Physics 2220 – Module 05 Homework
01.
1.0 × 1020 electrons flow through a cross section of a 2.0-mm-diameter iron wire in 5.0 s. What is the
electron drift speed?
From a table in the book, the electron number density of iron is:
n = 8.5 × 1028 electrons / m 3
The number of charge carriers (electrons) in a volume element is:
N e = nA Δ x = nAv d Δ t
The area of the cross section (assuming circular) put in terms of the diameter:
2
A =π r =
π d2
4
Substitute in the number of charge carriers equation and solve for the electron drift speed:
π d2
vd Δ t
4
4 Ne
4 (1.0 × 1020 electrons)
vd =
=
= 7.5 × 10−5 m/s
2
28
3
2
n πd Δt
(8.5 × 10 electrons/m ) π (0.002 m) (5.0 s)
N e = nAvd Δ t = n
02.
1.0 × 1016 electrons flow through a cross section of silver wire in 320 μs with a drift speed of 8.0 × 10 -4
m/s. What is the diameter of the wire?
From a table in the book, the electron number density of silver is:
n = 5.8 × 1028 electrons / m3
The number of charge carriers (electrons) in a volume element is:
N e = nA Δ x = nAv d Δ t
The area of the cross section (assuming circular) put in terms of the diameter:
2
A =π r =
π d2
4
Substitute in the number of charge carriers equation and solve for the diameter:
2
N e = nAvd Δ t = n
d=
d=
03.
√
16
√
πd
vd Δ t
4
4Ne
π n vd Δ t
4 (1.0 × 10 electrons)
= 9.3 × 10−4 m
3
−4
−6
π (5.8 × 10 electrons/m ) (8.0 × 10 m/s) (320 × 10 s)
28
The current in a 100 watt lightbulb is 0.85 A. The filament inside the bulb is 0.25 mm in diameter. What
is the current density in the filament?
Use the definition of current density:
J=
J=
I
I
=
=
A
π r2
0.85 A
0.00025 m
π
2
(
2
)
I
d
π
2
2
()
= 1.73 × 107 A/m2
04.
2.0 × 1013 electrons flow through a transistor in 1.0 ms. What is the current through the transistor?
Start with the definition of current:
I=
Δ Q (nAv d Δ t) q
=
Δ t
Δ t
The number of charge carriers (electrons) in a volume element is:
N e = nA Δ x = nAvd Δ t
Substitute into current equation and solve for current:
I=
05.
N e q (2.0 × 10 13 electrons) (1.60 × 10−19 C)
=
= 3.2 × 10−3 A = 3.2 mA
−3
Δ t
1.0 × 10 s
A 3.0-mm-diameter wire carries a 12 A current when the electric field is 0.085 V/m. What is the wire's
resistivity?
E
J =σ E = ρ
Use Ohm's Law and the definition of current density:
J=
I
A
Combine and solve for the resistivity:
E
I
ρ = A
2
EA
E π d 2 (0.085 V/m) π (0.003 m)
ρ=
=
=
= 5.0 × 10−8 Ω m
I
4I
4 (12 A)
06.
A 0.50-mm-diameter silver wire carries a 20 mA current. What are:
Note the resistivity of silver and number density:
σ = 6.2 × 107 Ω−1 m−1
n = 5.8 × 10 28 electrons/m3
(a)
The electric field in the wire
Use Ohm's Law and the definition of current density:
J =σ E
J=
I
A
Combine and solve for the electric field:
4 (20 × 10−3 A)
J
I
4I
E= σ =
=
=
= 1.64 × 10− 3 V/m
2
2
7
−1
−1
A σ π d σ π (0.00050 m) (6.2 × 10 Ω m )
(b)
The electron drift speed in the wire?
Start with the definition of current:
n vd q π d2
Δ Q (nAv d Δ t) q
I=
=
= nAv d q =
Δ t
Δ t
4
Solve for the electron drift speed:
4 I
nq π d 2
4 (20 × 10−3 A)
= 1.1 × 10−5 m/s
3
−19
2
electrons/m ) (1.6 × 10 C / electron) π (0.00050 m)
vd =
vd =
(5.8 × 10 28
07.
A 1.5 V battery provides 0.50 A of current.
(a)
At what rate (C/s) is charge lifted by the charge escalator?
The rate charge is lifted by the charge escalator is a fancy way of saying current.
I=
(b)
Δ Q
= 0.50 C/s = 0.50 Amps
Δ t
How much work does the charge escalator do to lift 1.0 C of charge?
Think the battery (external force) is doing the work.
W = Δ U = q Δ V = (1.0 C) (1.5 V ) = 1.5 J
(c)
What is the power output of the charge escalator?
Use the definition of power:
P=
08.
(a)
W
Δ U
QΔ V
=
=
= I Δ V = (0.50 A) (1.5 V) = 0.75 W
Δ t
Δ t
Δ t
How long must a 0.60-mm-diameter aluminum wire be to have a 0.50 A current when connected
to the terminals of a 1.5 V flashlight battery?
Note for a aluminum wire:
σ = 3.5 × 107 Ω−1 m−1
ρ = 2.8 × 10−8 Ω m
Use Ohm's Law and the definition of resistance.
ρ L
A
I ρ L 4I ρ L
Δ V=
=
2
A
πd
Δ V = IR
R=
Solve for the length of the wire:
(1.5 V ) π (0.00060 m)2
Δ V π d2
L=
=
= 30 m
4I ρ
4 (0.50 A) (2.8 × 10−8 Ω m)
(b)
What is the current if the wire is half the length?
Use the same equation and solve for the new current
I=
09.
(1.5 V) π (0.00060 m)2
Δ V π d2
=
= 1.0 A
4L ρ
4 (15 m) (2.8 × 10−8 Ω m)
You need to design a 1.0 A fuse that “blows” if the current exceeds 1.0 A. The fuse material in your
stockroom melts at a current density of 500 A/cm2. What diameter wire of the materiel will do the job?
Use current density and solve for the diameter:
J=
d=
√
√
I
4I
=
A
π d2
4 (1.0 A)
4I
−4
=
= 5.0 × 10 m = 0.50 mm
2
2
π J
π (500 A/cm ) (100 cm/m)