Q 12-08-188JEE MainJEE Main 2020 (3 Sep, Shift 2)Medium
The electric field of a plane electromagnetic wave propagating along the $x$ direction in vacuum is $\vec E = E_0\hat j\cos(\omega t - kx)$. The magnetic field $\vec B$, at the moment $t = 0$ is:
Answer: (C) $\vec B = E_0\sqrt{\mu_0\varepsilon_0}\cos(kx)\,\hat k$
$B_0 = \dfrac{E_0}{c} = E_0\sqrt{\mu_0\varepsilon_0}$, and $\vec B$ is in phase with $\vec E$.
$\vec E\times\vec B$ must be along $+\hat i$: $\hat j\times\hat k = \hat i$, so $\vec B$ is along $\hat k$.
At $t = 0$: $\cos(-kx) = \cos kx$, so $\vec B = E_0\sqrt{\mu_0\varepsilon_0}\cos(kx)\,\hat k$.
Solution by Sreeraj P, M.Sc Physics