The electric field associated with an electro | vedclass

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(D) The given electric field is $\vec{E} = 30(2 \hat{x} + \hat{y}) \sin \left[2 \pi \left(5 	imes 10^{14} t - \frac{10^7}{3} zight)ight]$.Compari

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$A, D$

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(D) The given electric field is $\vec{E} = 30(2 \hat{x} + \hat{y}) \sin \left[2 \pi \left(5 	imes 10^{14} t - \frac{10^7}{3} zight)ight]$.Comparing this with the standard wave equation $\vec{E} = \vec{E}_0 \sin(\omega t - kz)$,we get $\omega = 2 \pi 	imes 5 	imes 10^{14} 	ext{ rad s}^{-1}$ and $k = 2 \pi 	imes \frac{10^7}{3} 	ext{ m}^{-1}$.The speed of the wave in the medium is $v = \frac{\omega}{k} = \frac{5 	imes 10^{14}}{10^7 / 3} = 1.5 	imes 10^8 	ext{ m s}^{-1}$.The refractive index $\mu = \frac{c}{v} = \frac{3 	imes 10^8}{1.5 	imes 10^8} = 2$. Thus,option $(D)$ is correct.The magnetic field is given by $\vec{B} = \frac{1}{v} (\hat{k} 	imes \vec{E})$.$\vec{B} = \frac{1}{1.5 	imes 10^8} \left[ \hat{k} 	imes 30(2 \hat{x} + \hat{y}) ight] \sin \left[2 \pi \left(5 	imes 10^{14} t - \frac{10^7}{3} zight)ight]$.$\vec{B} = \frac{30}{1.5 	imes 10^8} (2 \hat{y} - \hat{x}) \sin \left[2 \pi \left(5 	imes 10^{14} t - \frac{10^7}{3} zight)ight] = 2 	imes 10^{-7} (- \hat{x} + 2 \hat{y}) \sin \left[2 \pi \left(5 	imes 10^{14} t - \frac{10^7}{3} zight)ight]$.So,$B_x = -2 	imes 10^{-7} \sin(\dots)$ and $B_y = 4 	imes 10^{-7} \sin(\dots)$. Thus,option $(A)$ is correct and $(B)$ is incorrect.The polarization direction is along $(2 \hat{x} + \hat{y})$,so $	an 	heta = \frac{E_y}{E_x} = \frac{1}{2} = 0.5$. Thus,$	heta = 	an^{-1}(0.5) \approx 26.57^{\circ}$,not $30^{\circ}$. Thus,option $(C)$ is incorrect.Therefore,the correct options are $(A)$ and $(D)$.

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