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Chapter 31: Problem 56

Suppose a \(35-\mathrm{kW}\) radio station emits EM waves uniformly in all directions. (a) How much energy per second crosses a \(1.0-\mathrm{m}^{2}\) area \(1.0 \mathrm{~km}\) from the transmitting antenna? (b) What is the rms magnitude of the \(\overrightarrow{\mathbf{E}}\) field at this point, assuming the station is operating at full power? What is the rms voltage induced in a \(1.0-\mathrm{m}\) -long vertical car antenna (c) \(1.0 \mathrm{~km}\) away, \((d) 50 \mathrm{~km}\) away?

Short Answer

Expert verified
(a) 0.00277 W/m虏, (b) 1.02 V/m, (c) 1.02 V, (d) 0.0204 V at 50 km.

Step by step solution

01

Understanding Power Distribution

The radio station emits power uniformly in all directions, which forms a spherical pattern around the source. Therefore, at a distance, the power is distributed over the surface area of a sphere with radius equal to the distance from the source. The surface area of a sphere is calculated as: \[ A = 4\pi r^2 \] where \( r \) is the radius in meters.
02

Calculating Power per Unit Area (Intensity)

The intensity \( I \) is defined as the power per unit area, calculated by dividing the total power by the surface area of the sphere:\[ I = \frac{P}{A} = \frac{35,000}{4\pi(1000)^2} = \frac{35,000}{4\pi \times 10^6} \]Evaluate this expression to find the intensity at 1 km away.
03

Evaluating Energy per Second through 1.0 m虏

To find the energy crossing a 1.0 m虏 area per second, use the intensity from Step 2 since intensity has units of power per unit area:\[ I \times 1 \,m^2 = I \]Calculate \( I \) from Step 2 to get the energy per second.
04

Calculating rms Electric Field (E)

The relationship between intensity and the rms electric field \( E_{rms} \) in an electromagnetic wave is given by:\[ I = \frac{1}{2} \epsilon_0 c E_{rms}^2 \]Solve for \( E_{rms} \):\[ E_{rms} = \sqrt{\frac{2I}{\epsilon_0 c}} \]where \( \epsilon_0 = 8.85 \times 10^{-12} \, F/m \) and \( c = 3 \times 10^8 \, m/s \).Use the intensity from Step 2 to calculate \( E_{rms} \).
05

Calculating rms Induced Voltage at 1 km

The rms voltage \( V_{rms} \) induced in a 1.0 m antenna by the electric field \( E_{rms} \) is calculated using:\[ V_{rms} = E_{rms} \times l \]where \( l = 1.0\, m \). Use the \( E_{rms} \) found in Step 4 to compute \( V_{rms} \).
06

Re-calculate for 50 km Distance

At 50 km, repeat the process from Steps 1 to 5, substituting \( r = 50,000 \, m \) instead of 1,000 m. This changes the surface area \( A \) and results in a different intensity \( I \) at this new distance. Subsequently, recalculate \( E_{rms} \) and \( V_{rms} \) for this distance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Intensity
Intensity refers to the power of the electromagnetic (EM) wave per unit area, which in this case can be understood through the radio station's emission. Imagine the radio station as a point source of energy, radiating waves uniformly in all directions. This energy expands like a balloon around the source, forming a spherical surface.
  • Intensity \(I\) is given by the formula: \[ I = \frac{P}{A} \] where \(P\) is the power and \(A\) is the area over which the power is distributed.
  • For a spherical distribution at a distance \(r\), the area \(A\) is \(4\pi r^2\).
The challenge is to determine how much of the power reaches a 1.0 m虏 area 1 km from the source. By substituting the given values into the intensity formula, you can calculate it. This measure of intensity tells us the concentration of the wave's energy at that specific location.
RMS Electric Field
The Root Mean Square (RMS) Electric Field \(E_{rms}\) is a measure of the electric component's strength within an electromagnetic wave over a specific area. It's crucial for understanding how strong the electric field is in any given area where the wave passes through.
  • The relationship between intensity and RMS Electric Field is given by: \[ I = \frac{1}{2} \epsilon_0 c E_{rms}^2 \]
  • Here, \( \epsilon_0 \) is the permittivity of free space \(8.85 \times 10^{-12} \, F/m\), and \(c\) is the speed of light \(3 \times 10^8 \, m/s\).
By rearranging and solving this equation for \(E_{rms}\), you can determine the strength of the electric field at a given location from the intensity derived earlier. This calculation is particularly valuable for assessing the potential impact of the field on nearby objects.
Induced Voltage
Induced Voltage, in this context, is the voltage that appears across a conductor placed in the path of an electromagnetic wave's electric field. This is particularly pertinent for antennas.
  • To find the RMS Voltage \(V_{rms}\) induced in a 1 m antenna, use: \[ V_{rms} = E_{rms} \times l \]
  • Where \(l = 1.0 m\) (the length of the antenna).
By multiplying the RMS Electric Field \(E_{rms}\) (which you've calculated from the intensity) by the length of the antenna, you obtain the induced RMS Voltage. This induced voltage is crucial for understanding how radio waves interact with conductive materials like antennas, allowing us to harness these waves for communication.
Energy Distribution
Energy distribution in electromagnetic waves is vital to comprehend how energy diminishes with distance from the radio station. As these waves travel, they cover increasing areas, reducing the intensity.
  • The energy is initially concentrated nearest to the source and spreads over a wider area as distance increases.
  • At a new distance, calculate the effect by using the formula for intensity but adjusting the radius \(r\), reflecting the greater distance.
As you increase the distance, such as moving from 1 km to 50 km, the surface area increases dramatically. This reduces the energy per unit area because the total energy remains the same while the area over which it is spread becomes larger. These calculations are essential for designing systems that rely on EM waves, ensuring proper power and data reception across varying ranges.

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Most popular questions from this chapter

(II) An electromagnetic wave has an electric field given by \(\overrightarrow{\mathbf{E}}=\hat{\mathbf{i}}(225 \mathrm{~V} / \mathrm{m}) \sin \left[\left(0.077 \mathrm{~m}^{-1}\right) z-\left(2.3 \times 10^{7} \mathrm{rad} / \mathrm{s}\right) t\right]\) (a) What are the wavelength and frequency of the wave? (b) Write down an expression for the magnetic field.

(I) How long does it take light to reach us from the Sun, \(1.50 \times 10^{8} \mathrm{~km}\) away?

Imagine that a steady current \(/\) flows in a straight cylindrical wire of radius \(R_{0}\) and resistivity \(\rho\) . \((a)\) If the current is then changed at a rate \(d I / d t\) , show that a displacement current \(I_{\mathrm{D}}\) exists in the wire of magnitude \(\epsilon_{0} \rho(d I / d t) .\) (b) If the current in a copper wire is changed at the rate of 1.0 \(\mathrm{A} / \mathrm{ms}\) , deter- mine the magnitude of \(I_{\mathrm{D}}\) (c) Determine the magnitude of the magnetic field \(B_{\mathrm{D}}\) (T) created by \(I_{\mathrm{D}}\) at the surface of a copper wire with \(R_{0}=1.0 \mathrm{mm}\) . Compare (as a ratio) \(B_{\mathrm{D}}\) with the field created at the surface of the wire by a steady current of 1.0 \(\mathrm{A}\) .

The metal walls of a microwave oven form a cavity of dimensions \(37 \mathrm{~cm} \times 37 \mathrm{~cm} \times 20 \mathrm{~cm}\). When 2.45 -GHz microwaves are continuously introduced into this cavity, reflection of incident waves from the walls set up standing waves with nodes at the walls. Along the 37 -cm dimension of the oven, how many nodes exist (excluding the nodes at the wall) and what is the distance between adjacent nodes? [Because no heating occurs at these nodes, most microwaves rotate food while operating.

(I) What is the range of wavelengths for (a) FM radio \(\begin{array}{llll}(88 \mathrm{MHz} & \text { to } 108 \mathrm{MHz}) & \text { and }(b) \text { AM radio }(535 \mathrm{kHz} \text { to }\end{array}\) \(1700 \mathrm{kHz}) ?\)
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