Chapter26
7
(a) The magnitude of the current density is given by
J
=
nqv
d
, where
n
is the number of
particles per unit volume,
q
is the charge on each particle, and
v
d
is the drift speed of the
particles.
The particle concentration is
n
= 2
:
0
£
10
8
cm
¡
3
= 2
:
0
£
10
14
m
¡
3
, the charge is
q
= 2
e
= 2(1
:
60
£
10
¡
19
C) = 3
:
20
£
10
¡
19
C, and the drift speed is 1
:
0
£
10
5
m
=
s. Thus
J
= (2
£
10
14
m
¡
3
)(3
:
2
£
10
¡
19
C)(1
:
0
£
10
5
m
=
s) = 6
:
4 A
=
m
2
:
(b) Since the particles are positively charged, the current density is in the same direction as their
motion, to the north.
(c) The current cannot be calculated unless the crosssectional area of the beam is known. Then
i
=
JA
can be used.
17
The resistance of the wire is given by
R
=
½L=A
, where
½
is the resistivity of the material,
L
is the length of the wire, and
A
is the crosssectional area of the wire. The crosssectional area
is
A
=
¼r
2
=
¼
(0
:
50
£
10
¡
3
m)
2
= 7
:
85
£
10
¡
7
m
2
. Here
r
= 0
:
50 mm = 0
:
50
£
10
¡
3
m is the
radius of the wire. Thus
½
=
RA
L
=
(50
£
10
¡
3
)(7
:
85
£
10
¡
7
m
2
)
2
:
0 m
= 2
:
0
£
10
¡
8
¢
m
:
19
The resistance of the coil is given by
R
=
½L=A
, where
L
is the length of the wire,
½
is the
resistivity of copper, and
A
is the crosssectional area of the wire. Since each turn of wire has
length 2
¼r
, where
r
is the radius of the coil,
L
= (250)2
¼r
= (250)(2
¼
)(0
:
12 m) = 188
:
5 m. If
r
w
is the radius of the wire, its crosssectional area is
A
=
¼r
2
w
=
¼
(0
:
65
£
10
¡
3
m)
2
= 1
:
33
£
10
¡
6
m
2
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '98
 Heckman
 Physics, Charge, Current

Click to edit the document details