Q 12-08-142JEE MainJEE Main 2022 (27 Jul, Shift 1)Medium
A beam of light travelling along $X$-axis is described by the electric field $E_y = 900\sin\omega\left(t - \dfrac xc\right)$. The ratio of electric force to magnetic force on a charge $q$ moving along $Y$-axis with a speed of $3\times10^7\ \text{m s}^{-1}$ will be :
[Given speed of light $= 3\times10^8\ \text{m s}^{-1}$]
Answer: (C) $10 : 1$
The magnetic field is along $z$ with $B = E/c$; the velocity along $y$ is perpendicular to it.
$$\frac{F_E}{F_B} = \frac{qE}{qvB} = \frac{c}{v} = \frac{3\times10^8}{3\times10^7} = 10$$
Ratio $10 : 1$.
Solution by Sreeraj P, M.Sc Physics