31.6.1 Energy and Intensity

Waves transfer energy. Ocean waves erode beaches, sound waves set your eardrums vibrating, and light from the sun warms the earth. The energy flow of an electromagnetic wave is described by the Poynting vector $\vec{S}$ , defined as

\vec{S} \equiv \frac{1}{μ_{0}} \vec{E} \times \vec{B}

(31.39)

The Poynting vector, shown in FIGURE 31.24, has two important properties:

The Poynting vector points in the direction in which an electromagnetic wave is traveling. You can see this by looking back at Figure 31.23.
It is straightforward to show that the units of S are $W / m^{2}$ , or power (joules per second) per unit area. Thus the magnitude S of the Poynting vector measures the rate of energy transfer per unit area of the wave front.

The figure illustrates a Poynting vector in the x y z coordinate plane.

FIGURE 31.24 Full Alternative Text

Because $\vec{E}$ and $\vec{B}$ of an electromagnetic wave are perpendicular to each other, and $E = c B$ , the magnitude of the Poynting vector is

S = \frac{E B}{μ_{0}} = \frac{E^{2}}{c μ_{0}} = c ϵ_{0} E^{2}

The Poynting vector is a function of time, oscillating from zero to $S_{max} = E_{0}^{2} / c μ_{0}$ and back to zero twice during each period of the wave’s oscillation. That is, the energy flow in an electromagnetic wave is not smooth. It “pulses” as the electric and magnetic fields oscillate in intensity. We’re unaware of this pulsing because the electromagnetic waves that we can sense—light waves—have such high frequencies.

Of more interest is the average energy transfer, averaged over one cycle of oscillation, which is the wave’s intensity I. In our earlier study of waves, we defined the intensity of a wave to be $I = P / A$ , where P is the power (energy transferred per second) of a wave that impinges on area A. Because $E = E_{0} sin [ 2 π (x / λ - f t) ]$ , and the average over one period of ${sin}^{2} [ 2 π (x / λ - f t) ]$ is $\frac{1}{2}$ , the intensity of an electromagnetic wave is

I = \frac{P}{A} = S_{avg} = \frac{1}{2 c μ_{0}} E_{0}^{2} = \frac{c ϵ_{0}}{2} E_{0}^{2}

(31.40)

Equation 31.40 relates the intensity of an electromagnetic wave, a quantity that is easily measured, to the amplitude $E_{0}$ of the wave’s electric field.

The intensity of a plane wave, with constant electric field amplitude $E_{0}$ , would not change with distance. But a plane wave is an idealization; there are no true plane waves in nature. You learned in Chapter 16 that, to conserve energy, the intensity of a wave far from its source decreases with the inverse square of the distance. If a source with power $P_{source}$ emits electromagnetic waves uniformly in all directions, the electromagnetic wave intensity at distance r from the source is

I = \frac{P_{source}}{4 π r^{2}}

(31.41)

Equation 31.41 simply expresses the recognition that the energy of the wave is spread over a sphere of surface area $4 π r^{2}$ .

EXAMPLE 31.4 Fields of a cell phone

A cell phone broadcasts a 0.60 W signal at a frequency of 1.9 GHz. What are the amplitudes of the electric and magnetic fields at a distance of 10 cm, about the distance to the center of the user’s brain?

MODEL Treat the cell phone as a point source of electromagnetic waves.

SOLVE The intensity of a 0.60 W point source at a distance of 10 cm is

I = \frac{P_{source}}{4 π r^{2}} = \frac{0.60 W}{4 π {(0.10 m)}^{2}} = 4.78 W / m^{2}

We can find the electric field amplitude from the intensity:

\begin{matrix} E_{0} & = & \sqrt{\frac{2 I}{c ϵ_{0}}} & = & \sqrt{\frac{2 (4.78 W / m^{2})}{(3.00 \times 10^{8} m / s) (8.85 \times 10^{- 12} C^{2} / N m^{2})}} \\ = & 60 V / m \end{matrix}

The amplitudes of the electric and magnetic fields are related by the speed of light. This allows us to compute

B_{0} = \frac{E_{0}}{c} = 2.0 \times 10^{- 7} T

REVIEW The electric field amplitude is modest; the magnetic field amplitude is very small. This implies that the interaction of electromagnetic waves with matter is mostly due to the electric field.