Saddle River, NJ, 3rd edition, 1999.