24.4

The Energy Carried by Electromagnetic Waves

Electromagnetic waves, like water waves or sound waves, carry energy. The energy is carried by the electric and magnetic fields that comprise the wave. In a microwave oven, for example, microwaves penetrate food and deliver their energy to it, as Figure 24-14 illustrates. The electric field of the microwaves is largely responsible for delivering the energy, and water molecules in the food absorb it. The absorption occurs because each water molecule has a permanent dipole moment; that is, one end of a molecule has a slight positive charge, and the other end has a negative charge of equal magnitude. As a result, the positive and negative ends of different molecules can form a bond. However, the electric field of the microwaves exerts forces on the positive and negative ends of a molecule, causing it to spin. Because the field is oscillating rapidly—about times a second—the water molecules are kept spinning at a high rate. In the process, the energy of the microwaves is used to break bonds between neighboring water molecules and ultimately is converted into internal energy. As the internal energy increases, the temperature of the water increases, and the food cooks.

Figure 24-14 A microwave oven. The rotating fan blades reflect the microwaves to all parts of the oven.

The physics of a microwave oven.

The energy carried by electromagnetic waves in the infrared and visible regions of the spectrum plays the key role in the greenhouse effect that is a contributing factor to global warming. The infrared waves from the sun are largely prevented from reaching the earth’s surface by carbon dioxide and water in the atmosphere, which reflect them back into space. The visible waves do reach the earth’s surface, however, and the energy they carry heats the earth. Heat also flows to the surface from the interior of the earth. The heated surface in turn radiates infrared waves outward, which, if they could, would carry their energy into space. However, the atmospheric carbon dioxide and water reflect these infrared waves back toward the earth, just as they reflect the infrared waves from the sun. Thus, their energy is trapped, and the earth becomes warmer, like plants in a greenhouse. In a greenhouse, however, energy is trapped mainly for a different reason—namely, the lack of effective convection currents to carry warm air past the cold glass walls.

The physics of the greenhouse effect.

A measure of the energy stored in the electric field of an electromagnetic wave, such as a microwave, is provided by the electric energy density. As we saw in Section 19.5, this density is the electric energy per unit volume of space in which the electric field exists:

| (19.12)

In this equation, the dielectric constant has been set equal to unity, since we are dealing with an electric field in a vacuum (or in air). From Section 22.8, the analogous expression for the magnetic energy density is

| (22.11)

The total energy density u of an electromagnetic wave in a vacuum is the sum of these two energy densities:

| (24.2a)

In an electromagnetic wave propagating through a vacuum or air, the electric field and the magnetic field carry equal amounts of energy per unit volume of space. Since , it is possible to rewrite Equation 24.2a for the total energy density in two additional, but equivalent, forms:

| (24.2b)

| (24.2c)

The fact that the two energy densities are equal implies that the electric and magnetic fields are related. To see how, we set the electric energy density equal to the magnetic energy density and obtain

But according to Equation 24.1, , so it follows that . Taking the square root of both sides of this result shows that the relation between the magnitudes of the electric and magnetic fields in an electromagnetic wave is

| (24.3)

In an electromagnetic wave, the electric and magnetic fields fluctuate sinusoidally in time, so Equations 24.2a, 24.2b and 24.2c give the energy density of the wave at any instant in time. If an average value for the total energy density is desired, average values are needed for E2 and B2. In Section 20.5 we faced a similar situation for alternating currents and voltages and introduced rms (root mean square) quantities. Using an analogous procedure here, we find that the rms values for the electric and magnetic fields, Erms and Brms, are related to the maximum values of these fields, E0 and B0, by

Equations 24.2a, 24.2b and 24.2c can now be interpreted as giving the average energy density , provided the symbols E and B are interpreted to mean the rms values given above. The average density of the sunlight reaching the earth is determined in the next example.

| Example 4 | | | The Average Energy Density of Sunlight

Sunlight enters the top of the earth’s atmosphere with an electric field whose rms value is . Find (a) the average total energy density of this electromagnetic wave and (b) the rms value of the sunlight’s magnetic field. Reasoning The average total energy density of the sunlight can be obtained from Equation 24.2b, provided the rms value is used for the electric field. Since the magnitudes of the magnetic and electric fields are related according to Equation 24.3, the rms value of the magnetic field is Brms = Erms/c.
Solution

a. | According to Equation 24.2b, the average total energy density is

b. | Using Equation 24.3, we find that the rms magnetic field is

As an electromagnetic wave moves through space, it carries energy from one region to another. This energy transport is characterized by the intensity of the wave. We have encountered the concept of intensity before, in connection with sound waves in Section 16.7. The sound intensity is the sound power that passes perpendicularly through a surface divided by the area of the surface. The intensity of an electromagnetic wave is defined similarly. For an electromagnetic wave, the intensity is the electromagnetic power divided by the area of the surface.
Using this definition of intensity, we can show that the electromagnetic intensity S is related to the energy density u. According to Equation 16.8 the intensity is the power P that passes perpendicularly through a surface divided by the area A of that surface, or . Furthermore, the power is equal to the total energy passing through the surface divided by the elapsed time t (Equation 6.10b), so that . Combining these two relations gives

Problem solving insight The concepts of power and intensity are similar, but they are not the same. Intensity is the power that passes perpendicularly through a surface divided by the area of the surface.

Now, consider Figure 24-15, which shows an electromagnetic wave traveling in a vacuum along the x axis. In a time t the wave travels the distance ct, passing through the surface of area A. Consequently, the volume of space through which the wave passes is ctA. The total (electric and magnetic) energy in this volume is

Using this result in the expression for the intensity, we obtain

| (24.4)

Thus, the intensity and the energy density are related by the speed of light, c. Substituting Equations 24.2a, 24.2b and 24.2c, one at a time, into Equation 24.4 shows that the intensity of an electromagnetic wave depends on the electric and magnetic fields according to the following equivalent relations:

| (24.5a)

| (24.5b)

| (24.5c)

Figure 24-15 In a time t, an electromagnetic wave moves a distance ct along the x axis and passes through a surface of area A.

If the rms values for the electric and magnetic fields are used in Equations 24.5a, 24.5b and 24.5c, the intensity becomes an average intensity, , as Example 5 illustrates.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 5 | | | Power and Intensity

Figure 24-16 shows a tiny source that is emitting light uniformly in all directions. At a distance of 2.50 m from the source, the rms electric field strength of the light is 19.0 N/C. Assuming that the light does not reflect from anything in the environment, determine the average power of the light emitted by the source.

Figure 24-16 At a distance of 2.50 m from the light source, the rms electric field of the light has a value of 19.0 N/C.

Reasoning Recall from Section 16.7 that the power crossing a surface perpendicularly is equal to the intensity at the surface times the area of the surface (see Equation 16.8). Since the source emits light uniformly in all directions, the light intensity is the same at all points on the imaginary spherical surface in Figure 24-16. Moreover, the light crosses this surface perpendicularly. Equation 24.5b relates the average light intensity at the surface to its rms electric field strength (which is known), and the area of the surface can be found from a knowledge of its radius.
Knowns and Unknowns The following data are available:

Description | Symbol | Value
Rms electric field strength 2.50 m from light source | Erms | 19.0 N/C
Distance from light source | r | 2.50 m
Unknown Variable |  |
Average power emitted by light source |  | ?

Modeling the Problem

| Step 1 Average Intensity
| According to the discussion in Section 16.7, the average power that passes perpendicularly through the imaginary spherical surface is equal to the average intensity times the area A of the surface, or . The area of a spherical surface is , where r is the radius of the sphere. Thus, the average power can be written as in Equation 1 at the right. The radius is known, but the average light intensity is not, so we turn to Step 2 to evaluate it.

| (1)

| Step 2 Average Intensity and Electric Field
| The average intensity of the light passing through the imaginary spherical surface is related to the known rms electric field strength at the surface by Equation 24.5b:

where c is the speed of light in a vacuum and ε0 is the permittivity of free space. We can substitute this expression into Equation 1, as indicated at the right.

Solution Algebraically combining the results of each step, we have

The average power emitted by the light source is

Related Homework: Problems 24, 26 and 57

Check Your Understanding 2

If both the electric and magnetic fields of an electromagnetic wave double in magnitude, how does the intensity of the wave change? (a) The intensity decreases by a factor of four. (b) The intensity decreases by a factor of two. (c) The intensity increases by a factor of two. (d) The intensity increases by a factor of four. (e) The intensity increases by a factor of eight. (The answer is given at the end of the book.)
Background:The intensity of an electromagnetic wave is the energy per second per square meter that the wave carries through a surf ace that is perpendicular to the direction of travel of the wave. It depends on the magnitudes of both the electric and magnetic fields.
For similar questions (including calculational counterparts), consult Self-Assessment Test 24.1, which is described at the end of Section 24.6.

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