A) \[\frac{1}{c}(6\hat{k}+8\hat{i})cos[(6x-8z+10ct)]\]
B) \[\frac{1}{c}(6\hat{k}-8\hat{i})cos[(6x-8z-10ct)]\]
C) \[\frac{1}{c}(6\hat{k}+8\hat{i})cos[(6x+8z-10ct)]\]
D) None of these
Correct Answer: D
Solution :
| \[\overrightarrow{E}=10\hat{j}\cos (6x+8z-10ct)\] |
| \[{{B}_{o}}=\frac{{{E}_{o}}}{C}=\frac{10}{C}\] |
| \[W=10C\] |
| \[\because \] \[\hat{E}\times \hat{B}=\hat{C}\] |
| \[\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 0 & 1 & 0 \\ {{B}_{x}} & {{B}_{y}} & {{B}_{z}} \\ \end{matrix} \right|=\frac{6\hat{i}+8\hat{j}}{10}\]\[\Rightarrow \]\[{{B}_{z}}\hat{i}-0\hat{j}-{{B}_{x}}\hat{k}=\frac{3}{5}\hat{i}+\frac{4}{5}\hat{j}\] |
| \[{{B}_{z}}=\frac{3}{5},{{B}_{y}}=0,{{B}_{z}}=\frac{4}{5}\] |
| \[\therefore \]\[\overrightarrow{B}=\frac{1}{C}(-8\hat{i}+6\hat{k})cos(6x+8z+10ct)\] |
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