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
±
= 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
NI
2
π r
.
Solution:
In a toroid, all the ²ux is confned
to the inside oF the toroid
B
=
μ
0
2
.
So, the ²ux through the loop oF wire is
Φ
B
1
=
±
B dA
=
μ
0
0
2
π
sin(
)
±
b
+
R
R
adr
r
=
μ
0
0
2
π
a
sin(
) ln
²
b
+
R
R
³
.
Applying ±araday’s law, the induced emF can
be calculated as Follows
E
=

N
±
d
Φ
B
1
dt
=

N
±
μ
0
0
2
π
ω a
ln
²
b
+
R
R
³
cos(
)
=
E
0
cos(
)
where
ω
=2
πf
was used.
The maximum magnitude oF the induced
emf
,
E
0
, is the coe³cient in Front oF cos(
).
E
0
=

N
±
d
Φ
B
1
dt
=

N
±
μ
0
0
ω
2
π
a
ln
´
b
+
R
R
µ
=

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

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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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
=
±
d
+
a
d
d
Φ
=
μ
0
4
π
(2
bI
)
±
d
+
a
d
dx
x
=
μ
0
4
π
(2
) ln
d
+
a
d
= (1
×
10

7
N
/
A
2
) 2 (0
.
15 m) (28 A)
×
ln
²
0
.
03 m + 0
.
05 m
0
.
03 m
³
=
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.
The rails are connected at one end by a load
resistor of 51 mΩ.
The acceleration of gravity is 9
.
8m
/
s
2
.
200 mT
200 mT
200 mT
35 cm
760 g
a
51 m
Ω
99 kg
What is the magnitude of the terminal ve
locity (
i.e.,
the eventual steadystate speed
v
∞
) reached by the bar?
Correct answer: 77
.
52 m
/
s.
Explanation:
Let :
m
= 99 kg
,
M
= 760 g
,
'
= 35 cm
,
B
= 200 mT
,
and
R
= 51 mΩ
.
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 Spring '10
 Turner
 Magnetic Field

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