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PROBLEM SOLUTIONS
20.1
The angle between the direction of the constant F
eld and the normal to the plane of the loop is
θ
=
0°, so
Φ
B
BA
==
( ) ×
( )
×
( )
⎡
⎣
−−
cos
.
.
0
5
0
8
01
0
1
21
0
22
T
m
m
⎤
⎦
=×
⋅=
−
cos
.
.
04
8
1
0
4
8
3
°
T
m
m
W
b
2
20.2
The magnetic ﬂ
ux through the loop is given by
Φ
B
BA
=
cos
, where
B
is the magnitude of the
magnetic F
eld,
A
is the area enclosed by the loop, and
is the angle the magnetic F
eld makes
with the normal to the plane of the loop. Thus,
Φ
B
BA
×
()
⎛
⎝
−
−
cos
.
.
500 10
200
5
c
m
10 m
1 cm
2
2
⎜
⎞
⎠
⎟
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
⋅
−
2
7
100 10
cos
.
cos
θθ
Tm
2
(a)
When
B
ur
is perpendicular to the plane of the loop,
=
0° and
Φ
B
⋅
−
7
.
T
m
2
(b)
If
×
⋅
( )
−
30 0
1 00
10
30 0
7
..
c
o
s
.
°, then
T m
2
Φ
B
°
⋅
−
866 10
8
T
m
2
(c)
If
×
⋅
( )
−
90 0
1 00
10
90 0
7
c
o
s
.
°, then
T m
2
Φ
B
°
=
0
20.3
Φ
B
BA
B
r
cos
(
)cos
θπ
2
where
is the angle between the direction of the F
eld and the
normal to the plane of the loop.
(a)
If the F
eld is perpendicular to the plane of the loop,
=
0°, and
B
r
B
=
( )
=
×⋅
( )
−
Φ
πθ
π
2
3
2
800 10
012
0
cos
.
.c
o
s
m
2
°
=
0 177
T
(b)
If the F
eld is directed parallel to the plane of the loop,
=°
90 , and
Φ
B
BA
BA
=
cos
cos
90
0
°
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Chapter 20
20.4
The magnetic ﬁ
eld lines are tangent to the surface of the cylinder, so that no magnetic ﬁ
eld lines
penetrate the cylindrical surface. The total ﬂ
ux through the cylinder is
zero .
20.5
(a)
Every ﬁ
eld line that comes up through the area
A
on one side of the wire goes back down
through area
A
on the other side of the wire. Thus, the net ﬂ
ux through the coil is
zero .
(b)
The magnetic ﬁ
eld is parallel to the plane of the coil, so
θ
=
90 0
.°. Therefore,
Φ
B
BA
BA
==
=
cos
cos
.
90 0
0
°
20.6
(a)
The magnitude of the ﬁ
eld inside the solenoid is
Bn
I
N
I
⎛
⎝
⎜
⎞
⎠
⎟
=×
⋅
( )
−
μμ
π
00
7
41
0
400
0 360
l
TmA
m
.
⎛
⎝
⎜
⎞
⎠
⎟
( )
=
−
5 00
6 98
10
6 98
3
..
.
A
T
m
T
(b)
The ﬁ
eld inside a solenoid is directed perpendicular to the crosssectional area, so
=°
0
and the ﬂ
ux through a loop of the solenoid is
Φ
B
BA
B
( )
( ) ×
−−
cos
cos
θπ
r
2
3
6 98 10
3 00
10
T
2
2
5
0
1 97 10
mT
m
2
( )
⋅
−
cos
.
°
20.7
(a)
The magnetic ﬂ
ux through an area A may be written as
Φ
B
BA
B
= (
)
=
cos
component of
perpendicular to
AA
( ) ⋅
Thus, the ﬂ
ux through the shaded side of the cube is
Φ
Bx
=⋅=
( )
⋅×
( )
⋅
50
25 10
31 10
2
2
3
.
T
m
T
m
2
(b)
Unlike electric ﬁ
eld lines, magnetic ﬁ
eld lines always form closed loops, without beginning
or end. Therefore, no magnetic ﬁ
eld lines originate or terminate within the cube and any
line entering the cube at one point must emerge from the cube at some other point. The net
ﬂ
ux through the cube, and indeed through any
closed
surface, is
zero .
20.8
ε
()
=
−
×
−
ΔΦ
Δ
Δ
Δ
B
t
t
cos
15
0
16 10
3
2
m
⎡
⎣
⎤
⎦
°
×
=
−
−
cos
0
120 10
10 10
010
3
4
s
V
m
V
20.9
With the constant ﬁ
eld directed perpendicular to the plane of the coil, the ﬂ
ux through the coil is
Φ
B
BA
BA
cos 0°
. As the enclosed area increases, the magnitude of the induced emf in the
coil is
⎛
⎝
⎜
⎞
⎠
⎟
= (
) ×
( )
=
−
ΔΦ
Δ
Δ
Δ
B
t
B
A
t
030
50 10
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 Spring '10
 ASD

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