Version 051 – MT3 – chiu – (58655)
This print-out should have 17 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001
10.0 points
N
R
b
7. The induced magnetic field is left to right
(Binduced =⇒) and the induced current flows
from a through R to b (I −→).
8. The induced magnetic field is left to right
(Binduced =⇒) and the induced current flows
from b through R to a (←− I).
A coil is suspended around an axis which is
co-linear with the axis of a bar magnet. The
coil is connected to a resistor with ends labeled a and b. The bar magnet moves from
left to right with North and South poles labeled in the figure.
a
1
S
v
Using Lenz’s law, what is the direction of
the induced magnetic field in the coil and
the direction of the induced current in the
resistor R when the bar magnet is moving left
to right?
1. The induced magnetic field is zero and
the induced current is zero.
2. The induced magnetic field is zero and
the induced current flows from a through R
to b (I −→).
3. The induced magnetic field is right to left
(⇐= Binduced ) and the induced current flows
from b through R to a (←− I). correct
4. The induced magnetic field is right to
left (⇐= Binduced ) and the induced current is
zero.
9. The induced magnetic field is left to right
(Binduced =⇒) and the induced current is
zero.
Explanation:
The induced magnetic field depends on
whether the flux is increasing or decreasing.
The magnetic flux through the coil is from
right to left. When the magnet moves from
left to right, the magnetic flux through the
coils decreases.
The induced current in the coil must produce an induced magnetic field from right
to left (⇐= Binduced ) to resist any change of
magnetic flux in the coil (Lenz’s Law).
The helical coil when viewed from the bar
magnet winds around the solenoid from terminal b counter-clockwise.
Since the induced field is right to left (⇐=
Binduced ), the induced current flows from b
through R to a (←− I); i.e., clockwise.
002 10.0 points
A long wire carries a current I1 = 1 A in
the −x̂-direction, and a rectangular loop of
height h = 0.2 m and width L = 0.5 m carries
a current I2 = 9 A counter-clockwise as shown
in Figure below. The loop is a distance d =
7 cm away from the long wire. The long
wire and the rectangular loop are in the same
plane.
I1
5. The induced magnetic field is zero and
the induced current flows from b through R to
a (←− I).
6. The induced magnetic field is right to left
(⇐= Binduced ) and the induced current flows
from a through R to b (I −→).
d
I2
h
y
x
L
Version 051 – MT3 – chiu – (58655)
Find the magnitude of the net magnetic
force exerted by the long wire on the rectangular loop.
1. 1.06e-05
2. 0.0002
3. 1.84e-05
4. 3.17e-05
5. 9.52e-06
6. 5.28e-06
7. 8.57e-05
8. 0.000333
9. 2.9e-05
10. 1.17e-05
A magnetic dipole is falling in a conducting
metallic tube. Consider the induced current
in an imaginary current loop when the magnet
is moving away from the upper loop and when
the magnet is moving toward the lower loop.
?
z
v
N
dipole
magnet
S
Spole
?
Ibelow
y
let : I1 = 1 A ,
I2 = 9 A ,
d = 7 cm = 0.07 m ,
h = 0.2 m , and
w = 0.5 m .
The forces on the vertical segments cancel
each other. The force on the top segment is:
~ top = I2 L B = I2 L µ0 I1 ŷ .
F
2πd
The force on the right bottom segment is:
µ0 I 1
FB = I2 L
(−ŷ) .
2 π (d + h)
The net force exerted by the long wire on
the loop is
µ0 I 1
µ0 I 1
− I2 L
2πd
2 π (d + h)
1
1
= I 2 L µ0 I 1
−
2 π d 2 π (d + h)
= (1 A)(9 A)(0.5 m)(2 × 10−7 N · A−2 )
1
1
−
0.07 m 0.07 m + 0.2 m
= 9.52 × 10−6 N .
Iabove
Npole
x
Correct answer: 9.52 × 10−6 N.
Explanation:
2
Determine the directions of the induced
currents Iabove and Ibelow in an imaginary loop
shown in the figure, as viewed from above,
when the loop is above the falling magnet and
when the loop is below the falling magnet.
1. Iabove = counter-clockwise and
Ibelow = counter-clockwise
2. no current flow
3. Iabove = clockwise and
Ibelow = clockwise
Fnet = I2 L
003
10.0 points
4. Iabove = counter-clockwise and
Ibelow = clockwise correct
5. Iabove = clockwise and
Ibelow = counter-clockwise
Explanation:
When the falling magnet is below the up→
per loop, −
µ ind must be up to attract the
falling magnet and slow it down; i.e., counterclockwise as viewed from above.
Version 051 – MT3 – chiu – (58655)
→
Before reaching the lower loop, −
µ ind must
be down to oppose the falling magnet; i.e.,
clockwise as viewed from above.
3
4. The charge is positive and with a constant
speed.
5. The charge is neutral and slowing down.
6. The charge is positive and speeding up.
x
z
v
Iabove
7. The charge is negative and with a constant speed.
Npole
8. The charge is negative and speeding up.
N
dipole
magnet
9. The charge is neutral and with a constant
speed.
S
Spole
10. None of these
y
Ibelow
004
Explanation:
We know that when a charged particle
moves in a uniform magnetic field with a
constant speed, it undergoes a circular motion with the centripetal force provided by
the magnetic force, namely
10.0 points
A static uniform magnetic field is directed
into the page. A charged particle moves in
the plane of the page following a counterclockwise spiral of decreasing radius as shown.
m
v2
= qvB,
r
so we know that the radius is in fact proportional to the speed,
B
r=
B
Neglect the effect due to gravity.
What is a reasonable explanation?
1. The charge is neutral and speeding up.
2. The charge is positive and slowing down.
correct
3. The charge is negative and slowing
down.
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 negative r̂ direction, the particle has a
positive charge.
005 10.0 points
Consider the setup shown, where a capacitor
with a capacitance C is connected to a battery
with emf V and negligible internal resistance.
The capacitor is fully charged. Denote the
energy stored by the capacitor by UC .
Version 051 – MT3 – chiu – (58655)
4
S
V
C
κ
The switch S is opened, disconnecting the
battery from the capacitor. Next, a dielectric
slab of dielectric constant κ is inserted between the capacitor plates, completely filling
the gap. Once the slab is completely inserted,
how much energy UC′ is stored by the capacitor?
1. UC′ = UC
Figure above shows 3 circuits labeled A, B
and C. All the light bulbs, batteries and the
capacitors are identical. Denote the charges
on the capacitors by QA (t), QB (t) and QC (t).
Choose the correct option that completes each
statement:
I. At some finite time t after the circuits
are closed but before the capacitors are fully
charged, the charges on the capacitors obey
the following relation:
2. UC′ = UC κ
3. UC′ = UC κ2
Ia. QC (t) > QA (t) > QB (t)
Ib. QC (t) = QA (t) = QB (t)
Ic. QC (t) < QA (t) < QB (t)
UC
κ2
UC
5. UC′ =
correct
κ
4. UC′ =
Explanation:
We are trying to find UC′ , the total energy
stored after the insertion of the dielectric, in
terms of UC , the energy stored before. The
capacitance increases from C to κ C, but the
voltage decreases from V to V /κ. We can
calculate
1
1
UC′ = C ′ V ′2 = κ C
2
2
11
=
CV 2
κ2
UC
.
=
κ
V
κ
2
II.Once all of the capacitors are fully
charged (i.e. let t → ∞), the charges on
them obey the following relation:
IIa. QC (∞) > QA (∞) > QB (∞)
IIb. QC (∞) = QA (∞) = QB (∞)
IIc. QC (∞) < QA (∞) < QB (∞)
1. Ib, IIc
2. Ia, II
3. Ic, IIc
4. Ic, IIa
5. Ic, IIb
6. Ia, IIb correct
7. Ia, IIc
006
10.0 points
8. Ib, IIb
Version 051 – MT3 – chiu – (58655)
9. Ib, IIa
Explanation:
In order to answer question I, we must know
the time constant for each circuit. Since the
time constant τ = RC and all the capacitors
are identical, the value of τ for circuits A, B,
and C differs only in the value of R. Since
the bulbs in B are in series while the bulbs in
C are in parallel, the total resistance of each
circuit is described by the inequality: RC <
RA < RB . Therefore, we have τC < τA < τB .
The time constant determines how quickly
each capacitor will charge. A charging capacitor obeys the equation Q(t) = Qmax(1 −
e−t/τ ). Examining this equation, we can see
that a smaller time constant leads to a faster
charging rate. Therefore, at any time t where
the capacitors are not yet fully charged, we
must have: QC (t) > QA (t) > QB (t).
For question II, once the capacitors are fully
charged, current no longer flows and the potential across each capacitor is equal to the
battery emf. Since all the capacitors are
identical and have the same potential potential difference, they all hold the same charge:
QC (t) = QA (t) = QB (t).
007 10.0 points
A 29 µF capacitor is first charged to a potential 41 V, then connected in parallel with a
5 µF capacitor that had been charged to the
half the initial potential of the other capacitor, i.e. 20.5 V. The capacitors are connected
positive plate to positive plate and negative
plate to negative plate.
What is the the final potential difference Vf
across the capacitors?.
1. 26.8667
2. 43.6429
3. 17.9412
4. 40.0
5. 39.0676
6. 35.8333
7. 15.1136
8. 24.2308
9. 37.9853
10. 45.1429
5
Correct answer: 37.9853 V.
Explanation:
Let :
C1 = 29 µF ,
C2 = 5 µF , and
V1 = 41 V .
Initially, before C2 is connected in paralle
to C1 , the total charge in the system is
Qi = C1 V1 + (0.5 C2 ) V1 = V1 (C1 + 0.5 C2 ) .
When they are connected in parallel, the
potential Vf is the same on both capacitors,
so by conservation of charge
Qi = Qf
V1 (C1 + 0.5 C2) = (C1 + C2 ) Vf .
C1 + 0.5 C2
C1 + C2
29 µF + 0.5 ∗ 5 µF
= (41 V)
29 µF + 5 µF
Vf = V1
= 37.9853 V .
008
10.0 points
In the circuit above, the battery has emf
E, the capacitor has capacitance C, and all of
the resistors have the same resistance R. A
“long time” after the switch S is closed, what
is the voltage VC across the capacitor?
3
1. VC = E correct
7
Version 051 – MT3 – chiu – (58655)
2. VC =
3. VC =
4. VC =
5. VC =
6
usual assumption that these plates are large
and that the charge on the surfaces of these
plates is distributed uniformly.
+Q
−3 Q
3
E
4
1
E
3
2
E
5
1
E
2
y
x
P
6. VC = E
Explanation:
A “long time” after the switch is closed,
the capacitor C is fully charged, so we may
analyze the circuit as though it were not there.
To find the voltage across the capacitor, we
must: determine the net current in the circuit;
use E − IR with the net current to calculate
the voltage at the upper capacitor lead.
To find net current, we calculate the equivalent resistance of the circuit. There is one
resistor in series with a parallel connection of
one resistor and three resistors.
−1
1
1
Req = R +
+
R 3R
−1
4
=R+
3R
7
3
= R + R = R.
4
4
The net current is then
E
4E
=
.
Req
7R
We may now determine VC :
VC = E − Inet R
4E
=E−
R
7R
3
= E.
7
009 (part 1 of 2) 5.0 points
Two charged conducting plates are parallel to
each other. They each have area A. Plate
#1 has a total positive charge Q while plate
#2 has a total charge −3 Q. We make the
#1
#2
What are the charges on the outer surfaces
of the two plates (the left side of Plate #1 ,
the right side of Plate #2)? Hint: Consider
what Gauss’ Law can tell you about charges
on surfaces of conductors.
1. Q, −3Q
2. 0, −2Q
3. −Q, −Q correct
4. 0, 0
5. Q/2, −3Q/2
Explanation:
The electric field to the left of Plate 1 is
the superposition of the electric field of two
uniformly charged sheets. The field due to
plate # 1 points to the left while the field due
to plate # 2 points to the right. The net field
will point to the right with a magnitude of
Enet = 3Q/(2Aǫ0) − Q/(2Aǫ0 ) = Q/(Aǫ0 )
Place a gaussian box so that one end is in
the conductor of plate # 1 and so it extends
out of left surface of plate # 1. The electric
field is zero on the end of the box inside the
conductor. On the opposite end the electric
field points perpendicularly inward so there is
a flux on this end of the box of
−Enet Abox = −Q/(Aǫ0 )Abox .
The flux on all the other sides is zero since
the electric field is parallel to their surfaces.
Using Gauss’ law,
−Q/(Aǫ0 )Abox = Qlef t,#1 /ǫ0 (Abox /A).
Version 051 – MT3 – chiu – (58655)
Qlef t,#1 = −Q
The charge on the right side of plate # 1
then must be 2Q so that the total charge on
the plate is Q.
Place a second gaussian box so that one
end is in plate #1 and the other end is in
plate #2. There is no flux out of this box so
the net charge inside must be zero. Thus the
charge on the left side of plate #2 is −2Q.
This leaves a charge of −Q for the right side
of plate #2.
Alternative explanation: Imagine pushing
the two plates together, resulting in a single
“effective” plate of charge −2Q. By symmetry, each outer surface of this “effective” plate
has charge −Q. Note that, in the process of
pushing the plates together, the electrostatic
force on a charge q on the outer surface of the
left plate will either remain constant or increase slightly, and this force will be directed
outward and normal to the surface; any increase will be balanced by the normal force of
the surface on the charge (it remains bound
to the metal). There will be no forces tangential to the surface, so any charge q on the
surface will experience no net force, and the
charge distribution on the outer surface will
not change.
Now let us separate the “effective” plate
into its original parts. From the argument
above, there is no net force on the exterior
surface charges, so the charges on the exterior
surfaces must remain −Q. Charge conservation then requires that the left plate have a
charge +2Q on its inner surface, while the
right plate must have charge −2Q, producing
the same answer. This is a useful doublecheck procedure.
010 (part 2 of 2) 5.0 points
What is the magnitude of the electric field at
point P in between the two plates?
1. 2Q/(Aǫ0) correct
4. 3Q/(2Aǫ0 )
5. Q/(Aǫ0 )
Explanation:
The electric field between the plates is the
superposition of the electric field of two uniformly charged sheets. The field due to plate
# 1 points to the right and the field due to
plate # 2 points to the right. The net field
will point to the right with a magnitude of
Enet = Q/(2Aǫ0 ) + 3Q/(2Aǫ0 ) = 2Q/(Aǫ0 )
Digression: There is a well known theorem in electrostatics based on Gauss’ law.
It states that, at any given point immediately above a conducting surface, the electric
field is always perpendicular to the surface
with E⊥ = σsurf /ǫ0 , where σsurf = dqsurf /dA
is the local surface charge density. The reader
should notice that this theorem is satisfied
both at the left surface and at the right surface within the gap.
011
10.0 points
A neutral copper bar oriented horizontally
moves upward through a region where there
is a magnetic field out of the page. Assuming
the bar has reached electrostatic equilibrium,
which diagram best illustrates the distribution of charge on the bar?
1.
2.
2. 3Q/(Aǫ0)
3. Q/(2Aǫ0)
7
3.
Version 051 – MT3 – chiu – (58655)
4.
correct
2. Va > Vb
and
3. Va > Vb
and
4. Vb = Va
and
5. Vb > Va
and
6. Va > Vb
and
7. Va > Vb
and
8. Vb = Va
and
9. Vb > Va
rect
and
10. Vb = Va
and
5.
Explanation:
Within the conductor, the magnetic force
on a (positive) test charge q is given by
~ = q ~v × B.
~ This magnetic force leads to a
F
separation of positive and negative charge. At
equilibrium, the uniform magnetic force must
be balanced out by a uniform electric force
|Fe | = qE = qvB due to the surface charge
distribution, so |E| = vB. (Surface charges
will automatically adjust their positions until
such an equilibrium state is reached.) Qualitatively, in order to produce a uniform E, the
surface charge is expected to decrease monotonically from the positive to the negative
end. See the discussion related to Fig. 19-18
in M&I vol. II.
012 (part 1 of 3) 3.0 points
The resistance of the rectangular current loop
is R, and the metal rod is sliding to the left.
The length of the rod is d, while the width
of the rails is ℓ. a and b are the contact
points where the rod touches the rails, and
d > ℓ . Consider the relationship between the
potentials Vb and Va and the direction of the
induced magnetic field.
a
B
R
v
B
b
Which statement is correct?
1. Vb > Va
d
m
and
b is down
B
ℓ
8
b is down
B
b is out of the page.
B
b is out of the page.
B
b is out of the page.
B
b is up
B
b is into the page.
B
b is up
B
b is into the page. corB
b is into the page.
B
Explanation:
The part of the rod which extends past the
rails does not have a bearing on the answers.
Lenz’s law states that the induced current
appears such that it opposes the change in the
magnetic flux. In this case the magnetic flux
through the rectangular loop is decreasing
(since the area of the loop is decreasing) with
the direction of the flux into the page, so that
the induced magnetic field must point into the
page in order to keep the flux through the loop
constant. This corresponds to an induced
current which flows in a clockwise direction,
so if you look at the potential drop across the
resistor, you can see that the potential at b is
greater than the potential at a.
As a double-check, an analysis based on
motional emf shows that the motion of the
bar causes a polarization with positive charge
on the bottom and negative charge on the top,
corresponding to Vb > Va and a CW current.
013 (part 2 of 3) 3.0 points
What is the magnitude of the induced current
around the loop?
Bdv
R2
Bℓv
2. |Iind | =
correct
R
1. |Iind | =
Version 051 – MT3 – chiu – (58655)
Bℓv
R2
B d v2
4. |Iind | =
R2
B d2 v
5. |Iind | =
R
Bdv
6. |Iind | =
R
B ℓ v2
7. |Iind | =
R
B ℓ v2
8. |Iind | =
R2
B ℓ2 v
9. |Iind | =
R
B d v2
10. |Iind | =
R
Explanation:
The rate of change of the area of the rectangular loop is
3. |Iind | =
dx
dA
=ℓ
.
dt
dt
Then from Faraday’s law, the magnitude of
the induced emf is given by
E = Bℓv.
From Ohm’s law, E = I R so the magnitude
of the induced current is
Iind =
E
Bℓv
=
.
R
R
9
2 2
~ k = B ℓ v , up
4. kF
R
2
2 2
~ k = B d v , up
5. kF
R
2 2
~ k = B ℓ v , down correct
6. kF
R
2
~ k = B d ℓ v , up
7. kF
R
2
2 2
~ k = B ℓ v , down
8. kF
R
2
B d2 v
~
9. kF k =
, up
R
2
~ k = B d ℓ v , down
10. kF
R
Explanation:
The magnitude of the magnetic force exerted on the metal rod is given by
F = I ℓB,
where I is the induced current found in Part
2.
~ = 0 outside the rectanℓ was used since B
gle.
Substituting in the expression for the induced current yields
B 2 ℓ2 v
.
R
The direction of the induced current is down
through the metal rod as explained in Part 1.
F =
015
10.0 points
014 (part 3 of 3) 4.0 points
The magnitude of the force exerted on the
metal rod by the magnetic field, and the direction of the current through the metal bar
are given respectively by
2 2
~ k = B d v , down
1. kF
R
2
2 2
~ k = B ℓ v , up
2. kF
R
2
B d2 v 2
~
, down
3. kF k =
R
In the circuit above: B1 is a small, thin
filament bulb with electron density n, crosssectional area A, mobility u, and length L;
Version 051 – MT3 – chiu – (58655)
B2
B2 is a large, thick filament bulb made of the
same material, but with cross-sectional area
2A and length 3L.
When the switch S is closed, how many
electrons flow off the negative terminal of the
battery each second?
5
E
1. nAu correct
3
L
E
4
2. nAu
3
L
E
3. nAu
L
E
4. 3nAu
L
2
E
5. nAu
3
L
E
6. 2nAu
L
Explanation:
The number of electrons per second off the
negative terminal of the battery is just the
net electron current inet . In order to determine inet , we must first determine the current
through each bulb. To do so, we must make
use of the two independent loop equations:
E
L
E
E − E1 3L = 0 → E2 =
3L
.
E − E1 L = 0 → E1 =
Substituting these values for E1 and E2 into
the electron current equations for each bulb,
we obtain:
i1 = nAu
E
L
E
i2 = n2Au
3L
.
The net current is given by the node equation:
inet = i1 + i2 . Using the above expressions,
we find that
E
5
inet = nAu .
3
L
016
10.0 points
10
θ
B1
h
B1
w
y
θ
B2
x
The figure shows a region of width w and
height h where the magnetic field along the
~ 1 along h and B
~2
boundaries is known to be B
along w. The angle between B2 and w is θ.
Calculate the current I that passes through
the shaded region.
{Note: You may notice that this field configuration is unrealistic. We use it to illustrate
the application Ampere’s law.}
2
1. I =
(B1 w cos θ + B2 w cos θ), out of
µ0
page
2
B2 w cos θ, out of page
µ0
1
B1 h + B2 w cos θ, out of page
3. I =
µ0
2
4. I =
(B1 h + B2 w cos θ), into page
µ0
2
5. I =
B2 w cos θ, into page
µ0
2. I =
6. 0 correct
1
(B1 h + B2 w cos θ), into page
µ0
2
8. I =
(B1 h + B2 w cos θ), out of page
µ0
2
9. I =
(B1 w cos θ +B2 w cos θ), into page
µ0
7. I =
Explanation:
This problem is a straightforward application of Ampere’s law
I
~ • d~ℓ = µ0 Ienc .
B
Version 051 – MT3 – chiu – (58655)
The vertical portions make no contribution
~ ⊥ d~ℓ. Integrating about the loop in
since B
~ 2 • d~ℓ > 0
a clockwise fashion, we see that B
~ 2 • d~ℓ < 0 for the
for the upper wire and B
~ along
lower wire. Since the component of B
d~ℓ is the same magnitude in either case, these
contributions also sum to zero. We conclude
there is no current enclosed.
017
10.0 points
A Metal bar of mass M and length L slides
down with negligible friction but good electrical contact between two vertical metal posts.
The bar falls at a constant speed v. The bar
and posts have negligible electrical resistance,
but the posts are connected at the bottom
by a resistor of resistance R. The apparatus
sits in a uniform magnetic field of unknown
magnitude B coming out of the page.
In terms of the given quantities and physical
constants, determine the magnitude of B.
r
M gR
1.
vL
M gR
2.
(vL)2
r
M gR
3.
correct
vL2
Mg
4.
ev
1 p
M gR
5.
vL
M gR
6.
IL
Explanation:
Since we are told the speed is constant,
the magnetic and gravitational forces must
balance, so
ILB = M g
11
and I flows counterclockwise. We need to
express I in terms of given quantities, and
V
Ohm’s law gives us I = . In this case, V is
R
supplied by the motional emf vBL. Combining everything and solving for B:
vBL
LB = M g
R
vL2 2
B = Mg
R
r
M gR
B=
.
vL2