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