Husain, Zeena – Homework 8 – Due: Mar 23 2004, 4:00 am – Inst: Sonia Paban
1
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printout
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have
19
questions.
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Central time.
001
(part 1 of 1) 10 points
A current
I
= 5
.
54 Amps flows through a wire
perpendicular to the paper and towards the
reader at
A
and back in the opposite direction
at
C
. Consider the wires below the plane at
A
and
C
to be semiinfinite.
In the figure,
L
1
= 3 m,
R
= 3
.
7 m, and
L
2
= 3 m and
there is an external uniform magnetic field
B
= 8
.
53 T, which is directed into the paper.
Being ’external’, this field is NOT due to the
current in the wire.
Caution:
It may be necessary to take
into account the contribution from the long
straight wire which runs
up to
and
down from
the underneath side of the page.
L
1
I
I
L
2
C
A
R
O
B
What is the magnitude of the magnetic field
at the center of the arc
O
due to the current
in the wire?
Correct answer: 2
.
62661
×
10

7
T.
Explanation:
The two straight current segments within
the plane of the paper do not contribute to the
magnetic field at point
O
, because they are
parallel to the radius vector from that point.
Therefore
ds
×
ˆ
r
= 0 on these segments. The
contribution from the curved part of the wire
is easy to find using the BiotSavart law
B
=
μ
0
I ds
×
ˆ
r
4
π
r
2
=
μ
0
I
π
2
4
π
R
=
μ
0
I
8
R
(

ˆ
k
)
,
but one must not forget to take into account
the contribution from the long straight wire
which runs
into
and
out of
the page. For the
wire above this is
B
1
=
μ
0
I
4
π
(
L
1
+
R
)
ˆ
ı ,
while for the wire below we have
B
2
=
μ
0
I
4
π
(
L
2
+
R
)
ˆ
.
These three must be added as vectors, giving
the magnitude of the field

B
O

=
μ
0
I
8
R
2
+
μ
0
I
4
π
(
R
+
L
1
)
2
+
μ
0
I
4
π
(
R
+
L
2
)
2
1
/
2
=
(8
.
26866
×
10

8
T)
2
+(8
.
26866
×
10

8
T)
2
+(2
.
35195
×
10

7
T)
2
1
2
= 2
.
62661
×
10

7
T
.
002
(part 1 of 2) 10 points
The loop in the figure carries a current 1
.
13 A.
The semicircular portion has a radius 5
.
73 cm.
A
5
.
73 cm
5
.
73 cm
2
×
5
.
73 cm
1
.
13 A
Determine the magnitude of the magnetic
field at A.
Correct answer: 1
.
17733
×
10

5
T.
Explanation:
Let :
I
= 1
.
13 A
,
R
= 5
.
73 cm
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Husain, Zeena – Homework 8 – Due: Mar 23 2004, 4:00 am – Inst: Sonia Paban
2
A
R
R
2R
I
Basic Concepts:
The BiotSavart Law is
given by
d B
=
μ
0
4
π
I d
×
ˆ
r
r
2
.
(1)
Solution:
Note:
The distance from a current
element on a circle to the center is a constant,
namely
r
, so we can pull this out of the inte
gral. Also, the current element,
I dl
, is always
perpendicular to ˆ
r
, so sin
θ
= 1.
Hence,
B
full circle
=
μ
0
4
π
I
r
2
d
=
μ
0
4
π
I
r
2
2
π
r
=
μ
0
I
2
r
,
where a halfcircular arc is onehalf this value.
Assume:
Out of the page to be positive.
The angle
θ
(between the vector
and the
vector
r
) is from the BiotSavart Law, Eq.
(1), for the magnetic field
B
a distance
a
away from a wire segment
B
=
μ
0
I
4
π
a
θ
2
θ
1
sin
θ
d
θ
=
μ
0
I
4
π
a
(cos
θ
1

cos
θ
2
)
,
where
a
=
R
for the sides of the semisquare
in figure.
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 Spring '10
 Turner
 Magnetic Field, Husain, Sonia Paban

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