The electric field associated with an electromagnetic wave propagating in a dielectric medium is given by E…

Speed of light in medium is $v=\frac{\omega }{k}$

$\Rightarrow v=\frac{2\pi \left(5\times {10}^{14}\right)}{2\pi \times {\displaystyle \frac{{10}^7}{3}}}$

$\Rightarrow v=1.5\times {10}^8 \mathrm{m} {\mathrm{s}}^{-1}$

Therefore, refractive index $\mu =\frac{c}{v}=\frac{3\times {10}^8}{1.5\times {10}^8}=2$

$\Rightarrow \mu =2$

Given $\overrightarrow{E}=30\left(2\stackrel{\wedge }{x}+\stackrel{\wedge }{y}\right)\sin \left(2\pi \left(5\times {10}^{14}t-\frac{{10}^7}{3}z\right)\right) \mathrm{V} {\mathrm{m}}^{-1}$

The wave is polarised in $xy-$plane with polarisation angle ${\tan }^{-1}\left(\frac{1}{2}\right)$ with respect to $x-$axis.

Amplitude of magnetic field, $B_0=\frac{E_0}{V}=\frac{30\sqrt{2^2+1^2}}{1.5\times {10}^8}=\frac{30\sqrt{5}}{1.5\times {10}^8}$

Now, direction of $\overrightarrow{B}$ is $\left(\overrightarrow{v}\times \overrightarrow{E}\right)$

$=\stackrel{\wedge }{k}\times \frac{\left(2{\displaystyle \stackrel{\wedge }{i}}+{\displaystyle \stackrel{\wedge }{j}}\right)}{\sqrt{5}}$

$=\left(\frac{-\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}}{\sqrt{5}}\right)$.

 $$\begin{gathered} \overrightarrow{B}=\left(\frac{-\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}}{\sqrt{5}}\right)\times \frac{30\sqrt{5}}{1.5\times {10}^8}\sin \left[2\pi \left(5\times {10}^{14}t-\frac{{10}^7}{3}z\right)\right] \mathrm{Wb} {\mathrm{m}}^{-2} \\ =\left(-\stackrel{\wedge }{\mathrm{i}}+2\stackrel{\wedge }{\mathrm{j}}\right)\times 2\times {10}^{-7}\sin \left[2\mathrm{\pi }\left(5\times {10}^{14}\mathrm{t}-\frac{{10}^7}{3}\mathrm{z}\right)\right] \mathrm{Wb} {\mathrm{m}}^{-2} \end{gathered}$$

Hence,

$B_x=-2\times {10}^{-7}\sin \left[2\mathrm{\pi }\left(5\times {10}^{14}\mathrm{t}-\frac{{10}^7}{3}\mathrm{z}\right)\right] \mathrm{Wb} {\mathrm{m}}^{-2}$

and $B_y=4\times {10}^{-7}\sin \left[2\mathrm{\pi }\left(5\times {10}^{14}\mathrm{t}-\frac{{10}^7}{3}\mathrm{z}\right)\right] \mathrm{Wb} {\mathrm{m}}^{-2}$