howard (cah3459) – ohw18 – turner – (56705)
1
This printout should have 15 questions.
Multiplechoice questions may continue on
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beFore answering.
001 (part 1 of 2) 10.0 points
A closed circuit consists oF two semicircles oF
radii 77 cm and 38 cm that are connected by
straight segments, as shown in the fgure.
A current oF
I
= 3 A ±ows around this
circuit in the clockwise direction.
The permeability oF Free space is 4
π
×
10

7
Tm
/
A.
O
38
cm
77 cm
ˆ
ı
ˆ
ˆ
k
ˆ
k
is upward
From the paper
I
= 3 A
I
What is the direction oF the magnetic feld
B
at point
O
.
1.
/
B
±
/
B
±
= +ˆ
2.
None oF these are true‘.
3.
/
B
±
/
B
±
= +ˆ
ı
4.
/
B
±
/
B
±
=

ˆ
5.
/
B
±
/
B
±
=+
ˆ
k
6.
/
B
±
/
B
±
=

ˆ
ı
7.
/
B
±
/
B
±
=

ˆ
k
correct
Explanation:
IF you curve the fngers on your right hand
in the direction oF the current, the magnetic
feld points along your thumb.
002 (part 2 of 2) 10.0 points
²ind the magnitude oF the magnetic feld at
point at point
O
.
Correct answer: 3
.
7042
μ
T.
Explanation:
Let :
r
1
= 77 cm = 0
.
77 m
,
r
2
= 38 cm = 0
.
38 m
,
I
= 3 A
,
and
μ
0
=4
π
×
10

7
/
A
.
The magnetic feld due to a circular loop at
its center is
B
=
μ
0
I
2
R
.
Because the loop is only halF oF a current
loop, the magnetic feld due to the arc 1 is
/
B
1
=

μ
0
I
4
r
1
ˆ
k,
and the magnetic feld due to
the arc 2 is
/
B
2
=

μ
0
I
4
r
2
ˆ
k.
Thus the total magnetic feld is
/
B
=
/
B
1
+
/
B
2
=

μ
0
I
4
±
1
r
1
+
1
r
2
²
ˆ
k
=

(4
π
×
10

7
/
A) (3 A)
4
×
±
1
0
.
77 m
+
1
0
.
38 m
²
ˆ
k
±
10
6
μ
T
1T
²
=(

3
.
7042
μ
T)
ˆ
k
±
/
B
±
=
1
.
003
10.0 points
The closed loop shown in the fgure carries a
current oF 20 A in the counterclockwise direc
tion. The radius oF the outer arc is 60 cm
,
that oF the inner arc is 40 cm
.
Both arcs ex
tent an angle oF 60
◦
.
The permeability oF Free space is 4
π
×
10

7
/
A.
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View Full Documenthoward (cah3459) – ohw18 – turner – (56705)
2
40 cm
C
B
60 cm
D
A
O
60
◦
y
z
Find the component of the magnetic ±eld
at point
O
along the
x
axis.
Correct answer:

1
.
74533
×
10

6
T.
Explanation:
Let :
r
i
= 40 cm = 0
.
4m
,
r
o
= 60 cm = 0
.
6m
,
I
= 20 A
,
and
μ
0
=4
π
×
10

7
Tm
/
A
.
The magnetic ±eld due to a circular loop at
its center is
B
=
μ
0
I
2
R
.
Because the loop is only onesixth of a current
loop, the magnetic ±eld due to the inner arc
is
/
B
i
=

μ
0
I
12
r
i
ˆ
ı,
and the magnetic ±eld due
to the outter arc is
/
B
o
=
μ
0
I
12
r
o
ˆ
ı.
Thus the total magnetic ±eld is
/
B
=
/
B
i
+
/
B
o
=
μ
0
I
12
±
1
r
o

1
r
i
²
ˆ
ı
=
(4
π
×
10

7
/
A) (20 A)
12
×
±
1
0
.

1
0
.
²
ˆ
ı
=
(

1
.
74533
×
10

6
T ) ˆ
ı
.
004
10.0 points
Given:
Three conductors (same length) have
the shape of an equilateral triangle (whose
sides are of length
.
3
), a ring (whose radius
is of length
r
), a square (whose sides are of
length
.
4
).
These conductors’ lengths are all equal
(perimeters are 3
.
3
=2
π r
.
4
).
All conductors carry the same counter
clockwise current
I
as shown in the ±gure
below.
.
3
I
P
.
4
I
P
r
I
P
Select the correct comparison for the mag
nitude of the magnetic ±eld at the center
points
P
of the current loops shown above.
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 Spring '10
 Turner
 Magnetic Field, Howard

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