Transcript Lecture 8

Electric Current
Chapter 27
Electric Current
Current Density
Resistivity – Conductivity - Resistance
Ohm’s Law (microscopic and macroscopic)
Power Dissipated
Motion of a Point Charge in an Electric Field
A particle of mass m and charge q, placed in an electric field E,
will experience a force F = q E
q
F=qE=ma
•
a
E
The particle will accelerate with acceleration: a = (q/m) E
In one dimension the motion of the particle is described by:
x = x0 + v0 t + a t2/2
v = v0 + a t
v2 = v02 + 2 a (x – x0)
Motion of a Point Charge in an Electric Field
Between the plates of a parallel plate capacitor (vacuum)
L
q
•
m
a
The charge accelerates with
a = (q/m) E = (q/m) (V/L)
V
If the particle (with charge q) starts at rest, and the
potential difference between the plates is V, then the
kinetic energy upon reaching the second plate will equal
the change in potential energy: K = m v2 / 2 = qV
(e)(1V)=1 eV = 1.6 x 10-19 J
ELECTRON VOLT
Electron motion in a conductor
E=0
You probably think of the conduction electrons as normally sitting
still unless pushed by an electric field. That is wrong.
Electrons are in constant motion.
(And it is fast – around 10% of the speed of light.)
But their motion is random – and constantly changing as they bounce
off of impurities. Because of the random motion no net flow occurs.
Electron motion in a conductor
E=0
Electron motion in a conductor
E=0
E=0
E
• An electric field accelerates the electrons (along -E) and
so modifies the trajectories of electrons between collisions.
• When E is nonzero, the electrons move almost randomly
after each bounce, but gradually they drift in the direction
opposite to the electric field.
• This flow of charge is called a current.
Electric current
I
dq passes through
a wire
in time dt
• We define the electric current as the movement of
charge, across a given area, per unit time:
I = dq / dt
• SI unit of current: 1 C/s = 1 Ampere (Amp)
• The direction of the current is the direction in which
positive charges would move.
• Electrons move opposite to the direction of the current.
Current density
I
Cross-sectional area A
• If current I flows through a surface A, the current
density J is defined as the current per unit area:
J = I/A
• After an electron collides with an impurity, it will
accelerate under an E field with a = e E / m.
• Suppose the average time between collisions is t.
Then the average velocity is vd = a t = e E t / m.
This velocity is called the electron drift velocity
(which turns out to be much less than the speed of
light)
Current density
A
vd
vdDt
Density of electrons: n
Number of electrons: N=n(AvdDt)
• Construct the above volume.
• In time Dt all the electrons in it move out
through the right end.
• Hence the charge per unit time (the current) is
I = (N e) / Dt = n e A vd Dt / Dt = n e A vd
• The current density is
J = I / A = n e vd = (n e2 t / m) E
Example: What is the drift velocity of electrons in a
Cu wire 1.8 mm in diameter carrying a current of 1.3 A?
In Cu there is about one conduction electron per atom.
28
3
n

8.49x10
m
The density of Cu atoms is
Example: What is the drift velocity of electrons in a
Cu wire 1.8 mm in diameter carrying a current of 1.3 A?
In Cu there is about one conduction electron per atom.
28
3
n

8.49x10
m
The density of Cu atoms is
Find vd from J=I/A=1.3A/(p(.0009m)2)=5.1x105 A/m2
Example: What is the drift velocity of electrons in a
Cu wire 1.8 mm in diameter carrying a current of 1.3 A?
In Cu there is about one conduction electron per atom.
28
3
n

8.49x10
m
The density of Cu atoms is
Find vd from J=I/A=1.3A/(p(.0009m)2)=5.1x105 A/m2
5
Now use

2
J
5.110 A /m
vd  
ne (8.49 10 28 /m 3 )(1.6 1019 C)
Example: What is the drift velocity of electrons in a
Cu wire 1.8 mm in diameter carrying a current of 1.3 A?
In Cu there is about one conduction electron per atom.
28
3
n

8.49x10
m
The density of Cu atoms is
Find vd from J = I/A = 1.3A/(p(.0009m)2) = 5.1x106 A/m2
5
Now use
2
J
5.110 A /m
vd  
ne (8.49 1028 /m 3 )(1.6 1019 C)
5
vd  3.8 10 m /s

Much less than one millimeter per second!
Ohm’s Law
• We found that J= (ne2t/m)E, that is, that the
current J is proportional to the applied electric
field E (both are vectors):
•
Ohm’s Law
• We found that J= (ne2t/m)E, that is, that the
current J is proportional to the applied electric
field E (both are vectors):
•
•
J=sE
Ohm’s Law
Ohm’s Law
• We found that J= (ne2t/m)E, that is, that the
current J is proportional to the applied electric
field E (both are vectors):
•
•
J=sE
Ohm’s Law
s is the Conductivity, s  J / E.
Units are (A/m2) divided by (V/m) = A/(Vm)
Ohm’s Law
• We found that J= (ne2t/m)E, that is, that the
current J is proportional to the applied electric
field E (both are vectors):
•
•
J=sE
Ohm’s Law
s is the Conductivity, s  J / E.
Units are (A/m2) divided by (V/m) = A/(Vm)
• It is useful to turn this around and define the
Resistivity as r = 1/s, so E= rJ.
Units of r are (V/A)m
Ohm’s Law
s and r are dependent only on the material,
- not its length or area.
However, consider a metal rod of resistivity r:
+V
r, area A
L
0 volts
Ohm’s Law
s and r are dependent only on the material,
- not its length or area.
However, consider a metal rod of resistivity r:
+V
r, area A
L
E= rJ
(V/L)  r (I/A)
V = (rL/A) I
0 volts
Ohm’s Law
s and r are dependent only on the material,
- not its length or area.
However, consider a metal rod of resistivity r:
+V
r, area A
0 volts
L
E= rJ
(V/L)  r (I/A)
V = (rL/A) I
V=IR
with R=rL/A
Ohm’s Law
The macroscopic form V = I R is the most
commonly used form of Ohm’s Law.
R is the Resistance
It depends on the material type and shape:
R= rL/A
Units: ohms (W).
As r = R A / L, common units for the
resistivity r are Ohm-meters.
Similarly, common units for the
conductivity s = 1 / r are (Ohm m)-1 or Mho/m
Ohm’s Law
J=sE
microscopic form
s = conductivity
r = 1/s = resistivity
s and r  1/ s are dependent only on the material,
(NOT on its length or area)
V = I R macroscopic form
R depends on the material type and shape
A
L
R = r L / A = resistance
Electrical Power Dissipation
• In traveling from a to b ,
energy decrease of dq is:
dU = dq V
•
•
•
•
V
a
dq
b
Resistance, R
Now, dq = I dt
Therefore, dU = I dt V
Rate of energy dissipation is dU / dt = I V
This is the dissipated power, P. (Watts, or Joules /sec)
Electrical Power Dissipation
• In traveling from a to b ,
energy decrease of dq is:
dU = dq V
•
•
•
•
V
a
dq
b
Now, dq = I dt
Resistance R
Therefore, dU = I dt V
Rate of energy dissipation is dU / dt = I V
This is the dissipated power, P. [Watts, or Joules /sec]
• Hence,
or
P=IV
P = I 2 R = V2 / R
Resistivities of Selected Materials
Material
Resistivity [W m]
Aluminum
Cooper
Iron
2.65x10-8
1.68x10-8
9.71x10-8
Water (pure)
Sea Water
Blood (human)
2.6x105
0.22
0.70
Silicon
640
Glass
Rubber
1010 – 1014
1013 – 1016
What is the resistance of a Cu wire, 1.8 mm in diameter,
and 1 m long ?.
R = r L / A  R = (1.68x10-8) 1 / p (0.0009)2 W
R = 6.6x10-3 W
What is the voltage difference between the extremes of a
Cu wire, 1.8 mm in diameter, and 1 m long, when the
current is 1.3 A ?.
V = I R = (1.3 A) 6.6x10-3 W = 8.6x10-3 V
What is the power dissipated in a Cu wire, 1.8 mm in
diameter, and 1 m long, when the current is 1.3 A ?.
P = I2 R = (1.3)2 6.6x10-3 W = 1.12x10-2 W