Thevalingam, Donald – Homework 28 – Due: May 31 2007, midnight – Inst: Eslami
1
This printout should have 34 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
The current in a wire decreases with time
according to the relationship
I
= (2
.
61 mA)
×
e

at
where
a
= 0
.
13328 s

1
.
Determine the total charge that passes
through the wire From
t
= 0 to the time the
current has diminished to zero.
Correct answer: 0
.
0195828 C.
Explanation:
I
=
dq
dt
q
=
Z
t
t
=0
I dt
=
Z
∞
t
=0
(0
.
00261 A)
e

0
.
13328 s

1
t
dt
= (0
.
00261 A )
e

0
.
13328 s

1
t

0
.
13328 s

1
f
f
f
f
f
∞
0
=
0
.
0195828 C
.
keywords:
002
(part 1 oF 1) 10 points
A copper wire carries 1 amp oF electric cur
rent.
What kind oF charge does the electron ±ow
create in the wire?
1.
No charge.
correct
2.
Negative
3.
Positive
Explanation:
The electrons are balanced out by an equal
number oF protons whether or not they are
moving.
keywords:
003
(part 1 oF 1) 10 points
In a ±uorescent tube oF diameter 4
.
4 cm, 1
.
1
×
10
18
1
/
s electrons and 1
×
10
17
1
/
s positive
ions (with a charge oF +
e
) ±ow through a
crosssectional each second.
What is the current in the tube?
Correct answer: 0
.
192 A.
Explanation:
Let :
n
electron
= 1
.
1
×
10
18
1
/
s
,
n
ion
= 1
×
10
17
1
/
s
,
q
electron
= 1
.
6
×
10

19
C
,
and
q
ion
= 1
.
6
×
10

19
C
.
Note that, while the positive and negative
charges ±ow in opposite directions, the total
current is their sum:
I
=
I
electron
+
I
ion
=
n
electron
q
electron
+
n
ion
q
ion
= (1
.
1
×
10
18
1
/
s) (1
.
6
×
10

19
C)
+ (1
×
10
17
1
/
s) (1
.
6
×
10

19
C)
= 0
.
176 A + 0
.
016 A
=
0
.
192 A
.
keywords:
004
(part 1 oF 5) 10 points
A current oF 25 A exists in a copper (Cu) wire
which has a diameter oF 1 mm.
Each atom oF copper has one conduction
band, and the average thermal speed
r
k T
m
oF an electron is 100000 m
/
s . The mass den
sity oF Cu is 8
.
92 g
/
cm
3
, its molar mass
is 63
.
5 g
/
mol, and Avogadro’s number is
6
.
02214
×
10
23
atoms
/
mole. The electron mass
is 9
.
10939
×
10

31
kg, and the resistivity oF
copper is 1
.
7
×
10

8
Ω
·
m.
What is the current density?
Correct answer: 3
.
1831
×
10
7
A
/
m
2
.
Explanation:
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View Full DocumentThevalingam, Donald – Homework 28 – Due: May 31 2007, midnight – Inst: Eslami
2
Let :
I
= 25 A
and
r
= 0
.
5 mm = 0
.
0005 m
.
The current density is given by
J
≡
I
A
=
I
π r
2
=
(25 A)
π
(0
.
0005 m)
2
=
3
.
1831
×
10
7
A
/
m
2
.
005
(part 2 of 5) 10 points
What is the density of conduction electrons
band of copper?
Correct answer: 8
.
45944
×
10
28
m

3
.
Explanation:
Let :
N
A
= 6
.
02214
×
10
23
atoms
/
mole
,
ρ
m
= 8
.
92 g
/
cm
3
= 8920 kg
/
m
3
,
and
M
= 63
.
5 g
/
mol = 0
.
0635 kg
/
mol
.
The density of the valence electrons is
n
e
=
(# of valence electrons)
volume
=
(# of kg)
m
3
(# of moles)
kg
(# of atoms)
mol
·
(1 conduction electron)
atom
=
N
A
ρ
m
M
=
6
.
02214
×
10
23
atoms
/
mole
0
.
0635 kg
/
mol
×
8920 kg
/
m
3
=
8
.
45944
×
10
28
m

3
.
006
(part 3 of 5) 10 points
What is the drift velocity of the electrons?
Correct answer: 0
.
00234854 m
/
s.
Explanation:
Let :
q
e
= 1
.
60218
×
10

19
C
.
Current density is
J
≡
I
A
=
n
e
q
e
v
d
v
d
=
J
n
e
q
e
=
3
.
1831
×
10
7
A
/
m
2
8
.
45944
×
10
28
m

3
×
1
1
.
60218
×
10

19
C
=
0
.
00234854 m
/
s
.
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 Spring '08
 Turner
 Physics, Electron, Work, Correct Answer, midnight – Inst, Thevalingam

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