The electric field in a plane electromagnetic wave is given by
$$E_z = 60\cos(5x + 1.5 \times 10^9 t)\ \text{V/m}.$$
Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field) :
Answer: (A) $B_y = 2 \times 10^{-7}\cos(5x + 1.5 \times 10^9 t)\ \text{T}$
Amplitude: $B_0 = \dfrac{E_0}{c} = \dfrac{60}{3 \times 10^8} = 2 \times 10^{-7}$ T.
The phase $(5x + \omega t)$ means the wave travels along $-x$. $\vec{B}$ must be perpendicular to both $\vec{E}$ (along $z$) and the direction of travel (along $x$), so $\vec{B}$ is along $y$.
Check the direction: $\vec{E} \times \vec{B}$ must point along $-x$, and $\hat{z} \times \hat{y} = -\hat{x}$. So $B_y$ is in phase with $E_z$ (same sign):
$$B_y = 2 \times 10^{-7}\cos(5x + 1.5 \times 10^9 t)\ \text{T}$$
(Also, $\omega/k = 1.5 \times 10^9/5 = 3 \times 10^8\ \text{m s}^{-1} = c$, as expected.)
Solution by Sreeraj P, M.Sc Physics