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164
TIMEVARYING FIELDS AND MAXWELL'S EQUATIONS
(a) At
t
=
0, the induced current is
1=0.
(b) At
t
=
10 ms, the induced current is
I
~

7 .83 mAo (Note that at
t
=
10 ms, since
dBz(t)
>
0
dt
Based on Lenz's law, the induced current
I
through the IO.Q resistor must flow in the clockwise
direction.
(c) At
t
=
100
IDS,
the induced current is
I
=
O.
(d) At t
=
1 s, the induced current is
I
=
O.
73. Two circular coils. (a) Using the result of Example 66, the magnetic flux density produced on
the axis of coil 1 with radius
al =
5 cm, current
II
=
l00cos(20007rt) A, and number of turns
NI
=
10, at a distance of
z
=
1 m away from its center can
be
calculated as
p.oN
I
l
l
aI
2
BIz
2( 2 2)3/2
al
+z
2
7
(411"
x 1O )(IO)[100cos(20007rt)](0.05)
~
1.56cos(20001l"t) fLT

'11')

2[(0.05)2
+
12]
The voltage induced across the terminals of coil 2
(a2
=
5 cm and
N2
=
100) due to the time
varying flux '1'12 produced by coil 1 linking coil 2 can
be
evaluated as
d'l'12
2
dBb
'Vind2
= 
dt
=
N2(1I"
a
2
);u;
~
(100)[11"(0.05)2][(1.56 x 10
6
)(200011") sin(20007rt)]
~7.72
x 10
3
sin(20001l"t) V
Therefore, the amplitude of the induced voltage'Vind2 is "'7.72 mY.
(b) Repeating the same calculations for the same current
II
oscillating at 10 kHz [i.e.,
II(t) =
l00cos(200007rt) A] results in an induced voltage'Vind2 of amplitude 1'V77.2 mY.
74. Two concentric coils. (a) From Example 74, we have
p.oNII
5
B
ctr
=
2
=
22.5
x
10
IT
2a
Substituting
NI =
30, we find the radius of the larger coil as
a
~
0.754 m.
(b) From Example 74, since
a
~
75.4
cm» 1 cm=b, the total flux linking the smaller coil can
be
approximated as
,
'1'12
=
N2
p.o~tl
(1I"b
2
)
=
(75)(2.5
X
10
5
1)[11"(10
2
)2]
~
5.89
X
10
7
1
(c) For
I(t)
= 10 cos(l
2011"t)
A, the induced voltage across the terminals of the smaller loop can
be
found as
'Vind
= 
d'l'12
=
(5.89 x 10
7
)[(10)(12011") sin(l201l"t)] V
~
2.22 sin(1201l"t) mV
dt
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TIMEVARYING FIELDS AND MAXWELL'S EQUATIONS
opposing the current driven by the source, consistent with Lenz's law. This process continues
until the velocity of the bar reaches a value
Vfinal
such that the induced emf is equal to the source
voltage making the net current in the loop zero. Beyond that point, there is no current, hence no
force, hence no acceleration, and the bar simply continues to move at velocity
Vfinal.
The terminal
velocity vo can be found by simply equating the motional emf to vo. We have
Vo
'Vemf
=
f
(Vfinal' X
Bo) . dI
=
vfinalBOl
=
Vo
~
Vfinal
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This note was uploaded on 11/16/2010 for the course ECE 305 taught by Professor Staff during the Summer '08 term at Michigan State University.
 Summer '08
 STAFF

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