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 :
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)You need to login to perform this action.
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