The magnetic field vector of an electromagnetic wave is given by $$B = B_0\frac{\hat{i}+\hat{j}}{\sqrt{2}}\cos kz - \omega t$$ where $$\hat{i}$$, $$\hat{j}$$ represents unit vector along x and y-axis respectively. At $$t = 0$$ s, two electric charges $$q_1$$ of $$4\pi$$ coulomb and $$q_2$$ of $$2\pi$$ coulomb located at $$\left(0, 0, \frac{\pi}{k}\right)$$ and $$\left(0, 0, \frac{3\pi}{k}\right)$$, respectively, have the same velocity of $$0.5c\hat{i}$$, (where $$c$$ is the velocity of light). The ratio of the force acting on charge $$q_1$$ to $$q_2$$ is:

The magnetic field of the plane electromagnetic wave is given as

$$\vec B \;=\;B_{0}\,\dfrac{\hat i+\hat j}{\sqrt2}\, \cos\!\bigl(kz-\omega t\bigr).$$

For a monochromatic wave travelling along the $$+z$$-direction, the electric field $$\vec E$$ is always perpendicular to $$\vec B$$, and the three vectors $$\vec E,\;\vec B,\;\hat k$$ form a right-handed triad. The magnitudes satisfy the relation

$$|\vec E| \;=\;c\,|\vec B|,$$

and the direction is obtained from the vector identity

$$\hat k \;=\;\dfrac{\vec E\times\vec B}{|\vec E\times\vec B|}.$$ Because $$\hat k=\hat k_z$$ in the present problem, we choose the electric field so that $$\vec E\times\vec B$$ points along $$\hat k_z$$. With $$\vec B$$ lying in the $$x$$-$$y$$ plane, the appropriate $$\vec E$$ is

$$\vec E \;=\;c B_0\,\dfrac{\hat i-\hat j}{\sqrt2}\, \cos\!\bigl(kz-\omega t\bigr).$$

At the instant $$t=0$$ we have

$$\vec B(z,0)=B_0\,\dfrac{\hat i+\hat j}{\sqrt2}\, \cos(kz),\qquad \vec E(z,0)=cB_0\,\dfrac{\hat i-\hat j}{\sqrt2}\, \cos(kz).$$

The two charges are situated at

$$z_1=\dfrac{\pi}{k}, \qquad z_2=\dfrac{3\pi}{k}.$$

Substituting these $$z$$-values,

$$\cos(kz_1)=\cos\!\bigl(k\cdot\tfrac{\pi}{k}\bigr)=\cos\pi=-1,$$

$$\cos(kz_2)=\cos\!\bigl(k\cdot\tfrac{3\pi}{k}\bigr)=\cos3\pi=-1.$$

Hence at both locations

$$\vec B = -\,B_0\,\dfrac{\hat i+\hat j}{\sqrt2},\qquad \vec E = -\,cB_0\,\dfrac{\hat i-\hat j}{\sqrt2}.$$

Both charges possess the same velocity

$$\vec v = 0.5\,c\,\hat i.$$

Now we invoke the Lorentz force formula

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

First we evaluate $$\vec v\times\vec B$$ (the subscript $$0$$ on $$B_0$$ is suppressed below for brevity):

$$$ \begin{aligned} \vec v\times\vec B &= \Bigl(0.5\,c\,\hat i\Bigr)\times\Bigl(\, -\,B_0\,\dfrac{\hat i+\hat j}{\sqrt2}\Bigr) \\[4pt] &= -\,\dfrac{0.5\,c\,B_0}{\sqrt2}\, \Bigl(\hat i\times\hat i+\hat i\times\hat j\Bigr) \\[4pt] &= -\,\dfrac{0.5\,c\,B_0}{\sqrt2}\, \Bigl(0+\hat k\Bigr) \\[4pt] &= -\,\dfrac{0.5\,c\,B_0}{\sqrt2}\,\hat k. \end{aligned} $$$

Adding the electric part, the bracket $$(\vec E+\vec v\times\vec B)$$ at either charge position becomes

$$$ \begin{aligned} \vec E + \vec v\times\vec B &= -\,cB_0\,\dfrac{\hat i-\hat j}{\sqrt2} \;-\;\dfrac{0.5\,c\,B_0}{\sqrt2}\,\hat k \\[6pt] &= -\,\dfrac{c\,B_0}{\sqrt2} \Bigl(\hat i-\hat j+0.5\,\hat k\Bigr). \end{aligned} $$$

This vector is identical at $$z_1$$ and $$z_2$$. Therefore, the magnitude of the force on each charge is directly proportional to the magnitude of the charge itself:

$$|\vec F_1| = |q_1|\; \bigl|\vec E+\vec v\times\vec B\bigr|,\qquad |\vec F_2| = |q_2|\; \bigl|\vec E+\vec v\times\vec B\bigr|.$$

Taking the ratio, the common factor cancels:

$$\dfrac{|\vec F_1|}{|\vec F_2|} \;=\;\dfrac{|q_1|}{|q_2|} \;=\;\dfrac{4\pi}{2\pi} \;=\;2.$$

Thus the force on $$q_1$$ is twice the force on $$q_2$$, i.e.

$$\vec F_1 : \vec F_2 = 2 : 1.$$

Hence, the correct answer is Option D.