Question 705035

From Maxwell's equations, $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$. Since $\vec{E} = E_0 \sin(\omega t - k_z z)\hat{j}$, we have $\nabla \times \vec{E} = \frac{\partial E_y}{\partial z} \hat{i} = -(-k_z) E_0 \cos(\omega t - k_z z) \hat{i} = k_z E_0 \cos(\omega t - k_z z) \hat{i}$. Integrating with respect to $t$, $\vec{B} = -\frac{k_z E_0}{\omega} \sin(\omega t - k_z z) \hat{i}$. The Poynting vector is $\vec{S} = \frac{1}{\mu_0} (\vec{E} \times \vec{B}) = \frac{1}{\mu_0} [E_0 \sin(\omega t - k_z z) \hat{j} \times (- \frac{k_z E_0}{\omega} \sin(\omega t - k_z z) \hat{i})] = \frac{1}{\mu_0} \frac{k_z E_0^2}{\omega} \sin^2(\omega t - k_z z) \hat{k}$.