A) \[k\frac{{{E}_{0}}}{C}\left( \frac{\widehat{i}-\widehat{j}}{\sqrt{2}} \right)\cos \left[ {{10}^{4}}\left( \frac{\widehat{i}-\widehat{j}}{\sqrt{2}} \right).\overrightarrow{r}-(3\times {{10}^{12}})t \right]\]
B) \[\frac{{{E}_{0}}}{C}\left( \frac{\widehat{i}-\widehat{j}}{\sqrt{2}} \right)\cos \left[ {{10}^{4}}\left( \frac{\widehat{i}+\widehat{j}}{\sqrt{2}} \right).\overrightarrow{r}-(3\times {{10}^{12}})t \right]\]
C) \[\frac{{{E}_{0}}}{C}\widehat{k}\cos \left[ {{10}^{4}}\left( \frac{\widehat{i}+\widehat{j}}{\sqrt{2}} \right).\overrightarrow{r}+(3\times {{10}^{12}})t \right]\]
D) \[\frac{{{E}_{0}}}{C}\frac{\left( \widehat{i}+\widehat{j}+\widehat{k} \right)}{\sqrt{3}}\cos \left[ {{10}^{4}}\left( \frac{\widehat{i}+\widehat{j}}{\sqrt{2}} \right).\overrightarrow{r}+(3\times {{10}^{12}})t \right]\]
Correct Answer: A
Solution :
| [a] Direction of B is , |
| \[=\widehat{K}\times \widehat{E}\] |
| \[=(\frac{\widehat{i}+\widehat{j}}{\sqrt{2}})\times \widehat{K}\] |
| \[=\frac{\widehat{i}+\widehat{k}}{\sqrt{2}}+\frac{\widehat{j}\times \widehat{k}}{\sqrt{2}}\] |
| \[=\frac{\widehat{j}}{\sqrt{2}}+\frac{\widehat{i}}{\sqrt{2}}=\frac{\widehat{i}-\widehat{j}}{\sqrt{2}}\] |
| So, ans is between (i) and (ii) |
| Propagation direction, \[\widehat{k}=\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}\] |
| \[\overrightarrow{B}\] wave will be |
| \[\Rightarrow \frac{{{E}_{0}}}{c}(\widehat{B})Cos[|k|\widehat{k}-wt]\] |
| \[\begin{matrix} \downarrow & \downarrow \\ \text{we know it is} & \text{we know it is} \\ \end{matrix}\] |
| \[\frac{\widehat{i}-\widehat{j}}{\sqrt{2}}\] \[\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}\] |
| Only (i) satisfies |
| Hence, correct answer is (i) |
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