The electric field associated with an electromagnetic wave...

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

$⇒ v = \dfrac{2\pi\left(5 \times {10}^{14}\right)}{2\pi \times \dfrac{{10}^{7}}{3}}$

$⇒ v = 1 . 5 \times {10}^{8}m{s}^{- 1}$

Therefore, refractive index $\mu = \dfrac{c}{v} = \dfrac{3 \times {10}^{8}}{1 . 5 \times {10}^{8}} = 2$

$⇒ \mu = 2$

Given $\overset{\to}{E} = 30\left(2\overset{∧}{x} + \overset{∧}{y}\right)\sin \left(2\pi\left(5 \times {10}^{14}t - \dfrac{{10}^{7}}{3}z\right)\right)V{m}^{- 1}$

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

Amplitude of magnetic field, ${B}_{0} = \dfrac{{E}_{0}}{V} = \dfrac{30\sqrt{{2}^{2} + {1}^{2}}}{1 . 5 \times {10}^{8}} = \dfrac{30\sqrt{5}}{1 . 5 \times {10}^{8}}$

Now, direction of $\overset{\to}{B}$ is \left(\overset{\to}{v} \times \overset{\to}{E}\right)$</p><p>$= \overset{∧}{k} \times \dfrac{\left(2\overset{∧}{i} + \overset{∧}{j}\right)}{\sqrt{5}}$</p><p>$= \left(\dfrac{- \overset{∧}{i} + 2\overset{∧}{j}}{\sqrt{5}}\right)$.</p><p>$\overset{\to}{B} = \left(\dfrac{- \overset{∧}{i} + 2\overset{∧}{j}}{\sqrt{5}}\right) \times \dfrac{30\sqrt{5}}{1 . 5 \times {10}^{8}}\sin \left[2\pi\left(5 \times {10}^{14}t - \dfrac{{10}^{7}}{3}z\right)\right]Wb{m}^{- 2} = \left(- \overset{∧}{i} + 2\overset{∧}{j}\right) \times 2 \times {10}^{- 7}\sin \left[2\pi\left(5 \times {10}^{14}t - \dfrac{{10}^{7}}{3}z\right)\right]Wb{m}^{- 2}$</p><p>Hence,</p><p>${B}_{x} = - 2 \times {10}^{- 7}\sin \left[2\pi\left(5 \times {10}^{14}t - \dfrac{{10}^{7}}{3}z\right)\right]Wb{m}^{- 2}$</p><p>and${B}_{y} = 4 \times {10}^{- 7}\sin \left[2\pi\left(5 \times {10}^{14}t - \dfrac{{10}^{7}}{3}z\right)\right]Wb{m}^{- 2}