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. Part A At a home 5.50 km away from the antenna, what average pressure does this wave exert on a totally reflecting surface? Express your answer in pascals

Part B At a home 5.50 km away from the antenna, what are the amplitudes of the electric and magnetic fields of the wave? Express the electric field amplitude in newtons per coulomb. Express the magnetic field amplitude in teslas. Enter your answers numerically separated by a comma.

Part C At a home 5.50 km away from the antenna, what is the average density of the energy this wave carries? Express your answer in joules per meter cubed.

Part D For the energy density in part (c), what percentage is due to the electric field? Express your answer as a percentage

Answer 1

The average pressure exerted by the wave on the reflecting surface is 9.0816 W/m^2, which can be expressed in pascals. The electric field amplitude of the wave is 6.316 x 10^-5 N/C and the magnetic field amplitude is 0.0234 T. The average density of the energy carried by the wave is 2.59 J/m^3. The percentage of the energy density due to the electric field is approximately 0%.

Step-by-step explanation:

In this case, we can calculate the average pressure that the wave exerts on a totally reflecting surface at a home 5.50 km away from the antenna using the inverse square law. The power of the sinusoidal radio signal is given as 777 kW. Since the wave spreads out uniformly into a hemisphere above the ground, half of the power will be spread over the area of a hemisphere. Therefore, the power received at the home is 777 kW / (2 * pi * (5.50 km)^2) = 777000 W / (2 * pi * (5.50x10^3 m)^2) = 4.5408 W/m^2. This is the intensity of the wave. To calculate the average pressure, we can use the formula P = 2 * I, where P is the pressure and I is the intensity. Therefore, the average pressure exerted by the wave on the reflecting surface is 2 * 4.5408 W/m^2 = 9.0816 W/m^2, which can be expressed in pascals.

To calculate the amplitudes of the electric and magnetic fields of the wave at the home 5.50 km away from the antenna, we can use the formulas E = sqrt(2 * I / (c * mu)) and B = sqrt(2 * I / (c * epsilon)), where E is the electric field amplitude, B is the magnetic field amplitude, I is the intensity, c is the speed of light, mu is the permeability of free space, and epsilon is the permittivity of free space. The speed of light is approximately 3 x 10^8 m/s, the permeability of free space is approximately 4 * pi * 10^-7 T*m/A, and the permittivity of free space is approximately 8.854 x 10^-12 F/m. Substituting these values and the calculated intensity of 4.5408 W/m^2 into the formulas, we have E = sqrt(2 * 4.5408 / (3 x 10^8 * (4 * pi * 10^-7))) = 6.316 x 10^-5 N/C and B = sqrt(2 * 4.5408 / (3 x 10^8 * 8.854 x 10^-12)) = 0.0234 T.

To calculate the average density of the energy carried by the wave at the home 5.50 km away from the antenna, we can use the formula density = P / c, where P is the power and c is the speed of light. Substituting the power of 777 kW into the formula, we have density = 777000 / (3 x 10^8) = 2.59 J/m^3.

Finally, to determine the percentage of the energy density due to the electric field, we can use the formula % = (E^2 / (E^2 + B^2)) * 100, where E is the electric field amplitude and B is the magnetic field amplitude. Substituting the calculated values of E and B into the formula, we have % = (6.316 x 10^-5^2 / (6.316 x 10^-5^2 + 0.0234^2)) * 100 = 0.0001704%. Therefore, the percentage of the energy density due to the electric field is approximately 0.

Answer 2

The average pressure exerted on a perfectly reflecting surface at 5.5 km is 4.01 x 10^-9 Pa.

The electric and magnetic field amplitudes at 5.5 km are 4.73 x 10^3 N/C and 1.58 x 10^-5 T, respectively.

The average energy density of the wave is 2.01 x 10^-8 J/m^3.

Approximately 50.1% of the energy density is due to the electric field.

Part A: Average Pressure on Reflecting Surface

Convert power to intensity:

Intensity I = Power / Area = 777 kW / (2π * (5.5 km * 1000 m/km)^2) = 6.01 W/m^2

Calculate radiation pressure (double for perfect reflection):

Pressure p = 2 * I/c = 2 * 6.01 W/m^2 / 3 x 10^8 m/s = 4.01 x 10^-9 Pa

Part B: Electric and Magnetic Field Amplitudes

Relate intensity to electric field amplitude (E):

I = c * ε_0 * E^2 / 2 = 6.01 W/m^2

E = sqrt(2 * I / (c * ε_0)) = 4.73 x 10^3 N/C

Relate electric and magnetic field amplitudes (B):

B = E / c = 4.73 x 10^3 N/C / 3 x 10^8 m/s = 1.58 x 10^-5 T

Part C: Average Energy Density

Energy density (u) from intensity:

u = I / c = 6.01 W/m^2 / 3 x 10^8 m/s = 2.01 x 10^-8 J/m^3

Part D: Percentage of Energy Due to Electric Field

Ratio of electric field energy density to total energy density:

% due to electric field = (ε_0 * E^2 / 2u) * 100% = (8.85 x 10^-12 F/m * (4.73 x 10^3 N/C)^2 / (2 * 2.01 x 10^-8 J/m^3)) * 100%

Calculate percentage:

% due to electric field = 50.1%