Version 022 – Exam 2 – hoffmann – (57505)
This print-out should have 20 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 10.0 points
A 19 V battery delivers 120 mA of current
when connected to a 78 Ω resistor.
Determine the internal resistance of the
battery.
1. 36.3333
2. 60.6139
3. 112.119
4. 80.3333
5. 81.5454
6. 62.7521
7. 41.1111
8. 102.991
9. 108.571
10. 65.6147
Find the equivalent capacitance between
points a and b.
1. 9.3
2. 10.6992
3. 11.4487
4. 13.4901
5. 12.5415
6. 12.0237
7. 15.1785
8. 16.7109
9. 17.2993
10. 14.128
Correct answer: 12.5415 µF.
Explanation:
Let : C1
C2
C3
C4
Correct answer: 80.3333 Ω.
Explanation:
Let :
= 4.4 µF ,
= 5 µF ,
= 5.6 µF and
= 5.5 µF .
C1
V = 19 V ,
I = 120 mA = 0.12 A ,
R = 78 Ω .
and
b
C3
C4
Eliminate junctions of zero capacitance:
V = I (R + r)
V
19 V
r=
−R=
− 78 Ω
I
0.12 A
= 80.3333 Ω .
C1
a
C2
C4
b
C3
C1 and C3 are connected in series, so
5 µF
002 10.0 points
Consider the capacitor network.
µF
4. 4
5.6 µF
C2
a
The internal resistance is in series with the
given resistor, so
a
1
1
1
1
C3 + C2
=
+
=
C23
C2 C3
C2 C3
(5 µF) (5.6 µF)
C2 C3
=
C23 =
C3 + C2
5.6 µF + 5 µF
= 2.64151 µF .
b
C1 , C23 and C4 are connected in parallel,
5. 5
µF
so
Cab = C1 + C4 + C23
Version 022 – Exam 2 – hoffmann – (57505)
= 4.4 µF + 5.5 µF + 2.64151 µF
The centripetal force Fc is provided by FB , so
= 12.5415 µF .
003 10.0 points
A particle with a positive charge q and mass
m is undergoing circular motion with speed
v. At t = 0, the particle is moving along
the negative x axis in the plane perpendicular
~ which points in the
to the magnetic field B,
positive z direction as shown in the figure
below.
y
~
B
~v
m v2
= qvB
r
v
qB
=
.
r
m
The period of oscillation is
which is intuitive since the particle traverses
a distance 2 π r in a revolution and, moving at
speed v, takes the time T to do so. We know
v
the ratio , so
r
T =
z
2qB
m
mB
2. T =
q
mB
3. T = 2 π
q
qB
4. T =
m
2m
5. T =
qB
2πm
6. T =
correct
qB
qB
7. T = 2 π
m
πqB
8. T =
m
m
9. T =
qB
2m B
10. T =
q
Explanation:
The force on a charge q in magnetic field is
1. T =
~ B = q ~v × B
~.
F
2πr
,
v
T =
x
Find the period of the circular motion; i.e.,
the time takes for the particle to complete one
revolution.
2
m
2πr
= 2π
.
v
qB
004 10.0 points
A resistor is constructed by forming a material of resistivity ρ into a shape of a hollow
cylindrical shell of length L with inner radius
ra and outer radius rb . Suppose a potential
difference is applied between the inner and
outer surfaces such that the current flows radially outward.
L
rb
ra
b
What is the total resistance R?
ra
ρ
ln
1. R =
2πL
r
b
ρ
rb
2. R =
ln 1 −
2πL
r
a
rb
ρ
correct
ln
3. R =
2πL
ra
ρ
rb
4. R =
ln 1 +
2πL
ra
Version 022 – Exam 2 – hoffmann – (57505)
ρ
ra
5. R =
ln 1 +
2πL
rb
ra
ρ
ln 1 −
6. R =
2πL
rb
Explanation:
The resistance due to the cylindrical element is the resistivity of the material times
the length of the element divided by the crosssectional area (the area “seen” by the current;
i.e., perpendicular to the direction of the current):
ρℓ
R≡
,
A
To visualize the setup, since the potential
difference is applied between the inner and
the outer surfaces, we may cut the cylinder
surface and partially straighten out the side
surface. Now the length ℓ here is the distance traveled by the current in crossing the
cylindrical element,
ℓ = ∆r ,
and the cross sectional area is that of the side
of the cylindrical element,
A = 2πrL.
Remember, L is the length of the cylinder,
while ℓ is the “length” traveled by the current
while crossing the area A. Thus we have
∆R =
ρ ∆r
.
2πrL
The total resistance is obtained by letting
these cylindrical elements be very thin (∆r →
dr) and integrating from ra to rb :
Z rb
ρ
dr
R=
2 π L ra r
ρ
[ln rb − ln ra ]
=
2πL ρ
rb
=
.
ln
2πL
ra
005 10.0 points
Consider the setup shown, where a capacitor
3
with a capacitance C is connected to a battery
with emf V and negligible internal resistance.
Before the insertion of the dielectric slab
with dielectric constant κ, the energy density
within the gap is u.
Now, keeping the battery connected, insert
the dielectric, which fills the gap completely.
d
V
C
Determine the energy density u′ within the
gap in the presence of the dielectric.
u
2κ
u
2. u′ = 2
κ
1. u′ =
3. u′ = u κ2
4. u′ = 2 u κ
u
5. u′ = √
κ
u
6. u′ =
κ
7. u′ = u
√
8. u′ = u κ
9. u′ = u κ correct
10. u′ =
uκ
2
Explanation:
We are trying to find u′ , the energy density
after the insertion of the dielectric, in terms
of u, the energy density before. Since V ′ = V ,
V′
and E ′ =
for a parallel plate capacitor, we
d
V
and thus E ′ = E. However,
also have E =
d
ǫ0 becomes ǫ = κ ǫ0 , so
u′ =
1
1
ǫ E ′2 = κ ǫ0 E 2 = κ u .
2
2
Version 022 – Exam 2 – hoffmann – (57505)
Alternately, one could calculate
1
1
U ′ = Q′ V ′ = κ Q V = κ U .
2
2
Since U changes and the volume doesn’t, u
must change by the same factor.
006 10.0 points
A conductor suspended by two flexible wires
as in the figure has a mass per unit length of
0.57 kg/m.
The acceleration of gravity is 9.8 m/s2 .
B in
What current must exist in the conductor in
order for the tension in the supporting wires
to be zero when the magnetic field is 1.5 T
into the page?
1. 1.51261
2. 3.724
3. 2.53448
4. 0.861212
5. 2.45
6. 1.02083
7. 1.47
8. 2.3765
9. 1.10645
10. 1.5925
4
To balance the wire, the magnetic
force
~ must be equal to the gravity F = m g, so
I=
0.57 kg/m 9.8 m/s2
mg
=
lB
1.5 T
= 3.724 A
where µ is the mass per unit length.
007 10.0 points
An ammeter reads 5 A at full-scale deflection.
The meter was constructed by inserting a resistor in parallel with a 39.3 Ω galvanometer
coil that deflects full-scale when the voltage
across it is 36.3 mV.
What is the value of the parallel resistance
that should be used?
1. 0.00498186
2. 0.00529997
3. 0.00790995
4. 0.00233351
5. 0.00332614
6. 0.00422296
7. 0.00258788
8. 0.0168748
9. 0.00392976
10. 0.00726134
Correct answer: 0.00726134 Ω.
Explanation:
Let : Itotal = 5 A
Rc = 39.3 Ω
V = 36.3 mV .
RC
Correct answer: 3.724 A.
Explanation:
Ic
Let :
B = 1.5 T ,
g = 9.8 m/s2 , and
m
µ=
= 0.57 kg/m .
ℓ
The magnetic force acting on a current carrying wire is
~ = I ~l × B
~.
F
a
b
Ip
RP
b
b
Itotal
The current through the coil at full scale
deflection is
V
0.0363 V
ic =
=
= 0.000923664 A .
Rc
39.3 Ω
Version 022 – Exam 2 – hoffmann – (57505)
5
Hence, the current that must exist in the
parallel branch is
Explanation:
~
~
~
The force is F = q ~v × B , B = B −k̂ ,
1
Ip = Itotal −Ic = 5 A−0.000923664 A = 4.99908 A .~v = √ v +k̂ + ı̂ , and q > 0 , so
2
Therefore, the parallel resistor has a value of
~ = +|q| ~v × B
~
F
h
i
V
0.0363 V
1
Rp =
=
= 0.00726134 Ω .
= +|q| √ v B +k̂ + ı̂ × (−̂)
Ip
4.99908 A
2
1
= +|q| √ v B +ı̂ − k̂
008 10.0 points
2
◦
A positively charged particle moving at 45
1
b
√
F
=
+ı̂
−
k̂
.
angles to both the z-axis and x-axis enters
2
a magnetic field (pointing into of the page),
as shown. ı̂ is in the x-direction, ̂ is in the
009 10.0 points
y-direction, and k̂ is in the z-direction.
A square loop of wire carries a current and is
×
×
×
x
located in a uniform magnetic field.
~ ×
×
B
The left side of the loop is aligned and
×
×
×
attached to a fixed axis (dashed line in figure).
y
z
×
×
← axis of rotation
×
×
×
~ = 0.11 T
B
×
×
~
B
×
×
×
×
×
+q
What is the initial direction of deflection?
b = √1 (+ı̂ − ̂)
1. F
2
1
b = √ (−ı̂ + ̂)
2. F
2
1 b
3. F = √
+̂ + k̂
2
b = √1
4. F
−̂ + k̂
2
1
b = √ (+ı̂ + ̂)
5. F
2
1
b
6. F = √
+ı̂ + k̂
2
1 b
√
7. F =
+ı̂ − k̂ correct
2
b = √1
8. F
+̂ − k̂
2
~
9. F = 0 ; no deflection
b = √1
−ı̂ + k̂
10. F
2
y
x
0.38 m
×
×
→ 3A →
v
×
0.38 m
~ = 0.11 T
B
When the plane of the loop is parallel to the
magnetic field in the position shown, what is
the magnitude of the torque exerted on the
loop about the axis of rotation, which is along
the left side of the square as indicated by the
dashed line in the figure?
1. 0.457229
2. 0.047652
3. 0.533293
4. 0.504074
5. 0.497178
6. 0.125773
7. 0.135808
8. 0.186732
9. 0.26825
10. 0.0865172
Correct answer: 0.047652 N m.
Version 022 – Exam 2 – hoffmann – (57505)
Explanation:
y
x
d
ℓ
→ I →
← axis of rotation
~
B
6. 1.04098
7. 1.10448
8. 1.80723
9. 0.5
10. 0.910448
Correct answer: 1.80723 A.
Explanation:
B
C
= 0.047652 N m .
15 V
E2
i1
r2
F
and
A
D
10.0 points
R
E
I
Let : E1 = 30 V ,
E2 = 15 V ,
r1 = 3.1 Ω ,
r2 = 2.1 Ω , and
R = 10.4 Ω .
~k
τ = k~d × F
= d×I ℓB
= (0.38 m) (3 A) (0.38 m) (0.11 T)
30 V
r1
i2
Only the right side of the loop contributes to
the torque. By definition, the torque is
010
E1
~
B
Let : d = 0.38 m ,
ℓ = 0.38 m ,
B = 0.11 T ,
I = 3 A.
6
From the junction rule, I = i1 + i2 .
Applying Kirchhoff’s loop rule, we obtain
two equations:
E1 = i1 r1 + I R
E2 = i2 r2 + I R
= (I − i1 ) r2 + I R
= −i1 r2 + I (R + r2 ) ,
3.1 Ω
2.1 Ω
Multiplying Eq. (1) by r2 , Eq. (2) by r1 ,
10.4 Ω
E1 r2 = i1 r1 r2 + r2 I R
E2 r1 = −i1 r1 r2 + I r1 (R + r2 )
Find the current through the 10.4 Ω lowerright resistor.
1. 1.26531
2. 0.968421
3. 0.636364
4. 0.762238
5. 1.0082
(1)
(2)
Adding,
E1 r2 + E2 r1 = I [r2 R + r1 (R + r2 )]
E 1 r2 + E 2 r1
r2 R + r1 (R + r2 )
(30 V) (2.1 Ω) + (15 V) (3.1 Ω)
=
(2.1 Ω) (10.4 Ω) + (3.1 Ω) (10.4 Ω + 2.1 Ω)
I=
= 1.80723 A .
Version 022 – Exam 2 – hoffmann – (57505)
7
011 10.0 points
The direction of the magnetic field in a certain
region of space is determined by firing a test
charge into the region with its velocity in
various directions in different trials.
The field direction is
1. the direction of the magnetic force.
2. the direction of the velocity when the
magnetic force is a maximum.
3. perpendicular to the velocity when the
magnetic force is zero.
4. one of the direction of the velocity when
the magnetic field is zero.
5. None of these correct
Explanation:
The magnetic field direction can be determined according to the path of a charge which
moves in the field. Since
Ignoring side effects, calculate the capacitance of this unit. The permittivity of a
vacuum is 8.8542 × 10−12 C2 /N · m2 .
1. 31.8426
2. 188.89
3. 128.233
4. 155.231
5. 58.0442
6. 98.2816
7. 55.9585
8. 51.8295
9. 168.469
10. 197.588
Correct answer: 98.2816 pF.
Explanation:
A = 11.1 cm2 = 0.00111 m2 ,
d = 0.8 mm = 0.0008 m , and
ǫ0 = 8.8542 × 10−12 C2 /N · m2 .
Let :
~ = q ~v × B
~ ,
F
The capacitance of a parallel plate capacitor
is proportional to the area of a plate and
inversely proportional to the plate separation:
the component of velocity along the magnetic
field doesn’t contribute to the magnetic force;
i.e., a charge moving along the magnetic field
will keep its motion status. On the contrary,
if the initial velocity in perpendicular to the
magnetic force, there will be a centripetal
force so that the charge will undergo circular
motion.
012 10.0 points
A capacitor is constructed of interlocking
plates as shown. The separation between adjacent plates is 0.8 mm, and the effective area
of one of the adjacent plates is 11.1 cm2 .
8 ǫ0 A
d
8 (8.8542 × 10−12 C2 /N · m2 )
=
0.0008 m
1 × 1012 pF
× (0.00111 m2 ) ·
1F
= 98.2816 pF .
C=
There are eight equal capacitors; the purpose
of this question is to count spaces, not plates.
013 (part 1 of 2) 10.0 points
An electron in an electron beam experiences
Version 022 – Exam 2 – hoffmann – (57505)
a downward force of 2.9 × 10−14 N while traveling in a magnetic field of 5.1 × 10−2 T west.
The charge on a proton is 1.60×10−19.
a) What is the magnitude of the velocity?
1. 10648100.0
2. 6402440.0
3. 1692710.0
4. 5092590.0
5. 2389710.0
6. 3553920.0
7. 12228300.0
8. 949519.0
9. 747951.0
10. 1598840.0
Correct answer: 3.55392 × 106 m/s.
Explanation:
Let : Fm = 2.9 × 10−14 N ,
B = 5.1 × 10−2 T , and
qe = 1.60 × 10−19 C .
The magnetic force is
Fm = q v B
Fm
v=
qB
Apply right-hand rule (for a negative
charge); force directed into the palm of the
hand, fingers in the direction of the field,
thumb in the direction of the velocity.
Palm faces up, fingers point west, so the
thumb points north.
015 10.0 points
Consider a galvanometer with an internal resistance of 55 Ω. If it deflects full-scale when
it carries a current of 0.8 mA, what is the
value of the series resistance that must be
connected to it if this combination is to be
used as a voltmeter having a full-scale deflection for a potential difference of 0.6 V?
1. 5428.0
2. 2090.86
3. 2145.0
4. 741.0
5. 695.0
6. 2117.67
7. 3178.0
8. 1938.0
9. 2914.0
10. 1103.86
Correct answer: 695 Ω.
Explanation:
Let : V = 0.6 V ,
I = 0.8 mA = 0.0008 A ,
r = 55 Ω .
2.9 × 10−14 N
=
(1.6 × 10−19 C) (0.051 T)
= 3.55392 × 106 m/s
2. South
3. None of these
4. East
5. North correct
Explanation:
and
The voltage drop across the terminals is
given by
014 (part 2 of 2) 10.0 points
b) What is its direction?
1. West
8
V =IR+Ir
= I (R + r)
So
V
−r
I
0.6 V
=
− 55 Ω
0.0008 A
= 695 Ω .
R=
016
10.0 points
Version 022 – Exam 2 – hoffmann – (57505)
Four resistors are connected as shown in
the figure.
c
A good rule of thumb is to eliminate junctions connected by zero resistance.
R1
Ω
98 V
b
73
c
R4
S1
Correct answer: 16.5181 Ω.
Explanation:
c
R1
The series connection of R2 and R3 gives
the equivalent resistance
R23 = R2 + R3
= 39 Ω + 63 Ω
= 102 Ω .
The total resistance Rab between a and b can
be obtained by calculating the resistance in
the parallel combination of the resistors R1 ,
R4 , and R23 ; i.e.,
1
1
1
1
+
+
=
Rab
R1 R2 + R3 R4
R4 (R2 + R3 ) + R1 R4 + R1 (R2 + R3 )
=
R1 R4 (R2 + R3 )
R1 R4 (R2 + R3 )
Rab =
R4 (R2 + R3 ) + R1 R4 + R1 (R2 + R3 )
R3
The denominator is
R2
b
R4
d
S1
R4 (R2 + R3 ) + R1 R4 + R1 (R2 + R3 )
= (73 Ω)[39 Ω + 63 Ω] + (27 Ω) (73 Ω)
+ (27 Ω) [39 Ω + 63 Ω]
= 12171 Ω2 ,
so the equivalent resistance is
Rab =
Let :
R1
R2
R3
R4
E
b
EB
d
EB
R3
d
Ω
Find the resistance between points a and b.
1. 12.9212
2. 14.0019
3. 10.9657
4. 16.2536
5. 15.3015
6. 8.83291
7. 16.5181
8. 16.8554
9. 9.53279
10. 14.5411
a
R2
a
39 Ω
a
63 Ω
27
9
= 27 Ω ,
= 39 Ω ,
= 63 Ω ,
= 73 Ω ,
= 98 V .
Ohm’s law is V = I R .
(27 Ω) (73 Ω) [39 Ω + 63 Ω]
(12171 Ω2 )
= 16.5181 Ω .
and
017 (part 1 of 2) 10.0 points
An electron is projected into a uniform mag~ = By ̂ + Bz k̂, where
netic field given by B
By = 4.1 T and Bz = 2 T.
Version 022 – Exam 2 – hoffmann – (57505)
The magnitude of the charge on an electron
is 1.60218 × 10−19 C .
z
2T
B
4.1 T
y
v = 3.8 × 105 m/s
x
electron
Find the direction of the magnetic force
when the velocity of the electron is v k̂, where
v = 3.8 × 105 m/s.
b = −k̂
1. F
b = √1
2. F
̂ + k̂
2
b = k̂
3. F
b = ı̂ correct
4. F
1 b
5. F = √
̂ − k̂
2
b = √1
6. F
k̂ − ̂
2
b = −̂
7. F
b = ̂
8. F
b = −ı̂
9. F
Explanation:
Let : q = 1.60218 × 10−19 C ,
By = 4.1 T , and
Bz = 2 T .
Basic Concepts: Magnetic force on a moving charge is given by
~ = q ~v × B
~.
F
Solution:
~ = (4.1 T) ̂ + (2 T) k̂
B
v = (3.8 × 105 m/s) k̂ for the electron.
10
Find: The vector expression for the force on
the electron. This solves both part 1 and part
2.
We will go through two methods of doing
the problem.
The first is more mathematically oriented
and the second uses more of a reasoning argument.
Method 1: The force acting on a charge q
with velocity ~v in the presence of an external
~ is given by
magnetic field B
~ = q ~v × B
~
F
~ we
Taking the cross product of ~v with B
obtain
~ = q ~v × B
~
F
ı̂ ̂
k̂
= q 0 0
v 0 By Bz n
= q [(Bz )(0) − (By )(v)] ı̂
− [(0)(0) − (By )(0)] k̂
o
+ [(v)(0) − (By )(0)] ̂
= −q By v ı̂
= −(−1.60218 × 10−19 C)(4.1 T)×
(3.8 × 105 m/s) ı̂
= (2.49619 × 10−13 N) ı̂ ,
and the direction is +k̂ .
Method 2: The other method is to realize that the only component of the magnetic
field which affects the electron is the component perpendicular to its velocity. Therefore,
~ = q v B⊥ with the direction
F = q |~v × B|
given by the right hand rule to be in the negative k̂ direction; but recalling to reverse the
direction because the electron has a negative
instead of positive charge.
F = q v B⊥
= (1.60218 × 10−19 C)×
(3.8 × 105 m/s) (4.1 T)
= 2.49619 × 10−13 N in the ı̂ direction.
Version 022 – Exam 2 – hoffmann – (57505)
018 (part 2 of 2) 10.0 points
What is the magnitude of this force?
1. 2.73972e-13
2. 2.81983e-13
3. 2.49619e-13
4. 2.69166e-13
5. 2.25266e-13
6. 2.37122e-13
7. 2.4305e-13
8. 3.12425e-13
9. 2.98325e-13
10. 2.92237e-13
Correct answer: 2.49619 × 10−13 N.
019 10.0 points
A crossed-field velocity selector has a magnetic field of magnitude 0.037 T.
The mass of the electron is 9.10939 ×
10−31 kg.
What electric field strength is required if
52 keV electrons are to pass through undeflected?
1. 1855830.0
2. 5003860.0
3. 7825010.0
4. 15963800.0
5. 719385.0
6. 217935.0
7. 1494240.0
8. 3015720.0
9. 2618500.0
10. 6773350.0
Correct answer: 5.00386 × 106 V/m.
qE = qvB,
so the electric field is
E = v Br
=B
2W
m s
= (0.037 T)
2 (52000 eV) (1.602 × 10−19 J/eV)
9.10939 × 10−31 kg
020
10.0 points
Consider the RC circuit shown. The emf
of the battery is V , the resistance R, and the
capacitance C.
C
R
E
S
i
Consider the following statements.
A1: I = 0
V
A2: I =
R
V −1/(RC)
e
A3: I =
R
B1: VC = 0
B2: VC = V
B3: VC = V e−1/(RC)
When the switch S is closed for a very long
time, what are the current through R and the
potential across the capacitor, respectively?
Explanation:
The speed of the electron is
r
2W
v=
,
m
where W is the energy of the electron. The
electric and magnetic forces must balance
= 5.00386 × 106 V/m .
Explanation:
See above.
Let : B = 0.037 T ,
W = 52 keV = 52000 eV ,
m = 9.10939 × 10−31 kg .
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and
1. A2, B3
2. A3, B1
3. A2, B2
4. A2, B1
Version 022 – Exam 2 – hoffmann – (57505)
5. A1, B2 correct
6. A1, B3
7. A3, B3
8. A1, B1
9. A3, B2
Explanation:
At t = ∞ ,
I = 0,
VC = V .
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