Version 026 – MT3 v3 – yao – (57465)
This print-out should have 18 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 (part 1 of 2) 10.0 points
In the figure below the switch S is initially
in position a .
b
R1
V
a
b
S
1
1. The current through R3 decreases.
2. The current through R3 increases.
3. The current through R2 is twice what it
was with R1 in the circuit. correct
4. The current through R2 and R3 are now
the same.
b
bb
b
c
b
5. The current through R2 is half what it
was with R1 in the circuit.
R3
R2
b
What happens to total current in the circuit
(i.e., the current flowing off the battery terminal) when the switch is moved to the open
position b? R1 = R2 = R, R3 = 2R. Neglect
the internal resistance of the battery.
1. The total current is reduced to 2/3 its
initial value.
2. The total current is reduced to 1/2 its
initial value. correct
3. The total current is unchanged.
6. The current through R2 remains the same
as when R1 was in the circuit.
Explanation:
The potential difference across R2 and R3
must be the same. Since R3 = 2R2 , the current through R2 must be double that through
R3 :
V = I3 R3 = I3 × 2R = 2I3 × R = I2 R2 .
When the switch was in position a, R2 was in
series with R1 , so the the branch had a total
resistance 2R and its current was identical
to I3 . Since I3 is the same in both circuits,
we conclude that the current through R2 is
double what it was in the initial circuit.
003
10.0 points
4. The total current increases to 3/2 its
initial value.
5. The total current increases to twice its
initial value.
Explanation:
The equivalent resistance of the initial circuit is R, so the total initial current is V /R.
When the switch is at b, the resistance of
the circuit is R3 = 2R, so the total current
becomes V /2R, 1/2 its initial value.
002 (part 2 of 2) 10.0 points
What happens when switch S is moved to
position c, leaving R2 and R3 parallel?
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 magnitude B
coming out of the page.
Version 026 – MT3 v3 – yao – (57465)
In terms of the given quantities and physical constants, determine the speed v of the
bar.
√
M gR
1.
LB
r
M gR
2.
LB
M gR
3.
correct
(LB)2
r
Mg
4. LB
R
gR
5.
LB
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. I =
6. M gR
mg
(LB)2
M g(LB)2
8.
R
Explanation:
Since we are told the speed is constant,
the magnetic and gravitational forces must
balance, so
Mg
ILB = M g ⇒ I =
LB
and I flows counterclockwise. Then, using the
expression
3. I =
7.
∆V = emf = vBL
for a “battery” with negligible internal resistance, we can apply a loop rule ∆V − IR = 0
and solve the resulting equation for v.
∆V = IR
Mg
vBL =
R
LB
M gR
v=
.
(LB)2
004
10.0 points
2
4. I =
5. I =
6. I =
1
(B1 h + B2 w cos θ), into page
µ0
2
(B1 h + B2 w cos θ), out of page
µ0
2
B2 w cos θ, into page correct
µ0
2
B2 w cos θ, out of page
µ0
2
(B1 h + B2 w cos θ), into page
µ0
7. 0
2
(B1 w cos θ +B2 w cos θ), into page
µ0
1
9. I =
B1 h + B2 w cos θ, out of page
µ0
8. I =
Explanation:
This problem is a straightforward application of Ampere’s law
I
~ • d~ℓ = µ0 Ienc .
B
The vertical portions make no contribution
~ ⊥ d~ℓ. By symmetry, the horizontal
since B
portions make an equal contribution, so the
path integral simplifies to
2B2 w cos θ = µ0 I
2
B2 w cos θ .
I=
µ0
Version 026 – MT3 v3 – yao – (57465)
~ • d~ℓ
Looking at the loop, we can see that B
is strictly positive when the path integral is
evaluated in a clockwise fashion. A clockwise
path defines an area vector that points into
~ • d~ℓ ≥ 0 at
the page, and the fact that B
every point on the loop means that current is
positive relative to this loop, so must flow in
the same direction that the area vector points:
into the page.
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.
Before the insertion of the dielectric slab
with dielectric constant κ, the charge on the
capacitor is Q = C V and the energy density
1
is u = ǫ0 E 2 .
2
Now, keeping the battery connected, insert
the dielectric, which fills the gap completely.
d
V
C
Suppose the battery remains connected
during the insertion of the slab. Determine
the energy density u′ within the gap in the
presence of the dielectric.
′
2
1. u = u κ
2. u′ = u κ correct
u
κ2
u
4. u′ =
κ
3. u′ =
5. u′ = u
Explanation:
We are trying to find u′ , the energy density
after the insertion of the dielectric, in terms
of u, the energy density before. The increase
of C to C ′ = κ C, with V unchanged, draws
more charge from the battery to the capacitor.
3
Then we can calculate
1
1
U ′ = C′ V 2 = κ C V 2 = κ U .
2
2
Since U changes and the volume doesn’t, u
must change by the same factor.
006 10.0 points
In the diagram below, suppose that VC − VB
= 2.2 V, and that VF − VC = 0.5 V. What is
the potential difference VB − VA ?
Bb
Cb
Loop
1
Db
Loop
2
b
b
b
A
F
E
1. 1.8
2. 0.6
3. -3.3
4. -0.9
5. -1.2
6. 6.0
7. 2.7
8. -2.7
9. -4.8
10. 0.9
Correct answer: −2.7 V.
Explanation:
The potential difference is given by the loop
equation
(VC − VB ) + (VF − VC ) + (VB − VA ) = 0
which may be rewritten as
VB − VA = −[(VC − VB ) + (VF − VC )]
= −(2.2 V + 0.5 V)
= −2.7 V .
007 10.0 points
Consider the situation in the figure below, in
which there is a uniform electric field in the y
Version 026 – MT3 v3 – yao – (57465)
4
direction and a uniform magnetic field in the
x direction.
y
−0.00028 V
b−
+b
Proton
~ (at rest)
E
b
~
B
Power
Supply
b− + b
x
z
Assume the proton begins at rest and is
released at t = 0. Notice that, for t > 0, vx
continues to be zero while vy and vz become
functions of time: ~v (t) = h0, vy (t), −vz (t)i.
What is the correct expression for the Lorentz
force vector?
1. Fe+ = e[(E − Bvy (t))ẑ + Bvz (t)ŷ]
2. Fe+ = e[(E + Bvz (t))ŷ + Bvy (t)ẑ]
3. Fe+ = e[(E − Bvy (t))ŷ + Bvz (t)ẑ]
4. Fe+ = e[(E − Bvz (t))ẑ + Bvy (t)ŷ]
5. Fe+ = e[(E + Bvy (t))ẑ − Bvz (t)ŷ]
6. Fe+ = e[(E−Bvz (t))ŷ−Bvy (t)ẑ] correct
7. Fe+ = e[(E + Bvz (t))ŷ − Bvy (t)ẑ]
Explanation:
The Lorentz force law tells us that
~ + ~v × B)
~
Fe+ = q(E
= q[Eŷ + (vy (t)ŷ − vz (t)ẑ) × Bx̂]
= q[Eŷ + Bvy (t)(−ẑ) + Bvz (t)(−ŷ)]
= q[(E − Bvz (t))ŷ − Bvy (t)ẑ] .
008 10.0 points
A slab made of conducting material is connected to a power supply as shown in the figure below. There is a uniform magnetic field
of 1.2 T pointing upward throughout this region (perpendicular to the horizontal slab).
Two voltmeters are connected to the slab and
read steady voltages as shown. The connections across the slab are carefully placed
directly across from each other.
7 cm
1.2 T
15
cm
1.1 cm
0.68 V
b−
+b
What is the drift speed v of the mobile
charges?
1. 0.00333333
2. 0.00247619
3. 0.003125
4. 0.00285714
5. 0.00666667
6. 0.00228571
7. 0.0017037
8. 0.00386364
9. 0.00357143
10. 0.00625
Correct answer: 0.00333333 m/s.
Explanation:
In the steady state, the transverse electric
and magnetic forces must balance, so
q EH = q v B sin 90◦ = q v B.
So EH = v B. The drift speed v is uniform
throughout the slab (current conservation and
constant cross-sectional area), and the magnetic field B is uniform throughout this region. So EH = v B must be uniform along the
7 cm path across the slab, and we can write
|∆V | = 0.00028 V
= EH w
= EH (7 cm)
0.00028 V
⇒ EH =
7 cm
= 0.004 V/m .
And
Version 026 – MT3 v3 – yao – (57465)
EH
B
0.004 V/m
=
1.2 T
= 0.00333333 m/s .
L
v=
C
Note that we have experimentally determined the drift speed v, independent of the
carrier charge q and the density of charge carriers n.
009 (part 1 of 2) 10.0 points
In the circuit shown, the switch is connected to position a for a long time, then
thrown to position b.
0.2 H
1 µF
14 V
S b
a
9Ω
Find the frequency of oscillation in the LC
circuit.
1. 375.132
2. 355.881
3. 237.254
4. 290.576
5. 425.359
6. 410.936
7. 479.87
8. 277.053
9. 562.698
10. 441.416
Correct answer: 355.881 Hz.
Explanation:
Let :
5
R = 9 Ω,
C = 1 µF = 1 × 10−6 F ,
L = 0.2 H , and
E = 14 V .
E
S b
a
R
The frequency of oscillation in the circuit is
1
√
f=
2π LC
1
p
=
2 π (0.2 H) (1 × 10−6 F)
= 355.881 Hz .
010 (part 2 of 2) 10.0 points
Find the maximum current in the inductor.
1. 0.0154919
2. 0.031305
3. 0.0339411
4. 0.0396863
5. 0.0129099
6. 0.033541
7. 0.0426028
8. 0.0183848
9. 0.0431161
10. 0.0189737
Correct answer: 0.031305 A.
Explanation:
When the current is maximum in the oscillating LC circuit, the charge on the capacitor
is zero; at this point, the total energy of the
circuit is stored in the magnetic field of the inductor. On the other hand, when the charge
reaches its maximum in the capacitor, the
current is zero in the circuit; at this point,
the total energy of the circuit is stored in the
electric field of the capacitor. Therefore, from
conservation of energy,
1
1
2
C E 2 = L Imax
2
2r
r
C
1 × 10−6 F
= 14 V
Imax = E
L
0.2 H
= 0.031305 A .
Version 026 – MT3 v3 – yao – (57465)
011 10.0 points
In an RL series circuit, an inductor of 4.57 H
and a resistor of 8.25 Ω are connected to a
22.7 V battery. The switch of the circuit
is initially open. Next close the switch and
wait for a long time. Eventually the current
reaches its equilibrium value.
At this time, what is the corresponding
energy stored in the inductor?
1. 45.9691
2. 33.2366
3. 16.7545
4. 32.4889
5. 43.6196
6. 26.1092
7. 8.84938
8. 31.7775
9. 17.2994
10. 8.74791
Correct answer: 17.2994 J.
Explanation:
Let : L = 4.57 H ,
R = 8.25 Ω ,
E = 22.7 V .
and
The current in an RL circuit is
E I=
1 − e−Rt/L .
R
The final equilibrium value of the current,
which occurs as t → ∞ , is
I0 =
E
22.7 V
=
= 2.75152 A .
R
8.25 Ω
The energy stored in the inductor carrying a
current 2.75152 A is
1
L I2
2
1
= (4.57 H) (2.75152 A)2
2
= 17.2994 J .
U=
012 (part 1 of 2) 10.0 points
6
Figure above shows 3 circuits labeled A,
B and C. All the light bulbs, batteries, and
capacitors are identical (model the light bulbs
as resistors). Denote the time constants of
the circuits by TA , TB and TC . Choose from
among the following statements:
TA
.
Ia. TB =
2
Ib. TB = TA .
Ic. TB = 2 TA .
TA
.
IIa. TC =
2
IIb. TC = TA .
IIc. TC = 2 TA .
1. Ic,IIa correct
2. Ia,IIa
3. Ic,IIc
4. Ia,IIc
5. Ib,IIb
6. Ib,IIc
7. Ic,IIb
8. Ia,IIb
9. Ib,IIa
Explanation:
For circuit A, TA = R C .
For circuitB, the two bulbs are in series. It
has a resistance 2 R. So TB = 2 R C.
Version 026 – MT3 v3 – yao – (57465)
For circuit C, the two bulbs are in parallel.
R
RC
It has a resistance . So TC =
.
2
2
Put all together, we have TB = 2 TA , and
TA
. So Ic and IIa are correct.
TC =
2
013 (part 2 of 2) 10.0 points
Once the capacitors are fully charged, denote
the potential differences across them by VA ,
VB , and VC . Choose from among the following
statements:
5. 1424.71
6. 3974.15
7. 3778.15
8. 535.026
9. 3178.79
10. 519.946
Correct answer: 1424.71 V.
Explanation:
Let :
1. VA < VB , VA < VC
2. VA < VB , VA = VC
3. VA > VB , VA = VC
4. VA = VB = VC correct
5. VA > VB , VA < VC
6. VA < VB , VA > VC
7. VA > VB , VA > VC
Explanation:
Once the capacitors are fully charged, current no longer flows in the circuits and the potential difference across all the resistors drops
to zero. In this case, the loop equation for
each circuit is identical,
V −
Q
= 0,
C
7
A = 0.11 m2 ,
f = 63 rev/s ,
B = 0.08 T , and
N = 409 turns .
Basic Concept:
Solenoid
Faraday’s Law for
d ΦB
dt
ΦB = B A cos θ
θ = ωt
E = N A B ω sin(ω t) .
E = −N
Solution: For generator, the maximum emf
is
Emax = N A B ω
= N A B (2 π f )
= (409 turns)(0.11 m2 )(0.08 T)
× 2 π (63 rev/s)
= 1424.71 V .
so it must be that
VA = VB = VC .
014 (part 1 of 2) 10.0 points
A loop of area 0.11 m2 is rotating at 63 rev/s
with the axis of rotation perpendicular to a
0.08 T magnetic field.
If there are 409 turns on the loop, what is
the maximum voltage induced in it?
1. 1529.56
2. 1980.71
3. 910.328
4. 2866.66
015 (part 2 of 2) 10.0 points
When the maximum induced voltage occurs,
what is the orientation of the plane of the loop
with respect to the magnetic field?
1. at 45 degrees to B
2. parallel to B correct
3. Not enough information is given.
4. perpendicular to B
Explanation:
Version 026 – MT3 v3 – yao – (57465)
The emf generated at time t is
3. v =
E(t) = N A B ω sin(ω t)
= N A B ω sinθ
5. v =
6. v =
keywords:
2
B
E
B
E
E
correct
B
E
B2
4. v =
π
|E| is maximum when |sinθ| = 1, or θ = ± ,
2
so the plane of loop is parallel to B.
8
7. v = E B
016 (part 1 of 2) 10.0 points
B2
E
E2
9. v =
B
2
E
10. v =
B
Explanation:
Since the electric and magnetic forces on
the ion are equal,
8. v =
A device (“source”) emits a bunch of negatively charged ions (particles) with a range
of velocities. Some of these ions pass through
the left slit and enter “Region I” in which
there is a vertical uniform electric field E (in
the −̂ direction) and a B 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.
+V
Region of
Magnetic
Field
B
q
m
r
y
d
qE = qvB
E
v= .
B
017 (part 2 of 2) 10.0 points
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.
If the magnitude of charge on each ion is
|q|, what is the mass of the ions?
1. m =
x
z
Region I
Region II
In “Region I”, the electric potential between the plates is V , the distance between
the plates is d, which gives rise to an electric
field E . The magnetic field in both “Regions
I and II” is B .
What is the velocity of the ion?
B
E2
√
2. v = E B
1. v =
2. m =
3. m =
4. m =
5. m =
6. m =
7. m =
|q| B r
E
|q| E r
B2
|q| B 2
rE
|q| E 2 r
B
|q| E
rB
|q| B 2 r
correct
E
|q| B r
E2
Version 026 – MT3 v3 – yao – (57465)
|q| E r
B
|q| E 2
9. m =
rB
|q| B
10. m =
rE
Explanation:
The radius of a circular path taken by a
charged particle in a magnetic field is given
by
8. m =
mv
r=
qB
q B2 r
Br
Br
=
m=q
=q
.
E
v
E
B
018
10.0 points
In the above figure, the radius of the
solenoid(r1 ) is 2 cm and the radius of the
ring (r2 ) is 10 cm.
The magnetic flux through the outer ring is
Ia) πr12 B
Ib) πr22 B
Now consider the case where I is increasing.
Direction of the induced current in the ring
with radius r2 , as viewed from the right is
IIa) clockwise
IIb) counterclockwise
Direction of the induced magnetic dipole of
the outer ring is
IIIa) to the right
IIIb) to the left
1. Ia, IIa, IIIa
2. Ia, IIb, IIIb
9
3. Ib, IIb, IIIb
4. Ib, IIa, IIIa
5. Ia, IIb, IIIa
6. Ib, IIa, IIIb
7. Ia, IIa, IIIb correct
8. Ib, IIb, IIIa
Explanation:
Ia: The magnetic flux encircled by the big
ring is: πr12B
IIa: With B increasing, Bind of the big loop
is pointing to the left, so from the RHR the
induced current is clockwise as viewed from
the right.
IIIb: The direction of the induced magnetic
dipole is to the left.