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Physics 2113
Jonathan Dowling
Lecture 42: FRI 04 DEC
Final Exam Review II
Final Exam
• 3:00PM – 5:00PM MON 07DEC
PHYS 2113-1 (Moreno) is scheduled to take their exam in Lockett 9
PHYS 2113-2 (Gaarde) is scheduled to take their exam in Lockett 10
PHYS 2113-3 (Dowling) is scheduled to take their exam in Lockett 6
PHYS 2113-4 (O'Connell) is scheduled to take their exam in Lockett 15
PHYS 2113-5 and 6 (Abdelwahab) is scheduled to take their exam in
Lockett 2
PHYS 2113-7 (Hansen) Last name starts with A - K: take their exam in
Lockett 10, last name starts with L - Z: take their exam in Lockett 2
Final Exam
• 100 PTS: CH 13, 21–30 / HW01-11
This part will be 11 multiple choice
questions one from each chapter.
• 100 PTS: CH 31–33 / HW12-14
This part will be three multiple choice
questions and three word problems, one
each from each chapter.
Displacement “Current”
Maxwell proposed it based on
symmetry and math — no experiment!
B
B!
B
i
i
E
Changing E-field Gives Rise to B-Field!
32.3: Induced Magnetic Fields:
Here B is the magnetic field induced along a
closed loop by the changing electric flux FE
in the region encircled by that loop.
Fig. 32-5 (a) A circular parallel-plate capacitor, shown in side view, is being charged by
a constant current i. (b) A view from within the capacitor, looking toward the plate at
the right in (a).The electric field is uniform, is directed into the page (toward the plate),
and grows in magnitude as the charge on the capacitor increases. The magnetic field
induced by this changing electric field is shown at four points on a circle with a radius r
less than the plate radius R.
Ba > Bc > Bb > Bd = 0
dF E d
dE
= EA = A
µ slope
dt
dt
dt
Example, Magnetic Field Induced by
Changing Electric Field:
Example, Magnetic Field Induced by Changing Electric
Field, cont.:
32.4: Displacement Current:
Comparing the last two terms on the right side of the above equation shows that the
term
must have the dimension of a current. This product is usually treated as
being a fictitious current called the displacement current id:
in which id,enc is the displacement current that is encircled by the integration loop.
The charge q on the plates of a parallel plate capacitor at any time is related to the
magnitude E of the field between the plates at that time by
in which A is
the plate area.
The associated magnetic field are:
AND
Example, Treating a Changing Electric Field as a Displacement Current:
The displacement current id = i is
distributed evenly over grey area.
So rank by i
enc
d
= amount
of grey area enclosed by each loop.
d =c>b>a
32.5: Maxwell’s Equations:
Mathematical Description of Traveling EM Waves
Electric Field:
Magnetic Field:
E = Em sin (kx - w t )
B = Bm sin (kx - w t )
Wave Speed:
c=
m0e 0
All EM waves travel a c in vacuum
Wavenumber: k =
EM Wave Simulation
1
w
=
c
2p
l
2p
Angular frequency: w =
T
Vacuum Permittivity: e 0
Vacuum Permeability:
Fig. 33-5
Amplitude Ratio:
Em
=c
Bm
Magnitude Ratio:
E (t )
=c
B (t )
m0
(33-5)
The Poynting Vector:
Points in Direction of Power Flow
Electromagnetic waves are able to transport energy from transmitter
to receiver (example: from the Sun to our skin).
The power transported by the wave and its
direction is quantified by the Poynting vector.
John Henry Poynting (1852-1914)
For a wave, since
E is perpendicular to B:
Units: Watt/m2
| S |=
1
m0
EB =
1
cm 0
E2
In a wave, the fields change
with time. Therefore the
Poynting vector changes
too!!
The direction is constant, but
the magnitude changes from
0 to a maximum value.
EM Wave Intensity, Energy Density
A better measure of the amount of energy in an EM wave is obtained by
averaging the Poynting vector over one wave cycle.
The resulting quantity is called intensity. Units are also Watts/m2.
I =S =
1
cm 0
___
2
E =
1
cm 0
__________ __
2
2
Em sin (kx - wt )
The average of sin2 over
one cycle is ½:
1
I=
Em 2
2cm0
Bm = Em / c
Both fields have the
same energy density.
1
1
1
B2
1 B2
2
2
uE = e 0 E = e 0 (cB) = e 0
=
= uB
2
2
2 e 0 m0 2 m0
The total EM energy density is then
u = e 0 E = B / m0
2
2
EM Spherical Waves
The intensity of a wave is power per unit area. If one has a
source that emits isotropically (equally in all directions) the
power emitted by the source pierces a larger and larger
sphere as the wave travels outwards: 1/r2 Law!
I=
Ps
4pr
2
So the power per
unit area decreases
as the inverse of
distance squared.
Example
A radio station transmits a 10 kW signal at a frequency of 100 MHz.
Assume a spherical wave. At a distance of 1km from the antenna, find the
amplitude of the electric and magnetic field strengths, and the energy
incident normally on a square plate of side 10cm in 5 minutes.
Ps
10 ´10 3 W
2
I=
=
=
0.8mW
/
m
4p r 2 4p (1´10 3m)2
1
2
I=
Em Þ Em = 2cm0 I = 0.775V/m
2cm0
Bm = Em / c = 2.58 nT
Radiation Pressure
Waves not only carry energy but also momentum. The effect is
very small (we don’t ordinarily feel pressure from light). If light
is completely absorbed during an interval Δt, the momentum
Transferred Δp is given by
Du
and twice as much if reflected.
Dp =
Newton’s law:
Dp
F=
Dt
c
F
A
I
Now, supposing one has a wave that hits a surface
of area A (perpendicularly), the amount of energy
transferred to that surface in time Δt will be
DU = IADt
Radiation
pressure:
IADt
D
p
=
therefore
c
IA
F=
c
I
2I
pr = (total absorption), pr =
(total reflection)
c
c
[Pa=N/m2]
The pressure p is independent
of the area A.
(a) The pressure remains the same.
The force F is proportional
to the area A.
(b) The force decreases.
EM waves: polarization
Radio transmitter:
If the dipole antenna
is vertical, so will be
the electric fields. The
magnetic field will be
horizontal.
The radio wave generated is said to be “polarized”.
In general light sources produce “unpolarized
waves”emitted by atomic motions in random
directions.
EM Waves: Polarization
Completely unpolarized light will have
equal components in horizontal and vertical
directions. Therefore running the light through
a polarizer will cut the intensity in half: I=I0/2
When polarized light hits a polarizing sheet,
only the component of the field aligned with the
sheet will get through.
E y = E cos(q )
And therefore:
I = I 0 cos 2 q
Example
Initially unpolarized light of
intensity I0 is sent into a system of
three polarizers as shown. What
fraction of the initial intensity
emerges from the system? What
is the polarization of the exiting
light?
• Through the first polarizer: unpolarized to polarized, so I1=½I0.
• Into the second polarizer, the light is now vertically polarized. Then, I2 =
I1cos2(60o)= 1/4 I1 = 1/8 I0.
• Now the light is again polarized, but at 60o. The last polarizer is
horizontal, so I3 = I2cos2(30o) = 3/4 I2 =3 /32 I0 = 0.094 I0.
• The exiting light is horizontally polarized, and has 9% of the original
amplitude.
Completely unpolarized light will have
equal components in horizontal and vertical directions.
Therefore running the light through first polarizer will cut the
intensity in half: I=I0/2
When the now polarized light hits second
polarizing sheet, only the component of the field
aligned with the sheet will get through.
(a) I 0 ® 12 I 0 ® 12 I 0 cos2 (0°) = 12 I 0
(b) I 0 ® 12 I 0 ® 12 I 0 cos2 (60°) = 18 I 0
(c) I 0 ® 12 I 0 ® 12 I 0 cos2 (90°) = 0
(d) I 0 ® 12 I 0 ® 12 I 0 cos2 (30°) = 83 I 0
I = I 0 cos2 q
First polarizer cuts intensity in half.
Second cuts by cos 2 q .
The q is angle between dashed lines.
More gets through when more aligned.
Less gets through when less aligned.
a>d >b>c=0
Reflection and Refraction
When light finds a surface separating two media (air and
water, for example), a beam gets reflected (bounces) and
another gets refracted (bends).
Law of reflection (Light Bounces): the
angle of incidence  1 equals the angle
of reflection  ’1.
 1 =  ’1
Law of Refraction (Light Bends): n2 sin q 2 = n1 sin q1 Snell's Law
n is the index of refraction of the medium.
In vacuum, n = 1. In air, n ~ 1. In all other media, n > 1.
33.8: Reflection and Refraction:
Toward ✔
Can’t go past normal
Away ✖
✖
In each case
going from
less to
greater.
Chromatic Dispersion
The index of refraction depends on the wavelength (color) of the light.