ECE329
Fall 2008
Homework 3
(Solution)
September 15, 2008
1. In this problem, we will study Faraday’s law,
˛
C
~
E
·
d
~
l
=

d
dt
ˆ
S
~
B
·
d
~
S,
which states that the electromotive force
E
=
¸
C
~
E
·
d
~
l
around any closed loop
C
equals the time rate
of change of the magnetic flux
Ψ
B
=
´
S
~
B
·
d
~
S
through the surface
S
bounded by the loop.
(a) If
~
B
= 0
at all times, the magnetic flux is also zero (
Ψ
B
= 0
), and therefore, according to Faraday’s
law, the electromotive force is zero (
E
= 0
) over any closed loop
C.
(b) If
~
B
6
=
0 then it is possible for
E 6
= 0
if the path
C
is disturbed or being displaced in such a way
that the magnetic flux
Ψ
B
=
´
S
~
B
·
d
~
S
varies in time.
(c) Let us define a closed loop
C
passing through the fixed points
P
1
and
P
2
(see the next figure).
P
1
P
2
dl
dl
path
A
path
B
C
Since
~
B
is timeindependent, the corresponding magnetic flux
Ψ
B
is also time independent, and
therefore,
˛
C
~
E
·
d
~
l
=

d
Ψ
B
dt
= 0
.
Breaking the closed path integral into two parts, we have
˛
C
~
E
·
d
~
l
=
ˆ
P
1
→
P
2
path
A
~
E
·
d
~
l
+
ˆ
P
2
→
P
1
path
B
~
E
·
d
~
l
= 0
.
Reversing the direction of integration of the second integral, it can be shown that
ˆ
P
1
→
P
2
path
A
~
E
·
d
~
l
=
ˆ
P
1
→
P
2
path
B
~
E
·
d
~
l.
In consequence, the line integral
´
12
~
E
·
d
~
l
does not depend on the path taken between
P
1
and
P
2
.
1
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ECE329
Fall 2008
2. Given
~
B
=
B
0
(
t
cos(
ωt
) ˆ
x
+ sin(
ωt
) ˆ
z
)
Wb
/
m
2
, we can apply Faraday’s law to compute the emf
E
around the following closed paths. Since the surfaces bounded by the closed paths are not varying in
time, and also since
~
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 Spring '08
 FRANKE
 Electromotive Force, Magnetic Field

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