mcconnell (kam2342) – homework24 – Turner – (60230)
1
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001
10.0 points
A toroid having a rectangular cross section
(
a
= 2
.
44 cm by
b
= 3
.
87 cm) and inner
radius 3
.
21 cm consists of
N
= 630 turns of
wire that carries a current
I
=
I
0
sin
ω t
, with
I
0
= 43 A and a frequency
f
= 42
.
1 Hz.
A
loop that consists of
N
ℓ
= 16 turns of wire
links the toroid, as in the figure.
b
a
N
R
N
l
Determine the maximum
E
induced in the
loop by the changing current
I
.
Correct answer: 0
.
442577 V.
Explanation:
Basic Concept:
Faraday’s Law
E
=

d
Φ
B
dt
.
Magnetic field in a toroid
B
=
μ
0
N I
2
π r
.
Solution:
In a toroid, all the flux is confined
to the inside of the toroid
B
=
μ
0
N I
2
π r
.
So, the flux through the loop of wire is
Φ
B
1
=
integraldisplay
B dA
=
μ
0
N I
0
2
π
sin(
ω t
)
integraldisplay
b
+
R
R
a dr
r
=
μ
0
N I
0
2
π
a
sin(
ω t
) ln
parenleftbigg
b
+
R
R
parenrightbigg
.
Applying Faraday’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
parenleftbigg
b
+
R
R
parenrightbigg
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 coefficient 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
bracketleftbigg
b
+
R
R
bracketrightbigg
=

(16 turns)
μ
0
(630 turns)
×
(43 A) (42
.
1 Hz) (2
.
44 cm)
×
ln
bracketleftbigg
(3
.
87 cm) + (3
.
21 cm)
(3
.
21 cm)
bracketrightbigg
=

0
.
442577 V
E
= 0
.
442577 V
.
002
10.0 points
A long straight wire carries a current 36 A. A
rectangular loop with two sides parallel to the
straight wire has sides 5 cm and 10 cm, with
its near side a distance 1 cm from the straight
wire, as shown in the figure.
The permeability of free space is 4
π
×
10

7
T
·
m
/
A.
1 cm
5 cm
10 cm
36 A
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mcconnell (kam2342) – homework24 – Turner – (60230)
2
Find the magnetic flux through the rectan
gular loop.
Correct answer: 1
.
29007
×
10

6
Wb.
Explanation:
Let :
I
= 36 A
,
a
= 5 cm = 0
.
05 m
,
b
= 10 cm = 0
.
1 m
,
d
= 1 cm = 0
.
01 m
,
and
μ
0
4
π
= 1
×
10

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

7
N
/
A
2
) 2 (0
.
1 m) (36 A)
×
ln
parenleftbigg
0
.
01 m + 0
.
05 m
0
.
01 m
parenrightbigg
=
1
.
29007
×
10

6
Wb
.
003
(part 1 of 4) 10.0 points
A bar of negligible resistance and mass of
95 kg in the figure 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 110 g.
The uniform
magnetic field has a magnitude of 630 mT,
and the distance between the rails is 45 cm.
The rails are connected at one end by a load
resistor of 91 mΩ.
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 Spring '10
 Turner
 Magnetic Field, Correct Answer, Magnet, McConnell

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