Physics 213
Homework 8 Solutions
Spring 2009
z
r
ˆ
r
a
dl
P
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
28.69.
According to the BiotSavart Law,
0
2
ˆ
4
μ Id
πr
lr
B=
.
If the current loop can be broken up into segments i=0,1,…,n, then
the integral can also be broken up:
0
2
0
ˆ
4
n
i
i
μ Id
.
In our case, there are 3 segments:
the straight parts and the
quarter circle.
The contribution from the straight segments is zero since
0.
d
This leaves only the curved
segment contributing to
B
.
For each interval on the curved segment,
ˆ
is perpendicular to
dr
l
and so the BiotSavart
Law gives
0
2
4
Idl
dB
R
in an upward direction.
(Note that the contributions from all intervals around the curved
segment are the same magnitude and direction: no cancellations, etc).
To calculate the total
B
, we integrate over all
dB
for ¼ of a circle.
In the expression for dB, everything is a constant except dl.
So the integral becomes just
2
/ 4
dl
R
, giving
00
1
4
28
μ I
μ I
B
RR
directed out of the page.
Note that in this expression, the
B
field from one
quarter of a circle is just ¼ of the B field from a whole circle.
(Because of symmetries, it is very simple to calculate
B
at
point
P
but it would be much more difficult to calculate
B
at other points.)
28.70.
The horizontal wire yields zero magnetic field since
0.
d
For the vertical current, the
setup is as shown on the right.
The magnetic field contribution
d
B
=
o
4π
I d
r
^
r
2
at point P due
to every current element I d
along this segment is directed out of the plane of the page, and the
magnitude
dB = d
B
 =
o
4π
I dz
r
2
cos
.
So the total magnetic field due to this straight segment is out of the page, and its
magnitude is
0
2
0
cos
4
I
dz
B
dB
r
.
B
1x
=B
1
sin
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View Full DocumentPhysics 213
Homework 8 Solutions
Spring 2009
We can evaluate the integral by making the usual change of variable to the angle
:
z
=
a tan
,
dz
=
a sec
2
,
r
2
=
a
2
+ z
2
=
a
2
(1 + tan
2
)
=
a
2
sec
2
,
to get:
/2
0
0
0
0
0
cos
sin
4
4
4
I
I
I
Bd
a
a
a
.
So we end up with a magnetic field which is
half
that an infinite wire. Actually, we could have predicted this ahead of time.
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 Spring '07
 PERELSTEIN,M
 Magnetism, Work, Heat, Magnetic Field

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