Answer:
The average outward power flow per unit area is given by the Poynting vector:
⃗ = 1/2Re(⃗ x ⃗ )
where Re denotes the real part.
Substituting the given electric field intensity, we get:
⃗ = 1/2Re((⃗ + ⃗ ) x ⃗ )
Expanding the cross product we get:
⃗ = 1/2Re(⃗ x ⃗ + ⃗ x ⃗ )
Since ⃗ x ⃗ is orthogonal to ⃗ and ⃗ , we have:
⃗ x ⃗ = -⃗ x ⃗
Therefore, simplifying the above expression, we get:
⃗ = -Re(⃗ x ⃗ )
Using the identity ⃗ x ⃗ = -⃗ x ⃗ , we can also write:
⃗ = Re(⃗ x ⃗ *)
where * denotes complex conjugation.
The outward power flow per unit area is then given by the magnitude of the Poynting vector:
|⃗ | = |Re(⃗ x ⃗ *)|
Substituting the given electric field intensity, we get:
|⃗ | = |Re((⃗ + ⃗ ) x (⃗ * + ⃗ *))|
Expanding the cross product, we get:
|⃗ | = |Re(⃗ x ⃗ * + ⃗ x ⃗ * + ⃗ x ⃗ * + ⃗ x ⃗ *)|
Simplifying, we get:
|⃗ | = |Re(2⃗ x ⃗ *)|
Using the identity ⃗ x ⃗ * = -⃗ * x ⃗ , we can also write:
|⃗ | = |-2Im(⃗ x ⃗ *)|
where Im denotes the imaginary part.
Substituting the given electric field intensity, we get:
|⃗ | = |-2Im((⃗ + ⃗ ) x (⃗ * + ⃗ *))|
Expanding the cross product, we get:
|⃗ | = |-2Im(⃗ x ⃗ * + ⃗ x ⃗ * + ⃗ x ⃗ * + ⃗ x ⃗ *)|
Simplifying, we get:
|⃗ | = |-4Im(⃗ x ⃗ *)|
Using the identity Im(⃗ x ⃗ *) = |⃗ ||⃗ *|sin(θ), where θ is the angle between ⃗ and ⃗ *, we can write:
|⃗ | = 4|⃗ ||⃗ *|sin(θ)
Substituting the given electric field intensity, we get:
|⃗ | = 4|⃗ ||⃗ *|sin(2πft)
where f is the frequency of the radiation.
Therefore, the average outward power flow per unit area of the antenna system is given by:
P = |⃗ |A = 4|⃗ ||⃗ *|A sin(2πft)
where A is the area over which the radiation is spread.