Q 12-08-134JEE MainJEE Main 2022 (26 Jul, Shift 1)Medium
The magnetic field of a plane electromagnetic wave is given by $\vec B = 2\times10^{-8}\sin(0.5\times10^3x + 1.5\times10^{11}t)\,\hat j\ \text{T}$. The amplitude of the electric field would be
Answer: (C) $6\ \text{V m}^{-1}$ along $z$-axis
$E_0 = cB_0 = 3\times10^8\times2\times10^{-8} = 6\ \text{V m}^{-1}$.
The phase $(kx + \omega t)$ means the wave travels along $-x$. $\vec E$ must be perpendicular to both $\vec B$ ($\hat j$) and the direction of travel, so it lies along the $z$-axis (indeed $\hat k\times\hat j = -\hat i$).
Solution by Sreeraj P, M.Sc Physics