Electromagnetic Euler Force

 $$\textbf{F}=\textbf{r}\times\frac{\partial\textbf{B}}{\partial t}=-\textbf{r}\times(\nabla\times\textbf{E})$$

Wave equation of a material particle

 $$E=\frac{1}{2}m\textbf{v}^2 + \frac{1}{2}I\boldsymbol{\omega}^2$$

If we assume

$$\textbf{v}=\textbf{v}(x,t),$$ $$\boldsymbol{\omega}=\boldsymbol{\omega}(x,t),$$

then

$$A\frac{\partial\textbf{v}}{\partial x}=-I\frac{\partial\boldsymbol{\omega}}{\partial t},$$ and $$B\frac{\partial\boldsymbol{\omega}}{\partial x}=-m\frac{\partial\textbf{v}}{\partial t},$$ we arrive at

$$\boxed{\frac{mI}{AB}\frac{\partial^2\textbf{v}}{\partial t^2}= \frac{\partial^2\textbf{v}}{\partial x^2}}$$

$$\boxed{\textbf{F}=m(\textbf{a}+\textbf{v}\times\boldsymbol{\omega})}$$

Capacitor - Inductor duality

We must look for a new effect in which, the energy of a capacitor/inductor, instead of 

$$E=\frac{1}{2}CV^2,$$ $$E=\frac{1}{2}LI^2,$$ 

is given by 

$$E \propto \dot{I},$$ $$E \propto \dot{V}.$$

This can only happen with AC currents.