Q 12-08-064JEE MainJEE Main 2024 (9 Apr, Shift 2)Medium
The magnetic field in a plane electromagnetic wave is $B_y = (3.5\times10^{-7})\sin(1.5\times10^3x + 0.5\times10^{11}t)\ \text{T}$. The corresponding electric field will be:
Answer: (D) $E_z = 105\sin(1.5\times10^3x + 0.5\times10^{11}t)\ \text{V m}^{-1}$
$E_0 = cB_0 = 3\times10^8\times3.5\times10^{-7} = 105\ \text{V m}^{-1}$.
The phase $(kx + \omega t)$ means the wave travels along $-x$. $\vec E$ must be perpendicular to $\vec B$ (along $y$) and to the direction of travel, so it is along $z$ (check: $\hat k\times\hat j = -\hat i$).
$$E_z = 105\sin(1.5\times10^3x + 0.5\times10^{11}t)\ \text{V m}^{-1}$$
Solution by Sreeraj P, M.Sc Physics