The magnetic field of an E.M. wave is given by $\vec B = \left(\dfrac{\sqrt3}{2}\hat i + \dfrac{1}{2}\hat j\right)30\sin\left[\omega\left(t - \dfrac{z}{c}\right)\right]$ (S.I. units). The corresponding electric field in S.I. units is
Answer: (A) $\vec E = \left(\dfrac{1}{2}\hat i - \dfrac{\sqrt3}{2}\hat j\right)30c\sin\left[\omega\left(t - \dfrac{z}{c}\right)\right]$
The phase $\omega(t - z/c)$ means the wave travels along $+\hat k$, and $\vec E$ must have the same phase. $|E_0| = cB_0 = 30c$.
$\vec E$, $\vec B$ and $\hat k$ satisfy $\vec E = c\,\vec B\times\hat k$:
$$\left(\frac{\sqrt3}{2}\hat i + \frac{1}{2}\hat j\right)\times\hat k = \frac{\sqrt3}{2}(-\hat j) + \frac{1}{2}\hat i = \frac{1}{2}\hat i - \frac{\sqrt3}{2}\hat j$$
$$\vec E = \left(\frac{1}{2}\hat i - \frac{\sqrt3}{2}\hat j\right)30c\sin\left[\omega\left(t - \frac{z}{c}\right)\right]$$
(Check: $\vec E\times\vec B$ points along $+\hat k$.)
Solution by Sreeraj P, M.Sc Physics