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Chapter 25: Problem 70

In reading the instruction manual that came with your garage-door opener, you see that the transmitter unit in your car produces a \(250 \mathrm{mW}\) signal and that the receiver unit is supposed to respond to a radio wave of the correct frequency if the electric field amplitude exceeds \(0.10 \mathrm{V} / \mathrm{m}\). You wonder if this is really true. To find out, you put fresh batteries in the transmitter and start walking away from your garage while opening and closing the door. Your garage door finally fails to respond when you're \(42 \mathrm{m}\) away. Are the manufacturer's claims true?

Short Answer

Expert verified
After the computation, if the calculated electric field amplitude value is found to be less than \( 0.10 V/m \), then the manufacturer's claim is valid.

Step by step solution

01

Understanding the formula

The intensity of an electromagnetic wave is given as the power per unit area. This can be expressed with the formula \( I = \frac{P}{A} \) where \( I \) is the intensity, \( P \) is the power of the wave and \( A \) is the area on which the power is distributed. Here, the area is the surface of the sphere with a radius equal to the distance of the transmitter from the garage.
02

Calculation of the intensity

Here, with given power \( P = 250 mW = 250 \times 10^{-3} W \), and distance \( d = 42 m \) we calculate the area of the sphere as \( A = 4 \pi d^2 \). Substituting these values into the intensity formula resulting in \( I = \frac{P}{A} = \frac{250 \times 10^{-3}}{4\pi(42)^2} \). Solve it for \( I \).
03

Calculating the electric field amplitude

The electric field \( E \) and intensity \( I \) are related by the equation \( I = \frac{1}{2} c \varepsilon_0 E^2 \), where \( c \) is the speed of light and \( \varepsilon_0 \) is the permittivity of free space. Solving for E yields \( E = \sqrt{\frac{2I}{c \varepsilon_0}} \). Substituting the obtained value of I, \( c \approx 3.00 \times 10^8 m/s \) and \( \varepsilon_0 \approx 8.85 \times 10^{-12} C^2/N\cdot m^2 \), calculate \( E \).
04

Comparsion with threshold

Lastly, the calculated value of electric field amplitude is compared with the threshold given in the manufacturer's claims, which is \( 0.10 V/m \). The claims will be valid if the calculated field strength is greater than or equal to the given threshold.

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Key Concepts

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

Intensity of Electromagnetic Waves
Electromagnetic waves, like those from your garage-door transmitter, carry energy through space. The intensity of these waves describes how much power flows through a given area and is calculated using the formula \( I = \frac{P}{A} \). Here, \( I \) represents intensity, \( P \) is the power in watts, and \( A \) is the area over which this power spreads.

Imagine spreading the wave's power over the surface of a sphere, with the transmitter at the center. The distance from the transmitter to any point on this sphere is the radius, and the area \( A \) of the sphere can be determined by the formula \( 4 \pi d^2 \), where \( d \) is the distance to the receiver. As you increase the distance, the area grows, causing the intensity to decrease. This is why the garage door's response changes as you walk away. Understanding these concepts helps in numerous everyday applications, demonstrating how power and area work together to influence wave intensity.
Electric Field Amplitude
The electric field amplitude, a vital characteristic of an electromagnetic wave, measures the strength of the wave at any given point. It鈥檚 closely related to the wave's intensity through the relationship \( I = \frac{1}{2} c \varepsilon_0 E^2 \). Here, \( c \) is the speed of light, \( \varepsilon_0 \) is the permittivity of free space, and \( E \) denotes the electric field amplitude.

By rearranging this equation, we find \( E = \sqrt{\frac{2I}{c \varepsilon_0}} \), allowing us to calculate the amplitude if the intensity is known. A strong electric field means a potent signal, necessary for devices like a garage door receiver to function properly.
  • A higher amplitude increases signal detection.
  • Exceeding the threshold value ensures proper operation.
The sensitivity of the receiver unit is calibrated to respond when the electric field amplitude exceeds a particular threshold, such as the \( 0.10 \mathrm{V}/\mathrm{m} \) described by the manufacturer.
Permittivity of Free Space
Permittivity of free space, denoted \( \varepsilon_0 \), is a fundamental constant vital to understanding electromagnetic waves. This constant reflects how electric fields interact with the vacuum of empty space. Its value is approximately \( 8.85 \times 10^{-12} \mathrm{C}^2/\mathrm{N}\cdot \mathrm{m}^2 \).

\( \varepsilon_0 \) plays a critical role in the relationship between electric field amplitude and intensity. It influences how wave energy spreads through free space and governs light's speed in a vacuum through the equation \( c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}} \), where \( \mu_0 \) is the permeability of free space.
  • Describes how easily electric fields can develop in free space.
  • Impacts the capacitance of capacitors in electrical circuits.
Understanding \( \varepsilon_0 \) is crucial for navigating topics in electromagnetism, offering insights into wave behaviors and helping refine technologies that rely on these principles.

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Most popular questions from this chapter

People immersed in strong unchanging magnetic fields occasionally report sensing a metallic taste. Some investigators suspect that motion in the constant field could produce a changing flux and a resulting emf that could stimulate nerves in the tongue. We can make a simple model to see if this is reasonable by imagining a somewhat extreme case. Suppose a patient having an MRI is immersed in a \(3.0 \mathrm{T}\) field along the axis of his body. He then quickly tips his head to the side, toward his right shoulder, tipping his head by \(30^{\circ}\) in the rather short time of 0.15 s. Estimate the area of the tongue; then calculate the emf that could be induced in a loop around the outside of the tongue by this motion of the head. How does this emf com- pare to the approximately \(15 \mathrm{mV}\) necessary to trigger an action potential? Does it seem reasonable to suppose that an induced emf is responsible for the noted effect?

The intensity of a polarized electromagnetic wave is \(10 \mathrm{W} / \mathrm{m}^{2} .\) What will be the intensity after passing through a polarizing filter whose axis makes the following angles with the plane of polarization? (a) \(\theta=0^{\circ}\) (b) \(\theta=30^{\circ}\) (c) \(\theta=45^{\circ}\) (d) \(\theta=60^{\circ},\) (e) \(\theta=90^{\circ}\)

Rod cells in the retina of the eye detect light using a photopigment called rhodopsin. \(1.8 \mathrm{eV}\) is the lowest photon energy that can trigger a response in thodopsin. What is the maximum wavelength of electromagnetic radiation that can cause a transition? In what part of the spectrum is this?

Which of the following changes would not produce a larger eddy current in the metal? A. Increasing the frequency of the oscillating current in the transmitter coil B. Increasing the magnitude of the oscillating current in the transmitter coil C. Increasing the resistivity of the metal D. Decreasing the distance between the metal and the transmitter

Some electronic devices such as phones and watches can be charged wirelessly. In the charger, there is a coil that generates a rapidly changing magnetic field. In the device, there is a second, 100-turn rectangular coil with side lengths of \(3.0 \mathrm{cm}\) and \(4.0 \mathrm{cm} .\) If the maximum rate of change of the magnetic field in this second coil is \(53 \mathrm{T} / \mathrm{s},\) what is the emf induced in this coil?
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