If the electric field intensity of a uniform plane electromagnetic wave is given as $\vec E = -301.6\sin(kz - \omega t)\hat a_x + 452.4\sin(kz - \omega t)\hat a_y\ \text{V m}^{-1}$, then the magnetic intensity $\vec H$ of this wave in $\text{A m}^{-1}$ will be [Given: speed of light in vacuum $c = 3\times10^8\ \text{m s}^{-1}$, permeability of vacuum $\mu_0 = 4\pi\times10^{-7}\ \text{N A}^{-2}$]
Answer: (C) $-0.8\sin(kz - \omega t)\hat a_y - 1.2\sin(kz - \omega t)\hat a_x$
The wave travels along $+z$ (phase $kz - \omega t$). Impedance of free space $\eta_0 = \mu_0c = 120\pi\approx377\ \Omega$, and $\vec H = \dfrac{1}{\eta_0}\hat a_z\times\vec E$.
$\hat a_z\times\hat a_x = \hat a_y$ and $\hat a_z\times\hat a_y = -\hat a_x$, with $\dfrac{301.6}{377} = 0.8$ and $\dfrac{452.4}{377} = 1.2$:
$$\vec H = -0.8\sin(kz - \omega t)\hat a_y - 1.2\sin(kz - \omega t)\hat a_x$$
Solution by Sreeraj P, M.Sc Physics