terry (ect328) – oldhomework 22 – Turner – (59130)
1
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001
10.0 points
Three very long wires are strung parallel to
each other as shown in the figure below. Each
wire is at a distance
r
from the other two, and
each wire carries a current of magnitude
I
in
the directions shown in the figure.
I
I
I
3
2
1
z
x
y
r
r
r
×
3
2
1
y
x
z
Crosssectional View
What angle does the net force on the upper
wire (wire 3) due to the other two wires make
with the positive
x
axis? (Measure your an
gles in the standard way:
counterclockwise
from the
x
axis.)
1.
180
◦
2.
60
◦
3.
90
◦
correct
4.
270
◦
5.
300
◦
6.
30
◦
7.
240
◦
8.
120
◦
9.
210
◦
10.
0
◦
Explanation:
The magnetic field due to a long straight
wire is
B
=
μ
0
I
2
π r
,
and the force per unit length between two
parallel wires is
F
ℓ
=
μ
0
I
1
I
2
2
π r
.
There are two ways to solve this problem
which are essentially the same. The first way
is to find the net magnetic field at the upper
wire from the two wires below (
vector
B
net
=
vector
B
1
+
vector
B
2
) and then find the force from
vector
F
=
I
vector
L
×
vector
B .
The crucial step here will be to add the
magnetic fields as
vectors
.
The second way
would be to use
vector
F
=
I
vector
L
×
vector
B
to find the net
force on the upper wire from the two lower
wires:
vector
F
net
=
vector
F
1
+
vector
F
2
,
where we must be
sure to add the forces as vectors. You should
recognize that the two methods are formally
identical.
Let’s do it the first way.
The
magnitude magnetic field from wire 1 is found
from Ampere’s law to be
B
1
=
μ
0
I
2
π r
.
Using the right hand rule the direction points
up and to the left of wire as shown in figure 2.
30
◦
30
◦
60
◦
60
◦
60
◦
B
1
B
B
2
×
3
2
1
ˆ
ˆ
ı
ˆ
k
Adding Magnetic Fields
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terry (ect328) – oldhomework 22 – Turner – (59130)
2
30
◦
30
◦
60
◦
60
◦
60
◦
F
13
F
F
23
×
3
2
1
ˆ
ˆ
ı
ˆ
k
Adding Forces
Its components will then be
vector
B
1
=
B
[sin (30
◦
)ˆ
−
cos (30
◦
)ˆ
ı
]
.
Similarly,
vector
B
2
points down and to the left of
wire 3; its components are given by
vector
B
2
=
B
[
−
sin (30
◦
)ˆ
−
cos (30
◦
)ˆ
ı
]
.
Notice that by symmetry the ˆ
(
y
) compo
nent of the magnetic field vanishes. The net
magnetic field is therefore:
vector
B
net
=
−
2
μ
0
I
cos (30
◦
)
2
π r
ˆ
ı .
The force is then
vector
F
=
I
vector
L
×
vector
B
=
I L
parenleftbigg
−
2
μ
0
I
cos (30
◦
)
2
π r
parenrightbigg
(
−
ˆ
k
×
ˆ
ı
)
=
I L
parenleftbigg
−
2
μ
0
I
cos (30
◦
)
2
π r
parenrightbigg
(
−
ˆ
)
vextendsingle
vextendsingle
vextendsingle
vextendsingle
F
L
vextendsingle
vextendsingle
vextendsingle
vextendsingle
=
2
μ
0
I
2
cos (30
◦
)
2
π r
Notice that the direction of
vector
L
is in the
−
ˆ
k
direction, since that is the direction of the
current.
After all the negative signs have
cancelled, we notice that the force is in the ˆ
direction. This is 90
◦
from the positive
x
axis.
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 Fall '09
 Turner
 Magnetic Field, Electric charge, Terry

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