physics 104 final examination, may13, 2004mcdonald/examples/ph104_2004/... · 2010. 11. 13. ·...

Print Name: _____________________________ Please circle your class/time/instructor below:

Problem Score 1 /252 /303 /254 /305 /106 /30Total /150

1: 9 am Trawick 5: 10 am Trawick 2: 9 am McDonald 6: 10 am Calaprice 3: 9 am Galbiati 7: 10 am Meyers 4: 9 am Itzhaki 8: 10 am Altshuler 9: 10 am Seiringer

Physics 104 FINAL EXAM

May 13, 2004 3 hours Instructions:

• When you are told to begin, check that this examination booklet contains all the numbered pages 2 through 20.

• You must show your work—the grade you get depends on how well the grader can understand your solution even when you write down the correct answer.

• Please write each final answer in the box provided. • Remember to include units in your calculations, and specify the directions of all vectors. • If a part of a problem depends on a previous answer you have not obtained, make an

educated guess and proceed. • Do not panic or be discouraged if you cannot do every problem; there are both easy and

hard parts in this exam. Keep moving and finish as much as you can. Rewrite and sign the Honor Pledge: “I pledge my honor that I have not violated the Honor Code during this examination.” _______________________________________________________________________ ______________________________________________________________________

Signature: ____________________________

Physics 104 Final Examination, May 13, 2004 Problem 1,

Problem 1.

R2

r

A cylindrical capacitor is formed by two coaxial(their axis coincide with the z-axis) infinitely longthin cylindrical shells of radii R1 and R2, R1<R2. The cylinders carry charges +λ and -λ per unit length. a) (7 points) Using Gauss’s law calculate the electric field E as a function of the distance fromthe cylinder axis r outside the big cylinder( ), inside the small cylinder (2Rr > 1Rr < ) and in the space between the two cylinders ( 21 RrR << ). Sketch the Gaussian surfa es on the picture.

21 RrR <<

1Rr <

2Rr >

According to the cylindrical symmetry of the problem the elecalong the vector r and depends only on r = | r |. For the csurfaces shown on the picture the electric flux equals to

2

210

1

0

0

Rr

RrRLRr

EAdAES

n

>

<<

<

==≡Φ ∫ ελ

where L is the length of the Gaussian cylinder, and A is itaccount that A=2πrL we obtain that

z

R1

+λ

-λ

c

tric field is directed ylindrical Gaussian

s area. Taking into

E = 0 ( ) 1Rr <

E = 02 επ

λr

( R 21 Rr << )

E = 0 ( ) 2Rr >


b) (6 points) Calculate the electric potential difference between the cylinders, ∆V = V2 –V1.

2 2

1 1

12 1

0 0

( ) ln2 2

R R

R R 2

RV V V E r dr drr R

λ λ∆πε πε

⎛ ⎞= − = − = − = ⎜ ⎟

⎝ ⎠∫ ∫

2

0 1

ln2

RVR

λ∆πε

⎛ ⎞= = − ⎜ ⎟

⎝ ⎠

c) (6 points) Assuming that V=0 at infinity, sketch the r-dependence of V on the diagram below.

V

2

0 1

ln2

RR

λπε

⎛ ⎞⎜ ⎟⎝ ⎠

R1 R2r


d) (3 points) What is the capacitance per unit length of the system? The capacitance is determined as

QCV∆

= ,

where Q = λL is the charge of each of the cylinders, and ∆V=(λ/2πε0) ln(R1/R2) is the potential difference. Therefore

0 0

2 2

1 1

2 2

ln ln

L LCR RR R

λ πε πε

λ= =

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

One can see that the capacitance is indeed proportional to the length of the cylinders. Capacitance per unit length equals to

0

2

1

2

ln

CL R

R

πε=

⎛ ⎞⎜ ⎟⎝ ⎠

e) (3 points) What is the energy stored in the capacitor per unit length? Express the answer in terms of λ, R1, and R2.

( )2 2 2 2

2 2 2Q LU LC C C L

λ λ= = =

Using the result for the capacitance per unit length one obtains

22

0 1

ln4

U RL R

λπε

⎛ ⎞= ⎜ ⎟

⎝ ⎠


Problem 2 a) (5 points) A long straight wire carries a current I1. Use Ampere’s law to predict the magnetic field at a point P located a distance x from the wire. Specify magnitude and direction.

B I1

x

P B The magnetic field at a given point is perpendicular to both the direction of the current and to the radius – vector. It means that the direction is tangential. It follows from the right hand rule that the direction of the field is as shown on the picture. The magnitude of the field depends only on the distance x. Using the dashed circle as an Ampere contour one can write the Ampere law as

102 IxBldBC

µπ ==∫rr

Therefore

xIB

πµ2

10=


b) (5 points) A square coil with side a is placed a distance a from the long wire, as shown. What is the net force on the square coil when it carries a clockwise current I2? Specify magnitude and direction.

Let both the wire and the square lie in the xz– plane and the dcurrent I1 be the positive z– direction. The magnetic field of this currplane is parallel to the y– axis. Above the wire the field directed in thdirection. Let us label the sides of the square as shown on the picturefrom the symmetry that the forces applied to the sides 2 and 4 cancel

. The magnetic field that acts on the sides 1 and 3 equals t2F = −r r

4F

0 11,3 1,3 1,3

1,3

; 2 ,2

IB j x a xx

aµπ

= =r

=r

The corresponding forces can be obtained from the equation IFdr

=

( ) 0 1 2 0 1 21,3 1 3

1,3

; ,2 4

I I a I I aF k j F i Fx a

µ µπ π

= ± × = −rr r rr r

The net force is thus directed in positive x– direction:

1 2netF F F= +r r r

I

a

a

a

x

y

1

irecente p. It

eaco

ldr

=

=

1

2
I2 3 4
tion of the in the xz– ositive y– follows h other:

Br

× :

0 1 2

2I I a i

aµ

πr

0 1 2

4I I a

aiµ

πr

z


c) (4 points) What is the magnetic moment of the square coil? What is the torque on the coil?

Magnetic dipole moment: nNIArr≡µ , where in our case N=1, I=I2, and A=a2 The

direction of the normal to the plane of the square has to be determined by the right-hand rule. It turns out to be negative y– direction: jn

rr−= . Therefore

jaIrr 2

2−≡µ

Torque: 0=×∝×= jjBrrrrr µτ . Indeed, the torque appears only when the

magnetic field has a component parallel to the plane of the loop

magnetic moment: jaIrr 2

2−≡µ torque: 0=τr

d) (5points) The source of the current I2 is now turned off, but the current I1 is still on and constant. Calculate the magnetic flux Φm through the square coil due to the magnetic field of the long wire.

By definition the magnetic flux through any contour is the integral over a surface, S,

∫=ΦS

nm dAB ,

bounded by this contour. Choosing the surface to be flat and taking into account that the magnetic field in the xz-plane in directed along the y-axis and depends only on x one can write the flux as

a

I1

a

a

( ) ( )2 2

0 1 0 1 0 1

0

2ln ln 22 2 2

a a a

ma a

I I a a I adz dx B x a dxx a

µ µ µΦπ π π

⎛ ⎞= = = =⎜ ⎟⎝ ⎠∫ ∫ ∫

( )0 1

ln 22m I aΦ µπ

=

e) (5 points) The current I1 now increases linearly at a rate dtdI1=α . Calculate the induced current in the square coil I2. Assume the square coil has resistance R.


Time dependence of the current in the long wire leads to the time-dependent magnetic flux through the square loop:

( ) ( )10 0

ln 2 ln 22 2

md dIa adt dtΦ µ µ α

π π= =

One can evaluate the electromotive force, Ε2,using the Faraday’s law:

Ε2 = ( )

0

ln 22

md adtΦ µ α

π− = −

The current can be calculated from the Ohm’s law: I2= Ε2 /R. Therefore

( ) 02

ln 22

aIR

µ απ

= −

f) (4 points) What is the direction of the force on the square coil as a result of the induced current?

One can determine direction of the induced current from the Lenz’s law: this current should oppose the changes of the current in a long wire. For example, in the latter current increases (α>0), then the induced current in the segment 3, which is closer to the wire should be in negative z-direction

g) (2 points) What is the mutual inductance M of the coil and wire?

By definition M=Φm/I1. Substitution of the equation for the flux Φm yields

( )0

ln 22

M aµπ

=

Direction: for positive the induced current is directed clockwise, i.e., it flows in the direction shown on the picture


Problem 3. A resistor and an inductor are connected to a battery through a switch, as shown.

I

a) (5 points) At time t=0 somebody closes the switch, allowing current to flow. In the space below, derive the current I(t) that flows through the inductor as a function of time. (Partial credit here for both the setup and for the final I(t), even if you are unable to actually solve the differential equation for I(t).) The total voltage V is distributed between the resistor and the inductor: V= VR+ VL. At a given current I the two voltage drops are VR=IR, VL=L(dI/dt). This leads to the equation:

dIL RIdt

V+ =

It is possible to separate the variables and revrite this equation as dIL d

V RIt=

−

Integrating both parts of this equation leads to

( ) ( ){ }ln lnL dI L Vt I RVR RIA

R= = − − +

−∫ ,

where is A a time-independent constant. This equation allows to determine I(t):

( ) exp ,V tI t AR τ

⎛ ⎞= + −⎜ ⎟⎝ ⎠


where τ=L/R. The constant A should be determined from the condition that I(t)=0 at t=0. As a result, A=-V/R and

( ) 1 expV tI tR τ

⎛ ⎞⎛ ⎞= − −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠


b) (4 points) On the axes below, draw graphs of the current I(t) through the inductor, and the voltage VL(t) across the inductor. Be sure to indicate the initial (t = 0) and final

)( ∞=t values of I and V.

I

t

LV

t

V/R V

In the following parts (c through e), instead of being connected to a battery, the inductor and resistor are connected to a voltage

( )tVV ωcos0= .

c) (7 points) The current through the inductor may be written as ( )δω −= tII cos0 . Find the magnitude and the phase 0I δ of the current through the inductor. The complex impedance of the sequentially connected reristor and inductor equals to

R LZ Z Z R i Lω= + = +

can be written as Z=|Z|exp(iδ) with

( )22 1; tan LZ R LR

ωω δ − ⎛ ⎞= + = ⎜ ⎟⎝ ⎠

The current through the inductor is the total current. It can be writtens as

( )( ) ( )0 0 0 0Re Re Re cosi t i t

i ti

V e V e V VI eZ Z e Z Z

ω ωω δ

δ tω δ−⎛ ⎞⎛ ⎞= = = =⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠−

Therefore

( )220I V R Lω= +

δ = tan-1(ωL/R)


d) (5 points) The voltage across the inductor may be written as ( )θω −= tVV LL cos0 . Find the magnitude VL0 and the phase θ of the voltage VL across the inductor. The voltage across the inductor equals to

( )( ) ( )( ) ( )( )/ 20 0 0Re Re Rei t i t i t

L LV Z I e i LI e LI eω δ ω δ ω π δω ω− −= = = + −

One can substitute the equation for I0 . As a result,

( )( )( )

( )/ 20 0

2 22 2Re cos

2i t

LLV LVV e t

R L R Lω π δω ω πω δ

ω ω

+ − ⎛ ⎞= = ⎜ ⎟⎝ ⎠+ +

+ −

The phase can be thus written as

1 1tan tan2 2

L RR L

π ω πθ δω

− −⎛ ⎞ ⎛ ⎞= − = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Therefore

( )0

0 22

1tan

LLVV

R LRL

ω

ω

θω

−

=+⎛ ⎞= − ⎜ ⎟⎝ ⎠

e) (4 points) What is the magnitude of the voltage VL0 across the inductor in the limit as 0→ω , and in the limit as ∞→ω ?

00

0 0

0 0L

L

LVVR

V V

ωωω

→ → →

→ ∞ →


Problem 4. a) (8 points) Lenses

An object of height 15 cm is placed 30 cm in front of a diverging lens of focal length f1 = -15 cm. A second lens (converging) of focal length f2 = +20 cm is placed 25 cm to the right of the first lens. Find the final image position and height by calculation (4 points) and by ray tracing on the figure below (4 points).

Scale: Each tic mark is 5 cm.

F1 F1` F2F2

Thin-lens equation 1 1 1s s f

+ =′

.

First lens is diverging, i.e., its focal length is negative; f1=-15 cm, s1 = 30 cm Therefore cm 1 10s′ = −

Second lens: f2=+20 cm, 2 125 35s s′= − = cm. From the equation

22 2 2

1 1 1 1 1 3 ; 46.720 35 140

ss f s

′= − = − = ≈′

cm

Magnification94

35303/14010

21

2121 −=

××−

=′′

==′

−=ssssmmm

ssm

1 21 2

1 2

10 35 1;30 140 / 3 4

s s sm m m ms s s′ ′ ′ − ×

= − = = = = −×

Image position: 46.7 cm Image height: -6.67 cm


b) (6 points) Total internal reflection. The sketch below shows a water-air interface. Sketch the direction of an incident light ray that can be totally internally reflected. Calculate the critical angle for total internal reflection. The index of the refraction of water is =1.33. wn

air

water θ θ

Total internal reflection: sin 1wn θ ≥ - Snell’s law for the refraction is not satisfied for any angle of refraction

Critical angle: 1 11 1sin sin 48.75

1.33cwn

θ − −⎛ ⎞ ⎛ ⎞= = ≈⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

o

1 1sin 48.75cwn

θ − ⎛ ⎞= ≈⎜ ⎟

⎝ ⎠o


c) (8 points) Magnifying mirror. Design a mirror that will give an upright image of your face with a magnification of two when your face is 25 cm from the mirror. Specify the type of mirror (concave or convex), the focal length f, the radius of curvature of the mirror R. Make a sketch of the mirror, and draw some rays that illustrate the object and image.

Mirror equation 1 1 1s s f

+ =′

. Magnification sms′

= − .

We need m=2 for s=25 cm. Therefore 1 1 1 1 50 2 10025 2 25 50

f r ff

= − = = = =×

Positive focal length means that the mirror is concave.

d) (8 points) Interference and diffraction

2525

object

F

image

A double slit system has slits with width b and slit separation 2b. Light of wavelength λ = b/2 is incident on the two slits and forms an interference pattern on the screen. Allowing for diffraction and interference effects, specify all fringes that appear on the screen. Double slit system: interference maxima appear at angles θm, such that

sin θm =mλ/2b=m/4, m=0,1,2,3,… . Single slit effect: zeros in the single slit diffraction pattern appear at mθ θ ′′= , where sin / / 2, 1,2,3,...m m b m mθ λ′′ ′ ′ ′= = = . Therefore the double slit bright fringes that correspond even values of m (except m=0) will not appear on the screen. One should expect bright fringes only at m=0,1,3., since sin θm can not exceed 1. m=0 and m=1 fringes are within the central peak of the diffraction. The fringe with m=3 should have rather low brightness.


Problem 5. A source with power P emits monochromatic light with wavelength λ equally in all directions. a) (6 points) Find I, the intensity of the light, and find the rms values of the electric and magnetic fields at a distance r away from the source. The power of the source is uniformly distributed over a sphere provided that the source is at the center of the sphere. The area of the sphere with a radius r is 4πr2. Therefore the intensity at this distance from the source is

( ) 24PI rrπ

=

The rms values of the electric and magnetic fields, Erms and Brms , are connected with the intensity by the equation

( ) ( ) ( )0

rms rmsE r B rI r

µ=

Since Erms = cBrms , each of the fields can be expressed through the intensity:

( ) ( ) ( ) ( )00 00

1 1;4 4rms rms

I rc P PE r c I r B rr c

µr c

µ µµπ π

= = = =

( )

( )

( )

2

0

0

41 ;

41

4

rms

rms

PI rr

c PE rr

PB rr c

πµπ

µπ

=

=

=


b) (2 points) Suppose that a small disc of radius b and mass m is located a distance r from the source. The light from the source is normally (perpendicularly) incident on the face of the disc. Find the acceleration a of the disc assuming that it absorbs light at that wave length completely. The disc absorbs the momentum of the light. The time derivative of the momentum of the disc dp/dt equals to the momentum absorbed per unit time. The latter equals to I/c times the area of the disc πb2:

22

24dp I Pbbdt c cr

π= =

According to the Newton’s law dp/dt is nothing but the force F aplied to the disc. Therefore, the acceleration a=F/m equals to

22

24I Pba b

mc mcrπ= =

c) (2 points) Explain qualitiatively how the acceleration will be affected if the ball is covered with a thin layer of a reflecting material? Assume that the mass of that thin layer is negligible compared to m. Because of momentum conservation the acceleration is increased by a factor of two.


Problem 6. A particle of mass m moves freely in the two-dimensional rectangular region defined by:

Lx ≤≤0 Ly 20 ≤≤ .

That is, the particle is in a two-dimensional infinite well, where 0=U inside the rectangular region, and outside that region. ∞=U a) (4 points) A wave function for a single particle in this potential must have the form

( ) )sin()sin(, 21 ykxkAyx =ψ

What are the possible values for and that are consistent with the boundary conditions? You must indicate allowable values of any quantum numbers (like “n”) you introduce in your answer.

1k 2k

The boundary conditions are that ( ) 0, =yxψ at the edges of the rectangle, and . (If we think of a standing wave, there must be nodes at the edges.) For this to be true, we must have

Lx = Ly 2=

Lnk π1

1 = , where K3,2,11 =n

Lnk2

22

π= , where K3,2,12 =n

b) (6 points) Use either the time-independent Schrödinger equation or the de Broglie relationship to derive the allowable energies for a single particle inside this well.

( ) ( ) ( ) ( ) ( )

( )

2 2 2

2 2

2 2 21 2 1 2

2 2

2 221 1 2 2 1

, , ,2

sin sin sin sin ,2 2 2

sin sin sin2 2 2

d dx y x y U x x E x ym dx dy

d n x n y d n x n yA A Em dx L L dy L L

n n x n y n n xA Am L L L L L

ψ ψ ψ ψ

π π π π ψ

π π π π π

⎡ ⎤− + + =⎢ ⎥

⎣ ⎦⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛− − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

h

h

h

x y=

( )

( ) ( ) ( )

( )

2

2 221 2

2 221 2

2 2 22 2 221 1 22 2

sin ,2

, , ,2 2

2 2

48 4 32

n y E x yL

n nx y x y E x ym L L

n nEm L L

h n hE n n nmL mL

π ψ

π πψ ψ ψ

π π

⎡ ⎤⎞ ⎛ ⎞ =⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞+ =⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞= +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

⎛ ⎞= + = +⎜ ⎟

⎝ ⎠

h

h


c) (5 points) Give the quantum numbers for the six lowest energy states for a single spinless particle in this well, starting with the ground state. Which, if any, of these states are degenerate?

Try some different values of and : 1n 2n

( ) ( ) 532

11432

1,1 2

222

2

2

⋅=+⋅=→mLh

mLhE

( ) ( ) 832

21432

2,1 2

222

2

2

⋅=+⋅=→mLh

mLhE

Apparently the six lowest energy states, in order from lowest to highest, are: (1, 1), (1, 2), (1, 3), (2, 1), (1, 4), and (2, 2). The states (1, 4) and (2, 2) have the same energy, so they are degenerate.

( ) ( ) 1332

31432

3,1 2

222

2

2

⋅=+⋅=→mLh

mLhE

( ) ( ) 2032

41432

4,1 2

222

2

2

⋅=+⋅=→mLh

mLhE

( ) ( ) 1732

12432

1,2 2

222

2

2

⋅=+⋅=→mLh

mLhE

( ) ( ) 2032

22432

2,2 2

222

2

2

⋅=+⋅=→mLh

mLhE

( ) ( ) 2532

32432

3,2 2

222

2

2

⋅=+⋅=→mLh

mLhE

( ) ( ) 2932

51432

5,1 2

222

2

2

⋅=+⋅=→mLh

mLhE

M d) (5 points) What is the ground state energy of six neutrons in this box? (Like electrons, neutrons are fermions and have two different spin states.)

Because the neutrons are fermions, each one must have a different state. There are two spin states for each spatial state, so the first three spatial states each hold two neutrons. The energy of this configuration is:

2

2

6

2

2

6

2

2

2

2

2

2

6

3,12,11,16

813

2632

2

1332

2832

2532

2

222

mLhE

mLhE

mLh

mLh

mLhE

EEEE

neutronsGS

neutronsGS

neutronsGS

neutronsGS

=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅+⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅⋅+⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅⋅=

++=

2, 2 1, 4

2, 1

1, 3

1, 2

1, 1

213h
28mL
E =


e) (4 points) A single particle is in the potential well, in its ground state. The wave function ),( yxψ for this particle is actually a function of both x and y, making it a challenge for most of

us to sketch it. So instead, consider ),( Lxψ along the line Ly = , right down the middle of the rectangular box. On the axes below, draw a graph of ),( Lxψ and a graph of along the line .

),(2 LxψLy =

0 L

ψ

x

0

0 L

2ψ

x

0


f) (6 points) Now suppose that the potential energy outside the well is lowered from ∞=U to some finite value, . Again, for a single particle in its ground state in this new finite potential well, draw a graph of

0UU =

),( Lxψ and a graph of along the line . ),(2 Lxψ Ly =

0

0 L

ψ

x 0 L

2ψ

x

0

How will lowering the potential outside the box from infinity to some finite value affect the probability of finding the particle between 0=x and 2/Lx = , if a measurement of its position is performed?

A) Higher probability for . ∞=UB) Higher probability for . 0UU =C) Same probability for both.

How will lowering the potential outside the box from infinity to some finite value affect the ground state energy of a single particle in the potential well?

A) Higher ground state energy for ∞=U . B) Higher ground state energy for 0UU = . C) Same ground state energy for both.

For the finite potential well, there is some non-zero probability that the particle will be found outside of the box if a measurement is made of its position. Thus, the probability that it will be found inside the box (specifically, on one particular side of the box) is smaller for the finite potential well ( 0UU = ), and larger for the infinite well ( ). ∞=U Comparing the drawings of ),( Lxψ in parts (d) and (e), the wavelength of the particle inside the box is clearly larger for the finite potential well (with outside the box). Essentially, lowering the potential outside the box has the same effect as making L bigger. Thus, the energy of the particle in its ground state is smaller for the finite potential well (

0UU =

0UU = ), and larger for the infinite well ( ). ∞=U

physics 104 final examination, may13, 2004mcdonald/examples/ph104_2004/... · 2010. 11. 13. ·...

Documents