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Final Exam Phys 248

May 11, 2009

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Problem Score

1 /30

2 /30

3 /30

4 /30

5 /30

Total /150

Useful constants: Planck constant h = 6.626 x 10-34 J s Reduced Planck constant =

�

! =1.054 x 10-34 J s Permittivity of free space:

�

!0

= 8.854 "10#12C2N

#1m

#2 hc = 1240 eV nm

�

!c = 197.3 eV nm c = 2.998 x 108 m/s 1 eV = 1.6 x 10-19 J ke2= 1.44 eV nm

Permeability of free space: µ0 = 4! "10#7T $ A

#1m

me = electron mass = 9.11 x 10-31 kg = 0.511 MeV/c2 e = electron charge = -1.6 x 10-19 C

Problem 1: Short Problems. (1) Damped oscillations: A 900.0 g object oscillates at the end of a vertical spring with spring constant k = 500.0 N/m. The damping coefficient due to the effect of air resistance is given by b = 2.0 N ·s/m. (a) What is the frequency of the damped oscillations? (b) In how much time does the energy of the system drop to 5% of its initial value? Solution: The frequency of the natural oscillations is:

�

!0

=k

m=

500

0.9= 23.6rad /s .

With damping:

�

! = !01"

b

2m!0

#

$ %

&

' (

2

= 23.6 ) 1"2

2 ) 0.9 ) 23.6#

$ %

&

' (

2

= 23.6rad /s

This is a case of weak damping, in fact b = 2 Ns/m << 2mω0

=42.48 Ns/m (b) The energy is proportional to the square of the amplitude:

�

E =1

2kA

2=1

2kA

0

2e!2bt

2m = E0e!bt

m

E

E0

= 0.05 = e!bt /m

" ln0.05 = !bt

m" t = !

m

bln0.05 =1.3s

(2) Radioactive decay: If initially there are N0=104 nuclei of a radioactive material with a half-life of t1/2, how many nuclei have decayed at time t=0.5 t1/2? Solution: The number of nuclei remaining are N(t) = N0e

!"t= N0e

! ln 2t /t1/2 , such

that at t=0.5 t1/2, we have N(0.5t1/2 ) = N0e!(ln 2)/2

=N0

2. The number of

nuclei that have decayed is N0!N0

2= .29N

0= 2.9 "10

3 .

(3) Standing waves: A tuning fork with a frequency 512 Hz is placed near the top of the pipe shown in the figure. The water level is lowered so that the first and second resonant modes are heard. Determine the values of L corresponding to these resonant modes. The velocity of sound at the temperature of your lab is 343 m/s. Solution: The water surface corresponds to a node of the standing wave since the air is not free to move longitudinally and the open end is an antinode. In a narrow tube open at one end

the resonance frequency is

�

f = nv

4L,n =1,3,5,...Assuming n = 1,3 the

lengths from the open end to the water level corresponding to

the first 2 resonant modes are

�

L1

=v

4 f=

343

4 ! 512=16.7cm and

�

L2

=3v

4 f=3! 343

4 ! 512= 50.2cm

L

(4) Electric energy and Potential: The figure below shows two plates, A and B. Plate A has a potential of VA = 0 V and plate B a potential of VB = 100 V. The dotted lines represent equipotential lines of 25, 50, and 75 V. A proton at point x is transferred to point z. Calculate the energy gained or expended by the test charge.

Solution

�

W = !"U = !q"V : Along the equipotential line the work is null since the potential is constant. Along the perpendicular path to the plates

�

W = !"U = !q"V =1.6 #10!19

# (75 ! 25) = 8.0 #10!18J

(5) Electromagnetic Waves: In an electromagnetic wave propagating in the vacuum, the electric field is

!E = z(150V m)cos[(2.00 !10

7m

"1)x "#t]. Calculate the numerical

value of ω and write expressions for the magnetic field and the Poynting vector of the wave. Solution The direction of propagation is the +x direction. The angular frequency is ! = ck = (2.998 "10

8m s)(2.0 "10

7m

#1) = 5.996 "10

15Hz . The B

field thus is

!B = ! y(5.0 "10

!7T )cos[(2.00 "10

7m

!1)x !#t] , and the Poynting

vector (

!S =

1

µ0

!E !!B) is

!S = x(59.72N !m

"1s"1)cos

2[(2.00 #10

7m

"1)x "$t] .

Problem 2: DC Circuit. In the circuit below, E1=50 V, E2=30 V, R1= R2=2 Ω, and R3= R4=4 Ω. a) Determine the current flowing through each resistor. Solution: reduce the circuit as follows: Here R34p=2 Ω. Kirchoff’s rules give

!1" I

0R1" (I

0+ I

1)R

34 p= 0

!2" (I

0+ I

1)R

34 p" I

1R2= 0

that can be solved to yield I0=35

3A, I

1=5

3A . Note that the current

through R3 and R4 is I0+ I

1

2=20

3A .

b) Determine the power dissipated in R2.

The power is I1

2R2=50

9W .

c) What is the potential difference Vb-Va between the points a and b indicated in the figure?

This is given by Vb !Va = "2+ I

0R1= "

1+ I

1R2= I

1R2+ (I

0+ I

1)R

34 p + I0R1 =160

3V .

E1 E2

R1

R3 R4

R2 a

b

R1

R2

R34p E2 E1

I0 I1

Problem 3: Static Electric and Magnetic Fields. 1) A long non-conducting cylindrical pipe of radius a has total charge per unit length λ uniformly distributed within its volume. The pipe is surrounded by a conducting cylindrical shell of inner radius b and outer radius c. The conducting shell has total charge per unit length λ’ distributed on its surface. Determine the electric field (magnitude and direction) in all regions of space delimited by the surfaces with r=a, b, and c. If λ’ =- λ/2, plot the E field as a function of the radius r. Head-on: The field is radially outward. For r<a, we have (defining the volume

charge density ρ) E(r)2!rl ="!r2l

#0

$ E(r) ="r

2#0

=%r

2!a2#0

. For a<r<b,

E(r)2!rl ="l

#0

$ E(r) ="

2!r#0

. For b<r<c, E=0. For r>c,

E(r)2!rl =(" + #" )l

$0

="l

2$0

% E(r) ="

4!r$0

. (In the plot below, a=1, b=2,

c=2.5, and λ=1.)

b

a c

−λ λ’+λ=λ/2

ρ

E(r)

r

2) Three long current-carrying wires are assembled as shown. Let I=1.5 A, a=30 cm, and b=40 cm. a) Compute the magnetic field (magnitude and direction) at P. Solution: label the wires as in the figure above. Then we have

!B1=

µ03I

2! (2b)y ,

!B2=

µ0I

2! (a2+ b

2)(ax " by) , and

!B3=

µ0I

2! (a2+ b

2)(ax + by) , such

that the total B field is

!B =!B1+!B2+!B3=µ0I

2!2a

(a2+ b

2)x +

3

2by

"#$

%&'. Putting in

numbers,

!B = (7.2 !10

"7T )x + (1.125 !10

"6T )y .

b) Suppose an additional wire with current I’=2I flowing out of the page (in the positive z direction) is placed at P. What is the force per unit length (magnitude and direction) on this wire? The force per unit length on the wire is

!F

l= 2Iz !

!Btot =

µ02I

2

2"2a

a2+ b

2y #

3

2bx

$%&

'(). Putting in values,

!F

l= (2.16 !10

"6N m)y + (3.375 !10

"6N m)x .

c) How do the answers to a) and b) change if the current direction in the wire with current 3I is reversed?

B changes as

!B =

µ0I

2!2a

(a2+ b

2)x "

3

2by

#$%

&'(, and F changes as

!F

l=µ02I

2

2!2a

a2+ b

2y +

3

2bx

"#$

%&'.

x

x

a

a

b

I

3I

I

b P

x

y

(1)

(2)

(3)

Problem 4: Magnetic Induction. 1) A long solenoid with n=250 turns per centimeter and radius a=10 cm carries a current I(t) as shown. The solenoid is surrounded by an N=10 turn circular coil of radius b=50 cm and resistance R=10 Ω. If I(t) = I

0e!4 t , where t is measured in seconds and I0=1 A, what is the

induced current I’ in the coil? (Be sure to indicate the direction of I’.) Solution: The B field of the solenoid is B = µ

0nI . The flux through the

coil is ! = Nµ0nI"a

2 , so the induced current as a function of time is

!I ="

R= #

d$ dt

R=4Nµ

0n%a

2I0e#4 t

R. Numerically, !I (t) = (3.95 "10

#3A)e

#4 t .

x

I

a

b

2) A rectangular loop with resistance R=5 Ω moves into a region of uniform magnetic field B=2.1 T (directed into the page) with velocity v=1.5 cm/s as shown. a) Taking x=0 and t=0 when the left edge of the loop enters the field, plot the flux, the induced current, and the external force required to keep the loop moving at a constant speed as a function of x. b) What is the direction of the current when the loop enters and exits the field? Solution: counterclockwise when it enters, clockwise when it exits c) What is the induced current (magnitude and direction) at t=0.1s? The induced current is I = Blv R = !3.15 "10

!4A .

x

x

x

x

x

x

x

x

x

x

x

x

x

x

3L

L=10 cm

L/2

x

Φ

x

Iind

x

x

Fext

v

Problem 5: Modern Physics. 1) A particle of mass m is in the first excited state of a one-dimensional harmonic oscillator with frequency ω. If the wave function

of the particle is

! (x) =1

2

m"#!

$%&

'()1/4

xe*m" x2 2! , determine the expectation

(average) value of x for this state.

The expectation value is

< x >= ! *(x)x! (x)

"#

#

$ dx =1

2

m%&!

x3e"m% x2 !

dx = 0

"#

#

$ .

2) The electron in a hydrogen atom is in the third excited energy level. a) What are the quantum numbers associated with this energy level? What is the number of states at this energy? The third excited state is n=4. Therefore, l=3,2,1,0 and m=-l,…l. The number of states (including spin) is 2n2=32.

b) If the electron undergoes a transition to the first excited state, what is the wavelength of the photon emitted in the transition? The transition is from n=4 to n=2. The wavelength of the photon is

given by the relation E4! E

2= !13.6eV

1

42!1

22

"#$

%&'=hc

(., such that

E4! E

2= " =

hc

13.6eV !1

42+1

22

#$%

&'(

=1240eV )nm13.6eV (3 /16)

= 486nm .

c) If the resulting wave function of the electron is given by

! (r) =1

4 2"a0

3 22 # r a

0( )e# r 2a0 , what is the probability of locating the

electron between r=a0 and r=1.05 a0 (a0 is the Bohr radius)? Since the range is small, we can approximate the probability by P(r) = 4!r2 |" (r) |2 #r . More precisely, we have

P(r = a0 ) = 4!a02|" (r = a0 ) |

20.05a0 =

4!a020.05a0

(16)(2! )a03e#1= 2.29 $10#3

.