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The vector potential $A$ is perpendicular to $B = \nabla \times A$, by definition, and hence, in a plane wave, it is either in the direction of $E$ or the direction of propagation. I suspect it is in the direction of propagation.

What is its direction?

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While for vectors $\vec{B}$ and $\vec{C}$, the cross product $\vec{B}\times\vec{C}$ is indeed perpendicular to both of the vectors, it is simply not the case that the curl of a vector field is orthogonal to the vector field. Do not read too much into the cross product notation.

In particular, you can add any constant vector field to $\vec{A}$ without changing the fields. So we can make it be nonorthogonal by adding a constant of our choice. When someone tells you a vector potential points in a particular direction they are simply making a gauge choice, and a different choice of gauge can result in the vector potential pointing in a different direction.

This means your question simply isn't well defined. We can find the direction of the electric field by seeing the force per unit charge of stationary charges, and we can find the magnetic field by finding the force on moving charges that move in three linearly independent directions. But there is no classical experiment to find the direction the vector potential points, so it isn't a scientific question.

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The vector potential is transverse to the direction of motion. To see this, perform a gauge transformation into the Coulomb gauge, such that $\nabla\cdot\mathbf{A}=0$. Using a plane wave solution $\mathbf{A}_0\cos(\mathbf{k}\cdot\mathbf{r}-\omega t)$ we then find $\mathbf{k}\cdot\mathbf{A}_0=0$, i.e. the potential is perpendicular to the direction of motion.

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  • $\begingroup$ In specifying that plane wave solution, aren't you choosing $\mathbf{A}$ to be perpendicular to $\mathbf{k}$? $\endgroup$ – Luke Burns Apr 1 '15 at 1:19
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    $\begingroup$ @LukeBurns Even if you select the plane wave solution for the electric and magnetic fields, there are still an infinite number of choices about the direction in which the vector potential can point. $\endgroup$ – Timaeus Apr 1 '15 at 1:22
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In general,

$$\mathbf E = -(\nabla \phi + \frac{\partial \mathbf A}{\partial t})$$

For the source free case, and in the transverse gauge with appropriate boundary conditions on $\phi$,

$$\mathbf E = -\frac{\partial \mathbf A}{\partial t}$$

Which can be checked by taking the curl of both sides

$$\nabla \times \mathbf E = -\nabla \times \frac{\partial \mathbf A}{\partial t} = -\frac{\partial}{\partial t}(\nabla \times \mathbf A) = -\frac{\partial \mathbf B}{\partial t} $$

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Magnetic vector potential is directed perpendicular to B. If B is along x axis it can have equal probabillity to be in any one of y or z direction. So there is no definite direction of A.

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  • $\begingroup$ Can you please add a few more details? Answers are usually expected to be a bit longer. Also consider using more precise language and avoiding typos. $\endgroup$ – user191954 Jul 12 '18 at 11:06
  • $\begingroup$ The vector potential isn't perpendicular to the magnetic field in general. $\endgroup$ – Chris Jul 12 '18 at 12:27

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