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
°
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document236
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
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 ASD
 Flux, Magnetic Field, ΔT, Δt Δt

Click to edit the document details