The magnetic field vector of an electromagnetic wave is given by $B=B_{0} \frac{\hat{i}+\hat{j}}{\sqrt{2}} \cos (k z-\omega t)$; where
$\hat{i}, \hat{j}$ represents unit vector along $x$ and $y$-axis respectively. At $\mathrm{t}=0 \mathrm{~s}$, two electric charges $\mathrm{q}_{1}$ of $4 \pi$ coulomb and $\mathrm{q}_{2}$ of $2 \pi$ coulomb located at $\left(0,0, \frac{\pi}{\mathrm{k}}\right)$ and $\left(0,0, \frac{3 \pi}{\mathrm{k}}\right)$, respectively, have the same velocity of $0.5 \mathrm{c} \hat{\mathrm{i}}$, (where $\mathrm{c}$ is the velocity of light). The ratio of the force acting on charge $\mathrm{q}_{1}$ to $\mathrm{q}_{2}$ is :-
Correct Option: , 3
$\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{V}} \times \overrightarrow{\mathrm{B}})$
$\overrightarrow{\mathrm{F}}_{1}=4 \pi\left[0.5 \operatorname{c} \hat{\mathrm{i}} \times \mathrm{B}_{0}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{2}\right) \cos \left(\mathrm{K} \cdot \frac{\pi}{\mathrm{K}}-0\right)\right]$
$\overrightarrow{\mathrm{F}}_{2}=2 \pi\left[0.5 \operatorname{ci} \times \mathrm{B}_{0}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{2}\right) \cos \left(\mathrm{K} \cdot \frac{3 \pi}{\mathrm{K}}-0\right)\right]$
$\cos \pi=-1, \quad \cos 3 \pi=-1$
$\therefore \frac{\mathrm{F}_{1}}{\mathrm{~F}_{2}}=2$
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