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Question

$\mathrm{B}_{\max }=\frac{\mathrm{E}_{\max }}{\mathrm{C}} \Rightarrow \mathrm{B}_{\max }=\frac{1.5 	imes 10^{6}}{3.0 	imes 10^{8}}=5.0 	imes 10^{-3

Vedclass · Accepted Answer

$\vec E = \hat k\, [1.5&times;10^6cos(5.93&times;10^5x+1.78&times;10^{14}t)]\,v/m$

$\vec B=\hat j\, [5.0&times;10^{-3}cos(5.93&times;10^5x+1.78&times;10^{14}t)]$

Vedclass · Answer

$\mathrm{B}_{\max }=\frac{\mathrm{E}_{\max }}{\mathrm{C}} \Rightarrow \mathrm{B}_{\max }=\frac{1.5 	imes 10^{6}}{3.0 	imes 10^{8}}=5.0 	imes 10^{-3} \mathrm{\,T}$

$\mathrm{K}=\frac{2 \pi}{\lambda} \Rightarrow \mathrm{K}=\frac{2 \pi}{10.6 	imes 10^{-6}} \Rightarrow \mathrm{K}=5.93 	imes 10^{5} \mathrm{\,rad} / \mathrm{m}$

$\omega=\mathrm{cK} \Rightarrow \omega=3 	imes 10^{8}\left(5.93 	imes 10^{5}ight)$

$=1.78 	imes 10^{14} \mathrm{\,rad} / \mathrm{sec}$

Direction of propagation $=\overrightarrow{\mathrm{E}} 	imes \overrightarrow{\mathrm{B}}$

$\Rightarrow-\hat{\mathrm{i}}=\overrightarrow{\mathrm{E}} 	imes \hat{\mathrm{A}} \Rightarrow \overrightarrow{\mathrm{A}}=\hat{\mathrm{j}}$

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