# 5.2 The energy of electromagnetic waves

The energy density of an electric field **E** is

Although we will not prove it in this course, a very similar result applies to magnetic fields. The energy density of a magnetic field **B** is

It follows that an electromagnetic wave has a certain energy density, and as the wave travels through space, this energy is transported with it. Energy transport is clearly an important feature of electromagnetic waves, and explains how we can benefit from the energy generated in the Sun, 1.5 × 10^{8} km away.

Let's compare the energy densities in the electric and magnetic waves in an electromagnetic wave. If *E* and *B* are the magnitudes of the electric and magnetic fields at a given point, we have

However, we know from Equations 7.28 and 7.29 that *E = cB* at any point in an electromagnetic wave. So the electric and magnetic waves have equal energy densities.

In a small time interval Δ*t*, the amount of energy transported across an area Δ*S*, perpendicular to the direction of propagation of the wave, is given by the energy in the shaded volume in Figure 13. Allowing for the equal energy densities of the electric and magnetic waves, this is

The rate of transfer of energy per unit area perpendicular to the direction of propagation of the wave is called the **energy flux**, so we have

Note: The energy flux is *per unit area*, and is a scalar field defined at each point in space. It has a different character from electric or magnetic fluxes which are not per unit area and are defined over specified surfaces.

The energy flux varies rapidly as peaks and troughs of the electromagnetic wave pass through the given area. Generally, we wish to know the average energy flux over one period of the wave. For a monochromatic, plane electromagnetic wave travelling in the *z*-direction, the electric field is proportional to *E*_{0} cos(*kz* − *ω**t*), so we need to average cos^{2}(*kz* − *ω**t*) over one period. Using the identity cos^{2}*θ* = (1 + cos 2*θ*) / 2, we have

Because *T* = 2/*ω*, the integral on the right-hand side is the integral of a cosine over a whole number of periods, and so is equal to zero. We therefore conclude that

## Exercise 8

At a receiver, a strong radio signal has an electric field of amplitude 0.01 V m^{−1}. What is the average energy flux associated with this signal?

### Answer

The average energy flux is

using the unit conversions 1 C = 1 A s, 1 A V = 1 W and 1 N m s^{−1} = 1 W.

*Comment*: The small value of this energy flux shows that amplification is an essential function of any radio receiver.

This is a suitable point at which to end this course. All of Maxwell's equations have been introduced, and you have seen that these equations permit electromagnetic waves to travel through empty space. Electric and magnetic fields are not just mathematical abstractions, but are real enough to transport energy from distant sources. You are bathed in various hues of light from the objects you see around you. Radio waves from radio and TV stations and a vast number of transmitting mobile phones are passing through you. In addition, there is a cosmic microwave background from the first minutes of the Universe and gamma rays from the most distant stars. No wonder Richard Feynman felt able to make the following judgement:

‘From a long view of the history of mankind – seen from, say, ten thousand years from now – there can be little doubt that the most significant event of the 19th century will be judged as Maxwell's discovery of the laws of electrodynamics.’