The magnetic field vector of an electromagnetic wave is given by $\vec B = B_0\dfrac{\hat i + \hat j}{\sqrt2}\cos(kz - \omega t)$, where $\hat i$, $\hat j$ represent unit vectors 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 $\left(0, 0, \dfrac\pi k\right)$ and $\left(0, 0, \dfrac{3\pi}{k}\right)$, 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:
Answer: (D) $2 : 1$
At $t = 0$: $\cos(k\cdot\frac\pi k) = \cos\pi = -1$ and $\cos(3\pi) = -1$, so $\vec B$ (and hence $\vec E$) is identical at both positions.
The force $\vec F = q(\vec E + \vec v\times\vec B)$ then has the same bracket for both charges, since their velocities are equal too.
$$\frac{F_1}{F_2} = \frac{q_1}{q_2} = \frac{4\pi}{2\pi} = 2 : 1$$
Solution by Sreeraj P, M.Sc Physics