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nguyen (jmn727) – oldhomework 24 – Turner – (59070)
1
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001
10.0 points
A toroid having a rectangular cross section
(
a
= 1
.
22 cm by
b
= 3
.
08 cm) and inner
radius 5
.
75 cm consists oF
N
= 230 turns oF
wire that carries a current
I
=
I
0
sin
ω t
, with
I
0
= 69
.
4 A and a Frequency
f
= 92
.
4 Hz. A
loop that consists oF
N
ℓ
= 13 turns oF wire
links the toroid, as in the fgure.
b
a
N
R
N
l
Determine the maximum
E
induced in the
loop by the changing current
I
.
Correct answer: 0
.
126091 V.
Explanation:
Basic Concept:
±araday’s Law
E
=

d
Φ
B
dt
.
Magnetic feld in a toroid
B
=
μ
0
N I
2
π r
.
Solution:
In a toroid, all the ²ux is confned
to the inside oF the toroid
B
=
μ
0
N I
2
π r
.
So, the ²ux through the loop oF wire is
Φ
B
1
=
i
B dA
=
μ
0
N I
0
2
π
sin(
ω t
)
i
b
+
R
R
adr
r
=
μ
0
N I
0
2
π
a
sin(
ω t
) ln
p
b
+
R
R
P
.
Applying ±araday’s law, the induced emF can
be calculated as Follows
E
=

N
ℓ
d
Φ
B
1
dt
=

N
ℓ
μ
0
N I
0
2
π
ω a
ln
p
b
+
R
R
P
cos(
ω t
)
=
E
0
cos(
ω t
)
where
ω
= 2
πf
was used.
The maximum magnitude oF the induced
emf
,
E
0
, is the coe³cient in Front oF cos(
ω t
).
E
0
=

N
ℓ
d
Φ
B
1
dt
=

N
ℓ
μ
0
N I
0
ω
2
π
a
ln
b
b
+
R
R
B
=

(13 turns)
μ
0
(230 turns)
×
(69
.
4 A) (92
.
4 Hz) (1
.
22 cm)
×
ln
b
(3
.
08 cm) + (5
.
75 cm)
(5
.
75 cm)
B
=

0
.
126091 V
E
= 0
.
126091 V
.
002
10.0 points
A long straight wire carries a current 28 A. A
rectangular loop with two sides parallel to the
straight wire has sides 5 cm and 15 cm, with
its near side a distance 3 cm From the straight
wire, as shown in the fgure.
3 cm
5 cm
15 cm
28 A
±ind the magnetic ²ux through the rectan
gular loop.
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View Full Document nguyen (jmn727) – oldhomework 24 – Turner – (59070)
2
The permeability of free space is 4
π
×
10

7
T
·
m
/
A.
Correct answer: 8
.
23897
×
10

7
Wb.
Explanation:
Let :
I
= 28 A
,
a
= 5 cm = 0
.
05 m
,
b
= 15 cm = 0
.
15 m
,
d
= 3 cm = 0
.
03 m
,
and
μ
0
4
π
= 1
×
10

7
N
/
A
2
.
d
a
b
x
dx
I
The magnetic Fux through the strip of area
dA
is
d
Φ =
B dA
=
μ
0
2
π
I
x
bdx
=
μ
0
4
π
2
bI dx
x
,
so the total magnetic Fux through the rectan
gular loop is
Φ
total
=
i
d
+
a
d
d
Φ
=
μ
0
4
π
(2
bI
)
i
d
+
a
d
dx
x
=
μ
0
4
π
(2
bI
) ln
d
+
a
d
= (1
×
10

7
N
/
A
2
) 2 (0
.
15 m) (28 A)
×
ln
p
0
.
03 m + 0
.
05 m
0
.
03 m
P
=
8
.
23897
×
10

7
Wb
.
003
(part 1 of 4) 10.0 points
A bar of negligible resistance and mass of
99 kg in the ±gure below is pulled horizon
tally across frictionless parallel rails, also of
negligible resistance, by a massless string that
passes over an ideal pulley and is attached
to a suspended mass of 760 g. The uniform
magnetic ±eld has a magnitude of 200 mT,
and the distance between the rails is 35 cm.
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This note was uploaded on 01/21/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
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