Summary
Lecture 24
In addition to carrying energy, electromagnetic waves also carry linear
momentum, Δp; hence, electromagnetic wave exerts some force and pressure.
If radiation is completely absorbed, the momentum
transferred:
Δp =
ΔU
c
If radiation is completely reflected, the momentum
transferred:
Δp =
2ΔU
c
Review for Exam 2
The radiation pressure if EM wave is fully absorbed:
The radiation pressure if EM wave is fully reflected:
P=
P=
I
c
2I
c
In amplitude modulation (AM), the amplitude of the carrier signal is made to
vary in proportion to the audio signal’s amplitude.
In frequency modulation (FM), the frequency of the carrier signal is made to
change in proportion to the audio signal’s amplitude.
Physics 112, Spring 2010, Mar 10, Lecture 24
Physics 112, Spring 2010, Mar 10, Lecture 24
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Announcement
♦ There is Exam 2 on Friday, March 12, at lecture hour, 12:10-1:00 pm, here, in
0005 Physics. You can come earlier, at ~11:50 am.
♦ For the exam please bring a pencil, a calculator, and your ID.
♦ Please do not bring or turn your cell phone off !
♦ Exam 1 is covered by the 14-24 lectures.
♦ There will be 10 problems (need some calculations) and 15 questions typical
to those solved and/or discussed at lectures.
Review for Exam 2
Done during Lecture 24
Please see additional problems below
♦This exam will use a computer form “bubble sheet” (answers from #26 to #50).
♦ You can bring and use one side of one page of handwritten notes and your
handwritten notes used for Exam 1.
♦ If have any questions, please ask me after lecture or send me E-mail.
Physics 112, Spring 2010, Mar 10, Lecture 24
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Physics 112, Spring 2010, Mar 10, Lecture 24
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1
Force on Charge Particle
Electromagnetic Waves
An electromagnetic wave has a wavelength of 650 nm. How would we
classify it and what is its frequency?
A negatively charged particles is moving to the left at constant speed as
shown. In which magnet position will there be a force on the particle
directed out of the page?
1) Visible light, blue
1) 1.95×1011 Hz
2) Infrared EM
2) 1.95×10-1 Hz
3) Ultraviolet EM
3) 4.6×1014 Hz
4) Visible light, red
4) 4.6×105 Hz
Visible electromagnetic waves
are from ~400 nm (blue) to
~750 nm (red).
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Physics 112, Spring 2010, Mar 10, Lecture 24
cε 0 A
Erms =
c=
P
cε 0 A
A = πr = (π / 4)d = (π / 4)(1.75 ×10 m)
2
−3
12.8 ×10−3W
= 1.42 ×103V / m
(3 ×10 m / s)(8.85 ×10 C 2 / N ⋅ m 2 )[(π / 4)(1.75 ×10−3 m) 2 ]
−12
8
Erms 1420 V / m
=
= 4.6 ×10−7 T
c
3 ×108 m / s
Physics 112, Spring 2010, Mar 10, Lecture 24
E = cB = (3 ×108 m / s)(17.5 ×10−9 T ) = 5.25V / m
In electromagnetic wave traveling west, the magnetic field oscillates
vertically and has a frequency of 80 kHz and an rms strength of 6.75x10-9
T. What is the frequency and rms strength of the electric field, and what is
its direction?
The frequency of the two fields must be the same: 80 kHz.
The rms strength of the electric field is:
Erms = cBrms = (3 ×108 m / s)(6.75 ×10−9 T ) = 2.03 V / m
4. The rms value of the magnetic field:
Brms =
E
B
2
3. The rms value of the electric field:
Erms =
6
Electromagnetic Waves
2. The area of the beam:
2
3 ×108 m / s
= 4.62 ×1014 Hz
650 × 10 −9 m
If the magnetic field in a traveling electromagnetic wave has a peak
magnitude of 17.5 nT at a given point, what is the peak magnitude of the
electric field?
1. The energy per unit area per unit time is:
P
λ
=
Electromagnetic Waves
A 12.8-mW laser puts out a narrow beam 1.75 mm in diameter. What are
the rms values of E and B fields in the beam?
2
=
Erms
c
Physics 112, Spring 2010, Mar 10, Lecture 24
Electromagnetic Waves
P
2
= cε 0 Erms
A
f =
The electric field is perpendicular to both the direction of travel and the magnetic
field, so the electric field oscillates along the horizontal north-south line.
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Physics 112, Spring 2010, Mar 10, Lecture 24
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