Version 057 – QZ2 – ditmire – (58565)
This print-out should have 30 questions.
Multiple-choice questions may continue on
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before answering.
001
1
constant speed, it undergoes a circular motion with the centripetal force provided by
the magnetic force, namely
m
10.0 points
A static uniform magnetic field is directed
out of the page. A charged particle moves
in the plane of the page following a counterclockwise spiral of decreasing radius as shown.
B
B
Neglect the effect due to gravity.
What is a reasonable explanation?
1. The charge is neutral and slowing down.
2. None of these
3. The charge is positive and speeding up.
4. The charge is positive and slowing down.
5. The charge is negative and speeding up.
6. The charge is negative and with a constant speed.
7. The charge is positive and with a constant
speed.
8. The charge is neutral and speeding up.
9. The charge is negative and slowing down.
correct
10. The charge is neutral and with a constant
speed.
Explanation:
We know that when a charged particle
moves in a uniform magnetic field with a
v2
= qvB,
r
so we know that the radius is in fact proportional to the speed,
r=
m
v.
qB
Since the particle follows a spiral of decreasing
radius, we can judge that it is slowing down.
~ = q ~v × B
~ must be
The magnetic force F
in the direction for the centripetal force −r̂
(pointed inward) of this particle in counter~ is in
clockwise circular motion. Since ~v × B
the positive r̂ direction, the particle has a
negative charge.
002 10.0 points
Calculate the average drift speed of electrons
traveling through a copper wire with a crosssectional area of 70 mm2 when carrying a
current of 90 A (values similar to those for the
electric wire to your study lamp). Assume
one electron per atom of copper contributes
to the current. The atomic mass of copper
is 63.5 g/mol and its density is 8.93 g/cm3 .
Avogadro’s number is 6.022 × 1023 and the
fundamental charge is 1.602 × 10−19 C.
1. 3.27594e-06
2. 0.000245696
3. 1.84272e-05
4. 2.45696e-05
5. 4.21193e-05
6. 4.42252e-06
7. 0.000103192
8. 0.000196557
9. 9.47684e-05
10. 3.68544e-05
Correct answer: 9.47684 × 10−5 m/s.
Explanation:
Let :
N = 1,
M = 63.5 g/mol ,
Version 057 – QZ2 – ditmire – (58565)
ρ = 8.93 g/cm3 ,
A = 70 mm2 ,
I = 90 A , and
q = 1.602 × 10−19 C , and
NA = 6.022 × 1023 /mol .
The number of current-carrying electrons
per unit volume in copper is
NA ρ
6.022 × 1023 /mol
=
M
63.5 g/mol
100 cm 3
3
× 8.93 g/cm
1m
28
3
= 8.46873 × 10 /m .
n=
so the drift speed in the copper wire is
vd =
=
I
nqA
90 A
(8.46873 × 1028 /m3 )(1.602 × 10−19 C)
2
1
1000 mm
×
70 mm2
1m
= 9.47684 × 10−5 m/s .
R2
1
=
R1
3
R2
=3
5.
R1
R2
3
6.
=
R1
2
R2
3
7.
=
R1
4
1
R2
= correct
8.
R1
2
R2
4
9.
=
R1
3
R2
10.
=4
R1
Explanation:
ρℓ
ℓ
ρℓ
=
∝ 2 when ρ1 = ρ2 , so
R=
2
A
πr
r
2
R2
ℓ 2 r1
1
(2 ℓ1) r12
=
=
.
=
2
R1
ℓ1 (2 r1)2
2
ℓ 1 r2
4.
004 (part 2 of 2) 10.0 points
What is the ratio of the currents through these
two conductors?
1.
003 (part 1 of 2) 10.0 points
Consider two cylindrical conductors made of
the same ohmic material.
I2
~E 2
1
I
~E 1
V2
V1
2.
3.
4.
b
b
ℓ1
r1
ℓ2
r2
If ρ2 = ρ1 , r2 = 2 r1 , ℓ2 = 2 ℓ1 , and V2 =
R2
V1 , find the ratio
of the resistances.
R1
1
R2
=
1.
R1
4
R2
=2
2.
R1
R2
2
3.
=
R1
3
2
5.
6.
7.
8.
9.
10.
I2
I1
I2
I1
I2
I1
I2
I1
I2
I1
I2
I1
I2
I1
I2
I1
I2
I1
I2
I1
=
1
4
=4
=
=
=
=
=
4
3
1
3
1
2
2
3
3
2
=3
= 2 correct
=
3
4
Version 057 – QZ2 – ditmire – (58565)
Explanation:
Since V = IR or I = V /R and the voltages
are the same through each conductor. The
ratio of the currents is the inverse of the ratio
of the resistances.
005
10.0 points
The switch S has been in position b for a
long period of time.
R3
R1
006 10.0 points
A resistor is made from a hollow cylinder of
length ℓ, inner radius a, and outer radius b.
The region a < r < b is filled with material of
resistivity ρ. Current runs along the axis of
the cylinder.
What is the resistance of this component?
1. R =
2. R =
C
R2
S b
E
a
3. R =
4. R =
When the switch is moved to position a,
find the characteristic time constant.
2
(R1 + R2 ) C
1
2. τ = √
R1 R2 C
R1 + R2
C
3. τ =
2
1. τ =
4. τ = R1 C
5. τ =
5. R =
6. R =
7. R =
8. R =
9. R =
1
(R1 + R2 ) C
10. R =
6. τ = R2 C
7. τ =
3
2ρ ℓ
π b2
ρa
πℓ b2
ρℓ
π b2
ρ ℓ b2
π a4
ρℓ
correct
2
π (b − a2 )
π b2 ρ
ℓ
2πρ ℓ
2
b − a2
ρℓ
π a2
ρ ℓ a2
π b4
π b2 − a 2 ρ
ℓ
Explanation:
1
R1 C
8. τ = (R1 + R2 ) C correct
9. τ =
R=ρ
1
R2 C
p
10. τ = R1 R2 C
Explanation:
In charging an R C circuit, the characteristic time constant is given by
τ = RC ,
where R is the equivalent resistance:
R = R1 + R2 .
ℓ
ρℓ
ℓ
=ρ
=
.
2
2
2
A
πb −πa
π (b − a2 )
007 (part 1 of 2) 10.0 points
A battery, a capacitor, a bulb, and a switch
are in the circuit as shown. Unlike most real
bulbs, the resistance of the bulb in this question does not change as the current through
it changes. The switch is initially open as
shown, and the capacitor is uncharged.
Version 057 – QZ2 – ditmire – (58565)
S
C
R
E
4
The capacitor will remain charged. Its potential difference will be equal to the emf of
the battery E. The potential difference across
the bulb will remain zero.
009 (part 1 of 2) 10.0 points
Which statement correctly describes the
bulb after the switch is closed? Assume the
battery is ideal (it has no internal resistance)
and the connecting wires have no resistance.
Two identical light bulbs A and B are connected in series to a constant voltage source.
Suppose a wire is connected across bulb B as
shown.
1. None of these is correct.
2. The bulb is bright and remains bright.
A
3. The bulb is dim and remains dim.
4. At first the bulb is dim and it gets brighter
and brighter until the brightness levels off.
E
B
Bulb A
1. will burn more brightly. correct
5. At first the bulb is bright and it gets
dimmer and dimmer until it goes off. correct
Explanation:
Right after S is closed, the capacitor has
not charged yet (its potential difference is
zero). At this moment, the potential difference across the bulb is E and the current in the
circuit is maximum. As the capacitor charges
up, the voltage across the bulb drops, and it
gets dimmer and dimmer. The bulb goes off
when the capacitor is fully charged.
008 (part 2 of 2) 10.0 points
Which of the following correctly describes
what happens after the switch has remained
closed for a long time?
1. None of these is correct.
2. The potential difference across the capacitor is steady and much smaller than E.
3. The bulb is permanently off. correct
4. The current in the circuit is steady.
5. The bulb continues is permanently on.
Explanation:
2. will burn less brightly.
3. will burn as brightly as before.
4. will go out.
Explanation:
Because the wire is added in parallel to
light bulb B, the equivalent resistance now
becomes just the resistance of light bulb A.
Since the resistance dropped, the current had
to increase which means that light bulb A will
get brighter.
010 (part 2 of 2) 10.0 points
and bulb B
1. will burn less brightly.
2. will burn more brightly.
3. will go out. correct
4. will burn as brightly as before.
Explanation:
Since the wire is the path of zero resistance,
all the current will flow through the wire and
none will flow through light bulb B. With no
current to light the light bulb, bulb B will go
Version 057 – QZ2 – ditmire – (58565)
out.
011 (part 1 of 2) 10.0 points
A long piece of wire with a mass of 0.268 kg
and a total length of 91.64 m is used to make
a square coil with a side of 0.395 m. The coil
is hinged along a horizontal side (the y axis),
carries a 3.14 A current, and is placed in a
magnetic field with a magnitude of 0.0196 T
in the vertical direction along the z axis as
shown in the figure below.
The acceleration of gravity is 9.8 m/s2 .
5
Let : L = 91.64 m ,
ℓ = 0.395 m ,
m = 0.268 kg ,
i = 3.14 A , and
B = 0.0196 T .
Look down the positive y-axis at the coil
(that is, from the right-hand side of the original figure).
θ
z
θ
z
φ
i = 3.14 A
0. 3
95
m
x
y
B
x
µ
0.395 m
B = 0.0196 T
B = 0.0196 T
mg
Determine the angle that the plane of the
coil makes with the z axis when the coil is in
equilibrium.
1. 37.4199
2. 49.6187
3. 4.24119
4. 26.3757
5. 11.328
6. 38.3737
7. 1.84659
8. 56.6515
9. 47.0353
10. 51.1637
Correct answer: 47.0353◦.
Explanation:
Let θ be the angle the plane of the loop
makes with the z axis as shown.
Then the angle the coil’s magnetic moment
µ makes with the z axis is φ = 90◦ − θ ; e.g.,
sin φ = cos θ and tan φ = cot θ . The number
of turns in the loop is
L
circumference
91.64 m
=
4 (0.395 m)
= 58 .
N=
The torque about the z-axis due to gravity
is
~
τg = ~r × F
ℓ
=
cos φ m g ,
2
where ℓ is the length of each side of the square
loop. This gravitational torque tends to rotate the loop clockwise.
Version 057 – QZ2 – ditmire – (58565)
The torque due to the magnetic force tends
to rotate the loop counterclockwise about the
z-axis and has magnitude τm = N B I A sin φ.
At equilibrium,
τm = τg
m g (ℓ cos φ)
N B I ℓ2 sin φ =
.
2
Thus
mg
2N BI ℓ
(0.268 kg) (9.8 m/s2 )
=
2 (58) (0.0196 T) (3.14 A) (0.395 m)
= 0.931365 .
tan φ =
Since tan φ = tan(90◦ − θ) = cot θ, the angle
the loop makes with the z axis at equilibrium
is
6
A 43.047 V emf is placed across a series combination of three resistors (10 Ω, 41.796 Ω,
and 58 Ω).
At what rate is heat generated in the
41.796 Ω resistor?
1. 11.2158
2. 1.21262
3. 24.1359
4. 16.5563
5. 0.138691
6. 12.2158
7. 0.819414
8. 6.42462
9. 1.2295
10. 9.6865
Correct answer: 6.42462 W.
Explanation:
θ = cot−1 (0.931365) = 47.0353◦ .
Let :
012 (part 2 of 2) 10.0 points
Find the torque acting on the coil due to the
magnetic force at equilibrium.
1. 0.00190419
2. 0.141419
3. 0.0714003
4. 0.00126754
5. 0.00557892
6. 0.379581
7. 0.00479968
8. 0.00821795
9. 0.00095664
10. 0.0447763
Correct answer: 0.379581 N · m.
Explanation:
At equilibrium,
τm = N B I ℓ2 sin φ
= N B I ℓ2 cos θ
= (58) (0.0196 T) (3.14 A) (0.395 m)2
× cos(47.0353◦ )
= 0.379581 N · m .
013
10.0 points
E
R1
R2
R3
= 43.047 V ,
= 10 Ω ,
= 41.796 Ω ,
= 58 Ω .
and
Since the three resistors are connected in series, the current
E
R1 + R2 + R3
43.047 V
=
10 Ω + 41.796 Ω + 58 Ω
= 0.392063 A .
i=
The heat generated in resistor R2 is
P2 = i2 R2
= (0.392063 A)2 (41.796 Ω)
= 6.42462 W .
014 10.0 points
A conductor suspended by two flexible wires
as in the figure has a mass per unit length of
0.5 kg/m.
The acceleration of gravity is 9.8 m/s2 .
Version 057 – QZ2 – ditmire – (58565)
7
α
4Ω
B in
α
Rc
4Ω
γ
β
What current must exist in the conductor in
order for the tension in the supporting wires
to be zero when the magnetic field is 4.8 T
into the page?
1. 1.56456
2. 0.382812
3. 1.28489
4. 5.30833
5. 2.49083
6. 0.17093
7. 0.306988
8. 0.943704
9. 2.45
10. 1.02083
Ra
2Ω
β
γ
Rb
If the resistances between the terminals α,
β, and γ in each of the two networks shown
above are equivalent, find Ra .
1. 1.53846
2. 0.666667
3. 0.75
4. 2.0
5. 1.25
6. 0.8
7. 1.71429
8. 1.15385
9. 1.33333
10. 1.2
Correct answer: 0.8 Ω.
Correct answer: 1.02083 A.
Explanation:
Let : B = 4.8 T ,
g = 9.8 m/s2 , and
m
= 0.5 kg/m .
µ=
ℓ
The magnetic force acting on a current carrying wire is
Explanation:
We are developing what is commonly called
a “△-Y” or “Delta-Star” transformation.
α
α
RC
I=
mg
0.5 kg/m 9.8 m/s2
=
lB
4.8 T
= 1.02083 A
where µ is the mass per unit length.
015
10.0 points
Rc
RA
γ
β
~ = I ~l × B
~.
F
To balance the wire, the magnetic
force
~ must be equal to the gravity F = m g, so
Ra
RB
β
Let : RA = 4 Ω ,
RB = 2 Ω ,
RC = 4 Ω .
Rb
γ
and
Let α Rβ , β Rγ , and α Rγ be defined by setting the expressions for the equivalent resistances between terminals α, β, and γ equal in
the △ and Y networks, as shown below
α Rβ
= Ra + Rb =
1
1
1
+
RC
RA + RB
R (R + RB )
= C A
RA + RB + RC
(1)
Version 057 – QZ2 – ditmire – (58565)
RA (RB + RC )
(2)
RA + RB + RC
RB (RA + RB )
. (3)
α Rγ = Ra + Rc =
RA + RB + RC
Then, rewriting, we have
+1 Ra +1 Rb +0 Rc = α Rβ (1)
+0 Ra +1 Rb +1 Rc = β Rγ (2)
+1 Ra +0 Rb +1 Rc = α Rγ (3)
Using determinants to solve for Ra , Rb , and
Rc , we have
α Rβ +1
0 β Rγ +1 +1 α Rγ
0 +1 Ra = +1
+1
0
0 +1 +1 +1
0 +1 β Rγ
= Rb + Rc =
Ra =
=
=
=
=
α Rβ + α Rγ − β Rγ
1+1
RC (RA + RB )
2 (RA + RB + RC )
RB (RA + RC )
+
2 (RA + RB + RC )
RA (RB + RC )
−
2 (RA + RB + RC )
2 RB RC
2 (RA + RB + RC )
RB RC
RA + RB + RC
(2 Ω) (4 Ω)
(2 Ω) + (2 Ω) + (4 Ω)
Explanation:
Let :
R1
R2
R3
R4
R5
= 2 Ω,
= 7 Ω,
= 3 Ω,
= 9 Ω,
= 4 Ω.
and
c
R1
a
R3
a
Ra
Rc
Rd
R2
d
where
10.0 points
4Ω
R1 R2
(2 Ω) (7 Ω)
=
R1 + R2 + R3
2Ω+7Ω+3Ω
= 1.16667 Ω
R1 R3
(2 Ω) (3 Ω)
Rc =
=
R1 + R2 + R3
2Ω+7Ω+3Ω
= 0.5 Ω and
(7 Ω) (3 Ω)
R2 R3
=
Rd =
R1 + R2 + R3
2Ω+7Ω+3Ω
= 1.75 Ω .
Ra =
b
3Ω
7Ω
Correct answer: 4.74863 Ω.
c
Consider the resistor combination below.
c
9Ω
2Ω
a
What is the resistance between a and b?
1. 4.10274
2. 4.66667
3. 3.84615
4. 4.04545
5. 3.97207
6. 4.17073
7. 3.71724
8. 4.80784
9. 4.37712
10. 4.74863
Use a Y ↔ ∇ transformation
= 0.8 Ω .
016
8
d
to redraw the circuit:
d
Version 057 – QZ2 – ditmire – (58565)
10. 13.5
c
Rc
R4
Ra
a
Correct answer: 21.5 Ω.
b
R5
Rd
d
Rc and R4 are in series as are Rd and R5 ,
with these two pairs of resistors in parallel, so
Req = Ra +
1
1
+
Rc + R4 Rd + R5
1
1
=
+
0.5 Ω + 9 Ω 1.75 Ω + 4 Ω
= 0.279176 Ω−1 ,
1
= 1.16667 Ω +
0.279176 Ω−1
= 4.74863 Ω .
017
Explanation:
Let : R1
R2
R3
R4
Req
= 70.0 Ω ,
= 17.0 Ω ,
= 17.0 Ω ,
= 70.0 Ω ,
= 65.0 Ω .
R3
R4
1
Since
1
1
+
Rc + R4 Rd + R5
Req
The equivalent resistance of the circuit in
the figure is Req = 65.0 Ω .
17 Ω
70 Ω
R
17 Ω
E
and
R1
R
R2
E
S
R1 and R2 are in series:
R12 = R1 + R2 = 70 Ω + 17 Ω
= 87 Ω
as are R3 and R4 :
R34 = R3 + R4 = 17 Ω + 70 Ω
= 87 Ω .
10.0 points
70 Ω
9
S
Then R12 and R34 are in parallel:
−1
1
1
+
R1234 =
R12 R34
−1
1
1
=
+
87 Ω 87 Ω
= 43.5 Ω
and R is then in series with R1234
Find the value of R.
1. 9.5
2. 17.0
3. 21.5
4. 6.5
5. 7.0
6. 18.0
7. 14.0
8. 8.5
9. 15.5
Req = R + R1234
R = Req − R1234 = 65 Ω − 43.5 Ω
= 21.5 Ω .
018 (part 1 of 4) 10.0 points
In the circuit shown below, A, B, C, and D
are identical light bulbs.
Version 057 – QZ2 – ditmire – (58565)
Assume that the battery maintains a constant potential difference between its terminals (i.e., the internal resistance of the battery
is assumed to be negligible) and the resistance
of each lightbulb remains constant.
E
B
D
A
C
What is the correct relationship for the
brightnesses of (the power consumed by) the
light bulbs?
10
PA > PD > PB = PC .
019 (part 2 of 4) 10.0 points
If the emf of the battery is 17 V and each
resistance is 2 Ω, what power is consumed by
bulb B?
1. 5.78
2. 0.64
3. 0.98
4. 5.12
5. 4.5
6. 0.333333
7. 1.0
8. 1.21
9. 0.853333
10. 0.49
Correct answer: 5.78 W.
1. PB = PC > PA > PD
Explanation:
2. PD > PA > PB = PC
Let :
3. PD > PB = PC > PA
4. PA > PB = PC > PD
E = 17 V
R = 2 Ω.
and
RB and RC are in series, so
5. PA > PB > PC > PD
RBC = RB + RC = 2 R .
RBC and RD are in parallel, so
6. PA > PD > PB = PC correct
Explanation:
The circuit diagram is as follows.
RBCD =
D
2R
RBC RD
=
.
RBC + RD
3
RA and RBCD are in series, so
A
C
RABCD = RA + RBCD =
B
E
Observing the currents IA = ID + IBC , so
IA > ID
The current in the battery (and through
E
3E
RA ) is IA =
=
.
RABCD
5R
RBC = 2 RD so ID = 2 IBC and
and
IBC =
2
PA = IA2 R > ID
R = PD .
IA
E
=
.
3
5R
Thus
Observing potential difference relationships
VD = VB + VC = 2 VB = 2 VC , so VD > VB ,
VB2
VD2
>
= PB = PC
PD =
R
R
5R
.
3
and
PB =
=
2
IBC
RB =
E
5R
2
R=
(17 V)2
= 5.78 W .
25 (2 Ω)
E2
25 R
Version 057 – QZ2 – ditmire – (58565)
020 (part 3 of 4) 10.0 points
Bulb D is then removed from its socket.
How does the brightness of bulb A change?
1. The brightness of bulb A cannot be determined.
11
In the figure below the battery has an emf
of 18 V and an internal resistance of 1 Ω .
Assume there is a steady current flowing in
the circuit.
18 V
1Ω
2. The brightness of bulb A remains the
same.
5Ω
8 µF
3. The brightness of bulb A increases.
4. The brightness of bulb A decreases. correct
Explanation:
When bulb D is removed from its socket,
the circuit becomes serial connection of bulb
A, B, and C.
RABC = RA + RB + RC = 3 R, so
E
3E
<
= IA ,
3R
5R
so the brightness of bulb A decreases.
IA′ =
021 (part 4 of 4) 10.0 points
How does the brightness of bulb B change
when bulb D is removed from its socket?
1. The brightness of bulb B increases. correct
2. The brightness of bulb B remains the
same.
3. The brightness of bulb B cannot be determined.
4. The brightness of bulb B decreases.
Explanation:
Since bulbs A, B, and C are in serial conE
′
nection, IB
= IA′ =
.
3R
E
Before the removal of bulb D, IB =
,
5R
′
so IB
> IB , and the brightness of bulb B
increases.
022
10.0 points
5Ω
Find the charge on the 8 µF capacitor.
1. 8.0
2. 79.0588
3. 65.4545
4. 20.8
5. 33.8824
6. 12.0
7. 42.0
8. 22.8571
9. 21.4286
10. 23.5789
Correct answer: 65.4545 µC.
Explanation:
Let :
R1
R2
rin
V
C
= 5 Ω,
= 5 Ω,
= 1 Ω,
= 18 V ,
= 8 µF .
and
The equivalent resistance of the three resistors
in series is
Req = R1 + R2 + rin
=5 Ω+5 Ω+1 Ω
= 11 Ω ,
so the current in the circuit is I =
V
, and
Req
the voltage across R2 is
R2
5Ω
V2 = I R2 =
V =
(18 V)
Req
11 Ω
= 8.18182 V .
Version 057 – QZ2 – ditmire – (58565)
Since R2 and C are parallel, the potential
difference across each is the same, and the
charge on the capacitor is
12
a
R1
a
R3
Q = C V2 = (8 µF) (8.18182 V)
⇒
RL
RR
= 65.4545 µC .
R2
023 (part 1 of 2) 10.0 points
The circuit has been connected as shown in
the figure for a long time. V = 13 V, R1 =
1 Ω, R2 = 3 Ω, R3 = 7 Ω, R4 = 2 Ω, and
C = 1.6 µF.
a
R3
R1
V
C
R2
R4
R4
b
b
where
RL = R1 + R2 = 1 Ω + 3 Ω = 4 Ω ,
RR = R3 + R4 = 7 Ω + 2 Ω = 9 Ω ,
V
13 V
IL =
=
= 3.25 A , and
RL
4Ω
V
13 V
IR =
=
= 1.44444 A .
RR
9Ω
Across R3
b
What is the electric potential across the
capacitor?
1. 9.0
2. 4.88889
3. 5.12121
4. 5.80556
5. 7.5
6. 6.86111
7. 6.5
8. 5.90909
9. 6.32727
10. 5.0
Correct answer: 6.86111 V.
Explanation:
Let :
V
R1
R2
R3
R4
= 13 V ,
= 1 Ω,
= 3 Ω,
= 7 Ω , and
= 2 Ω.
After a long time C is fully charged and the
circuit acts like
V3 = IR R3 = (1.44444 A)(7 Ω) = 10.1111 V
and across R1
V1 = IL R1 = (3.25 A)(1 Ω) = 3.25 V .
Since V3 and V1 are “measured” from the same
point a, ∆V across C must be
Vc = V3 −V1 = 10.1111 V−3.25 V = 6.86111 V .
024 (part 2 of 2) 10.0 points
If the battery is disconnected, how long does
it take for the capacitor to discharge to 0.092
of its initial voltage?
1. 4.08522
2. 21.2148
3. 5.30039
4. 23.4694
5. 7.80429
6. 3.02214
7. 4.8494
8. 4.91718
9. 11.7463
10. 6.13353
Correct answer: 11.7463 µs.
Version 057 – QZ2 – ditmire – (58565)
Explanation:
Let :
13
so the equation for discharge is
Vt = 0.092 V0
C = 1.6 µF .
and
With the battery removed, the circuit looks
like
R3
R1
Qt
= e−t/τ
Q0
Vt
= 0.092 = e−t/τ
V0
t
− = ln(0.092)
τ
t = −τ ln(0.092)
= −(4.92308 µs)(−2.38597) = 11.7463 µs .
C
R2
025
R4
10.0 points
The current
⇒
I = a t2 − b t + c
C
R1
R3
R2
R4
⇒
C
in a section of a conductor depends on time.
What quantity of charge moves across the
section of the conductor from t = 0 s to t =
t1 ?
a 3 b 2
t − t + c t1 correct
3 1 2 1
b
2. q = a t31 − t21 + ct1
2
1. q =
Ru
3. q = 3 a t21 − 2 b t1 + c
Rl
4. q = a t21 − b t1 + c
⇒
5. q = 2 a t21 − b t1
C
Req
where
Ru = R1 + R3 = 1 Ω + 7 Ω = 8 Ω ,
Rl = R2 + R4 = 3 Ω + 2 Ω = 5 Ω , and
−1
−1 1
1
1
1
+
+
=
Req =
Ru Rl
8Ω 5Ω
= 3.07692 Ω .
τ ≡ Req C = (3.07692 Ω)(1.6 µF) = 4.92308 µs ,
6. q =
a 3 b 2
t − t +c
3 1 2 1
7. q = 2 a t21 − 3 b t1 + c t1
8. q = a t31 − b t2 + c t1
9. q = 3 a t21 − b + c
10. q = 2 a t1 − b
Explanation:
The unit of current is Coulomb per second:
dq
or dq = I dt.
I=
dt
To find the total charge that passes through
the conductor, one must integrate the current
over the time interval.
Version 057 – QZ2 – ditmire – (58565)
q=
Z
t1
Z
t1
dq =
0
Z
027
t1
10.0 points
I dt
8.1 Ω
0
(a t2 − b t + c) dt
0
t1
a 3 b 2
=
t1 − t1 + c t 3
2
0
a 3 b 2
= t1 − t1 + c t1 .
3
2
=
14
I3
4.4 V
6.4 Ω
2.8 Ω
026 10.0 points
A proton moves perpendicularly to a magnetic field that has a magnitude of 1.55 × 10−2
T.
The charge on a proton is 1.60×10−19 C.
What is the speed of the particle if the
magnitude of the magnetic force on it is 1.60×
10−14 N?
1. 2900000.0
2. 3594320.0
3. 1541670.0
4. 4108800.0
5. 6451610.0
6. 5046380.0
7. 1409980.0
8. 13016100.0
9. 17268800.0
10. 2944960.0
I2
2.8 V
0.6 Ω
5.3 V
I1
Find the current I1 in the 0.6 Ω resistor
at the bottom of the circuit between the two
power supplies.
1. 0.587607
2. 1.04254
3. 1.08083
4. 0.858574
5. 0.808879
6. 0.491891
7. 1.13845
8. 1.18957
9. 0.481532
10. 1.45888
Correct answer: 0.808879 A.
Explanation:
RD
I3
Correct answer: 6.45161 × 106 m/s.
E3
RC
Explanation:
I2
Let : B = 1.55 × 10−2 T ,
Fm = 1.60 × 10−14 N ,
qe = 1.60 × 10−19 C .
RB
and
Fmagnetic = q v B
Fmagnetic
v=
qB
1.6 × 10−14 N
=
(1.6 × 10−19 C) (0.0155 T)
= 6.45161 × 106 m/s
E2
RA
E1
I1
At a junction (Conservation of Charge)
I1 + I2 − I3 = 0 .
(1)
Kirchhoff’s law on the large outside loop gives
(RA + RB ) I1 + RD I3 = E1 + E2 .
(2)
Kirchhoff’s law on the right-hand small loop
gives
R C I2 + R D I3 = E 3 .
(3)
Version 057 – QZ2 – ditmire – (58565)
D1
−81.81 V Ω
=
= 0.808879 A .
D
−101.14 Ω2
028 10.0 points
A metal rod having a mass per unit length of
0.0159 kg/m carries a current of 8.2 A . The
rod hangs from two wires (in the same plane
as the rod) in a uniform vertical magnetic
field as in the figure.
The acceleration of gravity is 9.8 m/s2 .
B
B
θ
7m
6 mm
b
←
8.
v
A
2
9.8 m/s2
If the wires make an angle of 21◦ with the
vertical when in equilibrium (v = 0 m/s),
determine the strength of the magnetic field.
1. 0.25448
2. 0.00729436
3. 0.206624
4. 0.0583
5. 0.0738857
6. 0.0917989
7. 0.432735
8. 0.00439691
9. 0.0393138
10. 0.120475
Correct answer: 0.00729436 T.
Explanation:
B
B
θ
ℓ
r
v
b
←
I
Let : g = 9.8 m/s2 ,
g
Let : RA = 0.6 Ω ,
RB = 2.8 Ω ,
RC = 6.4 Ω ,
RD = 8.1 Ω ,
E1 = 5.3 V ,
E2 = 2.8 V , and
E3 = 4.4 V .
Using determinants,
0
1
−1
E 1 + E2 0 RD E3
RC RD I1 = 1
1
−1 RA + RB
0 RD 0
RC RD Expanding along the first row, the numerator is
0
1
−1
D1 = E1 + E2 0 RD E3
RC RD E1 + E2 RD =0−1 E3
RD E1 + E2 0 + (−1) E3
RC = − [(E1 + E2 ) RD − E3 RD ]
− [RC (E1 + E2 ) − 0]
= RD (E3 − E1 − E2 ) − RC (E1 + E2 )
= (8.1 Ω) (4.4 V − 5.3 V − 2.8 V)
−(6.4 Ω) (5.3 V + 2.8 V)
= −81.81 V Ω .
Expanding along the first column, the denominator
is
1
1
−1 0 RD D = RA + RB
0
RC RD 0 RD = 1 RC RD 1
−1 +0
− (RA + RB ) RC RD = 0 − RC RD − (RA + RB ) (RD + RC )
= (6.4 Ω) (8.1 Ω)
−(0.6 Ω + 2.8 Ω) (8.1 Ω + 6.4 Ω)
= −101.14 Ω2 , and
I1 =
15
Version 057 – QZ2 – ditmire – (58565)
v = 0 m/s ,
r = 6 mm ,
ℓ = 7 m,
I = 8.2 A , and
λ = 0.0159 kg/m .
Since the rod is in equilibrium, the total forces
acting on it must be zero:
X
~ × B|
~ =0
Fx = T sin θ − I |L
X
Fy = T cos θ − mg = 0
or
T sin θ − I L B sin 90◦ = 0
T cos θ − λ L g = 0
Dividing these equations, we have
IB
tan θ =
, so
λg
λg
tan θ
B=
I
(0.0159 kg/m) (9.8 m/s2 )
tan (21◦ )
=
(8.2 A)
= 0.00729436 T .
16
8. 1510.48
9. 2063.29
10. 1139.17
Correct answer: 2063.29◦ C.
Explanation:
Let : R0 = 17 Ω ,
R = 174 Ω ,
T0 = 11◦ C , and
α = 0.0045 (◦ C)−1 .
Resistance is
R = R0 [1 + α ∆T ]
= R0 + R0 α (Tf − T0 )
R − R0
Tf − T0 =
R0 α
R − R0
+ T0
Tf =
R0 α
174 Ω − 17 Ω
=
+ 11◦ C
◦
−1
(17 Ω) [0.0045 ( C) ]
= 2063.29◦ C .
029 10.0 points
A certain lightbulb has a tungsten filament
with a resistance of 17 Ω when cold and 174 Ω
when hot.
If the equation
R = R0 [1 + α ∆T ]
can be used over the large temperature range
involved here, find the temperature of the filament when it is hot. Assume that α , the temperature coefficient of resistivity of tungsten,
is 0.0045 (◦ C)−1 and that the temperature of
the cold filament is 11◦ C.
1. 886.967
2. 962.866
3. 2562.68
4. 1037.0
5. 1067.5
6. 1079.44
7. 1578.67
030
10.0 points
A device (“source”) emits a bunch of
charged ions (particles) with a range of velocities (see figure). Some of these ions pass
through the left slit and enter “Region I” in
which there is a vertical uniform electric field
(in the −̂ direction, due to the potential difference 1100 V across the distance 1.2 cm) and
a 0.1 T uniform magnetic field (aligned with
the ±k̂-direction) as shown by the shaded
area. ı̂ is in the direction +x (to the right), ̂
is in the direction +y (up the page), and k̂ is
in the direction +z (out of the page).
The ions that make it into “Region II” are
observed to be deflected downward and then
follow a circular path with a radius of r =
0.26 m.
Version 057 – QZ2 – ditmire – (58565)
+1100 V
m
26
y
cm
1.2 cm
x
z
Region I
Region II
If the charge on each ion is 4.1 × 10−18 C ,
what is the mass of the ions? (There is magnetic field in both Region I and Region II.)
1. 2.1204e-25
2. 2.64776e-25
3. 7.29e-25
4. 1.05549e-24
5. 6.0912e-25
6. 1.16291e-25
7. 4.015e-25
8. 1.638e-24
9. 1.07931e-24
10. 2.01536e-25
Correct answer: 1.16291 × 10−25 kg.
Explanation:
Let : B = 0.1 T ,
V = 1100 V ,
d = 1.2 cm ,
r = 0.26 m = 0.26 m
q = 4.1 × 10−18 C .
E≡
and
V
1100 V
=
= 91666.7 N/C .
d
1.2 cm
To obtain a straight orbit, the upward and
downward forces need to cancel. The force on
a charged particle is
~ =F
~E + F
~ B = q (E
~ + ~v × B)
~ .
F
For the force to be zero, we need
~E + F
~B = 0
F
~ E = −F
~B
F
qE = qvB
91666.7 N/C
E
=
v=
B
0.1 T
= 9.16667 × 105 m/s .
Region of
Magnetic
Field
0.1 T
q
17
and the radius of a circular path taken by a
charged particle in a magnetic field is given
by
mv
.
qB
Br
m=q
v
r=
= (4.1 × 10−18 C)
(0.1 T)(0.26 m)
9.16667 × 105 m/s
= 1.16291 × 10−25 kg .