The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by  $\overrightarrow{E}=30(2\widehat{x}+\widehat{y})\sin \left[2\pi (5\times {10}^{14}t-\frac{{10}^7}{3}z)\right]v{m}^{-1}$. Which of the following option(s) is (are) correct? [Given: the speed of light in vacuum,  $c=3\times {10}^{8}m{s}^{-1}$]

$$\begin{gathered} \overrightarrow{E}=(60\widehat{x}+30\widehat{y})\sin \left[(2\pi ft-\frac{2\pi }{\lambda }z)\right]V{m}^{-1} \\ {E}_0=\sqrt{{60}^2+{30}^2}=30\sqrt{5}V\text{/}m \\ f=5\times {10}^{14}Hz\frac{1}{\lambda }=\frac{{10}^7}{3}{m}^{-1} \end{gathered}$$

Speed of light in dielectric medium,

$$\begin{gathered} v=\frac{5\times {10}^{14}}{{10}^7/3}=1.5\times {10}^{8}m\text{/}s \\ \mu =\frac{c}{v}=\frac{3\times {10}^8}{1.5\times {10}^8}=2 \end{gathered}$$

Also, $v=\frac{E_0}{B_0}$

$$\begin{gathered} \Rightarrow B_0=\frac{E_0}{v}=\frac{30\sqrt{5}}{1.5\times {10}^8}=2\sqrt{5}\times {10}^{-7}Wb{m}^{-2} \\ \tan \theta =\frac{30}{60}=\frac{1}{2}\Rightarrow \theta ={\tan }^{-1}\frac{1}{2} \end{gathered}$$

The wave is polarized in  xy-plane with polarization angle  ${\tan }^1\frac{1}{2}$ with respect to  x-axis

$$\begin{gathered} {\left(B_0\right)}_x=B_0\sin \theta =2\sqrt{5}\times {10}^{-7}\times \frac{1}{\sqrt{5}}=2\times {10}^{-7}Wb{m}^{-2} \\ {\left(B_0\right)}_y=B_0\cos \theta =2\sqrt{5}\times {10}^{-7}\times \frac{2}{\sqrt{5}}=4\times {10}^{-7}Wb{m}^{-2} \\ {B}_x=-2\times {10}^{-7}\sin \left[2\pi (5\times {10}^{14}t-\frac{{10}^7}{3})z\right]Wb{m}^{-2} \end{gathered}$$