ECE321 Spring 2008
Exam 3 Solution Outline
Problem 1
N
as
33
−
6
−
3
−
363 3
−
6
−
3
−
36
()
T
:=
N
slts
12
:=
P4
:=
N
slts
P
3
=
W
as
1
1
2
−
+
6
−
+
[]
:=
j2
1
2
..
:=
k1
1
1
..
:=
W
as
j
W
as
j1
−
N
as
−
−
:=
W
as
T
123456789
1
0
1
1
1
2
1
3
6
3
3
6
3
3
6
3
3
6
3
=
Problem 2
w
as
100 cos 4
φ
s
⋅
⋅
=
w
bs
250 sin 4
φ
s
⋅
⋅
=
i
as
5 cos
ω
e
t
⋅
π
8
+
⎛
⎝
⎞
⎠
⋅
=
B
1.2 cos
ω
e
t
⋅
π
8
+
4
φ
s
−
⎛
⎝
⎞
⎠
⋅
=
μ
0
4
π
⋅
10
7
−
⋅
:=
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View Full DocumentExpanding B we have
B
1.2 cos
ω
e
t
⋅
π
8
+
⎛
⎝
⎞
⎠
⋅
cos 4
φ
s
⋅
()
⋅
1.2 sin
ω
e
t
⋅
π
8
+
⎛
⎝
⎞
⎠
⋅
sin 4
φ
s
⋅
⋅
+
=
Also
B
F
g
μ
0
⋅
=
F
g
μ
0
B
=
Finally
Fw
as
i
as
⋅
w
bs
i
bs
⋅
+
=
so
500 cos
ω
e
t
⋅
π
8
+
⎛
⎝
⎞
⎠
⋅
cos 4
φ
s
⋅
⋅
i
bs
250
⋅
sin 4
φ
s
⋅
⋅
+
g
μ
0
1.2 cos
ω
e
t
⋅
π
8
+
⎛
⎝
⎞
⎠
⋅
cos 4
φ
s
⋅
⋅
1.2 sin
ω
e
t
⋅
+
⎛
⎝
⋅
+
⎛
⎝
⋅
=
which is satisfied if
g
500
μ
0
⋅
1.2
:=
g
5.236
10
4
−
×
=
i
bs
2 sin
ω
e
t
⋅
π
8
+
⎛
⎝
⎞
⎠
⋅
=
Problem 3
w
as
20
−
sin 3
φ
sm
⋅
⋅
=
B
cos 3
φ
sm
⋅
100 t
⋅
+
=
r
0.1
:=
l
0.1
:=
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 Spring '08
 Staff

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