ctance
1.1
Ohms law
Background:
When a marble drops in molasses, the drag it experiences is
proportional to the rate it falls (by a drag law due to Stokes.) For this case,
the net force on the marble is given by
F
net
= mg + F
d
(1)
= mg v
(2)
terminal velocity is reached very quickly as the marble drops. At terminal
velocity, the acceleration is zero and
F
net
= 0
(3)
v
=
1
mg
(4)
The important points of this situation are that (1) a terminal velocity is reached
very quickly, (2) the nal velocity is proportional to the applied force, (3) the
work done by the drag force must be of opposite sign as the work done by the
applied force. The power dissipated per unit time is therefore P = vmg.
The situation for charges (say electrons) moving in matter is similar to mar-
bles in molasses. Free charges in a conductor move by bouncing into hot nuclei
doing a kind of random walk through the material. If we are able to apply a
small additional force, then the random walk is modied ever so slightly.
The
collisions with hot nuclei act as a drag force proportional to the drift velocity
of the electrons. We have the situation that the drift velocity of the electrons
v
e
is directly proportional to the applied force F
e
:
v
e
= F
e
(5)
Actually, 5 is a bit too simple minded. In E&M we write
J = f
(6)
Here J is the local velocity of charge carriers times the local charge density. f
is the force per unit charge. Finally, is the conductivity of the material. We
will consider forces per unit charge f that are electrochemical, magnetic, and
electrical. The physics of 6 has little to do with the nature of the force applied
and everything to with a drag on the electrons proportional to the drift velocity.
1 If the force per unit charge is caused by an electric eld, then 6 becomes
J = E
(7)
which is a less familiar form of ohms Law.
1.2
Current density and current
To get to the more familiar form of ohms law, we need to introduce the con-
cept of current.
Current can replace the more exact current density J in a
conductor because of the fact that, in most conductors, there is no accumula-
tion of charge. Not every device behaves in this way (capacitors for instance,
accumulate charge.) However, even rather poor conductors do not like to build
up charge anywhere. This implies that charge per unit time passing any cross
section of a conductor is always the same:
J
I1=dQ1/dt
I2=dQ2/dt
dQ /dt=0
im plies
I1=I2
(8)
This is extremely handy, for it lets us describe what might be an enormously
complex current density J with a single quantity I.
Now suppose the only force acting on a charge is electrostatic, so ohms law
holds
J = E
(9)
In this case the potential dierence between two points is given by
V
=
a
b
E d
(10)
= 1 a
b
J d
(11)
As it turns out, the integral of 11 is always proportional to the current owing
from point a to b. The constant of proportionality depends on the geometry
of the conductor and is given the name resistance. Thus we have the familiar
form of Ohms law:
V = IR
(12)
SAMPLE PROBLEM: What is the resistance between two cocentric spheres of
radius a and b > a if the volume in between is lled with material of conductance
?
2 1.3
The free-electron model of conductivity
The free electron model of conductivity (due to Drude) helps us gain a physical
insight to the fact that J is proportional to f.
In this picture, a conductor
consists of a certain number per unit volume n
e
of charge carriers that move
a very high velocities. With a frequency , These electrons undergo collisions
with nuclei in the matter with a period . Each collision completely scrambles
the electrons original velocity. If a force per charge f is applied, the electrons
of charge e are accelerated, but only during the time between collisions. Thus
the average drift of the electrons due to the eld is given by
v
d
= e
2m f
(13)
The current density J is given by the current density = e n
e
multiplied by
this drift speed:
J = (e)
2
n
e 2m
f
(14)
If we assume is the average distance between nuclei in the conductor and
v
thermal
is the average speed of the electrons, we have
J =
e
2
n
e 2mv
thermal
f
(15)
Thus, according to the free-electron model, the conductivity of a material of
charge density n
e
and charge e is given by
=
e
2
n
e 2mv
thermal
= e
2
n
e 2m
(16)
Notice that the conductivity of a piece of material is independent of the sign
of the charge carrier. In a conductor, the conductivity tends to go down with
increasing temperature (because v
thermal
increases.)
In a semiconductor or
insulator, conductivity tends to increase with increasing temperature because
the number of charge carriers per unit volume (n
e
) increases.
1.4
Electromotive Force
In practice there are two types of forces on charge carriers: the rst is source
forces which we designate with a force per unit charge f
s
. These forces might be
chemical, mechanical, a temperature gradient, or magnetic. The second type
of force per charge are electrostatic forces E. The electrostatic forces prevent
charge from accumulating in any conductor and transfer charges from the source
to distance parts of the circuit. In any case, the total source per unit charge is
given by
f = f
s
+ E
(17)
3 The electromotive force in a circuit is not a force. Instead it is the force per
unit charge integrated about a closed loop: =
f d
(18)
=
f
s
d +
E d
(19)
=
f
s
d
(20)
There is a very important point in this denition of that must not be left out.
f is not the net force per unit charge. Instead it is the sum of electric forces
and those forces that do positive work on the electron.
It neglects resistive
forces! For a steady-state circuit, if resistive forces were not neglected, = 0
or the electrons could not go at a constant velocity.
1.4.1
Case 1: the ideal battery.
Consider an ideal battery that supplies power with no internal resistance (i.e.,
no heat loss) to a resistor in a simple circuit as shown below:
+
R
V
a
b
(21)
In ideal battery supplies charge without resistance. Thus may be taken as
innite, which implies for nite J, f = 0. Given f = 0, the source force and
electrostatic forces in the battery must cancel ( f
s
= E.) Thus, if we call one
terminal of the battery b and the other a, the potential at b with respect to a is
given by
V
=
b
a
E d
(22)
=
b
a
f
s
d
(23)
On the other hand, there are no source forces in the resistor, so
a
b
f
s
d = 0
(24)
nally we have
=
b
a
f
s
d +
a
b
f d = V
(25)
4 1.4.2
CASE 2: Motional emf
You might be thinking the last case was a bit cloudy.
We got to = V by
assuming is innite in the battery instead of by examining the nature of the
chemical forces that actually move the charges about.
The case of motional
, the source of energy is somebody actually pulling a wire and the forces are
much easier to see. The case of motional is also actually more common: it
is the way in which power generators produce the electricity that is delivered to
our houses.
We consider the simple case of a loop of wire being pulled through a region
of nonzero magnetic eld, as shown in the gure below:
R
B
out of
page
v
h
x
(26)
Motion through the magnetic eld causes the charges to feel a force per charge
of magnitude
f
s
= vB
(27)
that is directed from the top of the page to the bottom. The emf is therefore =
f
s
d
(28)
= vBh
(29)
QUESTION: What is the force required to move the loop at the constant
velocity v?
ANSWER: The power dissipated by the resistor is
2
/R = (vBh)
2
/R which
is in turn given by F
applied
v. Thus F
applied
=
B
2
h
2
R
v.
Notice that = vBh
(30)
=
d
B
dt
(31)
where B
=
B da through the loop
(32)
= Bhx.
(33)
The virtue of 31 is that it is true for any loop moving in an arbitrary direction
through an arbitrary eld. This is proved in your text.
5 1.5
Faradays law of induction
The concept of relativity is so ingrained in our teaching that you might not be
as inspired by the following experiment as the likes of Faraday and Einstein. If
you look at the situation too casually, you can almost miss the point.
Faraday did the following three experiments:
(1) He observed a current ow through a circuit because of motional emf
by pulling a wire loop through a magnetic eld as pictured in the last section.
This was not an unexpected result. The force of the magnetic eld on moving
charges was well understood at the time and the current owing was induced
simply by this force.
(2) The next thing he did was to move the magnetic eld rather that the
charges. AN INCREDIBLE THING HAPPENED! He observed a current in
the loop.
BUT WHY? the charges in the loop in this case are stationary.
Magnetic elds do not put forces on stationary charges.
(3) The nal thing he did was to reduce the value of the magnetic eld (using
an electromagnet.)
Again a current was induced in the loop.
BUT WHY?
Magnetic elds to not put forces on stationary charges.
What Faraday concluded was the beginning of the fusion of electricity and
magnetism. He concluded that a changing magnetic eld somehow creates an
electric eld such that
E d = d
B
dt
(34)
=
B
t
da
(35)
Using Stokes law, this can be written
× E = B
t
(36)
It is important to realize what an incredible dierence there is between the
rst experiment and the second two.
The rst experiment involved no great
departure from previous theories of electricity and magnetism:
A magnetic
eld was known to put forces on moving charges. The second two experiments,
however, were truly new results. The force in on the charges that create the
emf in this case are electric, not magnetic, in nature and are created because of
36.
In the end, the fact that the mechanisms for inducing t