CHAPTER 15
Solutions for Exercises
E15.1
If one grasps the wire with the right hand and with the thumb pointing
north, the fingers point west under the wire and curl around to point east
above the wire.
E15.2
If one places the fingers of the right hand on the periphery of the clock
pointing clockwise, the thumb points into the clock face.
E15.3
z
y
x
q
u
u
u
B
u
f
14
5
19
10
602
.
1
10
)
10
602
.
1
(
−
−
×
−
=
×
×
−
=
×
=
in which
u
x
,
u
y
,
and
u
z
are unit vectors along the respective axes.
E15.4
N
5
)
90
sin(
5
.
0
)
1
(
10
)
sin(
=
=
=
o
l
θ
B
i
f
E15.5
(a)
φ
mWb
927
.
3
)
05
.
0
(
5
.
0
2
2
=
=
=
=
π
π
r
B
BA
turns
mWb
39.27
=
=
φ
λ
N
(b)
V
27
.
39
10
10
27
.
39
3
3
−
=
×
−
=
=
−
−
dt
d
e
λ
More information would be needed to determine the polarity of the
voltage by use of Lenz’s law.
Thus the minus sign of the result is not
meaningful.
E15.6
T
10
4
10
2
20
10
4
2
4
2
7
−
−
−
×
=
×
×
=
=
π
π
π
μ
r
I
B
E15.7
By Ampère’s law, the integral equals the sum of the currents flowing
through the surface bounded by the path.
The reference direction for
the currents relates to the direction of integration by the righthand
rule.
Thus, for each part the integral equals the sum of the currents
flowing upward.
Referring to Figure 15.9 in the book,
we have
∫
=
⋅
1
Path
A
10
l
d
H
∫
=
−
=
⋅
2
Path
A
0
10
10
l
d
H
∫
−
=
⋅
3
Path
A
10
l
d
H
1
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E15.8
Refer to Figure 15.9 in the book.
Conceptually the lefthand wire
produces a field in the region surrounding it given by
T
10
2
10
2
10
10
4
2
5
1
7
−
−
−
×
=
×
×
=
π
π
π
μ
=
r
I
B
By the righthand rule, the direction of this field is in the direction of
Path 1. The field in turn produces a force on the righthand wire given by
N
10
2
)
10
)(
1
(
10
2
4
5
−
×
=
×
=
=
i
B
f
l
By the righthand rule, the direction of the force is such that the wires
repel one another.
E15.9
The magnetic circuit is:
The reluctance of the iron is:
4
7
2
iron
0
iron
iron
10
4
10
4
5000
10
27
−
−
−
×
×
×
×
×
=
=
π
μ
μ
A
R
r
l
3
iron
10
4
.
107
×
=
R
The reluctance of the air gap is:
4
7
2
gap
0
gap
gap
10
9
10
4
10
−
−
−
×
×
×
=
=
π
μ
A
R
l
6
gap
10
842
.
8
×
=
R
Then we have
mWb
45
.
0
10
9
5
.
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 Fall '08
 Styblinski
 Magnetic Field, Righthand rule, power loss, Magnetic Circuit, eddy currents, Riron

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