2814
Chapter 28
28.43.
IDENTIFY
and
SET UP:
Use the appropriate expression for the magnetic field produced by each current
configuration.
EXECUTE:
(a)
0
2
I
B
r
μ
π
=
so
2
6
7
0
22
(
2
.
0
0
1
0
m
)
(
3
7
.
2
T
)
3.72 10 A
41
0
T
m
/
A
B
I
ππ
μπ
−
−
×
==
=
×
×⋅
.
(b)
0
2
N
I
B
R
=
so
5
7
0
2
2(0.210 m)(37.2 T)
1.24 10 A
(100)(4
10 T m/A)
RB
I
N
−
=
×
.
(c)
0
N
B
I
L
=
so
7
0
(37.2 T)(0.320 m)
237 A
(4
10 T m/A)(40,000)
BL
I
N
−
=
.
EVALUATE:
Much less current is needed for the solenoid, because of its large number of turns per unit length.
28.44.
IDENTIFY:
Example 28.10 shows that outside a toroidal solenoid there is no magnetic field and inside it the
magnetic field is given by
0
.
2
μ
NI
B
π
r
=
SET UP:
The torus extends from
1
15.0 cm
r
=
to
2
18.0 cm.
r
=
EXECUTE:
(a)
r
= 0.12 m, which is outside the torus, so
B
= 0.
(b)
r
= 0.16 m, so
3
00
(250)(8.50 A)
2.66 10 T.
2
2 (0.160 m)
NI
B
r
μμ
−
=
×
(c)
r
= 0.20 m, which is outside the torus, so
B
= 0.
EVALUATE:
The magnetic field inside the torus is proportional to 1/
r
, so it varies somewhat over the cross
section of the torus.
28.45.
IDENTIFY:
Example 28.10 shows that inside a toroidal solenoid,
0
2
NI
B
r
=
.
SET UP:
0.070 m
r
=
EXECUTE:
3
(600)(0.650 A)
1.11 10 T.
2
2 (0.070 m)
NI
B
r
−
=
×
EVALUATE:
If the radial thickness of the torus is small compared to its mean diameter,
B
is approximately
uniform inside its windings.
28.46.
IDENTIFY:
Use Eq.(28.24), with
0
replaced by
m0
K
=
, with
m
80.
K
=
SET UP:
The contribution from atomic currents is the difference between
B
calculated with
and
B
calculated
with
0
.
EXECUTE:
(a)
0
(80)(400)(0.25 A)
0.0267 T.
2
(
0
.
0
6
0
m
)
NI
K
NI
B
rr
=
=
(b)
The amount due to atomic currents is
79
79
(0.0267 T)
0.0263 T.
80
80
BB
′
=
EVALUATE:
The presence of the core greatly enhances the magnetic field produced by the solenoid.
28.47.
IDENTIFY
and
SET UP:
2
K
NI
B
r
=
(Eq.28.24, with
0
replaced by
K
)
EXECUTE:
(a)
m
1400
K
=
2
7
0m
2
(2.90 10 m)(0.350 T)
0.0725 A
(2 10 T m/A)(1400)(500)
rB
I
KN
−
−
×
=
(b)
m
5200
K
=
2
7
2
0.0195 A
(2 10 T m/A)(5200)(500)
rB
I
−
−
×
=
EVALUATE:
If the solenoid were airfilled instead, a much larger current would be required to produce the same
magnetic field.
28.48.
IDENTIFY:
Apply
2
K
μ
NI
B
π
r
=
.
SET UP:
m
K
is the relative permeability and
mm
1
K
χ
=
−
is the magnetic susceptibility.
EXECUTE:
(a)
m
2
2 (0.2500 m)(1.940 T)
2021.
(500)(2.400 A)
π
rB
π
K
μ
NI
μ
=
(b)
1
2020.
K
=−
=
EVALUATE:
Without the magnetic material the magnetic field inside the windings would be
4
/2021
9.6 10 T.
B
−
=×
The presence of the magnetic material greatly enhances the magnetic field inside the windings.
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View Full DocumentSources of Magnetic Field
2815
28.49.
IDENTIFY:
The magnetic field from the solenoid alone is
00
.
B
μ
nI
=
The total magnetic field is
m0
.
B KB
=
M
is
given by Eq.(28.29).
SET UP:
6000 turns/m
n
=
EXECUTE:
(a)
(i)
13
0
(6000 m ) (0.15 A)
1.13 10
T.
B
μ
nI
μ
−−
==
=
×
(ii)
36
m
0
15
1
9
9
(1.13 10
T)
4.68 10 A m.
K
MB
μμ
−
−
×
=
×
(iii)
3
(5200)(1.13 10
T)
5.88 T.
BKB
−
×
=
(b)
The directions of
B
!
,
0
B
!
and
M
!
are shown in Figure 28.49. Silicon steel is paramagnetic and
0
B
!
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 Spring '06
 Buchler
 Physics, Current, Magnetic Field

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