toupal (rgt374) – Homework 08 – Chiu – (58295)
1
This
printout
should
have
15
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
A 0
.
9 m long wire weighing 0
.
072 N
/
m is
suspended directly above an infinitely straight
wire. The top wire carries a current of 47 A
and the bottom wire carries a current of 82 A
.
The permeablity of free space is 1
.
25664
×
10

6
N
/
A
2
.
Find the distance of separation between the
wires so that the top wire will be held in place
by magnetic repulsion.
Correct answer: 10
.
7056 mm.
Explanation:
Let :
I
1
= 47 A
,
I
2
= 82 A
,
λ
f
= 0
.
072 N
/
m
,
and
μ
0
= 1
.
25664
×
10

6
N
/
A
2
.
In order for the system to be in equilibrium,
the magnetic force per unit length
λ
F
on the
top wire must be equal to its weight per unit
length. Thus
λ
F
=
F
L
=
μ
0
I
1
I
2
2
π
d
so
d
=
μ
0
I
1
I
2
2
π λ
F
=
(1
.
25664
×
10

6
N
/
A
2
) (47 A) (82 A)
2
π
(0
.
072 N
/
m)
= 0
.
0107056 m
=
10
.
7056 mm
.
002
10.0 points
A rectangular loop with dimensions (hor
izontal = 0
.
11 m)
×
(vertical= 0
.
253 m), is
suspended by a string, and the lower hori
zontal section of the loop is immersed in a
magnetic field.
If a current of 6 A is maintained in the loop,
what is the magnitude of the magnetic field
required to produce a tension of 0
.
035 N in the
supporting string?
Assume:
Gravitational
force is negligible.
Correct answer: 0
.
0530303 T.
Explanation:
Let :
F
= 0
.
035 N
,
I
= 6 A
,
and
l
hor
= 0
.
11 m
.
The magnitude of the force on the lower leg is

F

=

I
hor
×
B

.
Therefore,
the magnitude of the magnetic
field due to this force is
B
=
F
I
hor
=
0
.
035 N
(6 A) (0
.
11 m)
= 0
.
0530303 T
.
003
(part 1 of 2) 10.0 points
Consider a current in the long, straight wire
which lies in the plane of the rectangular loop,
that also carries a current, as shown.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
toupal (rgt374) – Homework 08 – Chiu – (58295)
2
11 cm
13 cm
66 cm
4
.
3 A
→
15 A
→
x
y
Find the magnitude of the net force exerted
on the loop by the magnetic field created by
the long wire.
Correct answer: 4
.
1925
×
10

5
N.
Explanation:
Let :
c
= 0
.
11 m
,
a
= 0
.
13 m
,
= 0
.
66 m
,
I
1
= 4
.
3 A
,
and
I
2
= 15 A
.
c
a
I
1
→
I
2
→
x
y
The magnetic forces on the top and the
bottom segments of the rectangle cancel.
We now concern ourselves with the seg
ments of the rectangle which are parallel to
the long wire. Using Ampere’s law, the mag
netic fields from the long wire at distances
c
and (
c
+
a
) away are
B
c
=

μ
0
I
1
2
π
c
ˆ
k
B
ca
=

μ
0
I
1
2
π
(
c
+
a
)
ˆ
k .
The
forces
on
the
left
and
right
vertical
segments of the rectangle are, respectively,
F
c
=
I
2
B
c
and
F
ca
=
I
2
B
ca
.
The forces
are oppositely directed since the current
I
2
has a direction in the loop at distances
c
op
posite from the current
I
2
in the loop at a
distance
c
+
a
, away from the long wire.
The net force is
F
=
μ
0
I
1
I
2
2
π
+
1
c
+
a

1
c
ˆ
ı
=
μ
0
(4
.
3 A) (15 A) (0
.
66 m)
2
π
×
+
1
0
.
11 m + 0
.
13 m

1
0
.
11 m
ˆ
ı
=

4
.
1925
×
10

5
N ˆ
ı

F

=
4
.
1925
×
10

5
N
.
004
(part 2 of 2) 10.0 points
What is the direction of the net force?
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Turner
 Force, Magnetic Field, Electric charge, 0.9 m

Click to edit the document details