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Electromagnetic Theory I
Solution Set 9
Due: 31 Octber 2011
33. Consider the magnetic feld produced by a current
I
in an infnitely long wire lying on
the
z
axis
∞
< z <
∞
.
a) Use symmetry arguments and Amp`
ere’s law to obtain the
B
feld everywhere outside
the wire. Express the Cartesian components oF
B
as explicit Functions oF
x, y, z
.
Solution
:
Ampere’s law with a circular current loop centered on the
z
axis yields,
with
r
=
p
x
2
+
y
2
,
B
=
μ
0
I
2
πr
ˆ
ϕ
=
μ
0
I
2
πr
(

ˆ
x
sin
ϕ
+ ˆ
y
cos
ϕ
)
B
x
=

μ
0
Iy
2
π
(
x
2
+
y
2
)
,
B
y
=
μ
0
Ix
2
π
(
x
2
+
y
2
)
,
B
z
= 0
b) By direct integration oF each component oF
∇×
A
=
B
, fnd the vector potential
A
For
this
B
in Coulomb gauge,
∇·
A
= 0.
Solution
: We can fnd a vector potential with
A
x
=
A
y
= 0
∂
y
A
z
=
B
x
,
A
z
=

μ
0
I
4
π
ln(
x
2
+
y
2
) +
f
(
x, z
)
B
y
=

∂
x
A
z
=
μ
0
Ix
2
π
(
x
2
+
y
2
)
+
∂
x
f
(
x, z
)
,
∂
x
f
(
x, z
) = 0
We may set
f
=
constant
by fxing a gauge
∇ ·
A
=
∂
z
A
z
= 0.
We can choose this
constant to involve a distance
r
0
to make the argument oF the ln dimensionless
A
z
=

μ
0
I
4
π
ln
x
2
+
y
2
r
2
0
c) Since
∇×
B
= 0 “almost everywhere” we should be able to fnd a scalar potential such
that
B
=
∇
φ
“almost everywhere”. By explicitly integrating the components oF this
equation, fnd a candidate For
φ
as an explicit Function oF
x, y, z
.
Solution
:

∂φ
∂z
=
0
,

∂φ
∂x
=
B
x
,
φ
=
μ
0
I
2
π
tan

1
x
y
+
f
(
y
)

∂φ
∂y
=

f
0
(
y
) +
μ
0
Ix
2
π
(
x
2
+
y
2
)
=
B
y
,
f
0
(
y
) = 0
φ
=
μ
0
I
2
π
tan

1
x
y
(1)
1
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View Full Documentd) In view of the fundamental theorem of calculus
φ
(
y
)

φ
(
x
)
=
Z
y
x
d
l
·∇
φ
=

Z
y
x
d
l
·
B
,
(2)
explain how your result for part c) does not run afoul of Amp`
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 Current

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