Given electric field

$\vec E = E_0\,\frac{\hat i+\hat j}{\sqrt 2}\cos(kz+\omega t)$

For a plane electromagnetic wave, the magnetic field is perpendicular to both the propagation direction and the electric field.

Find direction of propagation

The phase is $kz+\omega t$.

A wave of the form $\cos(kz-\omega t)$ travels in $+z$ direction, while $\cos(kz+\omega t)$ travels in the negative $z$-direction.

So propagation is along $-\hat k$.

Electric field at the given instant and position

At $t=0$ and $z=\pi/k$,

$kz+\omega t = k\left(\frac{\pi}{k}\right)+0=\pi$

Hence, $\cos\pi=-1$

Therefore, $\vec E = -E_0\frac{\hat i+\hat j}{\sqrt 2}$

So the electric force on a positive charge is along $-\frac{\hat i+\hat j}{\sqrt 2}$

Magnetic field direction

For an electromagnetic wave, $\vec B = \frac{1}{c}\,\hat n\times \vec E$ where $\hat n$ is the propagation direction.

Here, $\hat n=-\hat k$ and $\vec E \parallel -\frac{\hat i+\hat j}{\sqrt 2}$$

Thus $\vec B$ lies in the transverse plane. But we do not even need its exact direction yet; we only need the magnetic force.

Magnetic force on the particle

The particle velocity at $t=0$ is $\vec v=v_0\hat k$

Lorentz force: $\vec F = q(\vec E + \vec v\times \vec B)$

Since $\vec B$ is transverse (in the $xy$ plane), $\vec v\times\vec B$ also lies in the $xy$ plane.

More specifically, for a plane wave, $\vec B = \frac{1}{c}\hat n\times \vec E$ with $\hat n=-\hat k$.

Then $\vec v\times\vec B = v_0\hat k\times \left(\frac{1}{c}(-\hat k\times \vec E)\right)$

Using vector identity, $\hat k\times(-\hat k\times \vec E)=\vec E$ because $\vec E\perp \hat k$.

Hence, $\vec v\times\vec B = \frac{v_0}{c}\vec E$

Therefore, $\vec F=q\left(\vec E+\frac{v_0}{c}\vec E\right)=q\left(1+\frac{v_0}{c}\right)\vec E$

So the total force is parallel to $\vec E$.

Direction of total force

Since $\vec E=-E_0\frac{\hat i+\hat j}{\sqrt 2},$ the force is along $-\frac{\hat i+\hat j}{\sqrt 2}$

i.e. antiparallel to $\frac{\hat i+\hat j}{\sqrt 2}$

Correct option

$\boxed{\text{C}}$