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Chapter 32: Problem 15

We can reasonably model a 75-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
(a) 331 W/m虏; (b) Electric field: 498 V/m, Magnetic field: 1.66 碌T.

Step by step solution

01

Determine the Total Surface Area

To find the surface area of the light-bulb, which is modeled as a sphere, use the formula for the surface area of a sphere: \[ A = 4 \pi r^2 \] Given the diameter is 6.0 cm, the radius \( r \) is 3.0 cm or 0.03 m. Substitute this into the formula:\[ A = 4 \pi (0.03)^2 \approx 0.0113 \text{ m}^2 \] This is the total surface area of the bulb.
02

Calculate Visible Power Output

Since only 5% of the total power is converted to visible light, calculate this portion of the 75 W:\[ P_{\text{visible}} = 0.05 \times 75 \text{ W} = 3.75 \text{ W} \]This represents the power contributing to visible light.
03

Compute Intensity of Visible Light

Intensity is the power per unit area. Use the formula:\[ I = \frac{P_{\text{visible}}}{A} \]Substitute the known values:\[ I = \frac{3.75 \text{ W}}{0.0113 \text{ m}^2} \approx 331 \text{ W/m}^2 \]This is the intensity of the visible light at the bulb's surface.
04

Find the Electric Field Amplitude

The intensity \( I \) is related to the amplitude of the electric field \( E_0 \) by the equation:\[ I = \frac{1}{2} c \varepsilon_0 E_0^2 \]where \( c \) is the speed of light \( (3 \times 10^8 \text{ m/s}) \) and \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \text{ F/m}) \). Solve for \( E_0 \):\[ E_0^2 = \frac{2I}{c \varepsilon_0} \]\[ E_0 = \sqrt{\frac{2 \times 331}{3 \times 10^8 \times 8.85 \times 10^{-12}}} \approx 498 \text{ V/m} \]This is the electric field amplitude at the surface.
05

Find the Magnetic Field Amplitude

The electric and magnetic fields are related by:\[ B_0 = \frac{E_0}{c} \]Substitute \( E_0 = 498 \text{ V/m} \):\[ B_0 = \frac{498}{3 \times 10^8} \approx 1.66 \times 10^{-6} \text{ T} \]This is the amplitude of the magnetic field at the bulb's surface.

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Key Concepts

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

Electric Field Amplitude
The electric field amplitude is a fundamental concept when dealing with electromagnetic waves, such as light. It represents the maximum strength of the electric field at a point in space. To find it, we use the relation between the intensity of light and the electric field amplitude. The formula that connects these two is: \[ I = \frac{1}{2} c \varepsilon_0 E_0^2 \] where:
  • \( I \) is the intensity of the light.
  • \( c \) represents the speed of light, approximately \(3 \times 10^8\) m/s.
  • \( \varepsilon_0 \) is the permittivity of free space, about \(8.85 \times 10^{-12}\) F/m.
  • \( E_0 \) is the electric field amplitude.
By manipulating this formula, we can solve for \( E_0 \). Given an intensity \( I \), as in the case with the light bulb's visible light, rearrange the equation to find \( E_0 \): \[ E_0 = \sqrt{\frac{2I}{c \varepsilon_0}} \] Plugging the values of \( I = 331 \) W/m虏, \( c \), and \( \varepsilon_0 \) gives \( E_0 \approx 498 \) V/m. This tells us the strength of the electric field at the light bulb's surface.
Magnetic Field Amplitude
The magnetic field amplitude is another critical component of electromagnetic waves. It's related to the electric field amplitude, \( E_0 \), through the speed of light, \( c \). The relationship can be given by: \[ B_0 = \frac{E_0}{c} \] where:
  • \( B_0 \) is the magnetic field amplitude.
  • \( E_0 \) is the electric field amplitude.
  • \( c \) is the speed of light, \( 3 \times 10^8 \) m/s.
By using the previously calculated electric field amplitude \( E_0 \approx 498 \) V/m, substitute it in the formula to find \( B_0 \): \[ B_0 = \frac{498}{3 \times 10^8} \approx 1.66 \times 10^{-6} \] T (Tesla). This is the amplitude of the oscillating magnetic field at the surface of the light bulb, emphasizing how these fields work together to transmit energy.
Surface Area of Sphere
Understanding the surface area of a sphere is essential for calculations involving objects like a light bulb. When we approximate an object as a sphere, we can apply the formula for the surface area of a sphere: \[ A = 4 \pi r^2 \] where:
  • \( A \) is the surface area.
  • \( r \) is the radius of the sphere.
For the light bulb in question, given the diameter is 6.0 cm, half of that gives us a radius \( r = 3.0 \) cm or 0.03 m. Substituting in the values: \[ A = 4 \pi (0.03)^2 \approx 0.0113 \] m虏. With this, we get the total surface area through which light is emitted. This information is crucial when dividing total emitted power by this area to determine light intensity. Calculating the surface area is a foundational step in assessing the distribution of energy across the light bulb's surface.

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Most popular questions from this chapter

The sun emits energy in the form of electromagnetic waves at a rate of \(3.9 \times 10^{26} \mathrm{W}\) . This energy is produced by nuclear reactions deep in the sun's interior. (a) Find the intensity of electromagnetic radiation and the radiation pressure on an absorbing object at the surface of the sun (radius \(r=R=6.96 \times 10^{5} \mathrm{km}\)) and at \(r=R / 2\) , in the sun's interior. Ignore any scattering of the waves as they move radially outward from the center of the sun. Compare to the values given in Section 32.4 for sunlight just before it enters the earth's atmosphere. (b) The gas pressure at the sun's surface is about \(1.0 \times 10^{4} \mathrm{Pa}\) ; at \(r=R / 2,\) the gas pressure is calculated from solar models to be about \(4.7 \times 10^{33} \mathrm{Pa}\) Comparing with your results in part (a), would you expect that radiation pressure is an important factor in determining the structure of the sun? Why or why not?

If the intensity of direct sunlight at a point on the earth's surface is \(0.78 \mathrm{kW} / \mathrm{m}^{2},\) find \((\mathrm{a})\) the average momentum density (momentum per unit volume) in the sunlight and (b) the average momentum flow rate in the sunlight.

A circular loop of wire can be used us a radio antenna. If a 18.0-cm-diameter antenna is located 2.50 \(\mathrm{km}\) from a 95.0 -MHz source with a total power of 55.0 \(\mathrm{kW}\) , what is the maximum emf induced in the loop? (Assume that the plane of the antenna loop is perpendicular to the direction of the radiation's magnetic field and that the source radiates uniformly in all directions.)

Interplanetary space contains many small particles referred to as interplanetary dust. Radiation pressure from the sun sets a lower limit on the size of such dust particles. To see the origin of this limit, consider a spherical dust particle of radius \(R\) and mass density \(\rho\) (a) Write an expression for the gravitational force exerted on this particle by the sun (mass \(M )\) when the particle is a distance \(r\) from the sun. (b) Let \(L\) represent the luminosity of the sun, equal to the rate at which it emits energy in electromagnetic radiation. Find the force exerted on the (totally absorbing) particle due to solar radiation pressure, remembering that the intensity of the sun's radiation also depends on the distance \(r .\) The relevant area is the cross-sectional area of the particle, \(n o t\) the total surface area of the particle. As part of your answer, explain why this is so. (c) The mass density of a typical interplanetary dust particle is about 3000 \(\mathrm{kg} / \mathrm{m}^{3}\) . Find the particle radius \(R\) such that the gravitational and radiation forces acting on the particle are equal in magnitude. The luminosity of the sun is \(3.9 \times 10^{26} \mathrm{W}\) . Does your answer depend on the distance of the particle from the sun? Why or why not? (d) Explain why dust particles with a radius less than that found in part (c) are unlikely to be found in the solar system. [Hint: Construct the ratio of the two force expressions found in parts (a) and (b).]

In a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency 110.0 \(\mathrm{MHz}\) in opposite directions inside a narrow cavity with reflectors at both ends, causing a standing wave pattern to occur. (a) How far apart are the nodal planes of the magnetic field? (b) If the standing wave pattern is determined to be in its eighth harmonic, how long is the cavity?
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