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PH202-NG Final Comprehensive Exam (August 4, 2009, 4:15PM-6:45PM)
[a]
•
You may not open the textbook nor notebook.
•
A letter size information may be used.
•
A calculator may be used. However, mathematics or physics formula programmed in a calculator may not be used.
•
Write down, reasoning, calculation and answer in the blank space after each problem. Use the backside of the sheet if necessary.
•
Answers without reasonable explanation or convincing mathematical derivation will receive no point even if your answers coincide with the correct ones.
Your reasoning must begin with basic physical principles. Substantial partial credit may be given to correct reasoning and mathematical procedures when your final answers are wrong. However, if your final answers are too obviously wrong, a partial credit may not be given.
•
Do not forget units. The lack of units may result in losing a few points.
Important Physical Constants:
• electron’s electric charge: e =
−
1 .
60
×
10 − 19 C
• electron’s mass: m e
= 9 .
11
×
10 − 31 kg
• proton’s mass: m p
= 1 .
67
×
10 − 27 kg
• permittivity of free space: ǫ
0 k = 8 .
99
×
10 9 N
· m 2 /C 2
= 1
4 πk
= 8 .
85
×
10 − 12 C 2 / ( N
· m 2 ) where
• permeability of free space: µ
0
= 4 π
×
10 − 7 T
· m/A
• acceleration due to gravity at the surface of the earth: g = 9 .
80 m/s 2
• speed of light in a vacuum : c = 2 .
99792458
×
10 8 m/s
•
Planck’s constant : h = 6 .
62606876
×
10 − 34 J
· s
1 2 3 4 5 6 7
Total
1

1.1
[5 opts] Consider a helium ion He + which has only one electron like a hydrogen atom. The electron in the n -th orbit has an energy of
E n
=
−
(2 .
18
×
10 − 18 )
Z 2 n 2
[ J ] where Z = 2 is the atomic number of helium. When the electron jumps from n = 3 to n = 1, it emits a photon. Find the frequency of the photon.
f =
E
3 −
E
1 h
=
−
(2 .
18
×
10 − 18 )(2 2 ) ¡ 1
3 2
6 .
626
×
10 − 34
−
1
1 2
¢
= 1 .
17
×
10
16
Hz
1.2
[5 pts.] An electron is moving with a speed of 2 .
4
×
10 5 of the electron.
m/s . Find the de Broglie wavelength p = h
λ
→
λ = h p
= h mv
=
6 .
626
×
10 − 34
(9 .
11
×
10 − 31 )(2 .
4
×
10 5 )
= 3 .
0
×
10 − 9 m or 3 .
0 nm
2

2.1
[5 pts.] Polonium 216 Po undergoes α decay to produce a daughter nucleus that itself
84 undergoes β decay. Which one of the following nuclei is the one that ultimately results? (Justify your answer.)
(a) 211
82
Pb, (b) 211
81
Tl, (c) 212
81
Tl, (d) 212
83
Bi , (e) 213
82
Pb
An α particle carries two protons and two neutrons, and a β particle have no proton nor neutron.
Therefore, the mass number is reduced by 4.
216
→
212.
An α carries a charge of +2 e and a β particle a charge of
− e . Therefore, in total + e is removed from the original nucleus.
84
→
83.
Hence, the final nucleus is 212
83
Bi .
2.2
[5 pts.] Two protons are moving directly toward one another. When they are very far apart, their initial speeds are 1 .
50
×
10 6 m/s . What is the distance of closest approach?
Initial Energy: E i
=
1
2 mv 2 +
1
2 mv 2 = mv 2 .
Final Energy: E f
= ke 2
R
.
Since energy must conserve,
E f
= E i → ke 2
R
= mv 2
→
R = ke 2 mv 2
=
(8 .
99
×
10 9 )(1 .
60
×
10 − 19 ) 2
(1 .
67
×
10 − 27 )(1 .
50
×
10 6 ) 2
= 6 .
12
×
10 − 14 m
3

3.1
[5 pts.] A charge particle enters a uniform magnetic field and follows the circular path shown in the drawing.
(a) Is the particle positively or negatively charged?
(b) The particle’s speed is 150 m/s , the magnitude of the magnetic field is 0 .
42 T , and the radius of the path is 960 m . Determine the mass of the particle, given that its charge has a magnitude of 8 .
2
×
10 − 4 C .
(a) Using RHR-1 (see Figure), the particle is negatively charged.
(b) Centripetal force : F c
= mv 2
R
Magnetic force: F
B
= qvB
Hence,
F c
= F
B
→ mv 2
R
= qvB
→ m = qvBR v 2
= qBR v
=
(8 .
2
×
10 − 4 )(0 .
42)(960)
150
= 2 .
2
×
10 − 3 kg
3.2
[5 pts.] A spotlight on a boat is y = 2 .
4 m above the water, and the light strikes the water at a point that is x =
6 .
6 m horizontally displaced from the spotlight (see the drawing). The depth of the water is 4 .
0 m and the index of refraction for water is 1.333. Determine the distance d , which locates the point where the light strikes the bottom.
Using the angles θ
1 and θ
2 defined in Figure, tan θ
1
= x y
→
θ
1
= tan − 1
µ x ¶ y
= tan − 1
µ 6 .
6 ¶
2 .
4
= 70 .
0 ◦
Applying the Snell’s law, n
1 sin θ
1
= n
2 sin θ
2
→ sin θ
2
= n
1 n
2 sin θ
1
=
1
1 .
333 sin 70 .
0 ◦ = 0 .
705
→
θ
2
= sin − 1 (0 .
705) = 44 .
8 ◦
Then, x ′ shown in Figure is x ′ = 4 .
0 tan 44 .
8 ◦ = 4 .
0.
Hence, d = x + x ′ = 6 .
6 + 4 .
0 = 10 .
6 m .
4

4
V
2
[10 pts.] An electric circuit with three batteries (
= 20 .
0 V , V
3
= 10 .
0 V ) and three resistors ( R
1
V
= 5
1
.
= 15 .
0 V ,
0 Ω, R
2
=
10 .
0 Ω, R
3
= 10 .
0 Ω) as shown in Figure.
(a) Assuming that electric currents ( I
1
, I
2
, and I
3
) flow in the directions specified in Figure, write down three equations that determine the currents using the Kirchhoff rules.
(b) Find the magnitude and direction (Up or Down) of the actual current in R
2
.
(a) Kirchhoff’s junction rule
I
3
= I
1
+ I
2
Kirchhoff’s loop rule for the left loop:
V
1
+ I
1
R
1
= V
2
+ I
2
R
2
→
5 .
0 I
1
−
10 .
0 I
2
= 5 .
0
→
I
1
−
2 .
0 I
2
= 1 .
0
Kirchhoff’s loop rule for the right loop:
V
2
+ I
2
R
2
+ I
3
R
3
= V
3 →
10 .
0 I
2
+ 10 .
0 I
3
=
−
10 .
0
→
I
2
+ I
3
=
−
1 .
0
(b) Substituting Eq. (1) to Eq. (3),
I
2
+ ( I
1
+ I
2
) =
−
1 .
0
→
I
1
+ 2 .
0 I
2
=
−
1 .
0
Substracting Eq. (2) from Eq. (4), we obtain
4 .
0 I
2
=
−
2 .
0
→
I
2
=
−
0 .
5 A
Since I
2 is negative, the direction is Up . The magnitude is 0 .
5 A .
(1)
(2)
(3)
(4)
5

5 [20 pts.] A parallel plate capacitor with a separation d =
2 .
00 mm is charged with a battery of V = 24 .
0 V as shown in Figure. At the same time, a uniform magnetic field is also applied to the area between two plates. When a positive charge is impinged to the capacitor in the + x direction with a speed of v = 42 .
0 m/s , it remains in the same straight line. (Assume that there is no gravity.)
(a) Find the direction of the electric field. (+ x ,
− x , + y ,
− y , Out of paper, or Into paper)
(b) Find the magnitude of the electric field.
(c) Find the direction of the magnetic field. (+ x ,
− x , + y ,
− y , Out of paper, or Into paper)
(d) Find the magnitude of the magnetic field.
(a) The upper plate is charged positively and the lower plate nagatively. Therefore, the electric field lines begin at the upper plate and end at the lower plate.
− y
(b)
E =
V d
=
24 .
0
2 .
00
×
10 − 3
= 1 .
2
×
10 4 N/C .
(c) In order to move in the straight line, the net force in the y direction must be zero. Since the particle has positive charge, the direction of force due to the electric field is the same as the direction of the electric field,
− y . Therefore, the firection of magnetic force must be + y . Using the RHR-1, the direction of magnetic field is Into paper .
(d) The magnitude of the electric force is F
E
F
B
= qvB . Since the net force is zero,
= qE and the magnitude of the magnetic force is
F
E
= F
B → qE = qvB
→
B =
E v
=
1 .
2
×
10 4
42 .
0
= 2 .
86
×
10
2
T .
6

6.
[20 pts.] A candle is placed 30 cm from a lens. An upright image is formed as shown in Figure.
The size of the image is one-third of the object.
(a) Is this lens a diverging or converging lens? Justify your answer with a few sentences.
(b) Find the location of focal points by drawing rays. Leave all lines used to find the focal points in the figure so that the instructor knows how you find them. [Note: There are two focal points.]
(c) Calculate the focal length of the lens.
(d) In order to increase the size of the image in which direction the object should be moved, toward the lens or away from the lens? Briefly justify your answer.
(a) The image is virtual and the magnification is smaller than one. Therefore, it is a diverging lens .
(b) The following figure shows two rays that determine the two focal points.
(c) The object distance is d o
= 30 distance from the magnification.
cm and the magnification is m = +1 / 3. We can find the image d i
=
− md o
=
−
1
3
×
30 =
−
10 cm
Using the thin lens equation,
1 f
=
1 d o
+
1 d i
=
1
30
+
1
−
10
2
=
−
30
1
=
−
15
Hence, f =
−
15 cm .
(d) As the object moves toward the lens, ray 2 in the figure moves upward. On the other hand, ray 1 does not change. Hence, the image size increases. Toward the lens
7

7 [20 pts.] Figure shows a copper wire (negligible resistance) bent into a circular shape with a radius of r = 0 .
42 m . The radial section BC has resistance R = 3 .
5 Ω and fixed in place, while the copper bar AC (negligible resistance) sweeps around at an angular speed of ω = 14 rad/s . The bar makes electrical contact with the wire at all times. A uniform magnetic field exists everywhere, is perpendicular to the plane of the circle, and has a magnitude of B = 5 .
5
×
10 − 3 T .
(a) Find change in the magnetic flux during t = 0 .
050 s .
(b) Find the magnitude of the current induced in the loop.
(c) Is the direction of the induced current in the copper bar “From C to A” or“From A to C”? (Do not forget a brief justification.)
(d) How much energy is dissipated in the resistor during
0 .
050 s ?
(a) The bar rotates θ = ωt = 14
×
0 .
050 = 0 .
70 rad . Then. the are of the loop ABC increases by
∆ A = r 2 θ
2
=
(0 .
42 m ) 2
×
(0 .
70 rad )
2
= 0 .
0617 m
2
Hence, the chage in the magnetic flux is
∆Φ = B ∆ A = (5 .
5
×
10 − 3
T )
×
(0 .
0617 m
2
) = 3 .
40
×
10 − 4
W b
(b) Emf induced by the motion of the bar is emf =
∆Φ t
=
3 .
40
×
10 − 4
0 .
050 s
W b
= 6 .
79
×
10 − 3 V
Hence. the current is
I = emf
R
=
6 .
79
×
10 − 3
3 .
5 Ω
V
= 1 .
94
×
10 − 3
A = 1 .
94 mA
(c) Since the area of the loop ABC is increasing, the induced current generates the magnetic field insdie the loop in the direction opposite to the original magnetic field. Using the right hand rule, the induced current flows from C to A .
(d)
P = IV = (1 .
94
×
10 − 3 A )(6 .
79
×
10 − 3 V ) = 1 .
32
×
10 − 5 W
W = P t = (1 .
32
×
10 − 5 )(0 .
050 s ) = 6 .
59
×
10 − 7 J
8