## Energy in Electromagnetic Waves

For simplicity, only electromagnetic waves at large distances from any antennas and with a single frequency are considered. That makes it possible to write down a relationship between the intensity and the squared electric field of the wave. As for all time-varying quantities, one can characterize the wave intensity either by its peak value or by its average. The average is taken over one period of the wave's oscillation. That is usually the most useful quantity to look at, for example when calculating how a focused light wave will raise the temperature of an object it strikes.

The text briefly mentions that there is something mysterious about the nature of light. Although we have just tried very hard to convince ourselves that light is an electromagnetic wave, there are also situations where the same light behaves more like a material object, i.e., a particle. In those situations one refers to light as made up of photons.

Fortunately, the language of energy we use in this section is universal, which means we can apply it no matter what we think the nature of light actually is. Energy can be transmitted by a stream of particles (photons) just as it can be transported by a wave. It also dilutes and concentrates the same way in both descriptions.

The reason this so-called wave particle duality crops up for light, but not for radio waves lies in the details of how light waves are generated and detected. There are no radio antennas small enough to create visible light because the wavelengths are billions of times shorter than for radio waves.

Instead, visible light (and waves of higher frequency) is generated in atoms, by processes that can only be described using the modern theories of matter, based on quantum physics. As the term "quantum" implies, energy is transmitted by these processes in packets that have many of the properties we typically associate with particles – and that is where the concept of the photon comes from.

As we move onto the next unit, it may help you to keep this photon idea in mind. We are going to investigate optics – the science of light – starting with a perspective that has a lot in common with the idea of light as a stream of particles.

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.

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

$I_{ave}=\frac{c\varepsilon _{0}E_{0}^{2}}{2}$ [Equation 24.18]

where $c$ is the speed of light, $\varepsilon _{0}$ 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 $\varepsilon _{0}=1/\mu _{0}c^{2}$, where $\mu _{0}$ is the permeability of free space. Algebraic manipulation produces the relationship

$I_{ave}=\frac{cB_{0}^{2}}{2\mu _{0}}$ [Equation 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

$I_{ave}=\frac{E_{0}B_{0}}{2\mu _{0}}$ [Equation 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, $I_{0}=2I_{ave}$.

#### 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

$I=\frac{P}{A}=\frac{1.00\:kW}{0.300\:m \times 0.400\:m}$ [Equation 24.21]

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

$I_{ave}=\frac{1000\:W}{0.120\:m^{2}}=8.33\times 10^{3}\:W/m^{2}$ [Equation 24.22]

Note that the peak intensity is twice the average:

$I_{0}=2I_{ave}=1.67\times 10^{4}\: W/m^{2}$ [Equation 24.23]

#### Solution for (b)

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

$E_{0}=\left ( \frac{2I_{ave}}{c\varepsilon _{0}} \right )^{1/2}$ [Equation 24.24]

Entering known values gives

$E_{0}=\sqrt{\frac{2(8.33\times 10^{3}\:W/m^{2})}{(3.00\times 10^{8}\:m/s)(8.85\times 10^{-12}C^{2}/N\cdot m^{2})}}$ [Equation 24.25]

$=2.51\times 10^{3}\:V/m$

#### 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

$B_{0}=\frac{E_{0}}{c}$ [Equation 24.26]

Entering known values gives

$B_{0}=\frac{2.51\times 10^{3}\:V/m}{3.0\times 10^{8}\:m/s}$ [Equation 24.27]

$8.35\times 10^{-6}T$

#### 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.

Source: Rice University, https://openstax.org/books/college-physics/pages/24-4-energy-in-electromagnetic-waves