A) \[\frac{{{E}_{0}}}{C}\frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}\cos \left[ {{10}^{4}}\frac{(\hat{i}+\hat{j})}{\sqrt{2}}.\vec{r}+(3\times {{10}^{12}})t \right]\]
B) \[\frac{{{E}_{0}}}{C}\frac{(\hat{i}-\hat{j})}{\sqrt{2}}\cos \left[ {{10}^{4}}\frac{(\hat{i}+\hat{j})}{\sqrt{2}}.\vec{r}-(3\times {{10}^{12}})t \right]\]
C) \[\frac{{{E}_{0}}}{C}\frac{(\hat{i}-\hat{j})}{\sqrt{2}}\cos \left[ {{10}^{4}}\frac{(\hat{i}-\hat{j})}{\sqrt{2}}.\vec{r}-(3\times {{10}^{12}})t \right]\]
D) \[\frac{{{E}_{0}}}{C}\hat{k}\cos \left[ {{10}^{4}}\frac{(\hat{i}+\hat{j})}{\sqrt{2}}.\vec{r}+(3\times {{10}^{12}})t \right]\]
Correct Answer: B
Solution :
[b]:\[\overline{E}\times \overline{B}\] gives direction of wave propagation. \[\Rightarrow \]\[\hat{k}\times \vec{B}||\frac{\hat{i}+\hat{j}}{\sqrt{2}}\] Now,\[\hat{k}\left( \frac{\hat{i}-\hat{j}}{\sqrt{2}} \right)=\frac{\hat{j}-(-\hat{i})}{\sqrt{2}}=\frac{\hat{i}+\hat{j}}{\sqrt{2}}\] Wave propagation vector should be along\[\frac{\hat{i}+\hat{j}}{\sqrt{2}}\] and direction of magnetic field is along \[\frac{\hat{i}-\hat{j}}{\sqrt{2}}\] .
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