A) \[\vec{E}=-{{B}_{0}}c\sin (kx+\omega t)\hat{k}V/m\]
B) \[\vec{E}={{B}_{0}}c\sin (kx-\omega t)\hat{k}V/m\]
C) \[\vec{E}={{B}_{0}}c\sin (kx+\omega t)\hat{k}V/m\]
D) \[\vec{E}=\frac{{{B}_{0}}}{c}\sin \,(kx+\omega t)\hat{k}V/m\]
Correct Answer: C
Solution :
\[C=\frac{{{E}_{0}}}{{{B}_{0}}}\] \[E=C{{B}_{0}}\] \[=C{{B}_{0}}\] \[=C{{B}_{0}}\sin (kx+\omega t)\hat{i}\]
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