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Electromagnetic Waves

Physics 6C

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

Electromagnetic (EM) waves are produced by an alternating current in a wire. As the charges in the wire oscillate back and forth, the electric field around them oscillates as well, in turn producing an oscillating magnetic field. This magnetic field is always perpendicular to the electric field, and the EM wave propagates perpendicular to both the E- and B-fields. This gives us a right-hand-rule relating the directions of these 3 vectors:

1) Point the fingers of your right hand in the direction of the E-field

2) Curl them toward the B-field.

3) Stick out your thumb - it points in the direction of propagation.

Electromagnetic Waves



Click here for an EM wave animation

http://www.phys.hawaii.edu/~teb/java/ntnujava/emWave/emWave.html

Like any other wave, we know the relationship between the wavelength and frequency, and the speed of propagation of the wave:

fvwave




fvwave

In the case of EM waves, it turns out that the wave speed is the speed of light.

So our formula for EM waves (in vacuum) is:

sm8

00

1031

c;fc




fvwave



It turns out that the speed of light is also the ratio of the strengths of the Electric and Magnetic fields in an EM wave. So we know that E=cB (in standard metric units)



sm8

00

1031

c;fc


fvwave



The continuum of various wavelengths and frequencies for EM waves is called the Electromagnetic Spectrum

It turns out that the speed of light is also the ratio of the strengths of the Electric and Magnetic fields in an EM wave. So we know that E=cB (in standard metric units)



sm8

00

1031

c;fc

Examples:

• Find the frequency of blue light with a wavelength of 460 nm.



Examples:


Hz105.6m10460

103cffc 14

9sm8



Examples:

Hz105.6m10460

103cffc 14

9sm8




• A cell phone transmits at a frequency of 1.25x108 Hz. What is the wavelength of this EM wave?

Examples:


Hz105.6m10460

103cffc 14

9sm8

• A cell phone transmits at a frequency of 1.25x108 Hz. What is the wavelength of this EM wave?

m4.2Hz1025.1

103

fc

fc8sm8

You will need to use this formula very often to convert back and forth between frequency and wavelength.



Energy and momentum in EM Waves

Electromagnetic waves transport energy. The energy associated with a wave is stored in the oscillating electric and magnetic fields.

We will find out later that the frequency of the wave determines the amount of energy that it carries. Since the EM wave is in 3-D, we need to measure the energy density (energy per unit volume).

2120

2212

021 BEBEu

00

Note that the energy can be written in a few equivalent forms. Each can be useful, depending on the information you know about the wave.




Electromagnetic waves transport energy. The energy associated with a wave is stored in the oscillating electric and magnetic fields.

We will find out later that the frequency of the wave determines the amount of energy that it carries. Since the EM wave is in 3-D, we need to measure the energy density (energy per unit volume).

2120

2212

021 BEBEu

00

Note that the energy can be written in a few equivalent forms. Each can be useful, depending on the information you know about the wave.



We can also talk about the intensity of an EM wave (for light we would think of it as brightness). Just as for sound, intensity is measured as average power/area.

cuArea

PowerI Just multiply the energy equation above

by the speed of light to get the intensity.

This is the energy per unit volume

Example: High-Energy Cancer Treatment

Scientists are working on a technique to kill cancer cells by zapping them with ultrahigh-energy pulses of light that last for an extremely short amount of time. These short pulses scramble the interior of a cell without causing it to explode, as long pulses do.

We can model a typical such cell as a disk 5.0 µm in diameter, with the pulse lasting for 4.0 ns with an average power of 2.0x1012 W. We shall assume that the energy is spread uniformly over the faces of 100 cells for each pulse.

a) How much energy is given to the cell during this pulse?

b) What is the intensity (in W/m2) delivered to the cell?

c) What are the maximum values of the electric and magnetic fields in the pulse?









Recall that power is energy/time. So 2.0x1012 W is 2.0x1012 Joules/sec.



J8000J108)s100.4()100.2(Energy 39sJ12

This is the total energy, which is spread out over 100 cells, so the energy for each individual cell is 80 Joules.







To get intensity, we need to divide power/area. The area for a cell is just the area of a circle:



211262 m100.2)m105.2(rArea







To get intensity, we need to divide power/area. The area for a cell is just the area of a circle:



211262 m100.2)m105.2(rArea

Now divide to get intensity:

2mW21

29

12

2100.1

m100.2

W100.2

r100

PowerIntensity

This is the total area of all 100 cells.







To get the field strengths, recall our formulas:



2120

2212

021 BEBEu

00 cuArea

PowerI







To get the field strengths, recall our formulas:



2120

2212

021 BEBEu

00 cuArea

PowerI

Since the power was stated as average power we should assume that is the rms value. So our field values should get multiplied by √2 to find the maximum.

mV11

0107.8

cI2

E

T109.2c

I2B 30


EM waves also carry momentum. This means that a ray of light can actually exert a force. To get the pressure exerted by a sinusoidal EM wave, just divide the intensity by the speed of light.

cI

essurePr

Radiation av



This is the same as the total energy absorbed by the surface.


EM waves also carry momentum. This means that a ray of light can actually exert a force. To get the pressure exerted by a sinusoidal EM wave, just divide the intensity by the speed of light.

cI

essurePr

Radiation av



Example: Solar Sails

Suppose a spacecraft with a mass of 25,000 kg has a solar sail made of perfectly reflective aluminized film with an area of 2.59x106 m. If the spacecraft is launched into earth orbit and then deploys its sail at right angles to the sunlight, what is the acceleration due to sunlight? Assume that at the earth’s distance from the sun, the pressure exerted by sunlight on an absorbing surface is 4.70x10-6 Pa.





Recall that Pressure = Force/Area. We can use this and F=ma to get our formula:

mF

aamF

APFAF

P

mAP

a

Now for the tricky part: When the pressure number was given above, that was for an absorbing surface. What happens when the sunlight reflects instead?





Recall that Pressure = Force/Area. We can use this and F=ma to get our formula:

mF

aamF

APFAF

P

mAP

a

Now for the tricky part: When the pressure number was given above, that was for an absorbing surface. What happens when the sunlight reflects instead?

Twice as much momentum is transferred!

2

2

sm4

4

26

mN6

1072.9kg105.2

m1059.2)107.4(2a

PolarizationThe Polarization of an EM wave is defined to be the direction of its Electric field vector.

EM waves (or light) can be passed through a filter (polarizer) to select for a particular polarization direction. This will cut down the intensity (brightness) of the light based on the following formula:

20 )cos(II

Polarizers can be placed in sequence to adjust the intensity and polarization of light.

The most obvious example is dark sunglasses, where 2 filters are placed at 90° to each other, blocking out most of the light (the formula would say all the light is blocked).



Polarization

Details about polarization:

Typical light sources are unpolarized, which means the EM waves are not oriented in any particular direction (sunlight behaves this way). When unpolarized light passes through a polarizer, half of its intensity is blocked, and the transmitted light is now polarized in the direction selected by the filter.

Example Problem• Sunlight passes through 2 polarizers which are oriented at 60° relative to each other. How much of the original sunlight intensity is transmitted?



Polarization



Example Problem• Sunlight passes through 2 polarizers which are oriented at 60° relative to each other. How much of the original sunlight intensity is transmitted?

Click this link for a java applet with polarizers.

sun812

21

sunfinal I)60cos(II

41



http://www.lon-capa.org/~mmp/kap24/polarizers/Polarizer.htm



Polarization



Reflected light is (at least partially) polarized parallel to the reflecting surface. A good example is sunlight reflecting from the water. Fishermen wear polarized sunglasses to block the reflected sunlight, giving them a better view of objects beneath the surface.

