The electric field of a plane polarized elect | vedclass

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Question

$\overrightarrow{\mathrm{E}}=10 \hat{\mathrm{j}} \cos (6 x+8 z-10 \mathrm{ct})$

$\mathrm{B}_{\mathrm{o}}=\frac{\mathrm{E}_{\mathrm{o}}}{\mathrm{C}}

Vedclass · Accepted Answer

$\frac{1}{c}\left( {6\hat k - 8\hat i} \right)\,\cos \,\left[ {\left( {6x + 8z - 10ct} \right)} \right]$

Vedclass · Answer

$\overrightarrow{\mathrm{E}}=10 \hat{\mathrm{j}} \cos (6 x+8 z-10 \mathrm{ct})$

$\mathrm{B}_{\mathrm{o}}=\frac{\mathrm{E}_{\mathrm{o}}}{\mathrm{C}}=\frac{10}{\mathrm{C}}$

$\mathrm{W}=10\, \mathrm{C}$

$\because \,\hat E 	imes \hat B = \hat C$

$\left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \ 
  0&1&0 \ 
  {{B_x}}&{{B_y}}&{{B_z}} 
\end{array}} ight| = \frac{{6\hat i + 8\hat j}}{{10}}$

$ \Rightarrow {B_z}\hat i + 0\hat j - {B_x}\hat k = \frac{3}{5}\hat i + \frac{4}{5}\hat j$

${B_z} = \frac{3}{5},\,{{	ext{B}}_{	ext{y}}} = 0,\,{{	ext{B}}_{	ext{z}}} = \frac{4}{5}$

$\,	herefore \overrightarrow {	ext{B}}  = \frac{1}{{	ext{C}}}( - 8\widehat {	ext{i}} + 6\widehat {	ext{k}})\cos (6{	ext{x}} + 8{	ext{z}} + 10{	ext{ct}})$

The electric field of a plane polarized electromagnetic wave in free space at time $t = 0$ is given by an expression

$\vec E(x,y) = 10\hat j\, cos[(6x + 8z)]$

The magnetic field $\vec B (x,z, t)$ is given by : ($c$ is the velocity of light)

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