A) \[\vec{E}=4.8\times {{10}^{2}}\cos \left( 2\times {{10}^{7}}z+6\times {{10}^{15}}t \right)\left( \hat{i}-2\hat{j} \right)\frac{V}{m}\]
B) \[\vec{E}=4.8\times {{10}^{2}}\cos \left( 2\times {{10}^{7}}z-6\times {{10}^{15}}t \right)\left( 2\hat{i}+\hat{j} \right)\frac{V}{m}\]
C) \[\vec{E}=4.8\times {{10}^{2}}\cos \left( 2\times {{10}^{7}}z-6\times {{10}^{15}}t \right)\left( -2\hat{j}+\hat{i} \right)\frac{V}{m}\]
D) \[\vec{E}=4.8\times {{10}^{2}}\cos \left( 2\times {{10}^{7}}z+6\times {{10}^{15}}t \right)\left( -\hat{i}+2\hat{j} \right)\frac{V}{m}\]
Correct Answer: A
Solution :
If we use that direction of light propagation will be along \[\vec{E}\times \vec{B}.\] Then option is correct. Detailed solution is as following. magnitude of E = CB \[E=3\times {{10}^{8}}\times 1.6\times {{10}^{-6}}\times \sqrt{5}\] \[E=4.8\times {{10}^{2}}\sqrt{5}\] \[\vec{E}\]and \[\vec{B}\] are perpendicular to each other\[\Rightarrow \vec{E}.\vec{B}=0\] \[\Rightarrow \] either direction of \[\vec{E}\] is \[\hat{i}-2\hat{j}\] or \[-\hat{i}+2\hat{j}\]from given option Also wave propagation direction is parallel to \[\vec{E}\times \vec{B}\]which is \[-\hat{k}\] \[\Rightarrow \]\[\vec{E}\]is along \[\left( -\hat{i}+2\hat{j} \right)\]You need to login to perform this action.
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