# Microscopic Physics of Ohm's Law by rt3463df

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```									         Current & Resistance

- Current and current density
- Ohm’s Law
- Resistivity
- Resistance
Electrical Current
CURRENT I is the amount of positive charge
flowing past a fixed point in the wire per unit time :
dQ
I       if charge dQ flows in time dt
dt
Units: 1 ampere (A) = 1 C/s

Direction: by convention, current is the direction
of movement of positive charge

+    +     +                     -       -   -
+    +     +                     -       -   -
I
I
Electron Velocities
• Random velocities of electrons are large (several km/s)
• Drift velocity is a slow, average motion parallel to E

no field                                       +
E                    F  (  e) E

end
end
start                    start
net displacement
Determining the current

+       +         +       +       +       +
E
+         +       +       +       +       +

L = vd Dt

Charge ΔQ in length L of wire passes through the
shaded disk of area A in time Δt :

ΔQ = (number of charge carriers/volume) x
(charge on each one) x volume
Charge: DQ = n q V
= n q (AL)
= n q A vd Dt (since L=vt)

Current: I = DQ/Dt = nqAvd Dt /Dt
So,    I = nqAvd

vd = average (“drift”) velocity of each charge
q = charge on each particle
n = number of charge carriers per unit volume
A = cross section area
L = length
AL = volume
Example The mobile charges in most metals are electrons,
with about one or two electrons per atom being
free to move. So there are about 1023 charges
per cm3 (or 10 29 m-3).

n  1029 electrons/m3
nq  ne  1.6  1010 C/m3

Take Area = 1 mm2, assume I = 1 A

 vdrift  0.06 mm/s

Which way are the electrons moving?
E = 0 inside ?
Note: in electrostatics, we had E=0 inside a conductor, if
not, charges would move, the conductor would not
be in equilibrium and there would be a current.

For a wire to carry a current, we must have an electric
field inside the conductor, which is caused by the
potential difference between the ends of the wire.
This is no longer electrostatics!
Example
A copper wire of cross sectional area 3x10-6 m2 carries a
current of 10A. Find the drift velocity of the electrons in
this wire. Copper has a density of 8.95g/cm3, and atomic
weight of 63.5g/mole. Avagadro’s number is 6.02x1023 atoms.
Assume each atom contributes one free electron.
Questions

•When you turn on a flashlight, how long does it take for
the electrons from the battery to reach the bulb?

• What happens to the wire as the electrons go through
it?

•If you double the electric field in the wire, does the
acceleration of the electrons double?

•If the electric field remains constant, how do the
electron kinetic energies change with time?
Current Density J

The current density is defined as the current per unit area
in a conductor, where A is the cross section of conductor.
The current density is a vector quantity, units: Amps/m2.

I
J              And since I=nqAvd;    J  nqv   d

A
now v is proportional to the electric field
d

Where  is a constant called the
so J   E           conductivity of the material.

The current density J and the electric field E are both
established in a conductor as a result of a potential
difference across the conductor.
Because J is proportional to the field, current in a wire is
proportional to the potential difference between the ends
of the wire.
L

A                        E

V

Uniform E        V  EL

J  E        I    V   V  I  L 
                       
 A 
 A      L
“Resistance”, R
RESISTIVITY: the inverse of conductivity.
1    (this depends on

    the type of material)

  L   IR
V I
 A 
            Ohm’s Law

L (Uniform wire, Length L,
where   R 
A cross-section area A)

volt
Unit of resistance R is: 1 ohm ()  1
amp
Ohm’s Law
•Current density  field:        J=E
•Current  potential difference: V = IR

 = “resistivity”, has units of  m

A
m2  1
 = “conductivity”    units,
V      m
m
Resistivities of a few materials

ρ (20°C)
(Ω·m)
Cu                     1.7 x 10-8
Al                     2.8 x 10-8
Graphite               3500 x 10-8
Si                     640
Quartz                 ~ 1018
Example
A copper wire, 2 mm in diameter and 30 m in length, has
a current of 5A. Find:
a) resistance         1.7 108   m for Cu
b) potential difference between the ends
c) electric field
d) current density
Quiz
The wire in the previous example is replaced with
a wire of the same length and half the diameter,
carrying the same current. By what factor will
each of the following change?

A)   resistance
B)   potential difference between the ends
C)   electric field
D)   electron number density
E)   electron drift speed
F)   current density
Summary

Current Density: J  I
A  nqvd
Conductivity:   :
    
J  E    (defines  )
1
Resistivity:  


L
Resistance: R  
A
Ohm’s Law: V  IR

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