Chapter 30
5
The magnitude of the magnetic field inside the solenoid is
B
=
¹
0
ni
s
,whe
re
n
is the number
of turns per unit length and
i
s
is the current. The field is parallel to the solenoid axis, so the
flux through a cross section of the solenoid is
©
B
=
A
s
B
=
¹
0
¼r
2
s
ni
s
,whe
re
A
s
(=
¼r
2
s
) is the
crosssectional area of the solenoid. Since the magnetic field is zero outside the solenoid, this is
also the flux through the coil. The emf in the coil has magnitude
E
=
Nd
©
B
dt
=
¹
0
¼r
2
s
Nn
di
s
dt
and the current in the coil is
i
c
=
E
R
=
¹
0
¼r
2
s
Nn
R
di
s
dt
;
where
N
is the number of turns in the coil and
R
is the resistance of the coil. The current
changes linearly by 3
:
0A in 50ms, so
di
s
=dt
=(3
:
0A)
=
(50
£
10
¡
3
s) = 60 A
=
s. Thus
i
c
=
(4
¼
£
10
¡
7
T
¢
m
=
A)
¼
(0
:
016 m)
2
(120)(220
£
10
2
m
¡
1
)
5
:
3
(60 A
=
s) = 3
:
0
£
10
¡
2
A
:
21
(a) In the region of the smaller loop, the magnetic field produced by the larger loop may be taken
to be uniform and equal to its value at the center of the smaller loop, on the axis. Eq. 29–26,
with
z
=
x
andmuchg
rea
te
rthan
R
,g
ives
B
=
¹
0
iR
2
2
x
3
for the magnitude. The field is upward in the diagram. The magnetic flux through the smaller
loop is the product of this field and the area (
¼r
2
)o
fthesma
l
le
rloop
:
©
B
=
¼¹
0
ir
2
R
2
2
x
3
:
(b) The emf is given by Faraday’s law:
E
=
¡
d
©
B
dt
=
¡
μ
¼¹
0
ir
2
R
2
2
¶
d
dt
μ
1
x
3
¶
=
¡
μ
¼¹
0
ir
2
R
2
2
¶μ
¡
3
x
4
dx
dt
¶
=
3
¼¹
0
ir
2
R
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View Full Documentthe field of the larger loop. It is counterclockwise as viewed from above, in the same direction
as the current in the larger loop.
29
Thermal energy is generated at the rate
E
2
=R
,wh
e
r
e
E
is the emf in the wire and
R
is the
resistance of the wire. The resistance is given by
R
=
½L=A
,where
½
is the resistivity of copper,
L
is the length of the wire, and
A
is the crosssectional area of the wire. The resistivity can be
found in Table 26–1. Thus
R
=
½L
A
=
(1
:
69
£
10
¡
8
¢
m)(0
:
500 m)
¼
(0
:
500
£
10
¡
3
m)
2
=1
:
076
£
10
¡
2
:
Faraday’s law is used to find the emf. If
B
is the magnitude of the magnetic field through the
loop, then
E
=
AdB=dt
,whe
re
A
is the area of the loop. The radius
r
of the loop is
r
=
L=
2
¼
and its area is
¼r
2
=
¼L
2
=
4
¼
2
=
L
2
=
4
¼
.Thu
s
E
=
L
2
4
¼
dB
dt
=
(0
:
500 m)
2
4
¼
(10
:
0
£
10
¡
3
T
=
s) = 1
:
989
£
10
¡
4
V
:
The rate of thermal energy generation is
P
=
E
2
R
=
(1
:
989
£
10
¡
4
V)
2
1
:
076
£
10
¡
2
=3
:
68
£
10
¡
6
W
:
37
(a) The field point is inside the solenoid, so Eq. 30–25 applies. The magnitude of the induced
electric field is
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 Fall '98
 Heckman
 Physics, Current, Magnetic Field, Faraday

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