Physics 54
TimeVarying Fields
The only cure for boredom is curiosity. There is no cure for curiosity.
— Dorothy Parker
Completion of Ampere’s Law
As an example of a timevarying situation, consider the Bfield
established by the current charging a capacitor. Shown is a
parallelplate capacitor with circular plates. We are interested
in the Bfield at point
P
, shown in the edgeon view.
This situation has axial symmetry, which we exploit to use
Ampere's law and calculate
B
. The field lines are circles around
the symmetry axis, so we choose as our path of integration a
circle through point
P
. We find
B
⋅
d
r
∫
=
2
π
rB
=
μ
0
I
linked
.
Here
I
linked
is the current passing through the area bounded by the
circular path. But what area? The simplest is a circular disk bounded by
the circle of radius
r
. Because there is
no
current through such a
disk, we would have
I
linked
=
0
, from which we predict
B
=
0
at
P
.
But experimentally the field is
not
zero at
P
. Ampere's law gives a
prediction that is wrong.
However, we could use a different surface bounded by the circular path.
Consider a bowlshaped surface with the circle as its rim; the bowl
surface can be chosen to go around the left plate so that the wire penetrates through it.
In that case we would conclude that
I
linked
=
I
, leading to
B
=
μ
0
I
2
π
r
.
This is in agreement with experiment.
These different choices of surface bounded by the path give contradictory results. How
do we know which surface to use? Ampere’s law itself gives no instruction on this
point. No law of nature can be this vague. It must be either modified or discarded.
I
a
I
P
r
PHY 54
1
Electrodynamics
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Now consider the flux of the Efield through these two surfaces bounded by the circle.
First use the circular disk surface. Outside the region between the plates the Efield is
negligible, so the flux of
E
through the disk surface is that through a circle of radius
a:
Φ
e
=
E
⋅
d
A
∫
=
π
a
2
E
=
π
a
2
Q
ε
0
π
a
2
=
Q
ε
0
.
(Here we have used
E
=
σ
/
ε
0
and
σ
=
Q
/
π
a
2
.) From this we find
ε
0
d
Φ
e
dt
=
dQ
dt
=
I
.
This shows that the quantity
ε
0
⋅
d
Φ
e
/
dt
has the dimensions of a current, which in this
case is the same as the current in the wires. Maxwell called it the
displacement current
.
The term “displacement” arose from a model Maxwell had in mind, and has no significance today.
He suggested that Ampere’s law needed to be completed by adding the displacement
current to the “true” current, so that the law reads
B
⋅
d
r
∫
=
μ
0
I
linked
+
ε
0
d
dt
E
⋅
d
A
∫
⎡
⎣
⎢
⎤
⎦
⎥
.
Now let us apply this to our capacitor situation to see how it resolves the ambiguity. If
we choose the disk surface, then
I
linked
=
0
but
ε
0
d
dt
E
⋅
d
A
∫
=
I
, to we find
B
=
μ
0
I
2
π
r
, as
we should. If we choose the bowl surface we have
I
linked
=
I
and
ε
0
d
dt
E
⋅
d
A
∫
=
0
, again
leading to
B
=
μ
0
I
2
π
r
. The choice of surface no longer matters.
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 Summer '09
 Thomas
 Physics, Current, Magnetic Field, Faraday, Bfield

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