A beam of light travelling along X -axis is described by the...

Given, ${E}_{y} = 900\sin \left(\omegat - \dfrac{\omegax}{c}\right)$

Here, ${E}_{0} = 900$

Electric field and electric force are related as ${F}_{E} = q{E}_{0}$

Magnetic force of a moving charge is given by ${F}_{B} = qv{B}_{0}$

Thus, the ratio $\dfrac{{F}_{E}}{{F}_{B}} = \dfrac{{E}_{0}}{v{B}_{0}}$

Using relation between magnetic and electric field, we have

$\dfrac{{F}_{E}}{{F}_{B}} = \dfrac{c}{v} = \dfrac{3 \times {10}^{8}}{3 \times {10}^{7}} = 10: 1$