For a steady current flowing through a straight wire, the magnetic field at a
point at a perpendicular distance
r
r
from the wire, has a value
B
=
μ
0
I
2
π
r
B
=
μ
0
I
2
π
r
If
we integrate around the wire in a circle, then clearly we get
∮
B
⃗
⋅
d
l
⃗
=
μ
0
I
2
π
r
∮
ⅆ
l
=
μ
0
I
2
π
r
2
π
r
=
μ
0
I
∮
B
⃗
⋅
d
l
⃗
=
μ
0
I
2
π
r
∮
ⅆ
l
=
μ
0
I
2
π
r
2
π
r
=
μ
0
I
This
is true for irregular paths around the wire

∮
B
⃗
⋅
d
l
⃗
=
∮
B
⋅
ⅆ
l
cos
θ
∮
B
⃗
⋅
d
l
⃗
=
∮
B
⋅
ⅆ
l
cos
θ
but
for small
d
l
d
l
d
l
cos
θ
=
r
d
φ
d
l
cos
θ
=
r
d
φ
∮
B
⃗
⋅
d
l
⃗
=
∮
B
r
ⅆ
φ
=
μ
0
I
2
π
∮
ⅆ
φ
=
μ
0
I
∮
B
⃗
⋅
d
l
⃗
=
∮
B
r
ⅆ
φ
=
μ
0
I
2
π
∮
ⅆ
φ
=
μ
0
I
In fact instead of current we use the surface integral of the current density
J
J
,
which is the current per unit area
∮
B
⃗
⋅
d
l
⃗
=
μ
0
∫
S
J
⃗
⋅
d
A
⃗
∮
B
⃗
⋅
d
l
⃗
=
μ
0
∫
S
J
⃗
⋅
d
A
⃗
Maxwell's great insight was to realize that this was incomplete. He
reasoned that
ⅆ
φ
B
ⅆ
t
ⅆ
φ
B
ⅆ
t
gives a
E
⃗
E
⃗
field so we should expect that
ⅆ
φ
E
ⅆ
t
ⅆ
φ
E
ⅆ
t
gives a
B
⃗
B
⃗
field.

Think
of a capacitor in a simple circuit. We can draw a surface such as shown in the
figure, with "surface 1" and take the line integral around the edge of the
surface. Now look at surface 2, this will have the same line integral, but
now the surface integral will be different. Clearly there is something
incomplete with Ampere's law as formulated above. Maxwell re wrote Ampere's
law

∮
B
⃗
⋅
d
l
⃗
=
μ
0
∫
S
(
J
⃗
+
ε
0
∂
E
⃗
∂
t
)
⋅
d
A
⃗
∮
B
⃗
⋅
d
l
⃗
=
μ
0
∫
S
(
J
⃗
+
ε
0
∂
E
⃗
∂
t
)
⋅
d
A
⃗
which
solves the problem.

Again it is left as an exercise to show that
∇
⃗
×
B
⃗
=
μ
0
(
J
⃗
+
ε
0
∂
E
⃗
∂
t
)
∇
⃗
×
B
⃗
=
μ
0
(
J
⃗
+
ε
0
∂
E
⃗
∂
t
)

Comments:"This book covers second year Physics at Rice University."