The electric fields of two plane electromagnetic waves in vacuum are given by $\vec{E}_1 = E_0\hat{j}\cos(\omega t - kx)$ and $\vec{E}_2 = E_0\hat{k}\cos(\omega t - ky)$. At $t = 0$, a particle of charge $q$ is at origin with a velocity $\vec{v} = 0.8c\,\hat{j}$ ($c$ is the speed of light in vacuum). The instantaneous force experienced by the particle is:
Answer: (D) $E_0q\left(0.8\hat{i} + \hat{j} + 0.2\hat{k}\right)$
For each wave $\vec{B}$ has magnitude $E/c$ and $\vec{E}\times\vec{B}$ points along the direction of travel.
Wave 1 travels along $+x$ with $\vec{E}$ along $\hat{j}$, so $\vec{B}_1$ is along $\hat{k}$ ($\hat{j}\times\hat{k} = \hat{i}$).
Wave 2 travels along $+y$ with $\vec{E}$ along $\hat{k}$, so $\vec{B}_2$ is along $\hat{i}$ ($\hat{k}\times\hat{i} = \hat{j}$).
At $t = 0$ at the origin both cosines are $1$:
$$\vec{E} = E_0(\hat{j} + \hat{k}),\qquad \vec{B} = \frac{E_0}{c}(\hat{i} + \hat{k})$$
$$\vec{v}\times\vec{B} = 0.8c\,\hat{j}\times\frac{E_0}{c}(\hat{i} + \hat{k}) = 0.8E_0(-\hat{k} + \hat{i})$$
$$\vec{F} = q(\vec{E} + \vec{v}\times\vec{B}) = E_0q\left(0.8\hat{i} + \hat{j} + 0.2\hat{k}\right)$$
Solution by Sreeraj P, M.Sc Physics