JEE Main 2021 — Electromagnetic Waves Question with Solution

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 = 0s, two electric charges q₁ of 4$$\pi$$ coulomb and q₂ 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 c $$\hat{i}$$, (where c is the velocity of light). The ratio of the force acting on charge q₁ to q₂ is :-