University Physics Volume 2

# 16.3Energy Carried by Electromagnetic Waves

University Physics Volume 216.3 Energy Carried by Electromagnetic Waves

### Learning Objectives

By the end of this section, you will be able to:
• Express the time-averaged energy density of electromagnetic waves in terms of their electric and magnetic field amplitudes
• Calculate the Poynting vector and the energy intensity of electromagnetic waves
• Explain how the energy of an electromagnetic wave depends on its amplitude, whereas the energy of a photon is proportional to its frequency

Anyone who has used a microwave oven knows there is energy in electromagnetic waves. Sometimes this energy is obvious, such as in the warmth of the summer Sun. Other times, it is subtle, such as the unfelt energy of gamma rays, which can destroy living cells.

Electromagnetic waves bring energy into a system by virtue of their electric and magnetic fields. These fields can exert forces and move charges in the system and, thus, do work on them. However, there is energy in an electromagnetic wave itself, whether it is absorbed or not. Once created, the fields carry energy away from a source. If some energy is later absorbed, the field strengths are diminished and anything left travels on.

Clearly, the larger the strength of the electric and magnetic fields, the more work they can do and the greater the energy the electromagnetic wave carries. In electromagnetic waves, the amplitude is the maximum field strength of the electric and magnetic fields (Figure 16.10). The wave energy is determined by the wave amplitude.

Figure 16.10 Energy carried by a wave depends on its amplitude. With electromagnetic waves, doubling the E fields and B fields quadruples the energy density u and the energy flux uc.

For a plane wave traveling in the direction of the positive x-axis with the phase of the wave chosen so that the wave maximum is at the origin at $t=0t=0$, the electric and magnetic fields obey the equations

$Ey(x,t)=E0cos(kx−ωt)Bz(x,t)=B0cos(kx−ωt).Ey(x,t)=E0cos(kx−ωt)Bz(x,t)=B0cos(kx−ωt).$

The energy in any part of the electromagnetic wave is the sum of the energies of the electric and magnetic fields. This energy per unit volume, or energy density u, is the sum of the energy density from the electric field and the energy density from the magnetic field. Expressions for both field energy densities were discussed earlier ($uEuE$ in Capacitance and $uBuB$ in Inductance). Combining these the contributions, we obtain

$u(x,t)=uE+uB=12ε0E2+12μ0B2.u(x,t)=uE+uB=12ε0E2+12μ0B2.$

The expression $E=cB=1ε0μ0BE=cB=1ε0μ0B$ then shows that the magnetic energy density $uBuB$ and electric energy density $uEuE$ are equal, despite the fact that changing electric fields generally produce only small magnetic fields. The equality of the electric and magnetic energy densities leads to

$u(x,t)=ε0E2=B2μ0.u(x,t)=ε0E2=B2μ0.$
(16.27)

The energy density moves with the electric and magnetic fields in a similar manner to the waves themselves.

We can find the rate of transport of energy by considering a small time interval $ΔtΔt$. As shown in Figure 16.11, the energy contained in a cylinder of length $cΔtcΔt$ and cross-sectional area A passes through the cross-sectional plane in the interval $Δt.Δt.$

Figure 16.11 The energy $uAcΔtuAcΔt$ contained in the electric and magnetic fields of the electromagnetic wave in the volume $AcΔtAcΔt$ passes through the area A in time $ΔtΔt$.

The energy passing through area A in time $ΔtΔt$ is

$u×volume=uAcΔt.u×volume=uAcΔt.$

The energy per unit area per unit time passing through a plane perpendicular to the wave, called the energy flux and denoted by S, can be calculated by dividing the energy by the area A and the time interval $ΔtΔt$.

$S=Energy passing areaAin timeΔtAΔt=uc=ε0cE2=1μ0EB.S=Energy passing areaAin timeΔtAΔt=uc=ε0cE2=1μ0EB.$

More generally, the flux of energy through any surface also depends on the orientation of the surface. To take the direction into account, we introduce a vector $S→S→$, called the Poynting vector, with the following definition:

$S→=1μ0E→×B→.S→=1μ0E→×B→.$
(16.28)

The cross-product of $E→E→$ and $B→B→$ points in the direction perpendicular to both vectors. To confirm that the direction of $S→S→$ is that of wave propagation, and not its negative, return to Figure 16.7. Note that Lenz’s and Faraday’s laws imply that when the magnetic field shown is increasing in time, the electric field is greater at x than at $x+Δxx+Δx$. The electric field is decreasing with increasing x at the given time and location. The proportionality between electric and magnetic fields requires the electric field to increase in time along with the magnetic field. This is possible only if the wave is propagating to the right in the diagram, in which case, the relative orientations show that $S→=1μ0E→×B→S→=1μ0E→×B→$ is specifically in the direction of propagation of the electromagnetic wave.

The energy flux at any place also varies in time, as can be seen by substituting u from Equation 16.23 into Equation 16.27.

$S(x,t)=cε0E02cos2(kx−ωt)S(x,t)=cε0E02cos2(kx−ωt)$
(16.29)

Because the frequency of visible light is very high, of the order of $1014Hz,1014Hz,$ the energy flux for visible light through any area is an extremely rapidly varying quantity. Most measuring devices, including our eyes, detect only an average over many cycles. The time average of the energy flux is the intensity I of the electromagnetic wave and is the power per unit area. It can be expressed by averaging the cosine function in Equation 16.29 over one complete cycle, which is the same as time-averaging over many cycles (here, T is one period):

$I=Savg=cε0E021T∫0Tcos2(2πtT)dt.I=Savg=cε0E021T∫0Tcos2(2πtT)dt.$
(16.30)

We can either evaluate the integral, or else note that because the sine and cosine differ merely in phase, the average over a complete cycle for $cos2(ξ)cos2(ξ)$ is the same as for $sin2(ξ)sin2(ξ)$, to obtain

$〈cos2ξ〉=12[〈cos2ξ〉+〈sin2ξ〉]=12〈1〉=12.〈cos2ξ〉=12[〈cos2ξ〉+〈sin2ξ〉]=12〈1〉=12.$

where the angle brackets $〈⋯〉〈⋯〉$ stand for the time-averaging operation. The intensity of light moving at speed c in vacuum is then found to be

$I=Savg=12cε0E02I=Savg=12cε0E02$
(16.31)

in terms of the maximum electric field strength $E0,E0,$ which is also the electric field amplitude. Algebraic manipulation produces the relationship

$I=cB022μ0I=cB022μ0$
(16.32)

where $B0B0$ is the magnetic field amplitude, which is the same as the maximum magnetic field strength. One more expression for $IavgIavg$ in terms of both electric and magnetic field strengths is useful. Substituting the fact that $cB0=E0,cB0=E0,$ the previous expression becomes

$I=E0B02μ0.I=E0B02μ0.$
(16.33)

We can use whichever of the three preceding equations is most convenient, because the three equations are really just different versions of the same result: The energy in a wave is related to amplitude squared. Furthermore, because these equations are based on the assumption that the electromagnetic waves are sinusoidal, the peak intensity is twice the average intensity; that is, $I0=2I.I0=2I.$

### Example 16.3

A Laser Beam The beam from a small laboratory laser typically has an intensity of about $1.0×10−3W/m21.0×10−3W/m2$. Assuming that the beam is composed of plane waves, calculate the amplitudes of the electric and magnetic fields in the beam.

Strategy Use the equation expressing intensity in terms of electric field to calculate the electric field from the intensity.

Solution From Equation 16.31, the intensity of the laser beam is

$I=12cε0E02.I=12cε0E02.$

The amplitude of the electric field is therefore

$E0=2cε0I=2(3.00×108m/s)(8.85×10−12F/m)(1.0×10−3W/m2)=0.87V/m.E0=2cε0I=2(3.00×108m/s)(8.85×10−12F/m)(1.0×10−3W/m2)=0.87V/m.$

The amplitude of the magnetic field can be obtained from Equation 16.20:

$B0=E0c=2.9×10−9T.B0=E0c=2.9×10−9T.$

### Example 16.4

Light Bulb Fields A light bulb emits 5.00 W of power as visible light. What are the average electric and magnetic fields from the light at a distance of 3.0 m?

Strategy Assume the bulb’s power output P is distributed uniformly over a sphere of radius 3.0 m to calculate the intensity, and from it, the electric field.

Solution The power radiated as visible light is then

$I=P4πr2=cε0E022,E0=2P4πr2cε0=25.00W4π(3.0m)2(3.00×108m/s)(8.85×10−12C2/N·m2)=5.77N/C,B0=E0/c=1.92×10−8T.I=P4πr2=cε0E022,E0=2P4πr2cε0=25.00W4π(3.0m)2(3.00×108m/s)(8.85×10−12C2/N·m2)=5.77N/C,B0=E0/c=1.92×10−8T.$

Significance The intensity I falls off as the distance squared if the radiation is dispersed uniformly in all directions.

### Example 16.5

Radio Range A 60-kW radio transmitter on Earth sends its signal to a satellite 100 km away (Figure 16.12). At what distance in the same direction would the signal have the same maximum field strength if the transmitter’s output power were increased to 90 kW?

Figure 16.12 In three dimensions, a signal spreads over a solid angle as it travels outward from its source.

Strategy The area over which the power in a particular direction is dispersed increases as distance squared, as illustrated in the figure. Change the power output P by a factor of (90 kW/60 kW) and change the area by the same factor to keep $I=PA=cε0E022I=PA=cε0E022$ the same. Then use the proportion of area A in the diagram to distance squared to find the distance that produces the calculated change in area.

Solution Using the proportionality of the areas to the squares of the distances, and solving, we obtain from the diagram

$r22r12=A2A1=90W60W,r2=9060(100km)=122km.r22r12=A2A1=90W60W,r2=9060(100km)=122km.$

Significance The range of a radio signal is the maximum distance between the transmitter and receiver that allows for normal operation. In the absence of complications such as reflections from obstacles, the intensity follows an inverse square law, and doubling the range would require multiplying the power by four.

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