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Name: ____________________________________________________
Student ID: ___________________________
Section #: _________
Physics 208 Exam 3
Nov. 28, 2006
Print your name and section clearly above. If you do not know your section number,
write your TA’s name.
Your final answer must be placed in the box provided. You must show all your work
to receive full credit. If you only provide your final answer (in the box), and do
not show your work, you will receive very few points.
Problems will be graded on reasoning and intermediate steps as well as on the final
answer. Be sure to include units, and also the direction of vectors.
You are allowed one 8½ x 11” sheet of notes and no other references. The exam lasts
exactly 90 minutes.
Planck’s constant
h = 6.626 x 10-34 m2 kg/s
Problem 1:
_______ / 20
Planck’s constant x velocity
of light hc = 1240 eV x nm
Problem 2:
_______ / 20
Problem 3:
_______ / 20
Problem 4:
_______ / 20
Problem 5:
_______ / 20
TOTAL:
_______ / 100
Bohr radius ao = 0.053 nm
electron mass: 9.11 x 10
melectron c2 = 511,000 eV
-31
kg
Reduced Planck constant:
h
h=
= 1.055 #10$34 J % s
2"
Speed of light in vacuum:
c = 3 "10 8 m /s
!
!
!
!
!
!
Permittivity of free space
"o = 8.85 #10$12 C 2 /N % m 2
1
k=
= 9 $10 9 Nm 2 /C 2
4 "#0
Permeability of free space
µo = 4 " #10$7 T % m / A
Magnitude of electron charge
e = 1.6 "10#19 C
2
1) [20 points, 4 points each]. Explain your reasoning for full credit.
Multiple choice/short answer questions.
i) How many turns of wire a solenoid needs to have to produce a uniform magnetic field
of 2.515 x 10-4 T inside it if the current in the turns is 1 A and its length is 5 m? Explain.
a. 200
b. 1000
c. 10000
d. not calculable because the diameter of the solenoid is not given
e. 100
Solution: b)
(5m)(2.515 "10#4 T )
B = µ0nI = N µ0I /L => N = LB/ (µ0I) =
= 1000turns
(4$ "10#7 T % m / A)(1A)
ii) A bar magnet is dropped through the conducting loop as shown in the figure. Which is
the direction of the induced current in the conducting loop (observed from the top) while
!
the magnet is falling towards
the loop and after it has passed through the loop and
moves away from it? Explain.
a. Counterclockwise when the bar is approaching the loop and clockwise when it is
falling away from the loop.
b. Clockwise when the bar is approaching the loop and counterclockwise when it is
falling away from the loop.
c. Counterclockwise always.
d. Clockwise always.
Solution: b)
The S pole of the magnet produces an upward B-field and the flux increases while the
magnet approaches the loop. As the magnet falls below the loop, the N pole also
produces a field upward, but it decreasing as the magnet falls. By Lenz’ law, the
induced current produces a field to oppose these changes, so when the bar is above
the loop the induced B-field is downward and the current is cw and when it leaves the
loop the induced field is upward, and hence ccw current.
3
iii) The light intensity incident on a metallic surface produces photoelectrons with a
maximum kinetic energy of 4 eV. The light intensity is doubled. Determine the maximum
kinetic energy of the photoelectrons (in eV):
a. 4
b. 2
c. √2
d. 3
e. it cannot be determined since the work function of the metallic surface is not
given.
Solution: a, the maximum kinetic energy does not change since the frequency is not
changed.
iv) The wavelength of a 45 kg boy moving at 5 m/s on a skateboard is about:
a. 5 x 10-37 m
b. 3 x 10-36 m
c. 7 x 10-34 m
d. 0.1 m
Solution: b, λ=h/(mv) =
6.626 "10#34 J $ s
= 2.94 "10#36 m
( 45kg)(5m /s)
v) A hydrogen atom is in the n=2 excited state. What is the minimum energy needed to
ionize the!atom?
a. -13.6 eV
b. -3.4 eV
c. 3.4 eV
d. 6.8 eV
e. -6.8 eV
Solution: c) E2 = 13.6/4 eV = +3.4 eV. This is the binding energy, so this much energy
must be supplied to ionize the atom.
4
2) [20 pts, 4 pts each] Short answer.
i) Your Madison oldies radio station broadcasts at 94.9 MHz (1 MHz = 106 Hz).
What is the wavelength?
Solution: The wavelength is given by " = c / f = ( 3 #10 8 m /s) /94.9 #10 6 s$1 = 3.16m
Value
Units
λ=
!
ii) A light wave with right-handed circular polarization (looking as it comes toward you,
as in the lab) is propagating in vacuum in the zˆ direction. It has a wavelength of
600 nm. At a particular instant in time to its electric field vector points in the xˆ
direction and has magnitude 10 V/m. What is the direction and magnitude of the
electric field vector at a later time t1 = t o + 5 "10#16 s ?
!
!
Solution: Since this is circular polarization, the E-field has constant magnitude, but
!
rotates in a plane perpendicular
to the direction of propagation. The period is
$9
8
T = " /c = 600 #10 m /3 #10 m /s = 2 #10$15 s . 5 "10#16 s is ¼ of the period, so the E-field
will have rotated ¼ of the way around. This means that it now points in the " yˆ direction
for clockwise rotation.
!
!
Magnitude
Direction Units
r
!
E=
iii) A particular metal has a work function!
of 1.5eV. What is the maximum wavelength
photon that will eject an electron from the metal?
Solution:
!
" = hc / # $
# = hc / " = 1240eV % nm /1.5eV = 827 nm
Value
λ=
Units
5
iv) An electron is accelerated through a potential difference of 25000 V. What is the de
Broglie wavelength of the electron?
Value
Units
λ=
Solution:
p2
h2
h
hc
1240
=
$#=
=
=
=
2
2
6
2m 2m#
2mK
2mc e"V
2 % 0.51%10 % 25000 %1.6 %10&19 /1.6 %10&19
= 7.76 %10&3 nm
K = e"V =
!
v) On a bright and sunny day, the intensity of solar radiation on the Earth’s surface is
1000 W/m2. If the average wavelength of the sunlight is 500 nm, how many photons
are incident on an area A =1 m2 of the Earth’s surface per second?
Solution:
(1000W /m )(1m )
2
N = IA /(hf ) = IA" /hc =
1.6 #10
$19
2
500nm
= 2.52 #10 21 ph /s
J /eV 1240eV % nm
Value
!
N=
6
3) [20 pts, 5 pts each] Three linear polarizers are arranged as shown, with
transmission axes indicated by the thick solid line. The incoming light is circularly
polarized, and propagating in the z-direction.
y
x
30˚
Circularly
polarized
light
z
90˚
Region 1
Region 2
Region 3
a) The incoming circularly polarized light has an electric field vector of magnitude E o .
The electric field vector of the linearly-polarized light in region 1 has an oscillating
electric field vector with amplitude (maximum value) of E1 . What is the ratio E1 / E o ?
Explain.
!
Solution: Polarizer 1 passes the y-component of the circularly polarized light. The ycomponent has magnitude Eo, since when the!rotating electric field!of the circularly
polarized light points along the y-axis, there is no x-component, only y. The ycomponent oscillates up and down, and has a maximum value of Eo. so the ratio
E1 / E o = 1
E1 / E 0 = Value
!
b) What is the time-average power per unit!area (intensity) of the incident circularly
polarized wave in the case that E o = 10 V/m ?
Solution: The instantaneous power per unit area is the magnitude of the Poynting vector
r r r
r
r
S = E " B / µo . Since !
E and B do not change magnitude, but only rotate, the
!
instantaneous power here does not depend on time. Since B = E /c , the average
power
per
unit
area
is
Then
Save = E 02 / µoc .
! (10V /m!) 2
Save =
= 0.27 W /m 2
$7
8
!
(4" #10 T % m / A)(3 #10 m /s)
!
Value
Units
Save =
!
!
7
c) What is the ratio E 3 / E1 of the maximum value of the electric field in region 3 (this is
the region to the right of the last polarizer) to that in region 1?
Solution: The max electric field in region 2 is E 2 = E1 cos 30˚ . This light is polarized at a
!
60˚ angle with respect to polarizer 3. The component of this light along the
transmission axis of polarizer III is E 2 cos60˚= E1 cos 30˚cos60˚ . This is the light that
gets through polarizer III, and so the ratio E 3 / E1 = cos 30˚cos60˚= 0.433
!
!
E 3 / E1 = Value
!
!
d) What fraction of the intensity incident on polarizer
3 is not absorbed?
r r
r
The instantaneous power in region II is S = E 2 " B2 / µo , and the time-averaged power is
!
Save = E 22 /2µoc since this is linearly-polarized light. The time-averaged power in
region 3 is Save = E 32 /2µoc . The ratio E 3 / E 2 = cos60 = 0.5 . Then the time-averaged
power in region 3 is ¼ that
! in region 2. This means that ¾ of the power was
absorbed by the polarizer. Or could say directly from Malus’ law, that the intensity in
region 3 is I2 cos2 60˚= I2 /4 . This says that ¼ of the power is transmitted, and the
!
!
remaining ¾ of the power is absorbed.
!
Value
I3 / I2 =
!
Units
8
4) [20 pts, 5 pts each]
A rectangular coil with resistance R = 10 Ω has N = 10 turns each of length l = 5 cm
And width w = 2 cm as shown in the figure below. The coil moves with constant velocity
v = 1 cm/s through a uniform magnetic field B = 2 T entering the page. The magnetic
field is 2 T only in the region shown, and is zero everywhere else.
!
w=2cm
B=2T
v=1 cm/s
l = 5cm
!
A) What is the induced current in the coil as it enters the magnetic field region?
Indicate its direction in the loop.
Solution:
When the coil enters the B-field region the flux increases and the current is ccw and
d#B
dx
NBwv 10 % 2 % 2 %10&2 %1%10&2
" =N
= NBw
$I=
=
= 0.4mA
dt
dt
R
10
!
w=2cm
l = 5cm
!
Value
I=
!
Units
Direction
9
What are the magnitude and the direction of the total magnetic force on the coils:
B) total force on the loop as it enters the magnetic field region;
Solution: F = NIBw=1.6e-4 N as in the figure. The magnetic forces on the horizontal
axes are equal and opposite and do not contribute to the net magnetic force.
B=0
w=2cm
I
F=0
F
l = 5cm
Value
!
Units
Direction
F=
C) total force on the loop as it moves within the field region;
Solution:
!
The net magnetic force is zero since the forces on the vertical sides and horizontal
sides are equal and opposite.
Value
F=
!
Units
Direction
10
D) total force on the loop as it leaves the field region.
Solution: the flux is into the page, and decreases with time. By Lenz’ law, the induced
current will produce a field into the page that opposes this decrease. So the current is
clockwise. The magnetic force is still 1.6e-4 N but opposite respect to part A). The
forces on the 2 horizontal sides do not contribute to the net magnetic force since they
are equal and opposite.
B=0
F=0
F
I
Value
F=
!
Units
Direction
11
5) [20 pts, 5 pts each]
Consider a hydrogen atom with a nuclear charge e and a single electron orbiting.
Answer these questions about the Bohr’s model on this atom:
A) Find the magnitude of the velocity of the electron in the circular orbit with n = 2
around the nucleus. [Hint: consider the attractive Coulomb force acting on the electron
due to the proton in the nucleus.]
Solution:
mvr = nh " r =
F=k
nh
mv
e2
v2
e 2 mv
e 2 9 #10 9 # (1.6 #10$19 ) 2
2
=
m
"
k
=
mv
"
v
=
k
=
= 1.1#10 6 m /s
r2
r
nh
nh
2 #1.055 #10$34
!
!
Value
Units
v=
B) Find the radius of this orbit.
!
Solution:
nh
2 #1.055 #10$34
mvr = nh " r =
=
= 2.1#10$10 m
$31
6
mv 9.11#10 #1.1#10
or
!
r = n2a0 = 4 x a0=2.1x10-10m
Value
r=
!
Units
12
C) Find the total energy of the electron in this orbit.
Value
Solution:
Units
E=
1
= "3.4eV
n2
p2
e2
e2
E=
" k = "k
2m
r
2r
E = "13.6eV
!
since F = k
!
e2
v2
=
m
r2
r
D) Find the wavelength of the photon emitted in the transition from the state with ni = 4
! to the one with nf = 2.
Solution
E f " Ei =
hc
hc
$#=
=
#
Ei " E f
hc
1240nmeV
=
= 486.3nm
%1
1 ( "13.6%' 1 " 1 (*
"13.6eV '' 2 " 2 **
&16 4 )
& ni n f )
Value
"=
!
!
Units