Answer:
A) P = 3.3 × 10^(-11) Pa
B) Amplitude of electric field = 1.931 N/C
Amplitude of magnetic field = 6.44 × 10^(-9) T
C) μ_av = 1.65 × 10^(-11) J/m³
D) 50% each for the electric and magnetic field
Explanation:
A) First of all let's calculate intensity.
I = P_av/A
We are given;
P_av = 777 KW = 777,000 W
Distance = 5 km = 5000 m
Thus;
I = 777000/(2π × 5000²)
I = 0.00495 W/m²
Now, the average pressure would be given by the formula;
P = 2I/C
Where C is speed of light = 3 × 10^(8) m/s
P = (2 × 0.00495)/(3 × 10^(8))
P = 3.3 × 10^(-11) Pa
B) Formula for the amplitude of the electric field is gotten from;
E_max = √[2I/(εo•c)].
Where εo is the Permittivity of free space with a constant value of 8.85 × 10^(−12) c²/N.mm²
I and c remain as before.
Thus;
E_max = √[(2 × 0.00495)/(8.85 × 10^(−12) × 3 × 10^(8))]
E_max = √3.72881355932
E_max = 1.931 N/C
Formula for amplitude of magnetic field is gotten from;
B_max = E_max/c
B_max = 1.931/(3 × 10^(8))
B_max = 6.44 × 10^(-9) T
C) Formula for average density is;
μ_av = εo(E_rms)²
Now, E_rms = E_max/√2
Thus;
E_rms = 1.931/√2
μ_av = 8.85 × 10^(−12) × (1.931/√2)²
μ_av = 1.65 × 10^(-11) J/m³
D) The energy density for the electric and magnetic field is the same. So both of them will have 50% of the energy density.
