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

University physics volume 2 Page 3 / 5

A laser beam

The beam from a small laboratory laser typically has an intensity of about

1.0 \times 10^{−3} {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 ε_{0} E_{0}^{2} .

The amplitude of the electric field is therefore

E_{0} = \sqrt{\frac{2}{c ε_{0}} I} = \sqrt{\frac{2}{(3.00 \times 10^{8} m/s) (8.85 \times 10^{−12} F/m)} (1.0 \times 10^{−3} {W/m}^{2})} = 0.87 V/m .

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

B_{0} = \frac{E_{0}}{c} = 2.9 \times 10^{−9} T .

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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.

Figure shows a light bulb in the centre illuminating a circular area around it. This area has a radius of 3 m.

Solution

The power radiated as visible light is then

\begin{matrix} I & = & \frac{P}{4 π r^{2}} = \frac{c ε_{0} E_{0}^{2}}{2}, \\ E_{0} & = & \sqrt{2 \frac{P}{4 π r^{2} c ε_{0}}} = \sqrt{2 \frac{5.00 W}{4 π {(3.0 m)}^{2} (3.00 \times 10^{8} m/s) (8.85 \times 10^{−12} C^{2} /N \cdot m^{2})}} = 5.77 N/C, \\ B_{0} & = & E_{0} / c = 1.92 \times 10^{−8} T . \end{matrix}

Significance

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

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Radio range

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?

A point is labeled radio source. A small square labeled A1 is in the path of the lines radiating from the radio source. The lines continue from the corners of A1 and reach A2, a slightly bigger square. A1 is at a distance r1 from the source and A2 is at a distance R2. — In three dimensions, a signal spreads over a solid angle as it travels outward from its source.

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 ε_{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{matrix} \frac{r_{2}^{2}}{r_{1}^{2}} & = & \frac{A_{2}}{A_{1}} = \frac{90 W}{60 W}, \\ r_{2} & = & \sqrt{\frac{90}{60}} (100 km) = 122 km . \end{matrix}

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.

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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 ε_{0} E_{0}^{2}}{2}

where I is the average intensity in ${W/m}^{2}$ 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 μ_{0}}

and in terms of both electric and magnetic fields as

I = \frac{E_{0} B_{0}}{2 μ_{0}} .

The three expressions for $I_{avg}$ 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 ${100 W/m}^{2}$ ) is can quickly produce a considerable change in temperature, but the light pressure (about $3.00 \times 10^{−7} {N/m}^{2}$ ) is much too small to notice.

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Source: OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3

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