PHYS 241
Ch. 25 Homework Solutions
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.
25.5.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
(
29
6
2
2
2
3
4.85 A
1.469
10
A/m
1.025
10
m
I
I
J
A
r
π
π

=
=
=
=
×
×
(
29
(
29
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

=
=
=
×
=
×
(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.
(
29
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.
25.13.
I
DENTIFY
:
E
J
ρ
=
, where
/
.
J
I
A
=
S
ET
U
P
:
For tungsten
8
5.25
10
m
ρ

=
×
Ω⋅
and for aluminum
8
2.75 10
m.
ρ

=
×
Ω⋅
E
XECUTE
:
(a)
tungsten:
8
3
3
2
(5.25 10
m)(0.820 A)
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 Fall '08
 Milsom
 Charge, Magnetism, Work, Volt, Potential difference, Electrical resistance, internal resistance

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