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Today’s agenda:
Electric Current.
You must know the definition of current, and be able to use it in solving problems.
Current Density.
You must understand the difference between current and current density, and be able to
use current density in solving problems.
Ohm’s Law and Resistance.
You must be able to use Ohm’s Law and electrical resistance in solving circuit problems.
Resistivity.
You must understand the relationship between resistance and resistivity, and be able to
use calculate resistivity and associated quantities.
Temperature Dependence of Resistivity.
You must be able to use the temperature coefficient of resistivity to solve problems
involving changing temperatures.
Current Density
When we study details of charge transport, we use the concept
of current density.
Current density is the amount of charge that flows across a unit
of area in a unit of time.
Current density: charge per area per time (current / area).
A current density J flowing through an infinitesimal area dA
produces an infinitesimal current dI.
dA
J
dI  J  dA
Current density is a vector. Its
direction is the direction of the
velocity of positive charge carriers.
The total current passing through A is just
I

surface
J  dA
No OSE’s on this page.
Simpler, less-general
OSE on next page.
Current density: charge per area per time.
J
A
If J is constant and parallel to dA (like in a wire), then
I
I   J  dA  J  dA  JA  J 
A
surface
surface
Now let’s take a “microscopic” view of current and calculate J.
vt
q
v
A
If n is the number of charges
per volume, then the number of
charges that pass through a
surface A in a time t is
number
 volume   n  vt A 
volume
The total amount of charge passing through A is the number of
charges times the charge of each.
vt
q
v
Q  nqvt A
A
Divide by t to get the current…
Q
I
 nqv A
t
…and by A to get J:
J  nqv .
To account for the vector nature of the current density,
J  nqv
Not quite
“official” yet.
and if the charge carriers are electrons, q=-e so that
Je  n e v.
Not quite
“official” yet.
The – sign demonstrates that the velocity of the electrons is
antiparallel to the conventional current direction.
Currents in Materials
Metals are conductors because they have “free” electrons,
which are not bound to metal atoms.
In a cubic meter of a typical conductor there roughly 1028 free
electrons, moving with typical speeds of 1,000,000 m/s…
…but the electrons move in random directions, and there is no
net flow of charge, until you apply an electric field.
Thanks to Dr. Yew San Hor for this slide.
E
electron “drift” velocity
inside a
conductor
just one
electron
shown, for
simplicity
The electric field accelerates the electron, but only until the
electron collides with a “scattering center.” Then the electron’s
velocity is randomized and the acceleration begins again.
Some predictions based on this model are off by a factor or 10
or so, but with the inclusion of some quantum mechanics it
becomes accurate. The “scattering” idea is useful.
A greatly oversimplified model, but the “idea” is useful.
Even though some details of the model on the previous slide
are wrong, it points us in the right direction, and works when
you take quantum mechanics into account.
In particular, the velocity that should be used in
J  n q v.
is not the charge carrier’s velocity (electrons in this example).
Instead, we should the use net velocity of the collection of
electrons, the net velocity caused by the electric field.
Quantum mechanics shows us how to deal
correctly with the collection of electrons.
This “net velocity” is like the terminal velocity of a parachutist;
we call it the “drift velocity.”
J  n q vd .
It’s the drift velocity that we should use in our equations for
current and current density in conductors:
J  n q vd
I  nqvd A
I
vd 
nqA
Example: the 12-gauge copper wire in a home has a crosssectional area of 3.31x10-6 m2 and carries a current of 10 A.
The conduction electron density in copper is 8.49x1028
electrons/m3. Calculate the drift speed of the electrons.
I
vd 
nqA
I
vd 
neA
10 C/s
vd 
(8.49 1028 m -3 )(1.60 1019 C)(3.31106 m 2 )
vd  2.22 104 m/s