Since, in a radio wave E → ⊥ B → \overrightarrow{E}\bot \overrightarrow{B} E ⊥ B and Poynting vector is defined by
S → = 1 μ 0 ( E → × B → ) ⟹ ∣ ∣ S → ∣ ∣ = 1 μ 0 ∣ ∣ E → ∣ ∣ ⋅ ∣ ∣ B → ∣ ∣ \overrightarrow{S}=\frac{1}{\mu_0}(\overrightarrow{E}\times\overrightarrow{B})\\
\implies ||\overrightarrow{S}||=\frac{1}{\mu_0}||\overrightarrow{E}||\cdot||\overrightarrow{B}|| S = μ 0 1 ( E × B ) ⟹ ∣∣ S ∣∣ = μ 0 1 ∣∣ E ∣∣ ⋅ ∣∣ B ∣∣ But, we also know that,
∣ ∣ B → ∣ ∣ = ∣ ∣ E → ∣ ∣ c ||\overrightarrow{B}||=\frac{||\overrightarrow{E}||}{c} ∣∣ B ∣∣ = c ∣∣ E ∣∣ Thus, we get
∣ ∣ S → ∣ ∣ = 1 c μ 0 ∣ ∣ E → ∣ ∣ 2 ||\overrightarrow{S}||=\frac{1}{c\mu_0}||\overrightarrow{E}||^2 ∣∣ S ∣∣ = c μ 0 1 ∣∣ E ∣ ∣ 2 Given, ∣ ∣ E → ∣ ∣ = E = 1 0 − 4 V / m ||\overrightarrow{E}||=E=10^{-4}V/m ∣∣ E ∣∣ = E = 1 0 − 4 V / m
Hence, on plugin the value we get
∣ ∣ S → ∣ ∣ = 2.655 × 1 0 − 11 W / m 2 ||\overrightarrow{S}||=2.655\times 10^{-11}W/m^2 ∣∣ S ∣∣ = 2.655 × 1 0 − 11 W / m 2