# Why does $\mathbf{A}(x) = \frac{1}{2}(\mathbf{B}(x) \times \mathbf{x})$ work?

In my textbook, the identity for a possible vector potential $$\mathbf{A}(x) = \frac{1}{2}(\mathbf{B}(x) \times \mathbf{x})$$ is used. How is this valid? If I compute the curl, I get $$\mathbf{B}(x) \equiv \nabla \times \mathbf{A}(x) = \frac{1}{2}\left({\bf{x}} \cdot \nabla {\bf{B}} - \left( {\nabla \cdot {\bf{B}}} \right){\bf{x}} + \left( {\nabla \cdot {\bf{x}}} \right){\bf{B}} - {\bf{B}} \cdot \nabla {\bf{x}} \right) = \frac{1}{2}\left({\bf{x}} \cdot \nabla {\bf{B}} + 3{\bf{B}} - {\bf{B}} \cdot \nabla {\bf{x}} \right)$$ but I can't see how the RHS equals $$\bf{B}$$?

• (1) Its true only for position independent B. so gradB is 0. (2) x is the component of the position vector that is perpendicular to the direction of B. Other component drops out when you take cross product in the first place. So B.grad x is 0. Commented Aug 31, 2023 at 16:07
• Just for reference, this is called the 'symmetric gauge'. Commented Aug 31, 2023 at 16:41
• @Mauricio Did not know that (just looked it up), thank you!
– EE18
Commented Aug 31, 2023 at 18:25

This identity holds only for a uniform magnetic field $$\vec B$$. Many terms cancel in your formula. It remains \eqalign{ \vec\nabla\times\vec A &={1\over 2}\big((\vec\nabla.\vec x)\vec B-(\vec B.\vec\nabla)\vec x\big)\cr &={1\over 2}\big(3\vec B-\vec B\big)\cr &=\vec B\cr } as expected.
• Thank you for your very nice answer! I just had one last question: why is ${\bf{B}} \cdot \nabla {\bf{x}} = {\bf{B}}$?
• @EE18. $\vec{B}\cdot\vec{x}$ is the directional derivative along the direction of $\vec{B}$ times the length of $\vec{B}$. The directional derivative of the position vector is just the unit vector along that direction (e.g., consider the directional derivative along the $y$-direction, i.e., $\frac{\partial}{\partial y}\vec{x}=\hat{y}$. Commented Aug 31, 2023 at 16:31
• $(\vec{B}\cdot \nabla )\vec{x}$ is meant I suppose? @march But that makes sense, thank you!