A plane electromagnetic wave travels in a medium of relative permeability $1.61$ and relative permittivity $6.44$. If magnitude of magnetic intensity is $4.5\times10^{-2}\ \text{A m}^{-1}$ at a point, what will be the approximate magnitude of electric field intensity at that point? (Given: permeability of free space $\mu_0 = 4\pi\times10^{-7}\ \text{N A}^{-2}$, speed of light in vacuum $c = 3\times10^8\ \text{m s}^{-1}$)
Answer: (C) $8.48\ \text{V m}^{-1}$
$E = vB = v\,\mu_0\mu_r H$ with $v = \dfrac{c}{\sqrt{\mu_r\varepsilon_r}} = \dfrac{3\times10^8}{\sqrt{1.61\times6.44}} = \dfrac{3\times10^8}{3.22}$.
$$E = \frac{3\times10^8}{3.22}\times4\pi\times10^{-7}\times1.61\times4.5\times10^{-2} = 3\times10^8\times\frac{4\pi\times10^{-7}\times4.5\times10^{-2}}{2} \approx 8.48\ \text{V m}^{-1}$$
Solution by Sreeraj P, M.Sc Physics