251
C
URRENT
,
R
ESISTANCE
,
AND
E
LECTROMOTIVE
F
ORCE
25.1.
I
DENTIFY
:
/
I
Q t
=
.
S
ET
U
P
:
1.0 h
3600 s
=
E
XECUTE
:
4
(3.6 A)(3.0)(3600 s)
3.89
10 C.
Q
It
=
=
=
×
E
VALUATE
:
Compared to typical charges of objects in electrostatics, this is a huge amount of charge.
25.2.
I
DENTIFY
:
/
I
Q t
=
.
Use
d
I
n q v A
=
to calculate the drift velocity
d
.
v
S
ET
U
P
:
28
3
5.8
10
m
.
n
−
=
×
19
1.60
10
C
q
−
=
×
.
E
XECUTE
:
(a)
2
420 C
8.75
10
A.
80(60 s)
Q
I
t
−
=
=
=
×
(b)
d
.
I
n q v A
=
This gives
2
6
d
28
19
3
2
8.75 10
A
1.78 10
m s.
(5.8 10
)(1.60
10
C)(
(1.3 10
m) )
I
v
nqA
π
−
−
−
−
×
=
=
=
×
×
×
×
E
VALUATE
:
d
v
is smaller than in Example 25.1, because
I
is smaller in this problem.
25.3.
I
DENTIFY
:
/
I
Q t
=
.
/
J
I
A
=
.
d
J
n q v
=
S
ET
U
P
:
2
(
/ 4)
A
D
π
=
, with
3
2.05
10
m
D
−
=
×
.
The charge of an electron has magnitude
19
1.60
10
C.
e
−
+
=
×
E
XECUTE
:
(a)
(5.00 A)(1.00 s)
5.00 C.
Q
It
=
=
=
The number of electrons is
19
3.12
10 .
Q
e
=
×
(b)
6
2
2
3
2
5.00 A
1.51
10
A/m .
(
/ 4)
(
/4)(2.05
10
m)
I
J
D
π
π
−
=
=
=
×
×
(c)
6
2
4
d
28
3
19
1.51
10
A/m
1.11
10
m/s
0.111 mm/s.
(8.5
10
m
)(1.60
10
C)
J
v
n q
−
−
−
×
=
=
=
×
=
×
×
E
VALUATE
:
(a)
If
I
is the same,
/
J
I
A
=
would decrease and
d
v
would decrease.
The number of electrons
passing through the light bulb in 1.00 s would not change.
25.4.
(a)
I
DENTIFY
:
By definition,
J = I/A
and radius is onehalf the diameter.
S
ET
U
P
:
Solve for the current:
I = JA = J
(
D
/2)
2
E
XECUTE
:
I
= (1.50
×
10
6
A/m
2
)(
π
)[(0.00102 m)/2]
2
= 1.23 A
E
VALUATE
:
This is a realistic current.
(b)
I
DENTIFY
:
The current density is
J = nqv
d
S
ET
U
P
:
Solve for the drift velocity:
v
d
= J/nq
E
XECUTE
:
Since most laboratory wire is copper, we use the value of
n
for copper, giving
6
2
d
(1.50
10
A/m )
v
=
×
/[(8.5
×
10
28
el/m
3
)(1.60
×
19
10
−
C) = 1.1
×
4
10
−
m/s = 0.11 mm/s
E
VALUATE
:
This is a typical drift velocity for ordinary currents and wires.
25.5.
I
DENTIFY
and
S
ET
U
P
:
Use Eq. (25.3) to calculate the drift speed and then use that to find the time to travel the
length of the wire.
E
XECUTE
:
(a)
Calculate the drift speed
d
:
v
(
)
6
2
2
2
3
4.85 A
1.469
10
A/m
1.025
10
m
I
I
J
A
r
π
π
−
=
=
=
=
×
×
(
)(
)
6
2
4
d
28
3
19
1.469
10
A/m
1.079
10
m/s
8.5
10
/ m
1.602
10
C
J
v
n q
−
−
×
=
=
=
×
×
×
3
4
d
0.710 m
6.58
10
s
110 min.
1.079
10
m/s
L
t
v
−
=
=
=
×
=
×
25
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
252
Chapter 25
(b)
d
2
I
v
r n q
π
=
2
d
r n q L
L
t
v
I
π
=
=
t
is proportional to
2
r
and hence to
2
d
where
2
d
r
=
is the wire diameter.
(
)
2
3
4
4.12 mm
6.58
10
s
2.66
10
s
440 min.
2.05 mm
t
⎛
⎞
=
×
=
×
=
⎜
⎟
⎝
⎠
(c) E
VALUATE
:
The drift speed is proportional to the current density and therefore it is inversely proportional to
the square of the diameter of the wire. Increasing the diameter by some factor decreases the drift speed by the
square of that factor.
25.6.
I
DENTIFY
:
The number of moles of copper atoms is the mass of
3
1.00 m
divided by the atomic mass of copper.
There are
23
A
6.023 10
N
=
×
atoms per mole.
S
ET
U
P
:
The atomic mass of copper is
63.55 g
mole, and its density is
3
8.96 g cm .
Example 25.1 says there are
28
8.5
10
×
free electrons per
3
m
.
E
XECUTE
:
The number of copper atoms in
3
1.00 m
is
3
6
3
3
23
28
3
(8.96 g cm )(1.00
10 cm
m )(6.023 10
atoms mole)
8.49
10
atoms m .
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '11
 Shaefer
 Statics, Resistance, Electromotive Force, ohm, Electrical resistance

Click to edit the document details