# Does a homogeneous oscillating electric field produce a magnetic field?

I am working on a homework problem that says an electron in a continuous laser field can be modeled as experiencing a homogeneous oscillating electric field $$\vec{E}(\vec{r},t)=\cos \omega t \ \hat {z}$$.

It then proceeds to say that a suitable gauge choice is magnetic potential $$\vec{A}=0$$ and electric potential $$\phi(\vec r,t) = -\cos \omega t \ \hat {z} \cdot \vec r$$.

This means that the magnetic field $$\vec B=0$$.

But Maxwell's equation $$\mathbf{\nabla \times B} = \mu_0 \mathbf{j} + \frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}$$ says that a changing electric field will produce a nonzero curl of the magnetic field so $$\vec B$$ cannot be zero.

What am I getting wrong here?

• Why would B = 0? – ggcg May 8 '19 at 15:34
• You are correct. You have just proven that the pre-supposed approach $\phi(\vec{r},t) = -\cos(\omega t)\hat{\vec{z}}\cdot\vec{r}$ and $\vec{A}(\vec{r},t)=\vec{0}$ is wrong. – Thomas Fritsch May 8 '19 at 16:10
• @ggcg I guess it is because $\vec{A}$ is zero and $B = \nabla \times A$. – Abhay Hegde May 8 '19 at 16:24

Basically, what you're saying is that a homogeneous electric field $$\vec{E}(\vec{r},t)=\cos \omega t \ \hat {z}$$ in vacuum requires you to have a magnetic field that satisfies $$\nabla \times \mathbf{B} =\frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t},$$ giving you a non-homogeneous magnetic field, which would then require the electric field to satisfy $$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t},$$ that therefore needs an inhomogeneous electric field, in conflict with your initial assumption? Or, in other words, you're saying that the electric field assumed by the question, $$\vec{E}(\vec{r},t)=\cos \omega t \ \hat {z}$$, is inconsistent with the Maxwell equations?