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 Planck’s constant x velocity of light hc = 1240 eV x nm Bohr radius ao = 0.053 nm electron mass: 9.11 x 10-31 kg melectron c2 = 511,000 eV Reduced Planck constant:

!

h =h

2"=1.055 #10

$34J % s

Speed of light in vacuum:

!

c = 3"108m /s

Permittivity of free space

!

"o

= 8.85 #10$12C2/N %m

2

!

k =1

4"#0

= 9 $109Nm

2/C

2

Permeability of free space

!

µo

= 4" #10$7

T %m /A Magnitude of electron charge

!

e =1.6 "10#19C

Problem 1: _______ / 20 Problem 2: _______ / 20 Problem 3: _______ / 20 Problem 4: _______ / 20 Problem 5: _______ / 20 TOTAL: _______ / 100

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)

B = µ0nI = N µ0I /L => N = LB/ (µ0I)

!

=5m( ) 2.515 "10#4T( )4$ "10#7T %m /A( ) 1A( )

=1000turns

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

#34J $ s

45kg( ) 5m /s( )= 2.94 "10#36m

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#108m /s( ) /94.9 #106 s$1 = 3.16m

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

!

ˆ z direction. It has a wavelength of 600 nm. At a particular instant in time to its electric field vector points in the

!

ˆ x direction and has magnitude 10 V/m. What is the direction and magnitude of the electric field vector at a later time

!

t1

= to

+ 5 "10#16s?

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

!

T = " /c = 600 #10$9m /3#10

8m /s = 2 #10

$15s.

!

5 "10#16s 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

!

" ˆ y direction for clockwise rotation. 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

λ = Value Units

!

r E =

Magnitude Units Direction

λ = Value Units

5

iv) An electron is accelerated through a potential difference of 25000 V. What is the de Broglie wavelength of the electron?

Solution:

!

K = e"V =p2

2m=

h2

2m#2$ # =

h

2mK=

hc

2mc2e"V

=1240

2 % 0.51%106% 25000 %1.6 %10

&19/1.6 %10

&19=

= 7.76 %10&3nm

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:

!

N = IA /(hf ) = IA" /hc =1000W /m

2( ) 1m2( )1.6 #10

$19J /eV

500nm

1240eV % nm= 2.52 #1021 ph /s

λ = Value Units

N = Value

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.

a) The incoming circularly polarized light has an electric field vector of magnitude

!

Eo.

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 y-

component 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 y-component oscillates up and down, and has a maximum value of Eo. so the ratio

!

E1/E

o=1

b) What is the time-average power per unit area (intensity) of the incident circularly

polarized wave in the case that

!

Eo

=10 V/m? Solution: The instantaneous power per unit area is the magnitude of the Poynting vector

!

r S =

r E "

r B /µ

o. Since

!

r E and

!

r 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

!

Save

= E0

2/µ

oc . Then

!

Save

=10V /m( )

2

4" #10$7T %m /A( ) 3#10

8m /s( )

= 0.27 W /m2

Circularly polarized light

x

y

z

Region 1 Region 2

90˚

30˚

Region 3

!

E1/E

0=

Value

!

Save

= Value Units

7

c) What is the ratio

!

E3/E

1 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

!

E2

= E1cos30˚ . 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

!

E2cos60˚= E

1cos30˚cos60˚ . This is the light that

gets through polarizer III, and so the ratio

!

E3/E

1= cos30˚cos60˚= 0.433

d) What fraction of the intensity incident on polarizer 3 is not absorbed? The instantaneous power in region II is

!

r S =

r E 2"

r B 2/µ

o, and the time-averaged power is

!

Save

= E2

2/2µ

oc since this is linearly-polarized light. The time-averaged power in

region 3 is

!

Save

= E3

2/2µ

oc . The ratio

!

E3/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

!

I2cos

260˚= I

2/4 . This says that ¼ of the power is transmitted, and the

remaining ¾ of the power is absorbed.

!

E3/E

1=

Value

!

I3/I2

=

Value 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. 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

!

" = Nd#

B

dt= NBw

dx

dt$ I =

NBwv

R=10 % 2 % 2 %10

&2%1%10

&2

10= 0.4mA

!

I = Value Direction Units

!

l = 5cm

w=2cm

v=1 cm/s B=2T

!

l = 5cm

w=2cm

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.

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.

F F = 0

!

F = Value Direction Units

B=0

!

F = Value Direction Units

!

l = 5cm

w=2cm I

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.

F I

B = 0 F = 0

!

F = Value Direction Units

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 =nh

mv

!

F = ke2

r2

= mv2

r" k

e2mv

nh= mv

2" v = k

e2

nh=9 #10

9# (1.6 #10

$19)2

2 #1.055 #10$34

=1.1#106m /s

B) Find the radius of this orbit. Solution:

!

mvr = nh" r =nh

mv=

2 #1.055 #10$34

9.11#10$31#1.1#10

6= 2.1#10

$10m

or r = n2a0 = 4 x a0=2.1x10-10m

!

v = Value Units

!

r = Value Units

12

C) Find the total energy of the electron in this orbit. Solution:

!

E = "13.6eV1

n2

= "3.4eV

E =p2

2m" k

e2

r= "k

e2

2r

since

!

F = ke2

r2

= mv2

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

"13.6eV1

ni2"1

n f

2

%

& ' '

(

) * *

=1240nmeV

"13.61

16"1

4

%

& '

(

) *

= 486.3nm

!

" = Value Units

!

E = Value Units