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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 the expression for the corresponding magnetic field is (here subscripts denote the direction of the field):
A
$B_y = 2 \times 10^{-7} \cos(5x + 1.5 \times 10^9 t)\text{ T}$
B
$B_x = 2 \times 10^{-7} \cos(5x + 1.5 \times 10^9 t)\text{ T}$
C
$B_z = 60 \cos(5x + 1.5 \times 10^9 t)\text{ T}$
D
$B_y = 60 \sin(5x + 1.5 \times 10^9 t)\text{ T}$
Explanation
Peak magnetic field is $B_0 = E_0 / c = 60 / (3 \times 10^8) = 2 \times 10^{-7}\text{ T}$. Since $\vec{E}$ is along $\hat{z}$ and wave travels along $-\hat{x}$, $\vec{B}$ is along $\hat{y}$.
Detailed Solution
The amplitude of the magnetic field is $B_0 = \frac{E_0}{c} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7}\text{ T}$. The wave argument is $(kx + \omega t)$, which indicates propagation along the $-\hat{x}$ direction. The direction of wave propagation is given by the Poynting vector $\hat{S} = \hat{E} \times \hat{B}$. Here $\hat{S} = -\hat{i}$ and $\hat{E} = \hat{k}$. Since $\hat{k} \times \hat{j} = -\hat{i}$, $\vec{B}$ must be oriented along the $y$-axis ($\hat{j}$). Therefore, $B_y = 2 \times 10^{-7} \cos(5x + 1.5 \times 10^9 t)\text{ T}$.