Finding Jso for Antenna Array in Vacuum

[robert25pl]

Antenna array consists of two infinite plane parallel sheets in the xy plane spaced half a wavelength apart and having current densities. Wave is propagating in vacuum.

$$J_{s1} = J_{so}cos(\omega t) \vec{i}$$ z = 0

$$J_{s2} = J_{so}sin(\omega t) \vec{i}$$ $$z = \frac{\lambda}{2}$$
The time average of the Poynting vector is required to e 500 W/m^2 to the right of second antenna ($$z > \frac{\lambda}{2}$$ )
I have to find surface current Jso is needed?
I think that I have to find total E = E1 + E2 and then H. Poynting vector is given but I don't know how to surface current Jso from that. Any suggestions. Thanks

 

[robert25pl]

Any help please

 

[thepatientmental]

To find the required surface current Jso for the antenna array in vacuum, we can use the time-averaged Poynting vector equation:

<S> = 0.5Re(E x H*)

First, we can calculate the electric field E by adding the individual electric fields from the two infinite plane parallel sheets:

E = E1 + E2

Since the electric fields are perpendicular to the sheets, we can use the superposition principle to add them. E1 is in the positive x direction and E2 is in the negative x direction, so the total electric field E is simply:

E = (E1 - E2)

Substituting the given expressions for E1 and E2:

E = Jso(cos(\omega t) - sin(\omega t))

Next, we can calculate the magnetic field H by using the relationship H = (1/\eta) x E, where \eta is the impedance of vacuum (377 ohms).

H = \frac{1}{377}(E1 - E2)

Substituting the expression for E:

H = \frac{1}{377}Jso(cos(\omega t) - sin(\omega t))

Finally, we can substitute the expressions for E and H into the Poynting vector equation and equate it to the given value of 500 W/m^2:

<S> = 0.5Re(E x H*) = 500 W/m^2

0.5Re(Jso(cos(\omega t) - sin(\omega t)) x \frac{1}{377}Jso(cos(\omega t) - sin(\omega t))) = 500 W/m^2

Simplifying and solving for Jso:

Jso = \sqrt{\frac{1000}{377}} = 5.31 A/m

Therefore, a surface current of Jso = 5.31 A/m is needed for the antenna array in vacuum to have a time-averaged Poynting vector of 500 W/m^2 at a distance greater than half a wavelength from the second antenna.