Answer:
[tex]\mathbf{S^{\to} (x,t) = \dfrac{{E_o} {B_o}}{\mu_o} sin^2 (kx -wt) \hat i }[/tex]
Explanation:
Consider:
[tex]E^{\to} =E_o \ Sin (kx - wt) \hat j[/tex]
[tex]B^{\to} =B_o \ Sin (kx - wt) \hat k[/tex]
The equation for the Poynting vector is given as:
[tex]S^{\to} (x,t) = \dfrac{E^{\to}\times B^{\to}}{\mu_o}[/tex]
[tex]S^{\to} (x,t) = \dfrac{E_o \ Sin(kx - wt) \hat j \times B_o sin (kx -wt) \hat k}{\mu_o}[/tex]
[tex]S^{\to} (x,t) = \dfrac{{E_o} {B_o}}{\mu_o} sin^2 (kx -wt) (\hat j \times \hat k)[/tex]
[tex]S^{\to} (x,t) = \dfrac{{E_o} {B_o}}{\mu_o} sin^2 (kx -wt) \hat i[/tex]
∴
[tex]\mathbf{S^{\to} (x,t) = \dfrac{{E_o} {B_o}}{\mu_o} sin^2 (kx -wt) \hat i }[/tex]
