JEE Main 2021 — Electromagnetic Waves Question with Solution

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