Version 052 – midterm 02 – chiu – (56565)
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001(part1of2)5.0points
Consider a current configuration shown be
low. A long (effectively infinite) wire segment
is connected to a quarter of a circular arc with
radius
a
. The other end of the arc is connected
to another long horizontal wire segment. The
current is flowing from the top coming down
vertically and flows to the right along the pos
itive
x
axis.
a
I
O
O
y
x
What is the direction of the magnetic field
at
O
due to this current configuration?
1.
perpendicular to and out of the page
cor
rect
2.
45
◦
counterclockwise from the +
x
axis
3.
along the negative
y
axis
4.
225
◦
counterclockwise from the +
x
axis
5.
along the positive
y
axis
6.
along the negative
x
axis
7.
135
◦
counterclockwise from the +
x
axis
8.
315
◦
counterclockwise from the +
x
axis
9.
along the positive +
x
axis
10.
perpendicular to and into the page
Explanation:
From the BiotSavart law we know that
d
vector
B
∝
I
d
vector
l
×
ˆ
r .
One should first verify that the magnetic field
at
O
contributed by any infinitesimal current
element along the current configuration given
is perpendicular to the page, which is coming
out of the page. Therefore the resulting mag
netic field must also be pointing out of the
page.
002(part2of2)5.0points
Let
I
= 2
.
2 A and
a
= 0
.
73 m.
What is the magnitude of the magnetic field
at
O
due to the current configuration?
1. 3.5708e06
2. 6.35602e06
3. 6.2846e06
4. 1.6724e06
5. 1.07613e06
6. 2.61433e06
7. 5.55457e06
8. 4.20094e06
9. 1.28945e06
10. 2.42814e06
Correct answer: 1
.
07613
×
10
−
6
T.
Explanation:
Let :
I
= 2
.
2 A
,
and
a
= 0
.
73 m
.
Consider first the onequarter of a circular
arc. Since each current element is perpendic
ular to the unit vector pointing to
O
, we can
write
B
arc
=
μ
0
4
π
I
integraldisplay
π/
2
0
a dθ
a
2
=
μ
0
4
π
I
π
2
a
=
μ
0
I
8
a
.
For the straight sections, we apply the formu
las derived from the figure below,
θ
1
θ
2
a
I
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Version 052 – midterm 02 – chiu – (56565)
2
vector
B
= ˆ
z
μ
0
I
4
π a
integraldisplay
θ
2
θ
1
dθ
sin
θ
= ˆ
z
μ
0
I
4
π a
(cos
θ
1
−
cos
θ
2
)
where ˆ
z
is the unit vector perpendicular to the
plane of the paper that points to the reader.
For the downward ydirected current,
θ
2
=
π
2
and
θ
1
= 0. For the horizontal xdirected
current
θ
2
=
π
and
θ
1
=
π
2
.
Thus we find the contribution at the point
O for the magnetic field from the long vertical
wire is same as in the long horizontal wire,
and the sum is equal to
vector
B
= ˆ
z
μ
0
I
4
π a
(1 + 1)
= ˆ
z
μ
0
I
2
π a
,
/noindent the same as in a long straight wire.
Adding the contributions from the straight
sections and the arc,
B
tot
=
μ
0
I
8
a
+
parenleftbigg
μ
0
I
2
π a
parenrightbigg
=
μ
0
I
2
a
parenleftbigg
1
4
+
1
π
parenrightbigg
=
(1
.
25664
×
10
−
6
T
·
m
/
A) (2
.
2 A)
2 (0
.
73 m)
×
parenleftbigg
1
4
+
1
π
parenrightbigg
=
1
.
07613
×
10
−
6
T
.
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 Fall '08
 Turner
 Physics, Current, Energy, Magnetic Field, Correct Answer, Electric charge

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