The electric field of a plane electromagnetic wave is given by $$\vec{E} = E_0 \frac{\hat{i}+\hat{j}}{\sqrt{2}}\cos(kz + \omega t)$$. At $$t = 0$$, a positively charged particle is at the point $$(x, y, z) = \left(0, 0, \frac{\pi}{k}\right)$$. If its instantaneous velocity at $$(t = 0)$$ is $$v_0 \hat{k}$$, the force acting on it due to the wave is:

We have the electric field of the plane electromagnetic wave written as

$$\vec E = E_0\,\frac{\hat i+\hat j}{\sqrt2}\,\cos(kz+\omega t).$$

For a monochromatic plane wave in free space the magnetic field is related to the electric field by Faraday’s law. Stating the result in vector form for a wave whose space-time dependence is $$\cos(\vec k\!\cdot\!\vec r+\omega t)$$, we get

$$\vec B = -\,\frac1\omega\,\vec k \times \vec E.$$

Here $$\vec k = k\,\hat k$$, so

$$\vec B = -\,\frac k\omega\,\hat k \times \vec E = -\,\frac1c\,\hat k \times \vec E,$$ because $$c=\dfrac\omega k.$

At the instant $$t=0$$ the particle is located at

$$z = $$\frac$$$$\pi$$ k,$$ so the phase of the cosine is

$$kz+$$\omega$$ t = k\!$$\left$$($$\frac$$$$\pi$$ k$$\right$$)+0=$$\pi$$.$$

Hence

$$$$\cos$$\!\bigl(kz+$$\omega$$ t\bigr)=$$\cos$$$$\pi$$=-1,$$ and therefore the electric field at the given point and time is

$$$$\vec$$ E(0) = E_0\,$$\frac{\hat i+\hat j}{\sqrt2}$$\,(-1) = -\,$$\frac{E_0}{\sqrt2}$$\,($$\hat$$ i+$$\hat$$ j).$$

To obtain the magnetic field we first calculate the cross-product $$$$\hat$$ k$$\times$$$$\vec$$ E$$:

$$$$\hat$$ k$$\times$$$$\vec$$ E = $$\hat$$ k $$\times$$\!$$\left$$[-\,$$\frac{E_0}{\sqrt2}$$($$\hat$$ i+$$\hat$$ j)$$\right$$] = -\,$$\frac{E_0}{\sqrt2}$$\bigl($$\hat$$ k$$\times$$$$\hat$$ i+$$\hat$$ k$$\times$$$$\hat$$ j\bigr).$$

Using the right-hand rule, $$$$\hat$$ k$$\times$$$$\hat$$ i=$$\hat$$ j$$ and $$$$\hat$$ k$$\times$$$$\hat$$ j=-$$\hat$$ i,$$ so

$$$$\hat$$ k$$\times$$$$\vec$$ E = -\,$$\frac{E_0}{\sqrt2}$$\,($$\hat$$ j-$$\hat$$ i) = $$\frac{E_0}{\sqrt2}$$\,($$\hat$$ i-$$\hat$$ j).$$

Substituting this in the magnetic-field relation, we get

$$$$\vec$$ B(0) = -$$\frac$$1c\,\bigl($$\hat$$ k$$\times$$$$\vec$$ E\bigr) = -\,$$\frac$$1c\,$$\frac{E_0}{\sqrt2}$$($$\hat$$ i-$$\hat$$ j) = $$\frac{E_0}{c\sqrt2}$$\,(-$$\hat$$ i+$$\hat$$ j).$$

The positively charged particle has instantaneous velocity

$$$$\vec$$ v = v_0\,$$\hat$$ k.$$

The Lorentz force formula, stated explicitly, is

$$$$\vec$$ F = q\bigl($$\vec$$ E+$$\vec$$ v$$\times$$$$\vec$$ B\bigr).$$

We now compute the magnetic part $$$$\vec$$ v$$\times$$$$\vec$$ B$$:

$$$$\vec$$ v$$\times$$$$\vec$$ B = v_0\,$$\hat$$ k $$\times$$\!$$\left$$[$$\frac{E_0}{c\sqrt2}$$\,(-$$\hat$$ i+$$\hat$$ j)$$\right$$] = $$\frac{v_0E_0}{c\sqrt2}$$\,\bigl($$\hat$$ k$$\times$$(-$$\hat$$ i)+$$\hat$$ k$$\times$$$$\hat$$ j\bigr).$$

As before, $$$$\hat$$ k$$\times$$(-$$\hat$$ i)=-$$\hat$$ j$$ and $$$$\hat$$ k$$\times$$$$\hat$$ j=-$$\hat$$ i,$$ giving

$$$$\vec$$ v$$\times$$$$\vec$$ B = $$\frac{v_0E_0}{c\sqrt2}$$\,(-$$\hat$$ j-$$\hat$$ i) = -\,$$\frac{v_0E_0}{c\sqrt2}$$\,($$\hat$$ i+$$\hat$$ j).$$

Adding the electric and magnetic contributions,

$$$$\vec$$ E + $$\vec$$ v$$\times$$$$\vec$$ B = -\,$$\frac{E_0}{\sqrt2}$$\,($$\hat$$ i+$$\hat$$ j) -\,$$\frac{v_0E_0}{c\sqrt2}$$\,($$\hat$$ i+$$\hat$$ j) = -\,$$\frac{E_0}{\sqrt2}$$\Bigl(1+$$\frac{v_0}{c}$$\Bigr)($$\hat$$ i+$$\hat$$ j).$$

Because the charge $$q$$ is positive, the force vector has exactly the same direction as the bracket above. Thus

$$$$\vec$$ F\propto -($$\hat$$ i+$$\hat$$ j),$$

i.e. the force is directed antiparallel to $$$$\frac{\hat i+\hat j}{\sqrt2}$$.$$

Hence, the correct answer is Option C.