# Can changing electric field produce static magnetic field.

As we know that the maxwells equation in free space are
$$\nabla\cdot \vec{E}=0$$ $$\nabla\cdot \vec{B}=0$$ $$\nabla\times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$$ $$\nabla \times \vec{B}= \frac{1}{c^2}\frac{\partial \vec{E}}{\partial t}$$
Now if we take $\nabla \times \vec{E}=0$, that is if $\frac{\partial \vec{B}}{\partial t}=0$, will the $\frac{\partial \vec{E}}{\partial t}$ be non zero. From here we can see that it can not represent a em wave equation. That's why I am curuius it will lead to $\frac{\partial \vec{E}}{\partial t}=0$ or it will still remain non zero. I have tried several ways to solve the equation taking $\nabla \times \vec{E}=0$ in the third equation but could not arrive at any conclution.

• Under the condition $\:\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{E}=\boldsymbol{0}\:$ differentianting the 4th equation with respect to $\:t\:$ we have $$\dfrac{\partial^{2} \mathbf{E}}{\partial t^{2}}=\boldsymbol{0} \quad \Longrightarrow \quad \dfrac{\partial\mathbf{E}}{\partial t}=\textbf{constant in time} \ne \boldsymbol{0} \quad \text{in general}$$ Dec 13, 2016 at 9:10

$\frac{\partial \vec{B}}{\partial t}=0$ does not necessarily mean $\frac{\partial \vec{E}}{\partial t}=0$.
Consider the whole space $\mathbb R^3$ and assume that therein the electric field varies in time like this $$\vec{E}(t,x,y,z) = tE \: \bf e_z$$ for some constant $E>0$. Let us also assume that the magnetic field $\vec{B}$ is instead stationary like this $$\vec{B}(t,x,y,z) = \frac{Ex}{c^2} \: \bf e_y\:.$$ We have $$\nabla\cdot \vec{E} = \frac{\partial }{\partial z}tE = 0$$ $$\nabla\cdot \vec{B}= \frac{\partial }{\partial y}Ex = 0$$ $$\nabla\times \vec{E}=0 =-\frac{\partial \vec{B}}{\partial t}$$ $$\nabla \times \vec{B}= \frac{1}{c^2} E {\bf e_z}=\frac{1}{c^2}\frac{\partial \vec{E}}{\partial t}$$ You see that Maxwell's equations are therefore satisfied. It is interesting to notice that redefining $\vec{B}$ by replacing ${\bf e_y}$ for every other unit vector normal to ${\bf e_z}$, we would obtain another solution, since the problem is invariant under rotations around $z$. This means that $\vec B$ can only be fixed by giving boundary conditions breaking this symmetry.
Why would $\nabla\times\vec{B}=0$ just because $\frac{\partial\vec{B}}{\partial{}t}=0$? Spatial and time derivatives are not the same thing. Neither side of equation (4) is necessarily zero.