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The magnetic field of a plane electromagnetic wave is given by $\vec{B}=3\times10^{-8}\cos(1.6\times10^{3}x+48\times10^{10}t)\hat{j}$, then the associated electric field will be:
Detailed Solution
$\frac{E_0}{B_0}=c \Rightarrow E_0=B_0c=3\times10^{-8}\times3\times10^{8}=9$ V/m. E and B are in phase. The wave travels along $-\hat{i}$, and $\hat{E}\times\hat{B}$ must point along propagation: $\hat{k}\times\hat{j}=-\hat{i}$, so $\hat{E}=\hat{k}$. Hence $\vec{E}=9\cos(1.6\times10^{3}x+48\times10^{10}t)\hat{k}$ V/m.