CHAPTER 8
8.1a. Find
H
in cartesian components at
P(
2
,
3
,
4
)
if there is a current filament on the
z
axis carrying 8 mA
in the
a
z
direction:
Applying the BiotSavart Law, we obtain
H
a
=
Z
∞
−∞
Id
L
×
a
R
4
πR
2
=
Z
∞
−∞
Idz
a
z
×
[2
a
x
+
3
a
y
+
(
4
−
z)
a
z
]
4
π(z
2
−
8
z
+
29
)
3
/
2
=
Z
∞
−∞
[2
a
y
−
3
a
x
]
4
2
−
8
z
+
29
)
3
/
2
Using integral tables, this evaluates as
H
a
=
I
4
π
±
2
(
2
z
−
8
)(
2
a
y
−
3
a
x
)
52
(z
2
−
8
z
+
29
)
1
/
2
²
∞
−∞
=
I
26
π
(
2
a
y
−
3
a
x
)
Then with
I
=
8 mA, we finally obtain
H
a
=−
294
a
x
+
196
a
y
µ
A
/
m
b. Repeat if the filament is located at
x
1,
y
=
2: In this case the BiotSavart integral becomes
H
b
=
Z
∞
−∞
a
z
×
[
(
2
+
1
)
a
x
+
(
3
−
2
)
a
y
+
(
4
−
z)
a
z
]
4
2
−
8
z
+
26
)
3
/
2
=
Z
∞
−∞
[3
a
y
−
a
x
]
4
2
−
8
z
+
26
)
3
/
2
Evaluating as before, we obtain with
I
=
8 mA:
H
b
=
I
4
π
±
2
(
2
z
−
8
)(
3
a
y
−
a
x
)
40
(z
2
−
8
z
+
26
)
1
/
2
²
∞
−∞
=
I
20
π
(
3
a
y
−
a
x
)
127
a
x
+
382
a
y
µ
A
/
m
c. Find
H
if both filaments are present: This will be just the sum of the results of parts
a
and
b
,or
H
T
=
H
a
+
H
b
421
a
x
+
578
a
y
µ
A
/
m
This problem can also be done (somewhat more simply) by using the known result for
H
from an
infinitelylong wire in cylindrical components, and transforming to cartesian components. The Biot
Savart method was used here for the sake of illustration.
8.2. A current filament of 3
a
x
A lies along the
x
axis. Find
H
in cartesian components at
−
1
,
3
,
2
)
:We
use the BiotSavart law,
H
=
Z
L
×
a
R
4
2
where
L
=
3
dx
a
x
,
a
R
=
[
−
(
1
+
x)
a
x
+
3
a
y
+
2
a
z
]
/R
, and
R
=
√
x
2
+
2
x
+
14. Thus
H
P
=
Z
∞
−∞
3
a
x
×
[
−
(
1
+
a
x
+
3
a
y
+
2
a
z
]
4
π(x
2
+
2
x
+
14
)
3
/
2
=
Z
∞
−∞
(
9
a
z
−
6
a
y
)dx
4
2
+
2
x
+
14
)
3
/
2
=
(
9
a
z
−
6
a
y
)(x
+
1
)
4
π(
13
)
√
x
2
+
2
x
+
14
³
³
³
∞
−∞
=
2
(
9
a
z
−
6
a
y
)
4
13
)
=
0
.
110
a
z
−
0
.
073
a
y
A
/
m
116
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View Full Document8.3. Two semiinfinite filaments on the
z
axis lie in the regions
−∞
<z<
−
a
(note typographical error in
problem statement) and
a<z<
∞
. Each carries a current
I
in the
a
z
direction.
a) Calculate
H
as a function of
ρ
and
φ
at
z
=
0: One way to do this is to use the field from an
infinite line and subtract from it that portion of the field that would arise from the current segment
at
−
a<z<a
, found from the BiotSavart law. Thus,
H
=
I
2
πρ
a
φ
−
Z
a
−
a
Idz
a
z
×
[
ρ
a
ρ
−
z
a
z
]
4
π
[
ρ
2
+
z
2
]
3
/
2
The integral part simplifies and is evaluated:
Z
a
−
a
Idzρ
a
φ
4
π
[
ρ
2
+
z
2
]
3
/
2
=
Iρ
4
π
a
φ
z
ρ
2
p
ρ
2
+
z
2
±
±
±
a
−
a
=
Ia
2
πρ
p
ρ
2
+
a
2
a
φ
Finally,
H
=
I
2
πρ
"
1
−
a
p
ρ
2
+
a
2
#
a
φ
A
/
m
b) What value of
a
will cause the magnitude of
H
at
ρ
=
1,
z
=
0, to be onehalf the value obtained
for an infinite filament? We require
"
1
−
a
p
ρ
2
+
a
2
#
ρ
=
1
=
1
2
⇒
a
√
1
+
a
2
=
1
2
⇒
a
=
1
/
√
3
8.4a.) A filament is formed into a circle of radius
a
, centered at the origin in the plane
z
=
0. It carries a
current
I
in the
a
φ
direction. Find
H
at the origin: We use the BiotSavart law, which in this case
becomes:
H
=
Z
loop
Id
L
×
a
R
4
πR
2
=
Z
2
π
0
Iadφ
a
φ
×
(
−
a
ρ
)
4
πa
2
=
0
.
50
I
a
a
z
A
/
m
b.) A filament of the same length is shaped into a square in the
z
=
0 plane. The sides are parallel to the
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 Spring '10
 Ferguson
 Cartesian Coordinate System, Electromagnet, René Descartes, Magnetic Field, AZ, DVD region code

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