If the magnetic field in a plane electromagnetic wave is given by $\vec B = 3\times10^{-8}\sin\left(1.6\times10^3x + 48\times10^{10}t\right)\hat j$ T, then what will be the expression for the electric field?
Answer: (B) $\vec E = 9\sin\left(1.6\times10^3x + 48\times10^{10}t\right)\hat k$ V/m
Check the speed: $\dfrac\omega k = \dfrac{48\times10^{10}}{1.6\times10^3} = 3\times10^8$ m/s.
Amplitude: $E_0 = cB_0 = 3\times10^8\times3\times10^{-8} = 9$ V/m.
The phase contains $kx + \omega t$, so the wave travels along $-x$. $\vec E\times\vec B$ must point along $-\hat i$; with $\vec B$ along $\hat j$, $\hat k\times\hat j = -\hat i$, so $\vec E$ is along $+\hat k$:
$$\vec E = 9\sin\left(1.6\times10^3x + 48\times10^{10}t\right)\hat k\ \text{V/m}$$
(The official answer key lists the $60$ V/m option, which does not satisfy $E_0 = cB_0$.)
Solution by Sreeraj P, M.Sc Physics