Correct Answer - Option 3 :
\(\frac{1}{120π} \)
Concept:
Poynting vector:
It states that the cross product of electric field vector (E) and magnetic field vector (H) at any point is a measure of the rate of flow of electromagnetic energy per unit area at that point that is
\( \vec P = \vec E \times \vec H\)
Where P = Poynting vector, E = Electric field and H = Magnetic field
The Poynting vector describes the magnitude and direction of the flow of energy in electromagnetic waves.
The unit of the Poynting vector is watt/m2
The Poynting vector will be defined as the multiplication of the magnitude and the unit direction vector.
Where,
The magnitude of the time-average Poynting vector:
\(\left| {\vec P} \right| = \frac{1}{2}η {H^2}\) [∵ |E|/|H| = η ]
The direction of the Poynting vector is the same as the direction of propagation.
Given:
|E| = 1 V/m
ϵr = 4
Analysis:
\(P = \frac{1}{2}\eta {\left| H \right|^2}\) ---(1)
\(\frac{{\left| E \right|}}{{\left| H \right|}} = \eta \)
\(\left| H \right| = \frac{1}{{120\pi }}\)
Putting in (1)
\(P = \frac{1}{2}\left( {120\pi } \right)\frac{1}{{120\pi \;\times \;120\pi }}\)
= \(\frac{1}{{240\pi }}~W/{m^2}\)
Time average power density vector is given by
\({P_{avg}} = \frac{{E_o^2}}{{2\eta }}\) where \(\eta = \sqrt {\frac{\mu }{\epsilon}} \)
\(\eta = \sqrt {\frac{{1{\mu _o}}}{{4{\epsilon_o}}}} = \frac{{{\eta _o}}}{2} = \frac{{120\pi }}{2} = 60\pi \)
\({P_{avg}} = \frac{{{{\left( 1 \right)}^2}}}{{2.\left( {60\pi } \right)}} = \frac{1}{{120\pi }}W/{m^2}\)
