# 4.3 Energy in electromagnetic waves  (Page 2/6)

 Page 2 / 6
${I}_{\text{ave}}=\frac{{\text{cB}}_{0}^{2}}{{2\mu }_{0}},$

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

One more expression for ${I}_{\text{ave}}$ 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

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

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, ${I}_{0}=2{I}_{\text{ave}}$ .

## 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 ${\text{W/m}}^{2}$ ? (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

$I=\frac{P}{A}=\frac{1\text{.}\text{00 kW}}{0\text{.}\text{300 m}\phantom{\rule{0.25em}{0ex}}×\phantom{\rule{0.25em}{0ex}}0\text{.}\text{400 m}}.$

Here $I={I}_{\text{ave}}$ , so that

${I}_{\text{ave}}=\frac{\text{1000 W}}{0\text{.}\text{120}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}}=8\text{.}\text{33}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}.$

Note that the peak intensity is twice the average:

${I}_{0}=2{I}_{\text{ave}}=1\text{.}\text{67}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{W}/{\text{m}}^{2}.$

Solution for (b)

To find ${E}_{0}$ , we can rearrange the first equation given above for ${I}_{\text{ave}}$ to give

${E}_{0}={\left(\frac{2{I}_{\text{ave}}}{{\mathrm{c\epsilon }}_{0}}\right)}^{1/2}.$

Entering known values gives

$\begin{array}{lll}{E}_{0}& =& \sqrt{\frac{2\left(8\text{.}\text{33}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}{\text{W/m}}^{2}\right)}{\left(3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\left(8.85×{\text{10}}^{–12}\phantom{\rule{0.25em}{0ex}}{\text{C}}^{2}/\text{N}\cdot {\text{m}}^{2}\right)}}\\ & =& 2.51×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{V/m}\text{.}\end{array}$

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

${B}_{0}=\frac{{E}_{0}}{c}.$

Entering known values gives

$\begin{array}{lll}{B}_{0}& =& \frac{2.51×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{V/m}}{3.0×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}\\ & =& 8.35×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{T}\text{.}\end{array}$

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.

## Section summary

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

where ${I}_{\text{ave}}$ 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
${I}_{\text{ave}}=\frac{{\text{cB}}_{0}^{2}}{{2\mu }_{0}}$

and in terms of both electric and magnetic fields as

${I}_{\text{ave}}=\frac{{E}_{0}{B}_{0}}{{2\mu }_{0}}.$
• The three expressions for ${I}_{\text{ave}}$ are all equivalent.

## Problems&Exercises

What is the intensity of an electromagnetic wave with a peak electric field strength of 125 V/m?

$\begin{array}{lll}I& =& \frac{{\mathrm{c\epsilon }}_{0}{E}_{0}^{2}}{2}\\ & =& \frac{\left(3.00×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\left(8.85×{\text{10}}^{\text{–12}}{\text{C}}^{2}\text{/N}\cdot {\text{m}}^{2}\right){\left(1\text{25 V/m}\right)}^{2}}{2}\\ & =& 20.{\text{7 W/m}}^{2}\end{array}$

Find the intensity of an electromagnetic wave having a peak magnetic field strength of $4\text{.}\text{00}×{\text{10}}^{-9}\phantom{\rule{0.25em}{0ex}}\text{T}$ .

Assume the helium-neon lasers commonly used in student physics laboratories have power outputs of 0.250 mW. (a) If such a laser beam is projected onto a circular spot 1.00 mm in diameter, what is its intensity? (b) Find the peak magnetic field strength. (c) Find the peak electric field strength.

(a) $I=\frac{P}{A}=\frac{P}{\pi {r}^{2}}=\frac{0\text{.}\text{250}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{W}}{\pi {\left(0\text{.}\text{500}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{m}\right)}^{2}}={\text{318 W/m}}^{2}$

(b) $\begin{array}{lll}{I}_{\text{ave}}& =& \frac{{\text{cB}}_{0}^{2}}{{2\mu }_{0}}⇒{B}_{0}={\left(\frac{{2\mu }_{0}I}{c}\right)}^{1/2}\\ & =& {\left(\frac{2\left(4\pi ×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}\text{T}\cdot \text{m/A}\right)\left(\text{318}\text{.}{\text{3 W/m}}^{2}\right)}{\text{3.00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}\right)}^{1/2}\\ & =& 1\text{.}\text{63}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{T}\end{array}$

(c) $\begin{array}{lll}{E}_{0}& =& {\text{cB}}_{0}=\left(3\text{.00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}\right)\left(\text{1.633}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{T}\right)\\ & =& 4\text{.}\text{90}×{\text{10}}^{2}\phantom{\rule{0.25em}{0ex}}\text{V/m}\end{array}$

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