24.4 Energy in Electromagnetic Waves - College Physics

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 can 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. If the frequency of the electromagnetic wave is the same as the natural frequencies of the system (such as microwaves at the resonant frequency of water molecules), the transfer of energy is much more efficient.

Connections: Waves and Particles

The behavior of electromagnetic radiation clearly exhibits wave characteristics. But we shall find in later modules that at high frequencies, electromagnetic radiation also exhibits particle characteristics. These particle characteristics will be used to explain more of the properties of the electromagnetic spectrum and to introduce the formal study of modern physics.

Another startling discovery of modern physics is that particles, such as electrons and protons, exhibit wave characteristics. This simultaneous sharing of wave and particle properties for all submicroscopic entities is one of the great symmetries in nature.

The propagation of two electromagnetic waves is shown in three dimensional planes. The first wave shows with the variation of two components E and B. E is a sine wave in one plane with small arrows showing the vibrations of particles in the plane. B is a sine wave in a plane perpendicular to the E wave. The B wave has arrows to show the vibrations of particles in the plane. The waves are shown intersecting each other at the junction of the planes because E and B are perpendicular to each other. The direction of propagation of wave is shown perpendicular to E and B waves. The energy carried is given as E sub u. The second wave shows with the variation of the components two E and two B, that is, E and B waves with double the amplitude of the first case. Two E is a sine wave in one plane with small arrows showing the vibrations of particles in the plane. Two B is a sine wave in a plane perpendicular to the two E wave. The two B wave has arrows to show the vibrations of particles in the plane. The waves are shown intersecting each other at the junction of the planes because two E and two B waves are perpendicular to each other. The direction of propagation of wave is shown perpendicular to two E and two B waves. The energy carried is given as four E sub u.

Figure 24.23 Energy carried by a wave is proportional to its amplitude squared. With electromagnetic waves, larger

E

-fields and

B

-fields exert larger forces and can do more work.

But there is energy in an electromagnetic wave, whether it is absorbed or not. Once created, the fields carry energy away from a source. If 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.

A wave’s energy is proportional to its amplitude squared ( $E^{2}$ or $B^{2}$ ). This is true for waves on guitar strings, for water waves, and for sound waves, where amplitude is proportional to pressure. In electromagnetic waves, the amplitude is the maximum field strength of the electric and magnetic fields. (See Figure 24.23.)

Thus the energy carried and the intensity $I$ of an electromagnetic wave is proportional to $E^{2}$ and $B^{2}$ . In fact, for a continuous sinusoidal electromagnetic wave, the average intensity $I_{ave}$ is given by

I_{ave} = \frac{{cε}_{0} E_{0}^{2}}{2},

24.18

where $c$ is the speed of light, $ε_{0}$ is the permittivity of free space, and $E_{0}$ is the maximum electric field strength; intensity, as always, is power per unit area (here in ${W/m}^{2}$ ).

The average intensity of an electromagnetic wave $I_{ave}$ can also be expressed in terms of the magnetic field strength by using the relationship $B = E / c$ , and the fact that $ε_{0} = 1 / μ_{0} c^{2}$ , where $μ_{0}$ is the permeability of free space. Algebraic manipulation produces the relationship

I_{ave} = \frac{{cB}_{0}^{2}}{{2μ}_{0}},

24.19

where $B_{0}$ is the maximum magnetic field strength.

One more expression for $I_{ave}$ in terms of both electric and magnetic field strengths is useful. Substituting the fact that $c \cdot B_{0} = E_{0}$ , the previous expression becomes

I_{ave} = \frac{E_{0} B_{0}}{{2μ}_{0}} .

24.20

Whichever of the three preceding equations is most convenient can be used, since they are really just different versions of the same principle: Energy in a wave is related to amplitude squared. Furthermore, since these equations are based on the assumption that the electromagnetic waves are sinusoidal, peak intensity is twice the average; that is, $I_{0} = 2 I_{ave}$ .

Example 24.4

Calculate Microwave Intensities and Fields

On its highest power setting, a certain microwave oven projects 1.00 kW of microwaves onto a 30.0 by 40.0 cm area. (a) What is the intensity in ${W/m}^{2}$ ? (b) Calculate the peak electric field strength $E_{0}$ in these waves. (c) What is the peak magnetic field strength $B_{0}$ ?

Strategy

In part (a), we can find intensity from its definition as power per unit area. Once the intensity is known, we can use the equations below to find the field strengths asked for in parts (b) and (c).

Solution for (a)

Entering the given power into the definition of intensity, and noting the area is 0.300 by 0.400 m, yields

I = \frac{P}{A} = \frac{1 . 00  kW}{0 . 300 m \times 0 . 400 m} .

24.21

Here $I = I_{ave}$ , so that

I_{ave} = \frac{1000 W}{0.120 m^{2}} = 8 . 33 \times 10^{3} {W/m}^{2} .

24.22

Note that the peak intensity is twice the average:

I_{0} = 2 I_{ave} = 1 . 67 \times 10^{4} W / m^{2} .

24.23

Solution for (b)

To find $E_{0}$ , we can rearrange the first equation given above for $I_{ave}$ to give

E_{0} = {(\frac{2 I_{ave}}{{cε}_{0}})}^{1/2} .

24.24

Entering known values gives

\begin{array}{l} E_{0} & = & \sqrt{\frac{2 (8.33 \times 10^{3} {W/m}^{2})}{(3.00 \times 10^{8} m/s) (8.85 \times 10^{- 12} C^{2} / N \cdot m^{2})}} \\ = & 2.51 \times 10^{3} V/m . \end{array}

24.25

Solution for (c)

Perhaps the easiest way to find magnetic field strength, now that the electric field strength is known, is to use the relationship given by

B_{0} = \frac{E_{0}}{c} .

24.26

Entering known values gives

\begin{array}{l} B_{0} & = & \frac{2.51 \times 10^{3} V/m}{3.0 \times 10^{8} m/s} \\ = & 8.35 \times 10^{- 6} T . \end{array}

24.27

Discussion

As before, a relatively strong electric field is accompanied by a relatively weak magnetic field in an electromagnetic wave, since $B = E / c$ , and $c$ is a large number.