Physics 2213
Homework 8 Solutions
Spring 2010
28.60.
We must find the vector sum of the magnetic fields due to each wire.
For a long straight wire
B
0
I
2
r
. The direction of
B
is given by the righthand rule and is
perpendicular to the line from the wire to the point where then field is calculated.
(a)
The magnetic field vectors are shown in Figure (a) below.
(b)
At a position on the
x
axis
0
0
0
net
22
2
2
2
2
2
sin
,
2
(
)
I
I
a
Ia
B
r
x
a
x
a
x
a
in the positive
x
direction.
(c)
The graph of
B
versus
x
/
a
is given below (Figure (b)).
(d)
The magnetic field is a
max at the origin,
x
= 0.
(e)
When
0
2
,.
Ia
x
a B
x
(f)
Ampere’s law states that
C
encl
I
l
d
B
0
.
The magnetic field about a cylindrical conductor
carrying a current should be constant on a circle centered
on the conductor and should be tangent to the circle.
(See the figure.)
So
Bdl
l
d
B
and the line integral
around
C
is
C
C
C
r
B
dl
B
l
d
B
)
2
(
.
Ampere’s law then
gives
I
I
r
B
encl
0
0
)
2
(
and then
r
I
B
2
0
.
B
1x
=B
1
sin
B
I
r
C
l
d
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentPhysics 213
Homework 8 Solutions
Spring 2009
#1. [Wire & Loop]
(a)
The magnetic field due to the current I
2
in each
segment of the loop points into the page
everywhere
along each of the other three segments.
The
directions of the forces on each segment due to the
other three segments are therefore:
Force on segment ab due to I
2
Force on segment bc due to I
2
Force on segment cd due to I
2
Force on segment da due to I
2
These forces try to expand
the loop.
The force on a wire segment of length L and current I due to magnetic field B is
F = IL× B
.
The magnetic field
B
1
due to the straight wire is pointing into the page everywhere along the
loop. The magnitude of the field at a distance d below the wire is B
1
(d) =
μ
o
I
1
2
d
.
The vertical sections of the loop (labeled abcd in the diagram at right) experience equal and
opposite forces:
F
da,1
= 
F
bc,1
,
and these do not contribute to the net force on the loop.
Segment ab of the loop experiences a force of magnitude:

F
ab,1
=
B
1
(y) I
2
D
=
μ
o
I
1
I
2
D
2
y
in
the up direction (
), and segment cd experiences a force of magnitude:

F
cd,1
=
B
1
(y+H) I
2
D
=
μ
o
I
1
I
2
D
2
(y+H)
in the down direction (
).
The net force on the loop has magnitude:
F
net
=

F
ab,1


F
cd,1
=
μ
o
I
1
I
2
D
2
1
y

1
y+H
=
μ
o
I
1
I
2
D H
2
y(y+H)
The direction of F
net
is
up
.
(b)
We need to find the distance y at which the magnetic force
F
net
from part (a) exactly
balances the weight of the loop
W
:
F
net
+
W
= 0
or
μ
o
I
1
I
2
D H
2
y(y+H)
=
mg.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 PERELSTEIN
 Work, Magnetic Field

Click to edit the document details