# 16.3 Energy carried by electromagnetic waves  (Page 3/5)

 Page 3 / 5

## A laser beam

The beam from a small laboratory laser typically has an intensity of about $1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}{\phantom{\rule{0.2em}{0ex}}\text{W/m}}^{2}$ . 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 [link] , the intensity of the laser beam is

$I=\frac{1}{2}c{\epsilon }_{0}{E}_{0}^{2}.$

The amplitude of the electric field is therefore

${E}_{0}=\sqrt{\frac{2}{c{\epsilon }_{0}}I}=\sqrt{\frac{2}{\left(3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\left(8.85\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-12}\phantom{\rule{0.2em}{0ex}}\text{F/m}\right)}\left(1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}{\phantom{\rule{0.2em}{0ex}}\text{W/m}}^{2}\right)}=0.87\phantom{\rule{0.2em}{0ex}}\text{V/m}.$

The amplitude of the magnetic field can be obtained from [link] :

${B}_{0}=\frac{{E}_{0}}{c}=2.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}\text{T}.$

## 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

$\begin{array}{ccc}\hfill I& =\hfill & \frac{P}{4\pi {r}^{2}}=\frac{c{\epsilon }_{0}{E}_{0}^{2}}{2},\hfill \\ \hfill {E}_{0}& =\hfill & \sqrt{2\frac{P}{4\pi {r}^{2}c{\epsilon }_{0}}}=\sqrt{2\frac{5.00\phantom{\rule{0.2em}{0ex}}\text{W}}{4\pi {\left(3.0\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{2}\left(3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\left(8.85\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-12}{\phantom{\rule{0.2em}{0ex}}\text{C}}^{2}\text{/N}·{\text{m}}^{2}\right)}}=5.77\phantom{\rule{0.2em}{0ex}}\text{N/C,}\hfill \\ \hfill {B}_{0}& =\hfill & {E}_{0}\text{/}c=1.92\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\phantom{\rule{0.2em}{0ex}}\text{T}.\hfill \end{array}$

## Significance

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

A 60-kW radio transmitter on Earth sends its signal to a satellite 100 km away ( [link] ). 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?

## 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 $I=\frac{P}{A}=\frac{c{\epsilon }_{0}{E}_{0}^{2}}{2}$ 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

$\begin{array}{ccc}\hfill \frac{{r}_{2}^{2}}{{r}_{1}^{2}}& =\hfill & \frac{{A}_{2}}{{A}_{1}}=\frac{90\phantom{\rule{0.2em}{0ex}}\text{W}}{60\phantom{\rule{0.2em}{0ex}}\text{W}},\hfill \\ \hfill {r}_{2}& =\hfill & \sqrt{\frac{90}{60}}\left(100\phantom{\rule{0.2em}{0ex}}\text{km}\right)=122\phantom{\rule{0.2em}{0ex}}\text{km}.\hfill \end{array}$

## 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.

## Summary

• The energy carried by any wave is proportional to its amplitude squared. For electromagnetic waves, this means intensity can be expressed as
$I=\frac{c{\epsilon }_{0}{E}_{0}^{2}}{2}$

where I is the average intensity in ${\text{W/m}}^{2}$ and ${E}_{0}$ is the maximum electric field strength of a continuous sinusoidal wave. This can also be expressed in terms of the maximum magnetic field strength ${B}_{0}$ as

$I=\frac{c{B}_{0}^{2}}{2{\mu }_{0}}$

and in terms of both electric and magnetic fields as

$I=\frac{{E}_{0}{B}_{0}}{2{\mu }_{0}}.$

The three expressions for ${I}_{\text{avg}}$ are all equivalent.

## Conceptual questions

When you stand outdoors in the sunlight, why can you feel the energy that the sunlight carries, but not the momentum it carries?

The amount of energy (about ${100\phantom{\rule{0.2em}{0ex}}\text{W/m}}^{2}$ ) is can quickly produce a considerable change in temperature, but the light pressure (about $3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}{\text{N/m}}^{2}$ ) is much too small to notice.

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