# What is the magnitude of the force on a charged particle due to electromagnetic radiation?

Suppose there is an electromagnetic wave moving forward in the $\mathbf{\hat{k}}$ direction. Its electric/magnetic field components are given by: $$\mathbf{E} = E_0 \sin(kz - \omega t) \mathbf{\hat{i}}$$ $$\mathbf{B} = B_0 \sin(kz - \omega t) \mathbf{\hat{j}}$$ If a particle of charge $q$ was lying on the wave's trajectory, the Lorentz force law says that the force is given $\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$. However, is an electromagnetic wave a combination of both E and B fields, requiring both fields to be plugged into the equation, or does the force on the electron only depend on one of the fields, and is an EM wave only either an electric field or a magnetic field at one instant? Edit: changed x to z in expression for EM wave.

Note. As indicated by user23660, the EM wave must be transverse which means the $x$'s in your phases should instead be $z$'s.
At a given time $t$ and spatial point $\mathbf x = (x,y,z)$, the electromagnetic wave you consider is a combination of both of the fields; \begin{align} \mathbf E(t,\mathbf x) &= E_0\sin(kz-\omega t) \hat{\mathbf x} \\ \mathbf B(t,\mathbf x) &= B_0\sin(kz-\omega t) \hat{\mathbf y} \end{align} There are some special points at which both of the fields vanish though. In particular, any time the argument of the $\sin$ is an integer multiple of $\pi$; \begin{align} kz-\omega t = n\pi, \qquad n\in\mathbb Z \end{align} As a result, a particle sitting in the wave will experience both of the fields at once, and both of these fields will have to be plugged into the Lorentz force equation. Explicitly, Newton's Second Law along with the Lorentz force equation with both fields plugged in gives us the following equation of motion: \begin{align} \ddot{\mathbf x} =\frac{q}{m}(E_0\sin(kz-\omega t) \hat{\mathbf x} +B_0\sin(kz-\omega t)\dot{\mathbf x}\times\hat{\mathbf y}). \end{align} In components, this can be written as the following system of coupled differential equations: \begin{align} \ddot x &= \omega_0\sin(kz-\omega t)(c - \dot z) \\ \ddot y &= 0 \\ \ddot z &= \omega_0\sin(kz-\omega t) \dot x \end{align} where I've used the relationship $E_0 = cB_0$ and I have defined \begin{align} \omega_0 = \frac{qB_0}{m}. \end{align} As far as I can tell, this is a pretty nasty system, and I'm not sure if the general solution can be written in closed form (although admittedly I haven't really tried very hard to figure that out.) It's actually not so bad since $y$ is completely decoupled from $x$ and $z$, and it's differential equation simply implies constant velocity in $y$. This leaves a pair of coupled equations for $x$ and $z$.