Download Selected Problems Tutorial #2

Selected Problems Tutorial # 2
1. One end of an Al wire (diameter 2.5mm) is welded
to one end of a Cu wire (diameter 1.8mm). The
composite wire carries current I=1.3A. What is
the current density in each wire?
Answer:
Cross-sectional area A of Al wire
A Al
1 2
 d  4.9110 6 m 2
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Current density (constant within wire except near
the junction)
I
5
2
J Al 
 2.6  10 A / m
A Al
For copper: ACu = 2.54 x 10-6 m2
JCu = 5.1 x 105 A/m2
2. What is the drift speed of the conduction
electrons in the copper wire of Problem 1?
Answer:
In Cu, there is nearly 1 conduction electron per atom
(on the average).
n = number of electrons per unit volume equals the
number of atoms per unit volume
n

NA

M
Avogadro’s number: NA = 6.02 x 1023 /mol
 = 9.0 x 103 kg/m3
density of copper:
molar mass of copper: M = 64 x 10-3 kg/mol
n = 8.47 x 1028 electrons/m3
J
51
.  105 A / m2
vd 

ne (8.47  1028 electrons / m3 )(16
.  1019 C / electron)
= 3.8 x 10-5 m/s = 14 cm/s
Electrons drift very slowly.
3. A strip of silicon has a width of 3.2mm and a
thickness of 250  m, and it carries a current
I=5.2mA. The silicon is an n-type semiconductor,
having been “doped” with a controlled phosphorus
impurity. The doping has the effect of greatly
increasing n, the number of charge carries per
unit volume, as compared with the value for pure
silicon. In this case, n=1.5 x 1023 m-3.
Q: a) What is the current density in the strip?
Answer:
w=width
t=thickness
I
J 
 6500 A / m 2
wt
Q: b) what is the drift speed?
Answer:
2
J
6500A / m
vd 

23
-3
19
ne (1.5  10 m )(16
.  10 C)
= 0.27m/s = 27cm/s
Note: The drift speed (0.27m/s) calculated for the
electrons in this doped silicon semiconductor
is much greater than the drift speed
(3.8 x 10-5 m/s) obtained in Problem 2 for the
conduction electrons of the metallic copper
conductor.
4. What is the strength of the electric field present in:
a) a copper conductor in which the current density is
J = 5.1  10 A / m .
Answer:
8
.  10   m
Resistivity of copper   169
5
2
E = J = (1.69  10-8   m)(5.1  105 A / m2 )
3
 8.6  10 V / m
b) an n-type silicon semiconductor in which the
current density is J=6500A/m2.
Answer:
Resistivity of n-type silicon semiconductor
  8.7  104   m
E = J = (8.7  10-4   m)(6500A / m2 )
=5.7V/m
c) comment on the results obtained in a) and b).
Answer:
The electric field in the silicon semiconductor
(5.7V/m) is considerably higher than that in the
copper conductor (8.6 x 10-3 V/m). This is due
to the much lower concentration of charge carriers
in silicon than in copper. For a given current density,
the fewer charge carriers in silicon must drift faster,
which means that the electric field acting on them
must be stronger.
5. a) What is the mean free time  between collisions
for the conduction electrons in copper?
Answer:
m
 
ne 2 q
n=8.47 x 1028 electrons/m3 (see example problems)
A
31
9.1  10 kg
14
A 

2
.
5

10
s
-17
3.66  10 kg / s
where units were converted as
C2   C2  V C2  J / C kg  m2 / s2 kg
 2
 2


2
2
m
m A m C / s
m /s
s
b) What is the mean free path  for these collisions?
Assume an effective speed veff of 1.6 x 106 m/s.
Answer:
  veff  (2.5  10
14
s)(1.6  10 m / s)
=4.0 x 10-8 m = 40nm
This is about 150 times the distance between nearest
neighbor ions in a copper lattice.
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