The electric field of a plane electromagnetic wave is given by $\vec E = E_0\dfrac{\hat i + \hat j}{\sqrt2}\cos(kz + \omega t)$. At $t = 0$, a positively charged particle is at the point $(x, y, z) = \left(0, 0, \dfrac\pi k\right)$. If its instantaneous velocity at $t = 0$ is $v_0\hat k$, the force acting on it due to the wave is:
Answer: (C) Antiparallel to $\dfrac{\hat i + \hat j}{\sqrt2}$
At the particle, $\cos(k\cdot\frac\pi k + 0) = \cos\pi = -1$, so
$$\vec E = -E_0\frac{\hat i + \hat j}{\sqrt2}$$
The phase $kz + \omega t$ means the wave travels along $-z$, so $\vec B = \dfrac1c(-\hat k)\times\vec E$.
Magnetic force:
$$q\vec v\times\vec B = \frac{qv_0}{c}\hat k\times(-\hat k\times\vec E) = \frac{qv_0}{c}\vec E$$
(using $\vec E\perp\hat k$).
Total force: $q\vec E\left(1 + \dfrac{v_0}{c}\right)$, which points along $\vec E$, i.e. antiparallel to $\dfrac{\hat i + \hat j}{\sqrt2}$.
(The official answer key lists 'parallel to $\frac{\hat i + \hat j}{\sqrt2}$', which ignores that $\cos\pi = -1$ at this point.)
Solution by Sreeraj P, M.Sc Physics