Question 59 of 165

A beam of light travelling along $X$-axis is described by the electric field ${E}_{y}=900\sin \omega \left(t-\dfrac{x}{c}\right)$. The ratio of electric force to magnetic force on a charge $q$ moving along $Y$-axis with a speed of $3\times {10}^{7}m{s}^{-1}$ will be :
[Given speed of light $=3\times {10}^{8}m{s}^{-1}$ ]

A$1:1$B$1:10$C$10:1$D$1:2$

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Answer: (C) $10:1$

Given, ${E}_{y}=900\sin \left(\omega t-\dfrac{\omega x}{c}\right)$
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
$\dfrac{{F}_{E}}{{F}_{B}}=\dfrac{c}{v}=\dfrac{3\times {10}^{8}}{3\times {10}^{7}}=10:1$