Partial differential equation with Faraday's equation

Question

We were asked to find what equation is satisfied by $\Psi(x,y,z,t)$ given that $\textbf{B} = \nabla \times (\textbf{z} \Psi)$ and $\textbf{E} = -\textbf{z} \frac{\partial \Psi}{\partial t}$ while making use of of Faraday's law $ \nabla \times \textbf{E} = -\frac{\partial \textbf{B}}{\partial t}$ (with z the unity vector in the z direction). I'm having trouble with calculating such an equation. When I substitute both sides and calculate the cross products, I arrive at the same expression on both sides namely: $\textbf{x}\frac{\partial^2 \psi}{\partial t \partial y} + \textbf{y}\frac{\partial^2 \psi}{\partial t \partial x}$ so that both sides cancel each other and I'm left with no equation. I don't really have an idea how to fix this and I'm starting to question whether the starting expressions are right. Because when our expressions for B and E where switched I do arrive at a satisfying solution, which is: $\frac{\partial^2 \psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2} + \frac{\partial^2 \psi}{\partial x \partial z}+\frac{\partial^2 \psi}{\partial y \partial z}= \frac{\partial^2 \psi}{\partial t^2}$. Is my assumption correct or am I missing something?

Answer 1

We are given that the magnetic flux density is given by

$$\begin{align} \vec B&=\nabla \times (\hat z\Psi) \end{align}$$

Then, using Faraday's Law, $\nabla\times \vec E=-\frac{\partial \vec B}{\partial t}$, we see that (under sufficient smoothness conditions)

$$\vec E=-\hat z\frac{\partial \Psi}{\partial t}$$

We know that in SI units, $\nabla \times \vec B=\mu_0\vec J+\frac1{c^2}\frac{\partial \vec E}{\partial t}$. Hence, we see that

$$\begin{align} \nabla \times \nabla \times \vec B&=\nabla \underbrace{\nabla \cdot \vec B}_{=0}-\nabla^2 \vec B\\\\ &=-\nabla \times \hat z\nabla^2\Psi\\\\ &=\mu_0\nabla \times \vec J+\frac1{c^2}\nabla \times \frac{\partial \vec E}{\partial t}\\\\ &=\mu_0\nabla \times \vec J+\frac1{c^2}\nabla \times \left(-\hat z \frac{\partial^2\Psi}{\partial t^2}\right) \end{align}$$

Therefore, we have

$$\nabla \times \left(\hat z\left(\nabla^2\Psi-\frac1{c^2}\frac{\partial ^2\Psi}{\partial t^2}\right)+\mu_0\vec J\right)=0$$

which implies that for any (smooth) scalar potential $\Phi$, we have

$$\nabla^2\Psi-\frac1{c^2}\frac{\partial ^2\Psi}{\partial t^2}=-\mu_0J_z+\frac{\partial \Phi}{\partial z}$$

with $\nabla_t\Phi=\vec J_t$ in terms of the transverse ("t") components of $\vec J$ and $\nabla \Phi$.