A) \[\vec{E}=\left( 60\sin (1.6\times {{10}^{3}}x+48\times {{10}^{10}}t)\hat{k}V/m \right)\]
B) \[\vec{E}=\left( 3\times {{10}^{-8}}\sin (1.6\times {{10}^{3}}x+48\times {{10}^{10}}t)\hat{i}V/m \right)\]
C) \[\vec{E}=\left( 9\sin (1.6\times {{10}^{3}}x+48\times {{10}^{10}}t)\hat{k}V/m \right)\]
D) \[\vec{E}=\left( 3\times {{10}^{-8}}sin(1.6\times {{10}^{3}}x+48\times {{10}^{10}}t)\hat{j}V/m \right)\]
Correct Answer: C
Solution :
[c] \[\vec{B}=3\times {{10}^{-8}}\sin \left( 1.6\times {{10}^{3}}x+48\times {{10}^{10}}t \right)\hat{j}T\] \[{{E}_{0}}={{B}_{0}}\times C=3\times {{10}^{-8}}\times 3\times {{10}^{8}}=9\,V/m\] \[\therefore \vec{E}=9\sin \left( 1.6\times {{10}^{3}}x+48\times {{10}^{10}}t \right)\hat{k}\,V/m\]
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