Solutions to Homework Questions 3

Physics 112 Homework 3 (solutions) (2004 Fall)
Solutions to Homework Questions 3
Chapt17, Problem-1: If a current of 80.0 mA exists in a metal wire, how many electrons flow past
a given cross section of the wire in 10.0 min? Sketch the directions of the current and the electrons’ motion.
Solution:
The charge that moves past the cross section is !Q = I (!t), so given the charge on a single electron, we
can calculate the number of electrons passing through a the given cross-section as being
( )
n = !Q e
= I !t
e
( ) 10.0 min
= 80.0 !10 "3 C s
[( )( 60.0 s min )]
1.60! 10 -19 C
= 3.00! 1020 electrons
Note that since we are asked for the number of electrons, we can ignore the fact that the electrons are
actually negatively charged.
For the sketch, simply remember that the negatively charged electrons move in the direction opposite to
the conventional current flow.
Chapt17, Problem-5: In the Bohr model of the hydrogen atom, an electron in the lowest energy
state moves at a speed of 2.19x10 6 m/s in a circular path having a radius of 5.29x10 –11 m. What is the
effective current associated with this orbiting electron?
Solution:
We are going to use the fact that current is amount of charge ‘flowing’ past in a given time. However the
key word to spot here is ‘circular’ – the electron can be thought of as orbiting the proton at the center of
the atom.. The period of the electron in its orbit is T = 2!r v (distance around the orbit divided by the
speed at which it is moving – review PHYS111 if you are unsure of Circular motion). So the current
represented by the orbiting electron is
I = !Q
!t = e T = ve
2"r
(
= 2 .19! 106 ms) ( 1.60! 10 "19 C )
= 1.05 !10 "3 Cs= 1.05 mA
2 # 5.29 !10 "11 m
( )
Again we can ignore the fact that the electrons are actually negatively charged since we are asked for the
current.
Chapt17, Problem-8: An aluminum wire with a cross-sectional area of 4.0x10 –6 m 2 carries a
current of 5.0 A. Find the drift speed of the electrons in the wire. The density of aluminum is 2.7 g/cm 3 .
(Assume that one electron is supplied by each atom.)
Solution:
Using the assumption (that the question tells us to make) that each aluminum atom contributes one
electron, then the density of charge carriers is the same as the number of atoms per cubic meter. This is
density
n =
mass per atom = !
= N A !
M N A
M ,
(
or n = 6.02 !1023 mol) [( 2.7 g cm 3
)10 ( 6 cm 3 1 m 3
)]
= 6.0 !10 28 m 3
26.98 g mol
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Physics 112 Homework 3 (solutions) (2004 Fall)
So we can now use the equation for the drift speed of the electrons to get
I
v d =
neA =
5.0 C s
( 6.0! 10 28 m 3
)1.60 ( ! 10 "19 C )( 4.0 !10 "6 m 2
) = 1.3!10 "4 ms
We can ignore the fact that the electrons are actually negatively charged since we are asked for a speed
Chapt17, Problem-11: A person notices a mild shock if the current along a path through the
thumb and index finger exceeds 80 µA. Compare the maximum allowable voltage without shock across the
thumb and index finger with a dry-skin resistance of 4.0x10 5 ! and a wet-skin resistance of 2 000!
Solution:
This is a straightforward application of Ohm’s Law. Let ("V) max be the maximum voltage before a person
feels the shock, then
(!V) max
= I max R = ( 80 "10 #6 A)R
Thus, if R = 4.0 ! 10 5 " , (!V) max
= 32 V ,
but the skin is wet, and R = 2000 !, (!V) max
= 0.16 V
Chapt17, Problem-12: Suppose that you wish to fabricate a uniform wire out of 1.00 g of copper.
If the wire is to have a resistance of R = 0.500!, and if all of the copper is to be used, what will be (a) the
length and (b) the diameter of this wire?
Solution:
This is a little harder. The question does not supply all the information required, and you are required to
realize what pieces of information are missing. .[I would give you all the information in a Quiz or Exam !]
However there is a strong hint in the question: you are given the mass of the material, and are required to
calculate the area & length – these last two give a volume, and mass and volume are related through
density. Furthermore resistance, cross-sectional area & length are related through the resistivity of a
material. Thus the missing information is actually the density of copper (which is given in Ex17.2 in the
text), and the resistivity of Copper (Table 17.1 – since no temperature is given, and the temperature
coefficient of resistivity is small, it is OK to assume room-temperature). So armed with these values, first
lets calculate the volume of copper we have at our disposal.
V =
m
density = 1.00! 10 "3 kg
8.92 !10 3 kg m 3 = 1.12 !10 "7 m 3
Now since, V = A!L , this gives A!L = 1.12 "10 #7 m 3 . Lets call that eqn(1)
(a) OK, well we know the resistance we require, and so can find another expression relating the length and
cross-sectional area using R = !L ,. Subsituting in the values, we find
A
"
A =
! % "
$ ' L =
1.70( 10)8 *+m %
# R&
$
' L = ( 3.40( 10 )8 m )L .
# 0.500 * &
So inserting this expression for A into eqn(1) gives
( 3.40 ! 10 "8 m)L 2 = 1.12! 10 "7 m 3 , which yields L = 1.82 m
(b) From eqn(1) we can simply get the area, A = !d2
4 = 1.12" 10#7 m 3
,
L
but note that it is actually the diameter that is asked for in the question, so
( )
d = 4 1.12 !10 "7 m 3
#L
= 2.80 ! 10 "4 m = 0.280 mm
( )
= 4 1.12 ! 10"7 m 3
# 1.82 m
( )
2

Physics 112 Homework 3 (solutions) (2004 Fall)
Chapt17, Problem-19: The breathing monitor shown
in the Figure girds the patient with a mercury-filled rubber tube
and measures the variation of the tube resistance. The tube has
an unstretched length of 1.25 m and an inside diameter of 2.51 mm.
The monitor is connected to a 100-mV power supply, and the total
resistance of the circuit is that due to the mercury plus 1.00! (an
internal resistance of the power supply). Determine the change of
current through the monitor as a patient draws in a breath and stretches
the hose by 10.0 cm. Take "(Hg) = 9.40x10 –7 ! m.
Solution:
When the tube is stretched, the cross-sectional area decreases.
Since the volume of mercury is constant, we have a simple
Expression relating the initial & final (stretched) dimensions, namely V = A f !L f = A i !L i ,
" !
i.e.
4 d 2 % "
$ f ' L
# & f = ! 4 d i 2 %
$ ' L
# & i . So this gives us d 2 f = d 2 i ( L i L f ).
The total resistance of the circuit is (the “plus 1.00!” just makes the problem more realistic)”
R = r + R Hg = 1.00 ! + " HgL
A
= 1.00 ! + 4" HgL
#d 2
The change in current through the monitor is (application of Ohm’s Law both before & after)
!I = !V " !V !V
!V
=
R f
R i
=
1.00 #+ 4$ "
2
HgL f
%d 2 i
L i
0.100 V
( ) 1.35 m
( ) 2 1.25 m
1.00 # + 4 9.4& 10-7 #' m
% 2 .51&10 "3 m
1.00 # + 4$ HgL i
%d i
2
( ) 2
( )
"
0.100 V
( ) 1.25 m
%( 2.51 & 10 "3 m) 2
1.00 # + 4 9.4&10 -7 #'m
which can be evaluated to finally give !I = " 2.5 #10 "3 A = a 2 .5 mA decrease
( )
Chapt17, Problem-28: A toaster rated at 1 050 W operates on a 120-V household circuit and has
a 4.00-m length of nichrome wire as its heating element. The operating temperature of this element is
320°C. What is the cross-sectional area of the wire?
Solution:
The resistance of the heating element when at its operating temperature is
( ) 2
( ) 2
R = !V
" = 120 V
1050 W = 13.7 #
So we have the resistance, and the length – so if we know the resistivity, we can calculate the crosssectional
area of the wire. Now we are also told an operating temperature, so that should hint that we need
to include the temperature dependence of resistivity (using values in Table 17.1)
R = R 0 [ 1 +! ( T " T 0 )]= # 0 L
[
A 1+ ! ( T " T 0 )], so the cross-sectional area is
A = ! 0 L
[
R 1 +" ( T # T 0 )]
( ) 4.00 m
= 150 $ 10#8 %&m
13.7 %
( )
ie. The requested area is A = 4.90! 10"7 m 2
[ 1+ ( 0.40 $10 #3 ( °C ) #1
)( 320°C# 20.0°C)
]
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Physics 112 Homework 3 (solutions) (2004 Fall)
Chapt17, Problem-32: The output power of the Sun is 4.0x10 26 W. Calculate at eight cents per
kilowatt-hour the cost of running the Sun for one second?
Solution:
From the definition of power, the energy produced by the Sun in 1.0 second is
E =!" t = 4.0 # 10 26 W ( )
( ) 1.0 s
Converting this to kilowatt-hours (Eqn 17.11) E= 4.0 ! 10 26 # 1 kW & # 1 h &
W"s%
$ 10 3 ( % ( = 1.1! 10 20 kWh
W'
$ 3600 s'
At a rate of 8.0¢ per kilowatt-hour, the cost of this energy is
cost = 1.1 !10 20 " $0.08%
( kWh)
$ ' =
# kWh &
8.8 ! 1018 dollars
Chapt17, Problem-37: What is the required resistance of an immersion heater that will increase
the temperature of 1.50 kg of water from 10.0°C to 50.0°C in 10.0 min while operating at 120 V?
Solution:
This is yet another question tht unfortunately required you to scramble around the book ssearching out a
piece of information not supplied by the question. In this case, the missing information was the specific
heat of water (which can be found in Table 11.1) and gives the energy required to raise the temperature of
1kg of water by 1 o C. Armed with this information, one can simply calculate how much energy we require to
raise the amount of water specified in the question by the 40 o C requested.
E = mc (!T
)= ( 1.50 kg) 4186 J kg "°C ( )= 2.51 $10 5 J
( ) 50.0°C# 10.0°C
and, if this is to be added in !t = 10.0 min = 600 s , the power input needed is
!= E "t = 2.51# 105 J
= 419 W
600 s
The power input to the heater may be expressed as != "V
( ) 2
( ) 2
R = !V
" = 120 V
419 W = 34.4 !
( ) 2 R , so the needed resistance is
Chapt17, Problem-41: We estimate that there are 270 million plug-in electric clocks in the United
States, approximately one clock for each person. The clocks convert energy at the average rate of 2.50 W.
To supply this energy, how many metric tons of coal are burned per hour in coal-fired electric-generating
plants that are, on average, 25.0% efficient? The heat of combustion for coal is 33.0 MJ/kg.
Solution:
The total power converted by the clocks is
!= ( 2 .50 W )( 270 "10 6
)= 6.75 " 10 8 W ,
and the energy used in one hour is
E =!" t = 6.75 # 10 8 W ( )= 2 .43# 10 12 J .
( ) 3600 s
The energy input required from the coal is
E
E coal =
efficiency = 2.43 !1012 J
= 9.72 !10 12 J
0.250
The required mass of coal is thus
E
m =
coal
heat of combustion = 9.72! 1012 J
33.0 ! 10 6 Jkg = 2.95 ! 105 kg
or
"
m = ( 2.95 !10 5 1 metric ton %
kg) $
# 10 3 kg
' = 295 metric tons
&
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Physics 112 Homework 3 (solutions) (2004 Fall)
Chapt17, Problem-42: The cost of electricity varies widely throughout the United States;
$0.120/kWh is one typical value. At this unit price, calculate the cost of (a) leaving a 40.0-W porch light on
for two weeks while you are on vacation, (b) making a piece of dark toast in 3.00 min with a 970-W toaster,
and (c) drying a load of clothes in 40.0 min in a 5 200-W dryer.
Solution:
E =!" t = ( 40.0 W )( 14.0 d)
24.0 h d
(a)
( )= 1.34# 10 4 Wh = 13.4 kWh
cost = E! ( rate)= ( 13.4 kWh )( $0.120 kWh)= $1.61
(b) E =!" t = ( 0.970 kW )( 3.00 min) ( 1 h 60 min)= 4.85 #10 $2 kWh
cost = E! ( rate)= ( 4.85 !10 -2 kWh) ( $0.120 kWh)= $0.00583 = 0.583 cents
(c) E =!" t = ( 5.20 kW) ( 40.0 min) ( 1 h 60 min)= 3.47 kWh
cost = E! ( rate)= ( 3.47 kWh )( $0.120 kWh )= $0.416 = 41.6 cents
Chapt17, Problem-52: . Birds resting on high-voltage power lines are a common sight. The
copper wire on which a bird stands is 2.2 cm in diameter and carries a current of 50 A. If the bird’s feet are
4.0 cm apart, calculate the potential difference across its body.
Solution:
The resistance of the 4.0 cm length of wire between the feet is
R = !L (
A = 1.7 " 10#8 $%m) ( 0.040 m )
& ( 0.011 m ) 2 = 1.79" 10 #6 $,
so the potential difference is
!V = IR = ( 50 A) ( 1.79 "10 #6 $ )= 8.9 " 10 #5 V =
89 µV
Chapt17, Problem-59: . (a) Determine the resistance of a lightbulb marked 100 W @ 120 V. (b)
Assuming that the filament is tungsten and has a cross-sectional area of 0.010 mm 2 , determine the length of
the wire inside the bulb when the bulb is operating. (c) Why do you think the wire inside the bulb is tightly
coiled? (d) If the temperature of the tungsten wire is 2 600°C when the bulb is operating, what is the length
of the wire after the bulb is turned off and has cooled to 20°C? (See Chapter 10, and use 4.5 _ 10 –6 /°C as the
coefficient of linear expansion for tungsten.)
Solution:
(
(a) From != "V ) 2
(!V) 2
( ) 2
, the resistance is R =
R
( b ) Solving R = !L
A
for the length gives
( )
( ) 0.010 mm 2
" = 120 V
100 W = 144 !
L = R! A
" = 144 #
% 1 m 2 (
5.6 $10 -8 '
#!m & 10 6 mm 2 * =
)
26 m
( c ) The filament is tightly coiled to fit the required length into a small space
( d ) From L = L 0 [ 1 +! ( T " T 0 )], where ! = 4.5 " 10 #6 °C
L
L 0 =
1+ !( T " T 0 ) = 26 m
1+ 4.5 #10 "6 °C
( ) #1 , the length at T 0 = 20°C is
( ( ) "1
)( 2600°C-20°C) = 25 m
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Physics 112 Homework 3 (solutions) (2004 Fall)
Chapt17, Conceptual-1: A coulomb is a very large unit of charge, and yet currents of several
Amperes are quite common. How is this possible?
Solution:
It is made possible by the extremely high density of charge carriers (~10 28 m -3 ) in metallic conductors. A
typical household copper wire is ~1x10 -6 m 2 in cross-sectional area, hence there are ~10 20 potential charge
carriers per cm.
Chapt17, Conceptual-3: Why don’t the free electrons in a metal fall to the bottom of the metal
due to gravity? And charges in a conductor are supposed to reside on the surface – why don’t the free
electrons all go to the surface?
Solution:
The gravitational force pulling the electron to the bottom of a piece of metal is much smaller than the
electrical repulsion pushing the electrons apart. Thus, free electrons stay distributed throughout the
metal. The concept of charges residing on the surface of a metal is true for a metal with an excess of
charge. The number of free electrons in an electrically-neutral piece of metal is the same as the number of
positive ions – the metal has zero net charge.
Chapt17, Conceptual-5: Newspaper articles often have statements such as “10 000 volts of
electricity surged through the victim’s body.” What is wrong with this statement?
Solution:
[Abridged/adapted version of answer in the back of the Text] A voltage is not something that “surges
through” a completed circuit. A voltage is a potential difference that is applied across a device or circuit.
Thus, it would be more correct to say “1 ampere of electricity surged through the victims body”. Although
this current would have disastrous results on the human body, it doesn’t sound that exciting to people who
have never studied physics. Perhaps the newspapers get it wrong out of ignorance – or perhaps they ‘choose’
to get it wrong for other reasons…
Chapt17, Conceptual-10: Some homes have light dimmers that are operated by rotation of a
knob. What is being changed in the electric circuit when the knob is rotated?
Solution:
The knob is connected to a variable resistor connected in series with the light bulb. As you increase the
magnitude of the resistance in the circuit, the current in the circuit is reduced (since the potential
difference of the total circuit remains constant – usually 120 V). The decrease in current results in a
reduction in the power delivered to (and hence radiated by) the light bulb.
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