Physics 1010:
The Physics of Everyday Life
TODAY
• EM waves
• Some exam questions
1
Today’s topics
• Electromagnetic waves; speed of light,
radio transmitters
• Go over some of the exam questions
2
Electromagnetic radiation
• Encoding sound on EM radiation
 AM
 FM
• The spectrum
c = 3 x 10^8 m/s
broadcasting antenna
3
Force due to electron?
A
60 m
direction of force on a “test” electron I stick at A. ?
a.
b.
c.
d.
e.
Cannot tell
4
Force due to electron?
A
60 m
direction of force on a “test” electron I stick at A. ?
a.
b.
c.
d.
e.
Cannot tell
5
Force due to electron that has moved?
move down
A
direction of force on a “test” electron I stick at A. ?
a.
b.
c.
d.
e.
Cannot tell
6
Force due to electron that has moved?
move down
A
direction of force on a “test” electron I stick at A. ?
a.
b.
c.
d.
e.
Cannot tell
It can be a or c; we need to know WHEN electron moved.
7
Moving electrons cause a wave
Stationary ions give attractive force
notice that antenna is neutral, + and - charges, but
only the minuses (electrons) move.
60 m
moving out with time at speed of light
c = 3 x 108 m/s
8
Moving electrons cause a wave
How long does it take radio wave to go 60 m?
a. .2 sec, b. .002 s, c. 2 x 10-6 s, d. 2 x 10-7 s, e. none of above
c = 3 x 108 m/s
9
Moving electrons cause a wave
How long does it take radio wave to go 60 m?
a. .2 sec, b. .002 s, c. 2 x 10-6 s, d. 2 x 10-7 s, e. none of above
c = 3 x 108 m/s
ans d. time = distance/speed = 60 m/(3 x 108m/s) = 2 x 10-7 sec
after electron in transmitter moves, receiver electron knows it
moved.
10
Combining forces gives a transverse wave
30 m
-
c = 3 x 10^8 m/s
+
-
combined force- antenna neutral charge, only
moving electrons.
magnified
vertically
drawing force arrows as a wave.
11
Wave motion?
30
m
+
c = 3 x 108 m/s
-
C
magnified
vertically
B
A
a very short time (much shorter than the period) after this “snapshot”, the
strength of the field at A
will be
a. more downward,
b. the same,
c. down but smaller,
d. zero,
e. large upward
12
Wave motion?
30
m
+
c = 3 x 108 m/s
-
C
magnified
vertically
B
A
a very short time (much shorter than the period) after this “snapshot”, the
strength of the field at A
will be
a. more downward,
b. the same,
c. down but smaller,
d. zero,
e. large upward
ans. c. - wave just to left of A moves over to position A.
13
EM waves can be used to transport energy!
Set up 100MHz transmitter. Hook flashlight bulb between
two halves of receiving antenna nearby.
Bulb will
a. light up if signal strong enough,
b. not light up because no current through it,
c. not light up because current oscillates up and down so fast.
a. lights up. current going
back and forth 100 M osc./s,
but always heats.
14
Variation with distance?
A
-
B -
signal strength (electric force field)
a. at A is stronger than at B
b. at A is same as at B
c. at A is weaker than at B
d. no way to tell
a) A is stronger than B. Energy in electromagnetic
wave spreads out. Total energy same, but field at
any point weaker.
Similar to sound!
15
Variation with distance because energy
spreads out
ans. A is stronger than B. Energy in
electromagnetic wave spreads out. Total
energy same, but field at any point weaker.
Amount of
energy in EM
wave spread
over sphere,
area 4πr2. So
signal gets
weaker as
1/(distance from
transmitter)2
A
-
B -
r
16
How should we orient the receiver?
signal strength (electric force field)
a. parallel to broadcast antenna is stronger than when perpendicular
b. parallel to broadcast antenna is same as when perpendicular
c. parallel to broadcast antenna is weaker than when perpendicular
17
Orient the receiver like the original antenna
Answer: a) electric force field in same direction as electrons move in broadcast
Antenna (up and down; “polarization”)
easy to move electrons large amount up and down in receiving antenna and
go through bulb if antenna up and down. Hard if antenna vertical.
same with receiving electronics of normal radio.
18
Demonstration
• http://www.colorado.edu/physics/phet/sim
ulations/emf/emf.jnlp
• Experiment with electromagnetic field
Static field
Waves
19
What do the arrows represent?
TRANSMITTING
ANTENNA
RECEIVING
ANTENNA
a. The velocity of the electrons that are at each of those points, moving due to
the electromagnetic wave.
b. The evenly spaced electrons (blue dots) moving up and down between the
two antennae.
c. The strength and direction of the force that would be exerted by the
electromagnetic wave on an electron.
d. The position of the electrons that are at each of those points, moving due to
the electromagnetic wave.
e. The force resulting from electrons moving off of the transmitting antenna
towards the receiving antenna following the curved path.
20
What do the arrows represent?
TRANSMITTING
ANTENNA
Here curve indicating,
large upward force at this time RECEIVING
ANTENNA
Here curve indicating
no force at this time
Answer is c. The strength and direction of the force that would be
exerted by the electromagnetic wave on an electron.
How does this work?
1) Electrons can exert a force on other electrons through electrostatic forces
(Coulombs law).
2) When electrons oscillate up and down the transmitting antenna, the force
that would be exerted on other electrons is constantly changing.
3) Horizontal component of force is canceled by horizontal component of
attractive force due to stationary positive charges in antenna, so resulting
force is up and down.
4) It takes some time for this force (electromagnetic signal) to radiate
(propagate) from the antenna outwards (velocity = speed of light).
21
How to measure the propagation speed of the wave?
TRANSMITTING
ANTENNA
RECEIVING
ANTENNA
Answer is a. How fast
The speed of the wave (signal) is measured as… peak moves toward
a. how fast this peak moves towards receiver.
receiving antenna.
b. how fast this peak moves up and down.
Speed of light (c) = 3 x
c. both a and b
108 m/s
22
Radios work at a particular frequency
To broadcast radiowaves:
Electrons oscillate up and down in antenna
(a rapidly oscillating electric current)
-
Wave traveling out at speed of light.
Each radio station broadcasts at a particular
frequency.
AM stations 530 to 1600 kHz
(530 kHz means 530,000 complete
oscillations of electron up and down
antenna per second)
23
electrons going up and down in antenna
(rapidly oscillating electric current)
Wavelength of radiowave = distance (horizontal)
between one maximum of signal to the next
-
Wave traveling out at speed of light.
If electrons oscillate up and down transmitting antenna more times per
second (higher frequency), the wavelength of the radio wave will:
a. Increase b. Decrease c. Stay the same.
b. Decrease. Signal travels away from transmitter at the speed of
light (c). At a higher frequency, there is less time between when each
maximum is produced so previous maximum hasn’t traveled as far.
24
Motion modified by frequency?
If frequency of the transmitting antenna is increased from 530 Hz
to 1060 Hz so that the electrons oscillate up and down in the
transmitting antenna more times per second, the electrons in
the receiving antenna will:
a. be unaffected by the change and continue to oscillate up and
down at the same frequency as before the change
b. oscillate up and down at a higher frequency than before the
change.
c. oscillate up and down in a full cycle 1060 times per second.
d. oscillate up and down at a lower frequency than before the
change.
Answer is e. Receiving electron always moves at same frequency
as transmitter frequency.
25
1. If the amplitude of motion in the transmitter is increased,
the wave will get to the receiver
a. sooner than the small amplitude wave,
b. same time,
c. faster
ans. b. all electromagnetic waves travel at c = speed of light.
26
2. If the amplitude of motion in the transmitter is increased, the
receiver electrons will….
a. move up and down with higher frequency
b. force that radiowave exerts on receiving electron will increase
c. move up and down with lower frequency
d. a and b
e. b and c
ans. b. frequency independent of amplitude.
Increasing amplitude of transmitting electron oscillation,
increases acceleration of electron.
Larger accelerations produce stronger electromagnetic waves.
(larger forces).
27
To broadcast radiowaves:
Electrons oscillate up and down in antenna
(a rapidly oscillating electric current)
-
Wave traveling out at speed of light.
So if we watch receiver antenna and find electrons oscillate up and
down at a frequency of 890 kHz, what is the wavelength of the
radiowave? (Speed of light = 3 x 108 m/s)
a. 330,000 m b. 337 m
c. 2670 m
d. 3.0 m e. 522 m
Speed of light = Wavelength of wave x Frequency OR c = λν
Speed of light (c) = 3 x 108 m/s
Wavelength = speed/frequency = (3 x 108 m/s) / (890000 Hz) = 337 m
Optimum antenna length for broadcast is (wavelength/4) = 84 m.
AM broadcast antennas are tall!
28
What about a higher frequency?
(FM band)
So if we watch receiver antenna and find electrons oscillate up and
down at a frequency of 100 MHz = 108 Hz, what is the wavelength of
the radiowave? (Speed of light = 3 x 108 m/s)
a. 300 m b. 30 m
c. 3 m
d. 0.3 m
Speed of light = Wavelength of wave x Frequency OR c = λν
Speed of light (c) = 3 x 108 m/s
Wavelength = speed/frequency = (3 x 108 m/s) / (108 Hz) = 3 m
29
Two steel balls each have 1,000 extra electrons on them. The balls are 0.5 m apart.
The force between them is
a.
b.
c.
d.
9.2×10-22 N and attractive
9.2×10-22 N and repulsive
9.2×10-19 N and attractive
9.2×10-19 N and repulsive
30
Two steel balls each have 1,000 extra electrons on them. The balls are 0.5 m apart.
The force between them is
a.
b.
c.
d.
9.2×10-22 N and attractive
9.2×10-22 N and repulsive
9.2×10-19 N and attractive
9.2×10-19 N and repulsive
F = k q2/r2 = 9*109 (1000*1.6*10-19)2 / 0.52 N = 9.2*10-22 N
31
In the circuit on the right, the voltage across the prongs of
the plug is 120V and all light bulbs have resistance 10 Ohms.
The voltage between F and I is:
a. 0 V
b. 40 V
c. 80 V
d. 120 V
32
In the circuit on the right, the voltage across the prongs of
the plug is 120V and all light bulbs have resistance 10 Ohms.
The voltage between F and I is:
a. 0 V
b. 40 V
c. 80 V
d. 120 V
Circuit in parallel
33
In the circuit on the right, the voltage across the prongs of
the plug is 120V and all light bulbs have resistance 10 Ohms.
The current between at point G is:
a.
b.
c.
d.
36. A
12. A
8.0 A
4.0 A
34
In the circuit on the right, the voltage across the prongs of
the plug is 120V and all light bulbs have resistance 10 Ohms.
The current between at point G is:
a.
b.
c.
d.
36. A
12. A
8.0 A
4.0 A
V=IR, so I = V/R = 120/10 A = 12A
35
In the circuit on the right, the voltage across the prongs of
the plug is 120V and all light bulbs have resistance 10 Ohms.
The current between the plug
and point E is:
a.
b.
c.
d.
36. A
12. A
8.0 A
4.0 A
36
In the circuit on the right, the voltage across the prongs of
the plug is 120V and all light bulbs have resistance 10 Ohms.
The current between the plug
and point E is:
a.
b.
c.
d.
36. A
12. A
8.0 A
4.0 A
There is a 12A current through each light bulb, and in
Circuits in parallel the voltage is the same but the currents add up.
37