Question

Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of 777 kW. Assume that the wave spreads out uniformly into a hemisphere above the ground. At a home 5.00 km away from the antenna,
(a) what average pressure does this wave exert on a totally reflecting surface,
(b) what are the amplitudes of the electric and magnetic fields of the wave, and
(c) what is the average density of the energy this wave carries?
(d) For the energy density in part (c), what percentage is due to the electric field and what percentage is due to the magnetic field?

Answer 1

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.