Version 040 – Midterm 02 – yao – (57465)
1
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
The figure shows a portion of two ribbons
of width L, each containing a large number
N of closely packed wires. Each wire in the
upper ribbon carries a current I into the page;
each wire in the lower ribbon carries a current
2I out of the page. Use Ampere’s law to
~ at point P . Assume the +x
determine B
direction is to the right.
We first work out the contribution to the
magnetic field due to the top ribbon alone.
Draw a rectangular Amperean loop (of width
ℓ and height h) as shown in the figure. Integrating clockwise about this path:
Along the sides of the path,
Z
~ · d~l = 0 ,
B
~ is perpendicular to d~l.
since B
Along the upper part of the path,
Z
~ · d~l = Bupperℓ .
B
1. B =
2. B =
3. B =
4. B =
5. B =
6. B =
7. B =
8. B =
µ0 N I
x̂
2L
µ0 N I
x̂
L
3µ0 N I
(−x̂)
L
3µ0 N I
(−x̂)
2L
3µ0 N I
x̂
L
µ0 N I
(−x̂)
L
µ0 N I
(−x̂) correct
2L
3µ0 N I
x̂
2L
Explanation:
From the diagram, we can see that any pair
of wires equidistant from but on either side of
P will generate magnetic fields whose vertical
components cancel. Therefore the direction of
the magnetic field must be to the right above
the top current ribbon and to the left below
(vice-versa for the lower ribbon).
Along the lower part of the path,
Z
~ · d~l = Blower ℓ = Bupperℓ ,
B
since Bupper = Blower by symmetry.
Therefore,
I
~ · d~l = 2Btop ℓ = µ0 Iinsidepath
B
N
= µ0
ℓI ,
L
where N/L is the current density in
wires/meter. So the top ribbon contributes
Btop =
µ0 N I
x̂ .
2L
at P .
A similar analysis for the lower ribbon reveals that its contribution at P is given by
Bbottom =
µ0 N I
(−x̂) .
L
Therefore the vector sum at P is equal to:
~ ) = µ0 N I (−x̂) .
B(P
2L
Version 040 – Midterm 02 – yao – (57465)
2
from C to D is given by
002
10.0 points
UD −UC = e(V (b)−V (a)) = k e q s
1
1
− 2
2
a
b
Intuitive reasoning on the sign of ∆U : Natural tendency of the motion is from high
potential energy to lower potential energy.
When the proton is released it should move
from D to C, so UD > UC .
Consider the setup in Figure above. What
is the change in potential energy ∆U = UD −
UC , in moving a proton from C to D?
1 1
1. k q s
−
a b
1 1
−
2. -k e q s
a b
1 1
−
3. -2k e q s
a b
1
1
−
4. -2k e q s
a 2 b2
1 1
−
5. 2k e q s
a b
1
1
−
correct
6. k e q s
a 2 b2
1
1
7. -k e q s
−
a 2 b2
1 1
−
8. k e q s
a b
1
1
−
9. 2k e q s
a 2 b2
1
1
−
10. k q s
a 2 b2
Explanation:
Letting the +x direction be to the right,
Z D
2kqs
VD − VC = −
(−x̂) • d~x
x3
C
Z −1
= 2kqs d
2 x2
1
1
= 2kqs − 2 +
.
(1)
2b
2 a2
Multiplying eq(1) by the proton charge e,
we arrive at the potential energy difference
003 10.0 points
A current-carrying solenoidal coil of length
L and radius R, L ≫ R, is uniformly wound
with N turns. Suppose the coil is now cut in
half, resulting in two new solenoids with half
the number of turns (N/2) as the original coil.
The coils are connected to separate circuits so
that current flows through them in the same
direction as through the original coil. When
brought close to each other so that the cut
ends face each other:
(Ia) The coils repel each other.
(Ib) The coils are attracted to each other.
(Ic) The coils do not interact magnetically.
If the magnetic field created inside of one
of the new coils (far from the ends) is B ′ , and
that created by the original coil is B (all other
parameters being the same), then which of
the following relations is true?
(IIa) B ′ = B
B
(IIb) B ′ =
2
B
(IIc) B ′ =
8
′
(IId) B = 2B
1. Ia, IId
2. Ia, IIb
3. Ia, IIa
4. Ic, IId
5. Ic, IIb
6. Ic, IIc
.
Version 040 – Midterm 02 – yao – (57465)
7. Ic, IIa
4. Southeast
8. Ib, IIc
5. West
9. Ib, IIa correct
6. East
10. Ib, IIb
7. Southwest
Explanation:
When carrying a current, each smaller coil
still acts like a magnetic dipole, so must have
a North and a South pole. With the current flowing the same direction as through
the original coil, the cut ends of the new coils
must be of opposite polarity — if they were
not, then one coil would have either two north
poles or two south poles, an impossibility. Ib
is the correct choice.
8. South
Since magnetic field strength is proportional to N/L and this ratio does not change
when the coil is divided, we still have B ′ = B.
Hence, IIa is correct.
3
Explanation:
At location 1, current flows to the left
(South) to make the compass deflect Northwest. Thus, at location 2, current flows to the
~ due to the current, at locaright (North). B
tion 2 above the wire, is downward toward the
bottom of the page (East). Thus the net magnetic field due to the Earth and the current
carrying wire is Northeast, so the compass
needle will point Northeast. The actual angle
will depend on the value of the current.
005
10.0 points
004 10.0 points
Consider the following diagram.
N
1
2
The wire rests below two compasses. When
no current is running, both compasses point
North (the direction shown by the gray arrows). When the current runs in the circuit,
the needle of compass 1 deflects as shown. In
what direction will the needle of compass 2
point?
1. Northeast correct
2. North
3. Northwest
Figure above shows a portion of long, negatively charged rod. You need to determine
the potential difference VB − VA due to the
charged rod. Use the convention that up is
along the +y direction.
Consider the following statements:
Ia. The sign of VB − VA is positive
Ib. The sign of VB − VA is negative
Now bring a test charge q from A to B.
Consider the following statements:
IIa. The sign of the potential difference
Version 040 – Midterm 02 – yao – (57465)
VB − VA due to the rod depends on the sign
of the test charge q.
IIb. The sign of the potential difference
VB − VA due to the rod does not depend on
the sign of the test charge q.
Choose the correct choice:
1. Ia, IIb correct
2. Ib, IIa
3. Ib, IIb
4
The z-component of the electric field is
∂V
= −(4 x − 10 z)
∂z
= 10(3.4 m) − 4(2 m) = 26 N/C .
Ez = −
007 10.0 points
The figure represents two long, straight, parallel wires extending in a direction perpendicular to the page and carrying currents of
equal magnitude. The current in the left wire
runs out of the page and the current in the
right runs into the page.
4. Ia, IIa
Explanation:
Ia is correct. Notice that with the source
charge on the rod being negative, it generates
a downward electric field. By inspection the
vector E is antiparallel to the path, so ∆V =
VB − VA > 0.
IIb is correct. The potential difference
is a property of E generated by the source
charges. It is independent of the sign or magnitude of the test charge q.
006 10.0 points
Consider the setup where the potential function is given by: V = 4xz + 2y − 5z 2 . Find
Ez at the point: h2 m, 2 m, 3.4 mi.
1. 63.0
2. 31.0
3. 47.0
4. 29.0
5. 26.0
6. 92.0
7. 8.0
8. 61.0
9. 25.0
10. 32.0
Correct answer: 26 N/C.
Explanation:
let : x = 2 m ,
y = 2 m , and
z = 3.4 m .
a
c
b
What is the direction of the magnetic field
created by these wires at location a, b and c?
(b is midway between the wires.)
1. down, down, up
2. up, zero, down
3. up, down, up
4. down, zero, up
5. up, up, down
6. down, up, down correct
Explanation:
By the right-hand rule the left wire has a
clockwise field and the right wire a counterclockwise field.
At the leftmost point, the field due to the
left wire points down, while that due to the
right wire points up. Since
Bwire =
µ0 I
,
2π r
The field due to the left wire is larger, so the
superposition of the fields produces a downward field. At the center point, both fields
point up. At the rightmost point, the field
due to the left wire points up, while that due
Version 040 – Midterm 02 – yao – (57465)
to the right wire points down. Since the right
wire is closer, it dominates, so the net field is
down.
008 10.0 points
You are given three parallel conducting
plates of cross-sectional area A that are perpendicular to the x-axis. They are labeled,
from left to right, as plates 1, 2 and 3, respectively. The corresponding plate charges are
Q1 = −q, Q2 = 3q and Q3 = −2q. The width
of the gap between both plates 1 and 2 and 2
and 3 is d.
Determine the ∆V = V3 − V1 .
(q/A)d
ǫ0
(q/A)d
2. −
correct
ǫ0
5(q/A)d
3.
ǫ0
5(q/A)d
4. −
ǫ0
3(q/A)d
5.
ǫ0
6(q/A)d
6.
ǫ0
4(q/A)d
7.
ǫ0
3(q/A)d
8. −
ǫ0
9(q/A)d
9. −
2ǫ0
Explanation:
One may regard the 3-plate system as a
composite system which involves two capacitor systems with the 12-capacitor followed by
the 23-capacitor.
The 12-capacitor has charges Q1 and Q2 +
Q3 , i.e charges of −q and q respectively.
The 23-capacitor has charges Q1 + Q2 and
Q3 , i.e charges of +2q and −2q respectively.
The potential difference is
1.
~ 12 • d(x̂) − E
~ 23 • d(x̂)
V3 − V1 = −E
(q/A)d 2(q/A)d
=
−
ǫ0
ǫ0
=−
5
(q/A)d
.
ǫ0
009 (part 1 of 2) 5.0 points
Three point charges +q are placed at corners
of a square with sides of length L.
+
+
O
L
+
A
B
What is the potential at point O?
√ kq
1. V = 2 , 2
L
1
kq
2. V = 2 + √
L
2
kq
3. V = 2
L
k q2
1
4. V = 2 + √
L
2
kq
5. V = 2 2
L
kq
1
6. V = 2 + √
2 L2
kq
7. V = 3
L
kq
8. V =
L
√ kq
9. V = 3 2
correct
L
kq
10. V = 3 √
2L
Explanation:
so
L
The distance from any vertex to O is √ ,
2
V =
X
i
Vi = 3
√ kq
kq
√ =3 2
.
L
(L/ 2)
010 (part 2 of 2) 5.0 points
Now place a −q charge at A and move the +q
charge at B to infinity.
Version 040 – Midterm 02 – yao – (57465)
+
+
O
L
−
B
How much work is required to bring the +q
charge from infinity to point O?
1. W
2. W
3. W
4. W
5. W
6. W
7. W
8. W
9. W
10. W
k q2
=−
4L
k q2
=− 2
L
k q2
= 3√
2L
kq
= −√
2 L2
√ k q2
= −3 2
L
2
kq
=√
2L
√ k q2
= 2
correct
L
√ k q2
=3 2
L
2
kq
=√
2 L2
√ k q2
=− 2
L
X
i
P6
P1
d
P5
θ
θ
eb+
θ
θ
P4
P2
P3
An proton is moving horizontally to the
left with speed 4 × 106 m/s. Each location
is d = 6 cm from the electron, and the angle
θ = 35 ◦ . Give the magnetic field at P6 using
the convention that out of the page is positive.
1. 1.02 × 10−17 T
2. −1.53 × 10−17 T
3. −2.04 × 10−17 T
4. 0.68 × 10−17 T
5. −1.02 × 10−17 T correct
6. 2.04 × 10−17 T
7. 1.53 × 10−17 T
8. −0.68 × 10−17 T
Explanation:
At point 0,
V =
6
Explanation:
√ kq √ kq
Vi = 2 2 − 2
L
L
√ kq
= 2
,
L
so the work to move the charge from infinity
to B is
√ k q2
.
WB = 2
L
011
10.0 points
Let : v = 4 × 106 m/s ,
q = 1.6 × 10−19 C ,
d = 6 cm = 0.06 m ,
θ = 35 ◦ .
and
The magnitude of the field at P3 is
µ0 q v sin(θ) B=
4π
d2
= (1 × 10−7 T · m/A)(1.6 × 10−19 C)
Version 040 – Midterm 02 – yao – (57465)
(4 × 106 m/s)sin(35 ◦ )
(0.06 m)2
T.
×
= 1.02 × 10−17
Using the right hand rule (and remembering
that q is positive), the magnetic field is into
the page.
012 10.0 points
A total current of 48 mA flows through an
infinitely long cylinderical conductor of radius
6 cm which has an infinitely long cylindrical
r
hole through it of diameter r centered at −
2
along the x-axis as shown.
y
x
What is the magnitude of the magnetic field
at a distance of 19 cm along the positive xaxis? The permeability of free space is 4 π ×
10−7 T · m/A . Assume the current density is
constant throughout the conductor.
1. 1.57158e-07
2. 1.24583e-07
3. 1.75238e-07
4. 1.76768e-08
5. 9.78884e-08
6. 1.98154e-07
7. 5.2823e-08
8. 6.98413e-08
9. 7.36744e-08
10. 4.12903e-08
Correct answer: 5.2823 × 10−8 T.
Explanation:
Basic Concepts: Magnetic Field due to a
Long Cylinder
µ0 I
B =
.
2πr
Principle of Superposition.
Our goal is to model the given situation,
which is complex and lacks symmetry, by
7
adding together the fields from combinations
of simpler current configurations which together match the given current distribution.
The combination of the currents in Fig. 2 will
do so if we choose Icyl and Ihole correctly.
y
Hole
r
r
y
2
Icyl
4
= I
3
+
x
x
1
Ihole = − I
3
Since the current is uniform, the current
I
density J =
is constant. Then
A
J=
Icyl
Ihole
=−
Acyl
Ahole
π r2
, so
Clearly, Acyl = π r 2 , and Ahole =
4
Icyl
Ihole = −
.
4
Note: The minus sign means Ihole is flowing
in the direction opposite Icyl and I, as it must
if it is going to cancel with Icyl to model the
hole.
We also require I = Icyl + Ihole . We then
4
1
have Icyl = I, and Ihole = − I. With these
3
3
currents, the combination of the two cylinders
in figure 2 gives the same net current and
current distribution as the conductor in our
problem.
The magnetic fields are
Bcyl
Bhole
4
I
µ0
3
=
2
πx 1
µ0 − I
3
,
=
2 π (x + r/2)
so the total magnetic field is
Btotal = Bcyl + Bhole
Version 040 – Midterm 02 – yao – (57465)
8


µ0 I  4
1 
q1 + q2
−
5. VC = k
r
6π x x+
c
√
2
q2


6. VC = 2 k
c
µ0 I  3 x + 2 r 
q
2
6π x x+ r
7. VC = k
c
2
−7
q1 + q2
q1
q1
(4 π × 10 T m/A) (48 mA)
8. VC = k
+k −k
R3
c
R3
6π


q1 + q2
q1
q1
9. VC = k
−k
+k
 3 (19 cm) + 2 (6 cm) 
R3
R2
R1

×
q
1

6 cm 
10. VC = k
(19 cm) 19 cm +
c
2
Explanation:
5.2823 × 10−8 T .
C is a point inside the outer shell. The
potential inside the shell is constant, so must
be equal to the potential at the outer surface
R3 . As a consequence,
=
=
=
=
keywords:
013 (part 1 of 2) 5.0 points
Consider a system of a metallic ball with
net charge q1 and radius R1 enclosed by a
spherically symmetric metallic shell with net
charge q2 , inner radius R2 and outer radius
R3 . If q2′′ is the charge on the outside surface
of the shell and q2′ the charge on its inside
surface, then q2′′ + q2′ = q2 .
q2
R1
R2
R3
q2′
q2′′
q1
O
B
C
Find the potential at C.
OC = c .
OB = b and
q1 + q2
correct
R3
q1
2. VC = 2 k
c
q1 − q2
3. VC = k √
2c
q1
q1
q1 + q2
−k
+k
4. VC = k
c
R2
b
1. VC = k
VC = k
q1 + q2
,
R3
since the potential at R3 is equal to that of a
point charge of charge q1 + q2 located at the
origin.
014 (part 2 of 2) 5.0 points
Determine the potential at B.
q1 − q2
1. VB = k √
2c
q1
q1
q1 + q2
−k
+k
correct
2. VB = k
R3
R2
R1
q1 + q2
q1
q1
3. VB = k
−k
+k
R3
R2
b
q1 + q2
4. VB = k
R3
q1 + q2
5. VB = k
b
q1 + q2
q1
q1
6. VB = k
−k +k
R3
b
R1
q1
7. VB = k
b
q1
8. VB = 2 k
b
√
q2
9. VB = 2 k
c
q2
10. VB = k
b
Explanation:
Version 040 – Midterm 02 – yao – (57465)
9
B is located inside the central sphere; again,
we know the potential inside the sphere is
constant and equal to the potential at its
surface. There are several ways to solve this
problem. We will consider two.
We can apply the superposition principle to the potentials due to each surface
charge distribution. The surface charges are:
q ′′ = q1 + q2 at R3 , q ′ = −q1 at R2 , and
q = q1 at R1 . Remember that the potential difference between any two points inside
a spherically symmetric shell of charge is zero,
so we must only consider the potential of each
charge distribution up to but not inside of is
radius. Consequently,
VB = Vq′′ (b) + Vq′ (b) + Vq (b)
q1 + q2
q1
q1
=k
−k
+k
.
R3
R2
R1
Alternatively, we can recognize that the
potential at B is just the sum of the potential
differences from ∞ → R3 and from R2 →
R1 , since the potential difference inside the
conductors is zero. ∆V from ∞ → R3 is just
the potential at R3 due to the total charge,
while ∆V from R2 → R1 is just due to the
charge q on the surface at R1 . Therefore,
In reality, each individual molecule of the
dielectric is polarized, becoming a dipole (see
the illustration Fig 17.42 in M&I v.II). The
sum of the dipole contributions gives the electric field due to the polarization of the dielectric Epol . The simple model shown above is
justified by assuming that the effects of the
polarized charges in the interior of the dielectric largely cancel, leaving only the outermost
layer of dipole charge at the top and bottom
surfaces to contribute.
Given a dielectric constant κ, plate charge Q,
and plate area A, determine the magnitude of
the polarized charge Qpol .
1.
VB = Vq′′ (R3) − Vq′′ (∞) + Vq (R1 ) − Vq (R2 )
q1
q1
q1 + q2
−0+k
−k
,
=k
R3
R1
R2
which is equal to the other method, as expected.
015 10.0 points
When a dielectric is present in a charged
capacitor, the dielectric is polarized. The
figure below shows a simple model: a layer
of polarized charge forms at the upper and
lower surfaces of the dielectric, leading to a
reduction of the “effective” plate charge and
~ ′.
a reduced field E
Q
κ
2. Q(κ − 1)
1
3. Q 1 −
correct
κ
4. κQ
1
5. Q 1 +
κ
κ−1
6. Q
κ+1
Explanation:
First, note that we can apply the superposition principle inside the dielectric:
~′ =E
~ Q +E
~ Q . We are told a layer of charge
E
pol
forms on the outer surfaces of the dielectric,
so we can treat the dielectric as though it were
Version 040 – Midterm 02 – yao – (57465)
composed of parallel plates with charge Qpol ;
we know the electric field for this charge distribution — it is (Qpol /A)/ǫ0 . Using this fact
together with E ′ = E/κ, we obtain:
~′ =E
~Q + E
~Q
E
pol
Q
Q
Q
pol
=
−
κAǫ0
Aǫ0
Aǫ0
Q
= Q − Qpol
κ
Q
1
Qpol = Q − = Q 1 −
.
κ
κ
016
10.0 points
The circuit shown above consists of a battery, a wire, and carries a conventional current
~
I. At the center of the semicircle, what is B?
You may assume L ≫ h and that the positive
direction is out of the page.
2
µ0 I
π+
1. −
2π R
h
µ0 I
2.
(1 + π)
4π R
2
µ0 I π
correct
+
3. −
4π R h
µ0 2I
4. −
(1 + π)
2π R
2
µ0 I
π+
5.
2π R
h
µ0 I π
2
6. −
+
2π R h
2
µ0 I π
+
7.
4π R h
2
µ0 I π
+
8.
2π R h
Explanation:
Examining the figure, we see that the semicircular section and the lower straight wire
contribute to |B|. The upper straight wire
portions have d~l k r̂, so do not contribute.
Likewise, the short segment of length h ≪ L
may be neglected. Consequently, we have a
10
superposition of the magnetic fields of a halfloop and a straight wire of length L,
1 µ0 2πI
µ0 2I
+
2 4π R
4π
h
2
µ0 I π
.
+
=
4π R h
~ =
|B|
By the RHR, the direction must be into the
page, so negative.
017 10.0 points
An isolated large-plate capacitor (not connected to anything) originally has a potential
difference of 1100 V with an air gap of 5 mm.
Then a plastic slab 3 mm thick, with dielectric constant 6, is inserted into the middle of
the air gap as shown in the figure below.
5 mm
+
+
+
+
+
+
+
+
1
2
1 mm 3 mm
Calculate V1 − V4 .
1. 550.0
2. 660.0
3. 750.0
4. 500.0
5. 416.0
6. 466.667
7. 540.0
8. 440.0
9. 600.0
10. 687.5
3
−
−
−
−
−
−
−
−
4
1 mm
Correct answer: 550 V.
Explanation:
Here we simply add the potential differences we’ve already found:
Version 040 – Midterm 02 – yao – (57465)
∆V14 = ∆V12 + ∆V23 + ∆V34
= (220 V) + (110 V) + (220 V)
= 550 V .
11