Download Chapter 3 Steady current

Chapter 3
Steady current
In the previous chapter, we have dealt with the electrostatic properties of
conductors, namely when there is no current flowing. In this chapter, we will
study the basic properties of the conductor with a steady current flowing,
which is caused by a gradient (or a difference) of the electrostatic potential.
2 Intensity of a current (or simply “current”) and current density
:
We begin by defining a few basic quantities:
Intensity of a current: I = total amount of charge passing through the cross
section of a conducting wire per unit time.
Unit of I = Coulomb /sec = A(mpère)
Current density: j = I/S =current per unit area
Current density in terms of the charge density:
When the charge ρ per unit volume moves with the velocity ~v , the amount of charge passing through a unit
area perpendicular to ~v is, from the figure, ρv. There- S
fore, we can write
dQ
= ρv · S = ρ · (Sv)
dt
~ = ρ~v
Vol.= Sv
I =
qqq
1
(3.1)
v
3.1
Ohm’s law
When one applies a voltage difference on a conductor, an electric field is produced and hence the conduction electrons inside the conductor get accelerated. However, this accelerated motion does not last long since the electrons
get scattered by the nuclei and other impurities in the conductor and get
slowed down. So, what happens is that the initial accelerated motion quickly
turns into a steady flow with a constant velocity called “terminal” or “final”
velocity . In this phenomenon, normally the magnitude of the electric
field and the intensity of the steady current are proportional to
each other with a characteristic constant of proportionality. This empirical
law is called the Ohm’s law.
3.1.1
Various forms of Ohm’s law
2 Maxwell’s expression :
Ohm’s law can be expressed in various forms. One basic form is the one
due to Maxwell, which expresses the above mentioned proportionality of the
electric field and the current (1864). It reads
~
~ = σ E
σ = electric conductivity
We will derive this expression from a microscopic point of view shortly.
2 Another basic form: V = RI and its derivation :
Consider a portion of a conducting wire as shown below:
A
l
B
φ(B)
φ(A)
2
From the figure, the voltage difference V is
Z
V
Z
B
= VAB = φ(A) − φ(B) =
B
E d` =
A
A
j
d`
σ
j
=
`
σ
where we used Maxwell’s form of the Ohm’s law in the last equality in the
first line. Let us rewrite this in terms of the current I = jS. Multiplying the
numerator and the denominator by S, we easily get
j
jS`
`
`=
=I
σ
σS
σS
Now we define the resistence R by
V
=
R≡
`
σS
This quantity does have the property expected of a resistence, namely it is
proportional to the length while inversely proportional to the cross sectional
area and the electric conductivity. Using this quantity, we obtain the wellknown and useful form of the Ohm’s law
V = RI
Unit of resistence： Ω (Ohm) = Volt/Ampère
Unit of the electric conductivity σ ： Since [R] = 1/[σ]L,
3.1.2
[σ] = (Ω m)−1
Origin of the Ohm’s law
Let us now understand the microscopic origin of the Ohm’s law. For this
purpose, we consider a simple model of a conductor, where a friction for
3
the electrons, which is proportional to the velocity, is produced from the
scattering with impurities. The friction coefficient is parametrized as m/τ ,
which has the correct dimension.
m
or
dv
m
= eE − v
dt
τ
dv
eE
1
=
− v
dt
m
τ
(3.2)
second term = friction due to scattering with crystal lattice
and/or impurities
m = “effective mass” of an electron
e = “effective charge” of an electron
τ = relaxation time (average time between collisions)
Remarks:
• The meaning of the “relaxation time” will become clear below.
• Carriers of the charges are actually not the bare electrons but “elementary
excitations”, which are electrons “dressed up ” with the effects of complex
many-body interactions. That is why we used the words “effective mass”
and “effective charge”.
Solution of the differential equation:
qqq
dv
dt
= −
eE
τ
v− mτ
µ
¶
eE
t
ln v −
τ
= − + const
m
τ
eE
τ = ce−t/τ
v−
qqq
m
4
(3.3)
Let us rewrite this result in terms of the initial velocity v0 at
t = 0. Substituting t = 0 in (3.3), we get
eE
τ +c
m
eE
c = v0 −
τ
mµ
¶
eE
eE
v =
τ + v0 −
τ e−t/τ
m
m
v0
v0 =
qqq
qqq
v
vf
τ
0
(3.4)
In particular, at t → ∞ the velocity approaches the final
velocity given by
vf =
eE
τ
m
Thus, after a time much longer than the relaxation time (t >> τ ), so that
many collisions have taken place, the velocity becomes a constant final
velocity vf , irrespective of the initial velocity.
If one is only interested in this final velocity, it is not necessary to solve the
differential equation. In fact evaluating the original equation at late time,
we get
dvf
m
= 0 = eE − vf
dt
τ
eE
vf =
τ
m
m
qqq
From this we obtain the Ohm’s law in the following form
qqq
j = ρvf =
eτ
σ =
ρ
m
³ eτ ´
ρ E
m
If we set n =number density = number of electrons per unit volume, then
ρ = ne and the conductivity σ is expressed as
σ=
ne2 τ
.
m
5
3.2
Joule’s heat （heat due to friction）
As we have seen above, when a current flows in a conductor, actually a lot
of collisions take place and the resistence is produced. In this process, a part
of the energy is converted to heat. This is called the Joule’s heat.
Since the electric field produces the motion of the electrons against the resistence, it performs a work. The work done per unit time on the charge ρ in
a unit volume is called the rate of work and is given by the time derivative
of the work. This is computed as
dw
d~x
~
= f~ ·
= f~ · ~v
(f~ = ρE)
dt
dt
~ · ~v = (ρ~v ) · E
~
= (ρE)
~
= ~ · E
(3.5)
~ we get
Substituting the Ohm’s law ~ = σ E
dw
1
= σE 2 = j 2
dt
σ
This is the energy converted to Joule’s heat per unit time.
Example:
Joule’s heat produced in a wire of resistence R per unit time:
I
S
1`
R =
σS
Volume = S`
j =
⇒
1
RS
=
σ
`
Therefore, the Joule’s heat per unit time of the entire wire is obtainedy by
multiplying by the volume:
1 2
RS I 2
Joule’s heat =
j (S`) =
S` = RI 2 = V I
2
σ
` S
(3.6)
This is the well-known formula for the Joule’s heat taught in high school.
6
Exercise 3.1
The electric conductivity of cupper is σ = 5.8×107 Ω−1 m−1 .
Suppose we let a current of 1A flow through a cupper wire of length 1m and
the cross section 1mm2 .
(1) What is the difference of the voltage between the end points ?
(2) How many Watts of Joule’s heat is produced along this wire ?
3.3
Conservation of electric charge
In E & M, a fundamental law which is as important as the Maxwell’s equations is the law of conservation of electric charge. (This corresponds to the
conservation of mass in hydrodynamics.)
Consider the total amount of charge going out per unit time from an arbitrary
boundary surface as shown in the figure.
n̂
~j
V
Surface Σ
By using the Gauss’ theorem, this is equal to
Z
Z
~ · ~
dS n̂ · ~ =
d3 x∇
Σ
V
If the conservation of the charge holds, then the charge inside must decrease
by the amount
Z
∂
−
d3 xρ(x) (> 0)
∂t V
which is positive. Therefore we have
¶
Z µ
∂ρ
~
∇ · ~ +
=0
∂t
V
7
Since this holds for an arbitrary region, the integrand must vanish and we
get
~ · ~ + ∂ρ = 0
∇
∂t
This is the expression of the conservation of electric charge.
In particular, for a steady current , for which ∂ρ/∂t = 0, it gets simplified to
~ · ~ = 0
∇
3.4
Electrical circuit made with resistors and
batteries
As a system where steady state currents are present, consider an electrical
circuit made of resistors and batteries, as in the figure. We wish to determine the currents flowing inside, given the information of the
extermal currents. This appears quite complicated, but the principle for
solving this problem consists only of two simple laws due to Kirchhoff.
I1ext
R1
R5
V2
I2ext
I4ext
R4
V1
V3
R3
R2
I3ext
8
1. Kirchhoff’s first law：= conservation of charge:
tion,
X
X
Ii +
Ijext = 0
i
At each junc-
j
Here the sign convention of the current is positive for those going
out of the junction.
2. Kirchhoff’s second law：=consistency of the distribution of electrostatic
potentials: For each closed circuit
X
X
Ri Ii =
Vk
i
k
This expresses the requirement that sum of the voltage drop due to
the resistors = sum of the voltage increases due to the batteries.
If we specify Ri , Vk , Ijext the distribution of the currents inside must be determined uniquely. This fact is intimately related to the famous Euler’s
theorem in topology.
9
Theorem : The number of equations expressing the
Kirchhoff’s first and the second laws is always one more
than the number of internal currents Ii .
Proof: Note the following facts
# of equations for 1st law = # of vertices of the graph ≡ V
# of equations for 2nd law = # of faces of the graph ≡ F
We can now invoke the Euler’s theorem for a plane figure with boundary.
V +F = L+1
V
= # of vertices
F = # of faces
L = # of edges (lines)
Examples
triangle
V
= 3,
F = 1,
L=3
n-gon
V
= n,
F = 1,
L=n
10
We can make a more complicated figure by gluing simpler figures along an edge, as shown in the figure. In this
process, focus on the change of V, F, L. Then, clearly
V
= V1 + V2 − 2
L = L1 + L2 − 1
F = F1 + F2
qqq
V + F = V1 + V2 − 2 + F1 + F2
= L1 + 1 + L2 + 1 − 2
V :4→2
L:2→1
= L1 + L2 = L + 1
Thus through this procedure the Euler’s theorem continues to hold and hence
it is valid for any plane figure.
Since L corresponds to the number of internal currents, the theorem is
proved.
The reason for the existence of one more equation than is necessary is that
it must ensure the consistency of the external currents, that is
X
Ijext = 0
j
So as long as we give such consistent currents, it is automatically satisfied.
Examples of simple circuits:
1. Series circuit and combined resistence
I
R1
R2
I
=
I
R
I
V1 = IR1 V2 = IR2
In this case, from the sum of the voltage drop, we easily find (without invoking the Kirchhoff’s laws)
qqq
V = V1 + V2 = (R1 + R2 )I = RI
(3.7)
R = R1 + R2 = combined resistence
(3.8)
2. Parallel circuit and combined resistence
11
R1
I1
I
I
I
=
R I
I2
R2
Now let us apply the Kirchhoff’s law honestly and find the currents I1 , I2 and
the combined resistence R.
(i) Current conservation at each junction: I = I1 + I2
(ii) Consistency of the voltage configuration: Since there is no batteries in
this case, the consistency is expressed as
R1 I1 − R2 I2 = 0
qqq
(3.9)
R1 I1 = R2 I2 = V = RI
(3.10)
Solving for the currents, we get
I1 =
V
,
R1
I2 =
V
,
R2
I=
V
R
(3.11)
Now using I1 + I2 = I we have
V
V
V
+
=
R1 R2
R
(3.12)
Dividing by V we get the famous formula for the combined resistence R for
the parallel circuit
1
1
1
=
+
R
R1 R2
(3.13)
3. Example of a circuit with batteries inserted
I3ext
I1
I3
V1
I1ext
R2
I2
12
R3
I2ext
(1)
I1 = I2 + I1ext
(2)
I2 = I3 + I2ext
(3)
I3 = I1 + I3ext
(4)
V1 = I2 R2 + I3 R3
From (1)+(2)+(3), we easily see that the condition for the extermal currents
I1ext + I2ext + I3ext = 0 is indeed necessary. The rest is to solve this simple set
of equations. For example, to get I2 substitute the relation obtained from
(1) and (3), namely I3 = I2 + I1ext + I3ext , into (4). Then we get
qqq
V1 = I2 (R2 + R3 ) + (I1ext + I3ext )R3
£
¤
1
I2 =
V1 − (I1ext + I3ext )R3
R2 + R3
Exercise 3.2
For the circuit shown below, express the intermal currents
Ii , i = 1 ∼ 5 in terms of J 、R and V .
3J
I2
I1
R
I3
2R
V
R
2J
2R
I4
I5
13
J