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A laser beam

The beam from a small laboratory laser typically has an intensity of about 1.0 × 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 = 1 2 c ε 0 E 0 2 .

The amplitude of the electric field is therefore

E 0 = 2 c ε 0 I = 2 ( 3.00 × 10 8 m/s ) ( 8.85 × 10 −12 F/m ) ( 1.0 × 10 −3 W/m 2 ) = 0.87 V/m .

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

B 0 = E 0 c = 2.9 × 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

I = P 4 π r 2 = c ε 0 E 0 2 2 , E 0 = 2 P 4 π r 2 c ε 0 = 2 5.00 W 4 π ( 3.0 m ) 2 ( 3.00 × 10 8 m/s ) ( 8.85 × 10 −12 C 2 /N · m 2 ) = 5.77 N/C, B 0 = E 0 / c = 1.92 × 10 −8 T .

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 = P A = 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

r 2 2 r 1 2 = A 2 A 1 = 90 W 60 W , r 2 = 90 60 ( 100 km ) = 122 km .

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 = 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 = c B 0 2 2 μ 0

and in terms of both electric and magnetic fields as

I = 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 × 10 −7 N/m 2 ) is much too small to notice.

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Questions & Answers

The properties of a system during a reversible constant pressure non-flow process at P= 1.6bar, changes from constant volume of 0.3m³/kg at 20°C to a volume of 0.55m³/kg at 260°C. its constant pressure process is 3.205KJ/kg°C Determine: 1. Heat added, Work done, Change in Internal Energy and Change in Enthalpy
Opeyemi Reply
please how do dey get 5/9 in the conversion of Celsius and Fahrenheit
Gwam Reply
what is copper loss
timileyin Reply
this is the energy dissipated(usually in the form of heat energy) in conductors such as wires and coils due to the flow of current against the resistance of the material used in winding the coil.
Henry
it is the work done in moving a charge to a point from infinity against electric field
Ashok Reply
what is the weight of the earth in space
peterpaul Reply
As w=mg where m is mass and g is gravitational force... Now if we consider the earth is in gravitational pull of sun we have to use the value of "g" of sun, so we can find the weight of eaeth in sun with reference to sun...
Prince
g is not gravitacional forcé, is acceleration of gravity of earth and is assumed constante. the "sun g" can not be constant and you should use Newton gravity forcé. by the way its not the "weight" the physical quantity that matters, is the mass
Jorge
Yeah got it... Earth and moon have specific value of g... But in case of sun ☀ it is just a huge sphere of gas...
Prince
Thats why it can't have a constant value of g ....
Prince
not true. you must know Newton gravity Law . even a cloud of gas it has mass thats al matters. and the distsnce from the center of mass of the cloud and the center of the mass of the earth
Jorge
please why is the first law of thermodynamics greater than the second
Ifeoma Reply
define electric image.obtain expression for electric intensity at any point on earthed conducting infinite plane due to a point charge Q placed at a distance D from it.
Mateshwar Reply
explain the lack of symmetry in the field of the parallel capacitor
Phoebe Reply
pls. explain the lack of symmetry in the field of the parallel capacitor
Phoebe
does your app come with video lessons?
Ahmed Reply
What is vector
Ajibola Reply
Vector is a quantity having a direction as well as magnitude
Damilare
tell me about charging and discharging of capacitors
Ahemen Reply
a big and a small metal spheres are connected by a wire, which of this has the maximum electric potential on the surface.
Bundi Reply
3 capacitors 2nf,3nf,4nf are connected in parallel... what is the equivalent capacitance...and what is the potential difference across each capacitor if the EMF is 500v
Prince Reply
equivalent capacitance is 9nf nd pd across each capacitor is 500v
santanu
four effect of heat on substances
Prince Reply
why we can find a electric mirror image only in a infinite conducting....why not in finite conducting plate..?
Rima Reply
because you can't fit the boundary conditions.
Jorge
what is the dimensions for VISCOUNSITY (U)
Branda
Practice Key Terms 1

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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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