Download Steady Current

Electromagnetism
Chapter 9: Steady Current
Pablo Laguna
Georgia Institute of Technology, USA
Fall 2012
Notes based on textbook: Modern Electrodynamics by A. Zangwill
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Electromagnetism
Let’s consider the class of problems where the source of E(r) is electric
charge in steady motion; that is, current density j = j(r) which yield a current
Z
I = dS · j
S
From the continuity equation we have that
∂ρ
+∇·j=∇·j=0
∂t
And the Maxwell equations still read
∇·E=
ρ
0
and
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∇×E=0
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The condition ∇ · j = 0 implies that the lines of current density are like the
electric field lines.
In some instance, we can define the convection current density as
j(r) = ρ(r)υ(r)
Current density in many systems obeys Ohm’s law,
j = σE.
Like P = 0 χE, Ohm’s law is a constitutive relation which describes the
material-dependent response of a many-particle system to a specified
stimulus.
Here, it is the conductivity σ which carries the material-dependent
information.
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If dS k j (Filamentary Wire Limit)
Z
Z Z
Z
Z
Z
j d 3r =
j (dS · d`) = (j · dS)
d` = I d`
Thus, for an arbitrary vector C(r)
Z
Z
d 3 r j × C → I d` × C.
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Since
∇ · j(r)
=
0
j(r)
=
σ(r)E(r)
then
σ(r) ∇ · E(r) + E(r) · ∇σ(r) = 0
If σ = constant, and given that ρ(r) = 0 ∇ · E(r), we have that
σ∇·E
ρ
σ
0
ρ
=
0
=
0
=
0
Thus ∇2 ϕ = 0
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Matching Conditions Ohmic matter
From
∇·j
=
0
j
=
σE
∇×B
=
0
we can derive
n̂2 · [j1 − j 2 ] = 0
∂ϕ1 ∂ϕ2 = σ2
σ1
∂n S
∂n S
or
and
n̂2 × [E1 − E2 ] = 0
or
ϕ1 |S = ϕ2 |S
Notice the similarity with the matching conditions at the interface between two
dielectrics.
If no current flows through a boundary surface separating an Ohmic medium
from a non-conducting medium,
∂ϕ = 0.
∂n S
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Electrical Resistance
"! # 0
"n
S
! #!A
! # !B
"! # 0
"n
S
R=
ϕA − ϕB
.
I
where j is found from
Z
dS · j
I=
S
and ϕ from ∇2 ϕ = 0
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An alternative to computing
R=
is to fix I from
ϕA − ϕB
I
Z
Z
dS · j = σ
dS
∂ϕ
= ±I
∂n
S
S
and we calculate the potential difference from the line integral of E from one
electrode to the other. Therefore,
RB
A
R= R
RB
d` · E
1
A
R
= ×
σ
dS · j
dS · E
d` · E
S
S
Consider calculating the resistance of two conducting electrodes are
embedded in an infinite medium with conductivity σ
RC =
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0
σ
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Joule Heating
The inelastic collisions remove kinetic energy from the current-carrying
charges.
That energy is dissipated in the form of Joule heat.
In an Ohmic medium, the moving charges have a constant drift velocity
with no change in potential energy.
From the 1st law of thermodynamics, the rate of Joule heating is equal
to the rate the electric field does work on the current carrying charges:
Z
N
d X
dW
=
qk E(rk ) · rk (t) = d 3 r j · E
dt
dt
k=1
But
Z
V
Thus
d 3r j · E = −
Z
V
d 3 r ∇ · (jϕ) = −
V
Z
dS · j ϕ = (ϕA − ϕB )I = I 2 R.
S
dW
= I2R
dt
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Electromotive Force
A steady current requires the presence of a source of energy to
compensate for Joule heating.
The source of energy must be non-electrostatic (chemical, thermal,
gravitational, nuclear, or even electromagnetic) to maintain the current.
The force associated with this source of energy is called electromotive
force (EMF).
One represents the effect of any source of EMF by a fictitious electric
field E0 such that
j(r) = σ[E(r) + E0 (r)].
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The voltage difference between two points 1 and 2 along a circuit as
Z2
ϕ1 − ϕ2 =
d` · E.
1
The EMF between the points 1 and 2 is
Z2
E12 =
d` · E0 .
1
Thus
1
σ
Z2
Z2
d` · j =
1
d` · [E + E0 ] = ϕ1 − ϕ2 + E12
1
and
I R12 = ϕ1 − ϕ2 + E12
H
For a closed circuit, d` · E = 0, and then E = IR where R is the
resistance of the entire circuit and the EMF of the entire circuit is
I
I
1
E = d` · E0 =
d` · F.
q
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Kirchhoff’s Laws
They are a consequence of ∇ · j = 0 and ∇ × E = 0 for filamentary wire
circuits.
From ∇ · j = 0 currents must flow into and out of each node in a circuit with
no accumulation of charge. Thus
X
Ik = 0
(Kirchhoff’s current law).
k
with Ik the current on each segment.
H
For each closed loo, ∇ × E = 0 implies that d` · E = 0. Therefore, if In is
the current which flows through resistor Rn , then
X
X
Ek =
I n Rn
(Kirchhoff’s voltage law).
k
n
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