(twb493) brown Ch19-h1 chiu (56565)
This print-out should have 24 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering.
001 10.0 points
Why does the the brightness of a bulb not
change noticeably when you use longer copper
wires to connect it to the battery?
emf = V1 + Vbulb + V2
The potential dierence across a connecting wire is V1 = E1 L1 . Because the cross
sectional area of the connecting wires is much
larger than the cross sectional area of the lament, then for each wire E1
Ebulb . Thus,
V1
Vbulb .
For (A), we need to write down the above
equation in terms of the energy equation,
U = q V . Thus,
q (emf) = q V1 + q Vbulb + q V2
Since V1
Vbulb , then the energy loss
of electrons through the connecting wire is
much smaller than through the bulb and q V1
is negligible. As a result, emf Vbulb .
C) is incorrect because E=0 necessarily implies I=0. (D) is incorrect because it violates
charge conservation. (E) is incorrect because
current is not used up.
Explanation:
(A) and (B) are correct.
For (B), when there are two connecting
wires, applying the Loop Rule gives
+
Correct answer: A,B.
002 (part 1 of 5) 10.0 points
The following questions correspond to the gure shown, consisting of two ashlight batteries and two Nichrome wires of dierent
lengths and dierent thicknesses as shown
(corresponding roughly to your own thick and
thin Nichrome wires).
+
A. Very little energy is dissipated in the
thick connecting wires.
B. The electric eld in the connecting wires
is very small, so emf Ebulb Lbulb .
C. The electric eld in the connecting wires
is zero, so emf = Ebulb Lbulb .
D. The current in the connecting wires is
smaller than the current in the bulb.
E. All the current is used up in the bulb, so
the connecting wires dont matter.
1
The thin wire is 60 cm long, and its diameter is 0.22 mm. The thick wire is 17 cm long,
and its diameter is 0.41 mm.
The emf of each ashlight battery is 1.3 V.
Determine the steady-state electric eld inside each Nichrome wire. Remember that in
the steady state you must satisfy both the
current node rule and energy conservation.
These two principles give you two equations
for the two unknown elds.
Find the electric eld in the thin wire rst.
Correct answer: 4.00649 V/m.
Explanation:
Apply the loop rule. Call the thin wire 1
and the thick wire 2.
2 emf E1 L1 E2 L2 = 0
Apply the node rule.
i1 = i2
nA1 uE1 = nA2 uE2
A1 E 1 = A2 E 2
A2
E1 =
E2
A1
(0.41 mm)2
=
E2
(0.22 mm)2
= 3.47314E2
Substitute into the loop equation.
brown (twb493) Ch19-h1 chiu (56565)
2 emf 3.47314E2 L1 E2 L2 = 0
2 emf E2 (3.47314L1 + L2 ) = 0
2 emf
3.47314L1 + L2
2(1.3 V)
=
(3.47314)(0.6 m) + (0.17 m)
= 1.15356 V/m
E2 =
So, E1 = 3.47314 E2 = 4.00649 V/m .
003 (part 2 of 5) 10.0 points
Find the electric eld in the thick wire.
Correct answer: 1.15356 V/m.
Explanation:
See Explanation 1.
004 (part 3 of 5) 10.0 points
How long does it take an electron to drift
through both Nichrome wires if electron mobility in Nichrome is 9 105 (m/s)(N/C).
Correct answer: 3301.4 s.
Explanation:
v = uE , so calculate drift speed and t
for each wire separately since v is dierent for
each wire.
m/ s
(4.00649 V/m)
V /m
= 0.000360584 m/s
v1 =
9 105
t 1 =
L1
v1
0. 6 m
0.000360584 m/s
= 1663.97 s
=
For wire 2,
2
v2 = uE2
m/ s
(1.15356 V/m)
V /m
= 0.000103821 m/s
L2
t 2 =
v2
0.17 m
=
0.000103821 m/s
= 1637.44 s
=
9 105
t = t 1 + t 2
= 1663.97 s + 1637.44 s
= 3301.4 s .
005 (part 4 of 5) 10.0 points
On the other hand, about how long did it take
to establish the steady state when the circuit
was rst assembled?
1. 0.00256667 s
2. 2.56667 106 s
3. 2.56667 s
4. 2566.67 s
5. 2.56667 109 s correct
Explanation:
The minimum time required to reach steady
state is given by the speed of light. Using
L = L1 + L2 = 0.77 m
L
t
L
t =
c
0.77 m
=
3 108 m/s
c=
= 2.56667 109 s .
Notice that this is much smaller than the
time interval for a mobile electron to travel
around the circuit.
006 (part 5 of 5) 10.0 points
brown (twb493) Ch19-h1 chiu (56565)
There are about 6 1028 mobile electrons per
cubic meter in Nichrome. How many electrons
cross the junction between the two wires every
second?
Correct answer: 8.22419 1017 electrons/s.
Explanation:
The electron current is
i1 = nA1 uE1
3
009 (part 3 of 3) 10.0 points
The electron current in the rst circuit
(Nichrome) is i1 . The electron current in the
second circuit (wire with the higher mobility)
is i2 . Which of the following statements is
true?
1. i2 > i1 correct
2. i2 = i1
(0.00022 m)2
4
m/ s
9 105
(4.00649 V/m)
V /m
= (6 1028 m3 )
= 8.22419 1017 electrons/s .
007 (part 1 of 3) 10.0 points
A Nichrome wire 63 cm long and 0.34 mm in
diameter is connected to a 1.9 volt ashlight
battery.
What is the electric eld inside the wire?
Correct answer: 3.01587 V/m.
Explanation:
The loop rule gives,
emf EL = 0
emf
E=
L
1. 9 V
=
0.63 m
= 3.01587 V/m
008 (part 2 of 3) 10.0 points
Next, the Nichrome wire is replaced by a wire
of the same length and diameter, and same
mobile electron density but with electron mobility 8 times as large as that of Nichrome.
Now what is the electric eld inside the wire?
3. i2 < i1
4. Not enough information is given to compare the two circuits
Explanation:
i = nAuE . i u so 8u gives 8i. As a
result, i2 > i1 .
010 10.0 points
Suppose that the plates of a mechanical battery are very large compared to the distance s
between the plates, and the area of each plate
is A. What is the emf of the battery if the
charge on one of the plates is Q?
1. emf =
2. emf =
3. emf =
4. emf =
5. emf =
Qs
2A o
Qs
correct
Ao
Q
2A o s
Q
A os
2Q s
Ao
Explanation:
Using Gauss law on one of the plates, we
can determine
|E |
Correct answer: 3.01587 V/m.
Explanation:
The electric eld remains the same. It only
depends on V and L. However, the current
will change.
|Q|/A
,
o
emf |E |s
|Q|s
.
oA
011 (part 1 of 2) 10.0 points
brown (twb493) Ch19-h1 chiu (56565)
4
In the diagram below, suppose that VC VF
= 5.2 V, and that VD VE = 4.5 V. What is
the potential dierence VC VD ?
B
C
B2
B1
D
P1
Loop
1
A
P2
B3
P3
Loop
2
F
E
P0
+
+
P4
Correct answer: 0.7 V.
Explanation:
The potential dierence is given by the loop
equation
(VC VF ) (VC VD ) (VD VE ) = 0
which may be rewritten as
To start the analysis of this circuit, you
must write energy conservation (loop) equations. Each equation must involve a roundtrip path that begins and ends at the same
location. Each segment of the path should
go through a wire, a bulb, or a battery (not
through the air). How many valid energy
conservation (loop) equations is it possible to
write for this circuit?
1. One
(VC VD ) = (VC VF ) (VD VE )
(VC VD ) = 5.2 V 4.5 V
(VC VD ) = 0.7 V
2. Five
3. Innite
4. Four
012 (part 2 of 2) 10.0 points
If the element between C and D is a battery,
is the (+) end of the battery at C or at D?
1. There is no positive end.
2. C correct
3. D
Explanation:
The positive end of the battery is at C,
since point C is at a higher potential.
013 (part 1 of 7) 10.0 points
Three identical light bulbs are connected to
two batteries as shown in the following gure.
5. Six
6. Three correct
7. None
8. Two
Explanation:
There are three valid loops for this circuit:
1. P0 , P1 , P2 , P4 , P0
2. P0 , P1 , P3 , P4 , P0
3. P2 , P3 , P2
014 (part 2 of 7) 10.0 points
Which of the following equations are valid
brown (twb493) Ch19-h1 chiu (56565)
energy conservation (loop) equations for this
circuit? E1 refers to the electric eld in bulb
1; L refers to the length of a bulb lament.
Assume that the electric eld in the connecting wires is small enough to neglect. List all
that apply, separated by commas. If none
apply, enter none.
A.
B.
C.
D.
E.
F.
G.
+E2 L E3 L = 0
E1 L E3 L = 0
+2emf E2 L E3 L = 0
E1 L E2 L = 0
+2emf E1 L E2 L = 0
+2emf E1 L E3 L = 0
+2emf E1 L E2 L E3 L = 0
5
5 106 m (its very thin!). The electron
mobility of tungsten is 0.0018 (m/s)(V/m).
Tungsten has 6 1028 mobile electrons per
cubic meter. Since there are three unknown
quantities, we need three equations relating
these quantities. Use any two valid energy
conservation equations and one valid charge
conservation equation solve to for E1 , E2 , i1 ,
and i2 . Start with i2 .
Correct answer: 1.13097 1018 s1 .
Explanation:
One way of solving this is as follows. Start
with the following loop equation and use the
general expression i = n A u E.
Correct answer: A, E, F.
Explanation:
The equations given by A corresponds to
loop 3, E to loop 1, F to loop 2 are valid
energy conservation (loop) equations for this
circuit.
015 (part 3 of 7) 10.0 points
It is also necessary to write charge conservation equations (node) equations. Each such
equation must related electron current owing into a node to electron current owing out
of a node. Using that i2 corresponds to the
electron current through B2 (and similar for
the other subscripts), which of the following
are valid charge conservation equations? List
all that apply, separated by commas. If none
apply, enter none.
A. i1 = i3
B. i1 = i2
C. i1 = i2 + i3
Correct answer: C.
Explanation:
The equations given by A and B are incorrect, because the total current cannot be equal
to the current in a single parallel branch.
016 (part 4 of 7) 10.0 points
Each battery has an emf of 1.6 V volts. The
length of the tungsten lament in each bulb
is 0.008 m. The radius of the lament is
2 emf
2 emf E1 L E2 L = 0
i1
i2
L
L=0
nAu
nAu
Also since E2 L = E3 L, we know that E2 =
E3 and i2 = i3 . The Current Node Rule gives
i1 = i2 + i3
= 2 i2
Substitute into the previous loop equation
and solve:
2 emf
(2 i2 ) L
i2 L
=0
nAu
nAu
3 i2 L
=0
2 emf
nAu
2(emf)n A u
3L
2(emf)n ( R2 ) u
=
3L
= 1.13097 1018 s1 .
i2 =
017 (part 5 of 7) 10.0 points
What is i1 ?
Correct answer: 2.26195 1018 s1 .
Explanation:
From the solution to part 4, we know that
brown (twb493) Ch19-h1 chiu (56565)
6
1
N
i1 = 2 i2
A
= 2.26195 1018 s1 .
+
C
2
018 (part 6 of 7) 10.0 points
Now nd E1 .
+
Correct answer: 266.667 V/m.
B
D
Explanation:
We can write
E1 =
i1
nAu
= 266.667 V/m .
019 (part 7 of 7) 10.0 points
Finally, nd E2
Correct answer: 133.333 V/m.
Explanation:
We can write
1
E1
2
= 133.333 V/m .
E2 =
020 (part 1 of 4) 10.0 points
A circuit is assembled that contains a thinlament bulb (bulb #1) and a thick-lament
bulb (bulb #2), as shown in the following gure, with four compasses placed underneath
the wires (were looking down on the circuit).
The thin and thick laments are the same
length. When the thin-lament bulb is unscrewed from its socket, compass C deects
19 degrees. When the thin-lament bulb is
screwed back in and the thick-lament bulb is
unscrewed, compass C deects 9 degrees.
With both bulbs screwed into their sockets,
is the magnitude of the deection of compass
A?
Correct answer: 9 degrees.
Explanation:
Bulb 2 has the thicker lament; as a result,
when it is connected by itself to the battery,
the current will be larger and the compass
next to the battery will deect more. Thus,
it is the thick-lament bulb that deects the
compass more. Lets write the loop equation
for a single bulb circuit for bulb 1.
1:
2emf E1 L = 0
2emf
E1 =
.
L
The current through bulb 1 is
i = n A u E1 = n A u
2emf
L
when connected by itself to the batteries. Now
we apply the loop equation for the two-bulb
circuit using a loop consisting of the batteries
and bulb 1:
2emf E1 L = 0.
Note that this is exactly this same as when
it was the only bulb in the circuit. Thus,
the current through the bulb in the two-bulb
circuit is the same when bulb 2 is unscrewed.
As a result, the compass next to bulb 1 will
deect the same amount, 9 .
021 (part 2 of 4) 10.0 points
brown (twb493) Ch19-h1 chiu (56565)
What is the magnitude of the deection in
compass B?
Correct answer: 19 degrees.
Explanation:
The exact same reasoning can be applied
to bulb 2, so the compass next to bulb 2 will
deect the same amount as when bulb 1 is
unscrewed, which is 19 .
022 (part 3 of 4) 10.0 points
What is the magnitude of the deection in
compass C?
Correct answer: 26.6892 degrees.
Explanation:
In the two-bulb circuit, applying the current node rule shows that the current through
the battery and connecting wires is equal to
the sum of the currents through the bulbs,
i = i1 + i2 , in accordance with conservation
of charge. If the current i1 creates a magnetic eld B1 and if the current i2 creates a
magnetic eld B2 , then the total current will
create a magnetic eld that is B = B1 + B2
(since B i for a wire). Note that both of
these elds are in the same direction, east, so
we can simple add their magnitudes. Compass C will deect an amount given by
Bwire
BEarth
B1 + B2
=
BEarth
BEarth tan 9 + BEarth tan 19
=
BEarth
= tan 9 + tan 19
= 0.502712
tan =
= tan1 (0.502712)
= 26.6892 .
023 (part 4 of 4) 10.0 points
What is the magnitude of the deection in
compass D?
Correct answer: 26.6892 degrees.
7
Explanation:
Since the current through the wire over
compass D is the same as the current through
the wire over compass C, their deections will
have the same magnitude. Note that the
direction of deection will be opposite, since
the currents are owing over the compasses in
opposite directions.
024 10.0 points
A solid metal sphere of radius R carries a
charge of +Q. Another solid metal sphere of
radius r carries a charge of q . The amount
of charge is not enough to cause breakdown
in air. The two spheres are very far apart
(distance
R and distance
r ). At t = 0 a
very thin wire of length L is connected to the
two spheres, as in the following gure.
+
+ +Q +
+
+
R
+
+
+
L
q
r
The mobility u of mobile electrons in this
wire is very small, and the wire conducts electrons so poorly that it takes about an hour for
the system to reach equilibrium. In a short
time t (a few seconds), how many electrons
leave the sphere of radius r ? There are n mobile electrons per cubic meter in the wire, and
the wire has a constant cross-sectional area A.
Qq
t correct
+
Rr
Qq
nAu
t
+
2.
4 0 ( L + R + r )
Rr
Qq
nAu
t
+
3.
4 0 ( L + R + r )
Rr
nAu
Qq
4.
t
4 0 L
Rr
qQ
nAu
t
5.
4 0 L
rR
nAu
Qq
6.
t
+
4 0 L
Rr
1.
nAu
4 0 L
brown (twb493) Ch19-h1 chiu (56565)
Explanation:
Since this system takes a very long time to
come to static equilibrium, we can consider it
to be in a quasi steady state. That is, at any
instant we can assume electron current i is
the same at all locations in the wire. We are
interested in the electron current in the wire
a short time t after the connection has been
made-long enough for the rearrangement of
surface change to have occurred (this takes
only nanoseconds), but short enough that the
charge on the two spheres has changed very
little.
Since i = n A u E , and we know n, A, and u,
we need to nd the electric eld E in the wire.
There are three sources of charge contributing
to E in the wire: the two charged spheres,
and the charge on the surface of the wire.
Although there is not a lot of charge on the
wire, this charge is very close to the wire,
and its contribution to E inside the wire is
signicant. Since we dont know the exact
amount and distribution of charge on the wire,
we cant calculate E directly from Coulombs
Law.
Therefore, we start (as usual) from a fundamental principle: V = 0 for a round trip
path. We choose a path that goes through
the wire, where we want to know E , and also
through the air, as in the following gure.
+
+ +Q +
+
+
R
+
+
+
L
q
r
Consider the part of the path indicated by
the dashed line. Along the sections of the
path inside the spheres, the electric eld is
zero, and hence V is zero.
Along the section of the path in the air, if
we assume that (1) we are suciently far from
the wire that the small amount of charge on
the wire contributes a negligible electric eld,
and (2) the charged spheres are suciently far
8
from each other that the electric eld of one
is negligible near the other, then we nd that
along the dotted line portion of the path:
Vair = VB VA
1
qQ
4 0
rR
So for a round trip, we have
V = 0
Vwire + Vair = Vwire +
1
4 0
qQ
+
rR
Sine the system is in a quasi steady state, E
must be uniform in the wire, so Vwire = E L
(it is positive, since we are traveling opposite
to the direction of E in the wire). Thus,
E=
qQ
+
rR
L
1
4 0
and
i = nAuE
1
4 0
qQ
+
rR
= nAu
.
L
Therefore, a number of electrons equal to
nAu
1
4 0
qQ
+
rR
t
L
leave the small sphere in a time t.