Answer:
0.09511 V/m
[tex]3.17033\times 10^{-10}\ T[/tex]
7389.7799 W
Explanation:
I = Intensity of signal = [tex]12\times 10^{-6}\ W/m^2[/tex]
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]4\pi \times 10^{-7}\ H/m[/tex]
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
A = Area of the hemisphere
Intensity of Electric field
[tex]I=\frac{E^2}{2\times \epsilon_0 c}\\\Rightarrow E=\sqrt{2 \epsilon_0 c I}\\\Rightarrow E=\sqrt{2\times 4\pi \times 10^{-7}\times 3\times 10^8\times 12\times 10^{-6}}\\\Rightarrow E=0.09511\ V/m[/tex]
Electric field of the wave is 0.09511 V/m
Magnetic field
[tex]B=\frac{E}{c}\\\Rightarrow B=\frac{0.09511}{3\times 10^8}\\\Rightarrow B=3.17033\times 10^{-10}\ T[/tex]
Magnetic field of the wave is [tex]3.17033\times 10^{-10}\ T[/tex]
Power
[tex]P=IA\\\Rightarrow P=I\times 2\pi r^2\\\Rightarrow P=12\times 10^{-6}\times 2\pi \times 9900^2\\\Rightarrow P=7389.7799\ W[/tex]
The transmitter radiates 7389.7799 W of power
