/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Fields from a Light Bulb. We can... [FREE SOLUTION] | 91Ó°ÊÓ

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Fields from a Light Bulb. We can reasonably model a \(75-\mathrm{W}\) incandescent light bulb as a sphere \(6.0 \mathrm{~cm}\) in diameter. Typically, only about \(5 \%\) of the energy goes to visible light; the rest goes largely to nonvisible infrared radiation. (a) What is the visible-light intensity (in \(\mathrm{W} / \mathrm{m}^{2}\) ) at the surface of the bulb? (b) What are the amplitudes of the electric and magnetic fields at this surface, for a sinusoidal wave with this intensity?

Short Answer

Expert verified
The visible-light intensity is approximately 332 W/m²; electric field amplitude is ~502 V/m; magnetic field amplitude is ~1.673 μT.

Step by step solution

01

Determine the visible power output

First, calculate the visible power output of the light bulb. Since only 5% of the total power is converted to visible light, apply the formula:\[ P_{\text{visible}} = 0.05 \times P_{\text{total}} = 0.05 \times 75 \, \text{W} = 3.75 \, \text{W} \]
02

Calculate the surface area of the bulb

Model the bulb as a sphere with a diameter of 6.0 cm. Calculate the surface area using the formula for the surface area of a sphere:\[ A = 4 \pi r^2 \]where the radius \( r \) is half of the diameter:\[ r = \frac{6.0 \, \text{cm}}{2} = 3.0 \, \text{cm} = 0.03 \, \text{m} \]Thus, the surface area is calculated as:\[ A = 4 \pi (0.03 \, \text{m})^2 \approx 0.0113 \, \text{m}^2 \]
03

Calculate the intensity of visible light

Using the visible power and the surface area, calculate the intensity using the formula:\[ I = \frac{P_{\text{visible}}}{A} = \frac{3.75 \, \text{W}}{0.0113 \, \text{m}^2} \approx 331.86 \, \text{W/m}^2 \]
04

Determine the electric field amplitude

The intensity \( I \) of an electromagnetic wave is related to the electric field amplitude \( E_0 \) by the equation:\[ I = \frac{1}{2} c \varepsilon_0 E_0^2 \]Where \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light and \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) is the permittivity of free space. Rearrange to solve for \( E_0 \):\[ E_0 = \sqrt{\frac{2I}{c\varepsilon_0}} \approx \sqrt{\frac{2 \times 331.86}{3 \times 10^8 \times 8.85 \times 10^{-12}}} \approx 501.96 \, \text{V/m} \]
05

Calculate the magnetic field amplitude

The magnetic field amplitude \( B_0 \) is related to the electric field amplitude \( E_0 \) by the formula:\[ B_0 = \frac{E_0}{c} \]Thus, the magnetic field amplitude is:\[ B_0 = \frac{501.96 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \approx 1.673 \times 10^{-6} \, \text{T} \]

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Key Concepts

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

Intensity of Light
The intensity of light, often denoted as \( I \), is a measure of how much energy a wave carries per unit area, per unit time. It is expressed in watts per square meter (W/m²). In the context of a light bulb, not all the provided power converts to visible light. For a 75 W light bulb, only 5% contributes to visible light. Therefore, calculating the visible light intensity involves determining how much of the bulb's power emerges in the visible spectrum.
The visible power \( P_{\text{visible}} \) is calculated as \( 0.05 \times 75 \, \text{W} = 3.75 \, \text{W} \).

Next, we use the surface area of the bulb (modeled as a sphere with a diameter of 6.0 cm) to find the intensity. The surface area \( A \) of a sphere is given by:
  • Formula: \( A = 4 \pi r^2 \)
  • Radius \( r = \frac{6.0 \, \text{cm}}{2} = 0.03 \, \text{m} \)
  • \( A = 4 \pi (0.03 \, \text{m})^2 \approx 0.0113 \, \text{m}^2 \)
Finally, we calculate the intensity \( I \) using:
  • Formula: \( I = \frac{P_{\text{visible}}}{A} \)
  • \( I = \frac{3.75 \, \text{W}}{0.0113 \, \text{m}^2} \approx 331.86 \, \text{W/m}^2 \)
Thus, the intensity of visible light at the bulb's surface is approximately 331.86 W/m².
Electric Field Amplitude
The electric field amplitude \( E_0 \) is a crucial aspect of electromagnetic waves, representing the strength of the electric field of the wave. This varies directly with the light wave's intensity. For electromagnetic waves, the intensity \( I \) is connected to the electric field amplitude by the equation:
  • \( I = \frac{1}{2} c \varepsilon_0 E_0^2 \)
  • \( c = 3 \times 10^8 \, \text{m/s} \), the speed of light
  • \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \), the permittivity of free space
Given the intensity \( I = 331.86 \, \text{W/m}^2 \), we can solve for \( E_0 \):
  • \( E_0 = \sqrt{\frac{2I}{c\varepsilon_0}} \)
  • \( E_0 = \sqrt{\frac{2 \times 331.86}{3 \times 10^8 \times 8.85 \times 10^{-12}}} \approx 501.96 \, \text{V/m} \)
This calculation reveals that the electric field amplitude at the surface of the bulb is approximately 501.96 V/m, indicating the intensity of the electric component of the wave.
Magnetic Field Amplitude
The magnetic field amplitude \( B_0 \) indicates the strength of the magnetic component of electromagnetic waves and is intrinsically linked to the electric field amplitude \( E_0 \). According to Maxwell's equations, these two components are interconnected through the speed of light \( c \). Therefore, given \( E_0 \), you can find \( B_0 \) using the simple relation:
  • \( B_0 = \frac{E_0}{c} \)
  • Substituting \( E_0 = 501.96 \, \text{V/m} \) and \( c = 3 \times 10^8 \, \text{m/s} \)
  • \( B_0 = \frac{501.96 \, \text{V/m}}{3 \times 10^8 \, \text{m/s}} \approx 1.673 \times 10^{-6} \, \text{T} \)
This result means that the magnetic field amplitude at the surface of the bulb measures approximately \(1.673 \times 10^{-6} \) T (tesla), capturing the wave's magnetic influence. Understanding the relationship between the electrical and magnetic fields is key to grasping how electromagnetic waves propagate through space.

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Most popular questions from this chapter

Consider each of the following electric-and magnetic-field orientations. In each case, what is the direction of propagation of the wave? (a) \(\vec{E}=E \hat{i}, \overrightarrow{\boldsymbol{B}}=-B \hat{j} ;\) (b) \(\overrightarrow{\boldsymbol{E}}=E \hat{\boldsymbol{J}}, \overrightarrow{\boldsymbol{B}}=B \hat{\boldsymbol{i}} ;\) (c) \(\vec{E}=\) \(-E \hat{k}, \vec{B}=-B \hat{i} ;(\) d) \(\vec{E}=E \hat{i}, \vec{B}=-B \hat{k}\).

Testing a Space Radio Transmitter. You are a NASA mission specialist on your first flight aboard the space shuttle. Thanks to your extensive training in physics, you have been assigned to evaluate the performance of a new radio transmitter on board the International Space Station (ISS). Perched on the shuttle's movable arm, you aim a sensitive detector at the ISS, which is \(2.5 \mathrm{~km}\) away. You find that the electric-field amplitude of the radio waves coming from the ISS transmitter is \(0.090 \mathrm{~V} / \mathrm{m}\) and that the frequency of the waves is \(244 \mathrm{MHz}\). Find the following: (a) the intensity of the radio wave at your location; (b) the magnetic-field amplitude of the wave at your location; (c) the total power output of the ISS radio transmitter. (d) What assumptions, if any, did you make in your calculations?

An electromagnetic wave with frequency \(5.70 \times 10^{14} \mathrm{~Hz}\) propagates with a speed of \(2.17 \times 10^{8} \mathrm{~m} / \mathrm{s}\) in a certain piece of glass. Find (a) the wavelength of the wave in the glass; (b) the wavelength of a wave of the same frequency propagating in air; (c) the index of refraction \(n\) of the glass for an electromagnetic wave with this frequency; (d) the dielectric constant for glass at this frequency, assuming that the relative permeability is unity.

(a) How much time does it take light to travel from the moon to the earth, a distance of \(384,000 \mathrm{~km}\) ? (b) Light from the star Sirius takes \(8.61\) years to reach the earth. What is the distance from earth to Sirius in kilometers?

A source of sinusoidal electromagnetic waves radiates uniformly in all directions. At \(10.0 \mathrm{~m}\) from this source, the amplitude of the electric field is measured to be \(1.50 \mathrm{~N} / \mathrm{C}\). What is the electric-field amplitude at a distance of \(20.0 \mathrm{~cm}\) from the source?

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