Version 003 – Test #2 – Antoniewicz – (57030)
This print-out should have 21 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
1
001 (part 1 of 2) 7.0 points
VD − VC
b
!
"
−1
= 2kqs
d
2 x2
a
!
"
1
1
= 2kqs
−
2 a 2 2 b2
"
!
1
1
−
> 0.
(1)
= kqs
a 2 b2
#
Alternative explanation:
Let the center of the dipole be at the origin.
At a distance x along the +x̂ direction
!
"
!
"
s
s
Vdipole (x) = Vq x +
+ V−q x −
2
2
Consider the setup as shown in the figure
above where the distances from the points C
and D to the center of the dipole are a and b
respectively. What is the change in potential
∆V = VD − VC , in moving an electron from
C to D?
"
!
1
1
−
1. 2k q s
a 2 b2
"
!
1
1
−
2. -2k q s
a 2 b2
!
"
1 1
3. -2k q s
−
a b
"
!
1
1
−
4. -k q s
a 2 b2
"
!
1 1
−
5. k q s
a b
"
!
1
1
−
correct
6. k q s
a 2 b2
!
"
1 1
7. 2k q s
−
a b
"
!
1 1
−
8. -k q s
a b
"
!
1
1
−
9. -k e s
a 2 b2
"
!
1 1
−
10. -k e s
a b
Explanation:
"
# b!
2kqs
dx
VD − VC = −
− 3
x
a
Vdipole (x) =
Vdipole (x) =
kq
k(−q)
+
x + s/2
x − s/2
(x − s/2) − (x + s/2)
kqs
≈ − 2
2
2
x − (s/2)
x
So, we have
VD (x) − VC (x) = − k q s
!
1
1
− 2
2
b
a
"
which on simplification yields
"
!
1
1
−
>0
VD (x) − VC (x) = k q s
a 2 b2
Check: E is along the -x̂ direction, VD is
expected to be higher than VC .
002 (part 2 of 2) 3.0 points
What is the change in potential energy ∆U =
UD − UC , in moving an electron from C to
D? The charge of the electron is denoted by
e = 1.6×10−19 C.
1. e (V (b) − V (a))
2. (V (b) − V (a))
3. −e (V (b) − V (a)) correct
Explanation:
Multiplying eq(1) by the electronic charge
−e, we arrive at the potential energy difference from C to D is given by
UD − UC = −e(V (b) − V (a))
Version 003 – Test #2 – Antoniewicz – (57030)
Intuitive reasoning on the sign of ∆U : Natural tendency of the motion is from high
potential energy to lower potential energy.
Since when the electron is released it should
move from C to D, so UC > UD , or UD has
lower potential energy compared to UC , i.e.
UD − UC < 0.
003 (part 1 of 2) 4.0 points
When you bring a current-carrying wire down
onto the top of a compass, aligned with the
original direction of the needle and 6 mm
above the needle, the needle deflects by 9
degrees, as in the figure below.
9
◦
BEarth
Wire
Assuming the compass needle was originally pointing toward the north, what direction is the conventional current traveling in
the wire, and what is the direction of the
force on the compass needle due to the magnetic field caused by the current in the wire?
1. North, West
2. South, West
3. South, East correct
4. North, East
Explanation:
We just need to think about how the compass needle responds to the presence of the
current-carrying wire to answer this question.
Since the compass needle deflects to the east,
that is the direction of the magnetic force on
the needle due to the current in the wire. To
decide which way current is traveling in the
wire, we use the right hand rule. Your fingers
should point toward the east, so in order for
your fingers to curl around the wire and point
2
toward the east, your thumb must point toward the south. So the current is traveling
toward the south.
004 (part 2 of 2) 6.0 points
Calculate the amount of current flowing in
the wire. The measurement was made at a
location where the horizontal component of
the Earth’s magnetic field is
BEarth = 2 × 10−5 T.
Use
µ0
= 1 × 10−7 Tm/A.
4π
1. 0.200722
2. 0.14879
3. 0.172047
4. 0.133975
5. 0.0950307
6. 0.123429
7. 0.0613923
8. 0.136066
9. 0.0705308
10. 0.106278
Correct answer: 0.0950307.
Explanation:
Ultimately, we want to use the expression
Bwire =
µ0 2 I
4π r
to find the current in the wire. We know r,
but not Bwire yet. To find Bwire , we can use
what we know about the Earth’s magnetic
field and the deflection of the compass needle.
Consider the following simple diagram:
" wire
B
" Earth
B
Needle
From this drawing, we can write down
Version 003 – Test #2 – Antoniewicz – (57030)
Bwire
tan θ =
BEarth
⇒ Bwire = BEarth tan θ
≈ (2 × 10−5 T) tan 9◦
= 3.16769 × 10−6 T .
Now we simply rearrange the expression
above to find the current:
Bwire r
! "
µ0
2
4π
(3.16769 × 10−6 T)(6 mm)
=
2(1 × 10−7 Tm/A)
I=
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 −q and +q respectively.
The potential difference is
V3 − V1 = Egap,12 d + Egap,23 d
V3 − V1 =
2(q/A)d
$0
4(q/A)d
2.
$0
9(q/A)d
3.
2$0
4. 0
6.
7.
8.
9.
3(q/A)d
correct
2$0
5(q/A)d
$0
7(q/A)d
2$0
5(q/A)d
2$0
6(q/A)d
$0
3(q/A)d
2$0
006 10.0 points
What is the magnetic dipole moment of an
orbiting electron, as shown in the figure? Here
the orbit is a circle with the radius R and the
tangential speed is v. The direction of the
electron’s motion is CCW.
1.
5.
(q/A)d
(q/A)d
+
$0
2$0
V3 − V1 =
= 0.0950307 A .
005 10.0 points
Given three parallel conducting plates
which are aligned perpendicular to the x-axis.
They are labeled, from left to right as plate 1,
2 and 3 respectively. The corresponding plate
charges are Q1 = −q, Q2 = 0 and Q3 = q.
The width of the gap between 1 and 2 is d,
and the width between plates 2 and 3 is d/2.
Determine the magnitude of the potential
difference across 1 and 3.
3
The magnitude is
Ia. µ =
1 eR
2 v
Ib. µ = e R v
Ic. µ =
eR
v
Version 003 – Test #2 – Antoniewicz – (57030)
Id. µ =
1
eRv
2
The direction is
IIa. into the page
IIb. out of the page
Choose the correct choice:
We solve this problem based on Ampere’s
Law.
The left hand side of Ampere’s law is given
by:
Ia) 2πrB
Ib) 2πRB
1. Ib, IIb
The right hand side of Ampere’s law is:
2. Id, IIa correct
IIa) µ0 I1 !
r2
R2
"
3. Id, IIb
IIb) µ0 I1
4. Ia, IIa
The field B at r is given by:
5. Ia, IIb
IIIa)
6. Ib, IIa
µ0 I 1
2πR !
r2
µ0 I 1
R2
IIIb)
2πr
7. Ic, IIa
8. Ic, IIb
4
"
1. Ia, IIb, IIIa
Explanation:
The current I is the measure of how much
charge passes one location per second. The
charge on the electron is −e, and the time
it takes the electron to go around is T =
2 π R/v, where v is the speed of the electron,
so
e
ev
I=
=
.
2 π R/v
2πR
Therefore the magnetic pole moment of a single orbiting electron would be
$
%
% 1
ev $
µ = I π R2 =
π R2 = e R v .
2πR
2
007
10.0 points
A long thick wire of radius R carries a
current I1 as shown in the figure above. We
would like to determine the magnetic field B,
inside the wire, a distance r from the center
of the wire, where r < R, near the midpoint
of the wire where the end effect at both ends
are negligible.
2. Ia, IIa, IIIb
3. Ia, IIb, IIIb correct
4. Ib, IIb, IIIb
5. Ib, IIa, IIIa
6. Ib, IIb, IIIa
7. Ib, IIa, IIIb
8. Ia, IIa, IIIa
Explanation:
Along the Amperian loop of radius r and
magnetic field B, the left hand side of Ampere’s Law is given by 2πrB. Hence the
answer is Ia.
The
! 2 "current encircled by the loop is
r
I1
. Thus, the right hand side of the
R2
! 2"
r
expression is given by µ0 I1
. Hence, the
R2
answer is IIb.
Version 003 – Test #2 – Antoniewicz – (57030)
The above two answers lead to
! 2"
r
µ0 I 1
R2
B =
2πr
Hence, the answer is IIIb.
008 (part 1 of 2) 5.0 points
Consider a parallel plate capacitor where
the charge on the plate is Q and the plate
area is A. One plate is fixed and the other
" which maintains
plate is pulled by a force F
an equilibrium with a gap width d.
Determine the magnitude of the force.
!
"
Q
1. Q
A$
! 0 "
Q
2. Q
4d2 $0
"
!
Q
correct
3. Q
2A$0
"
!
Q
4. Q
2d2 $
! 0"
Q
5. 2Q
d2 $0
"
!
Q
6. Q
4A$0
!
"
Q
7. Q
d2 $
! 0 "
Q
8. 2Q
A$0
Explanation:
The magnitude of the attractive force is
"
!
Q
"
"
|F | = Q|Eplate| = Q
2A$0
009 (part 2 of 2) 5.0 points
Now slowly increase the gap width by a distance ∆s. Find the increase of the energy
in the capacitor system ∆U . (Hint: consider
only the electric field that exerts a force on
the mobile plate.)
"
!
Q
∆s
1. Q
A$0
5
!
"
QA
2. Q
∆s
2$0
"
!
QA
(d + ∆s)
3. Q
$0
"
!
Q
d
4. Q
A$0
!
"
Q
5. Q
∆s correct
2A$0
"
!
Q
d
6. Q
2A$0
"
!
QA
∆s
7. Q
$0
"
!
Q
(d + ∆s)
8. Q
2A$0
!
"
QA
9. Q
(d + ∆s)
2$0
"
!
Q
(d + ∆s)
10. Q
A$0
Explanation:
The magnitude of the attractive force before the stretch is
(Q/A)
2$0
Energy principle implies that
"| = Q
|F
" ∆s cos θ
∆U = −W = −Q |E|
Notice that E and ∆s are in opposite directions, i.e θ = 180◦ . !
"
Q
Hence, ∆U = Q
∆s > 0
2A$0
010 10.0 points
A conductor consists of an infinite number
of adjacent wires, each infinitely long and
carrying a current I (whose direction is out-ofthe-page), thus forming a conducting plane.
A
C
Version 003 – Test #2 – Antoniewicz – (57030)
If there are n wires per unit length, what is
" measured at A or C?
the magnitude of B
1. B =
µ0 I
2
2. B = 2 µ0 I
3. B =
µ0 n I
4
6
gle has dimensions l and w. The net current
through the loop is n I l. Note that since there
" in the direction of w, we
is no component of B
are only interested in the contributions along
sides l
&
" · d"s = 2 B l = µ0 n l I
B
B=
4. B = µ0 I
5. B = 4 µ0 I
6. B =
011 10.0 points
In a television picture tube, electrons are
boiled out of a very hot metal filament placed
near a negative metal plate. These electrons
start out nearly at rest and are accelerated
toward a positive metal plate. They pass
through a hole in the positive plate on their
way toward the picture screen, as shown in
the diagram.
µ0 n I
correct
2
7. B = 4 µ0 n I
8. B = µ0 n I
9. B =
µ0 I
4
10. B = 2 µ0 n I
Explanation:
B
A
l
W
C
µ0 n I
.
2
B
By symmetry the magnetic fields are equal
and opposite through point A and C and horizontally oriented. Following the dashed curve
in
& a counter-clockwise direction, we calculate
" · d"s, which by Ampere’s law is proporB
tional to the current through the dashed loop
coming out of the plane of the paper. In
this problem this is a positive current. Hence
" along the horizontal legs points in the diB
rection in which we follow the dashed curve.
Ampere’s Law is
&
" · d"s = µ0 I .
B
To evaluate this line integral, we use the rectangular path shown in the figure. The rectan-
− Plate
E
−
−
−
F = eE
Hot filament
−
−
−
−
L
+ Plate
+
+
+
v=?
+
+
+
+
+
+
The high-voltage supply in the television
set maintains a potential difference of 16500 V
between the two plates, what speed do the
electrons reach? Use me = 9.11 × 10−31 kg
and qe = 1.6 × 10−19 C and assume that this
is not relativistic.
1. 74968000.0
2. 64924200.0
3. 79515500.0
4. 71367400.0
5. 67575200.0
6. 72587400.0
7. 66263000.0
8. 70126100.0
9. 76130300.0
10. 78403400.0
Correct answer: 7.61303 × 107 m/s.
Version 003 – Test #2 – Antoniewicz – (57030)
Explanation:
The net energy remains constant throughout the whole process. We can use the following train of logic:
∆E
∆U + ∆K
q ∆V + ∆K
(−e)∆V + ∆K
⇒ ∆K
=0
=0
=0
=0
= e ∆V
= (1.6 × 10−19 C)(16500 V)
= 2.64 × 10−15 J .
7
What is the charge on the spherical surface
at r = r2 ?
1. 2q2
2. q1 − q2
3. −2q2
4. −q1 − q2
5. −2q1
6. q2
7. 2q1
Then
∆K = Kf − Ki
1
= m vf2 − 0
2
'
2∆K
⇒ vf =
( m
=
2(2.64 × 10−15 J)
9.11 × 10−31 kg
= 7.61303 × 107 m/s .
012 (part 1 of 2) 4.0 points
8. q1
9. q1 + q2 correct
10. −q1 + q2
Explanation:
The charge q1 at the center will induce −q1
on the inner surface of r1 . Since the region R2
is inside the conducting shell, the net charge
there is zero. Therefore, an equal amount
of +q1 will be induced on the outer surface
of r2 . However there already exists a charge
of q2 , which contributes to the total. The
two contributions together yield a charge of
q = q1 + q2 .
013 (part 2 of 2) 6.0 points
What is the potential at r = R3 ?
k(q1 + q2 + q3 )
kq2
−
r4
R3
k(q1 + q2 + q3 )
k(q1 + q2 )
2. V =
−
+
r4
r3
k(q1 + q2 )
correct
R3
k(q1 + q2 )
3. V =
r3
1. V =
A point charge q1 is concentric with two
spherical conducting thick shells, as shown
in the figure below. The smaller spherical
conducting shell has a net charge of q2 and
the larger spherical conducting shell has a net
charge of q3 .
4. 0
5. V =
k(q1 + q2 )
k(q1 + q3 )
+
r4
R3
Version 003 – Test #2 – Antoniewicz – (57030)
6. V =
7. V =
8. V =
9. V =
10. V =
kq2
r4
k(q1 + q2 + q3 )
r4
k(q1 + q2 )
R3
k(q1 + q2 + q3 )
k(q1 + q2 )
−
r4
r3
k(q1 + q3 )
R3
Explanation:
Let us begin by noting that there are three
contributions that contribute to the potential
at r = R3 .
The first contribution arises when one considers the two charges q1 and q2 enclosed by a
Gaussian surface of radius R3 .
k(q1 + q2 )
V1 =
R3
The second contribution arises from the fact
that there is an induced charge of −(q1 + q2 )
on the inner Gaussian surface of radius r3 .
V2 = −
k(q1 + q2 )
r3
The outermost surface has a total charge enclosed of (q1 + q2 + q3 ) at r = r4 which gives
rise to the third contribution.
V3 =
k(q1 + q2 + q3 )
r4
By the principle of superposition, we obtain
V = V1 + V2 + V3
V =
k(q1 + q2 + q3 ) k(q1 + q2 ) k(q1 + q2 )
−
+
r4
r3
R3
014 (part 1 of 2) 5.0 points
A particular alnico (aluminum, cobalt,
nickel, and iron) bar magnet (magnet A) has
a mass of 10 grams.It produces a magnetic
field of magnitude 4.5 × 10−5 T at a location
0.19 m from the center of the magnet, on the
axis of the magnet. Approximately what is
8
the magnitude of the magnetic field of magnet A a distance of 0.4 m from the center of
the magnet, along the same axis?
Assume that the size of the magnet is negligible compared to the distance between the
magnet and the location where field is measured.
1. 3.221e-06
2. 6.616e-06
3. 6.039e-06
4. 7.764e-06
5. 5.203e-06
6. 4.375e-06
7. 4.823e-06
8. 2.416e-06
9. 8.125e-06
10. 6.47e-06
Correct answer: 4.823 × 10−6 T.
Explanation:
Let :
B1 = 4.5 × 10−5 T ,
r1 = 0.19 m , and
r2 = 0.4 m .
We know that the magnetic field due to a
dipole along the axis is given by
" dipole | =
|B
µ0 2µ
4π r 3
Now, we note that the value of µ remains the
same. Hence, the product Br 3 must be a
constant. Using this we have
B1 r13 = B2 r23
! "3
r1
B2 = B1
r2
!
"3
0.19 m
−5
B2 = (4.5 × 10 T)
0.4 m
B2 = 4.823 × 10−6 T
015 (part 2 of 2) 5.0 points
If you removed the original magnet and replaced it with a magnet made of the same material but with a mass of 50 grams (magnet
Version 003 – Test #2 – Antoniewicz – (57030)
B), approximately what would be the magnetic field at a location 0.19 m from the center
of the magnet, on the axis of the magnet?
1. 0.00025
2. 0.000175
3. 0.0004
4. 0.000225
5. 0.000375
6. 0.000425
7. 0.0002
8. 0.000275
9. 0.00015
10. 0.0003
Correct answer: 0.000225 T.
Explanation:
Increasing the mass by a factor of 5, causes
an increase in µ of a factor of 5. As a result,
the value of B also increases by a factor of
5, since B ∝ µ. Hence, the new value of the
magnetic field is
! #"
!
"
m
50 g
Bnew =
B1 =
(4.5 × 10−5 T)
m
10 g
Bnew = 0.000225 T
where the new mass is m# = 50 g and the
initial mass is m = 10 g.
9
Explanation:
Since VB − VA is a positive number, it is
clear that B is at the higher potential.
017 (part 2 of 2) 5.0 points
What is the magnitude and direction of the
field E inside the wire? (Since this is a 1D
problem, the sign of your answer will indicate
the direction.)
1. −336 V/m
2. 331.2 V/m
3. 204 V/m
4. −144 V/m
5. 136.8 V/m
6. 403.2 V/m
7. −192 V/m
8. −288 V/m
9. 201.6 V/m
10. −240 V/m correct
016 (part 1 of 2) 5.0 points
The potential difference from one end of a 1
cm-long wire to the other in a circuit is
∆V = VB − VA = 2.4 V,
as shown in the figure below.
A
B
Explanation:
It should be clear that E points toward the
left, since E points toward the lower potential.
For a constant electric field,
∆V = −
#
" · d"% = −E
" · ∆"%.
E
For a path from A to B,
1 cm
Which end of the wire is at the higher potential?
1. A
2. B correct
3. The ends are at the same potential.
∆"% = &∆x, 0, 0'.
∆V = −Ex ∆x
∆V
⇒ Ex = −
∆x
2.4 V
=−
0.01 m
= −240 V/m .
Version 003 – Test #2 – Antoniewicz – (57030)
018 (part 1 of 2) 7.0 points
A very long wire carrying a conventional current I is straight except for a circular loop of
radius R (see the figure below). Calculate the
magnitude of the magnetic field at the center
of the loop.
10
019 (part 2 of 2) 3.0 points
What is the direction of the magnetic field at
the center of the loop? Take the current to
be moving in the +x direction and the loop
extending outward in the +y direction, with
the +z direction coming out of the page.
1. −x̂
R
2. +x̂
I
3. −ẑ
4. +ŷ
µ0 2 I
4 R
µ0 2 I
(1 + π) correct
4π R
µ0 2 I
4π R
µ0 2 I
(1 − π)
4π R
µ0 2 I
(1 + 2π)
4π R
Explanation:
Using the superposition principle, we have
) )
)" )
1. )B
)=
) )
)" )
2. )B
)=
) )
)" )
3. )B
)=
) )
)" )
4. )B
)=
) )
)" )
5. )B
)=
5. +ẑ correct
6. −ŷ
Explanation:
Using the right hand rule, the magnetic
fields from the wire and the loop both point
in the same direction, the +z direction.
020 (part 1 of 2) 5.0 points
Bnet = Bwire + Bloop .
Now,
µ0 2 I
.
4π R
So we just need to find the field due to the
loop:
Bwire =
µ0 2 I A
4π R3
µ0 2 I π R 2
=
4π R3
µ0 2 I π
=
.
4π R
Adding the contributions together, we obtain
Bloop =
µ0 2 I π
µ0 2 I
+
4π R
4π R
µ0 2 I
=
(1 + π) .
4π R
Bnet =
A power cord consists of two parallel wires
that are a distance d apart and carry currents
that flow in opposite directions (but have the
same magnitude I). Let the magnitude of the
magnetic field created by the cord at a perpendicular distance of r (>> d) away from
the center of the cord on the plane containing the two wires be B. What will be the
magnitude of the field B # at a distance of 2r
away from the cord? Assume the cord has
(practically) infinite length.
1.
B
2
2. 4B
Version 003 – Test #2 – Antoniewicz – (57030)
B
8
B
correct
4.
4
3.
7. 9B
Explanation:
From the explanation in part1, we see that
for a fixed value of r, B is proportional to d.
5. B
3d
B ##
=
= 3
B
d
6. 2B
Explanation:
Let P be the point where we are measuring
the magnetic field. P is at a distance of r from
one of the wires and r + d from the other wire.
Since the two wires carry current in opposing
directions, the magnetic fields due to the two
wires point in opposing directions. The net
magnetic field can be calculated as follows
µ0 2I
µ0 2I
−
4π r
4π r + d
*
+
µ0 2I
2I
BP =
−
4π r
r+d
*
+
d
µ0 I
BP =
2π r(r + d)
BP =
BP ≈
µ0 I d
2π r 2
Clearly, the magnitude of the net field is proportional to r −2 . Therefore, if we increase the
B
distance to 2r, the field will go down to .
4
021 (part 2 of 2) 5.0 points
What will be the magnitude of the field B ##
at a distance r from the cord, if the distance
between the wires is increased from d to 3d?
1.
B
9
2. B
3.
B
6
4. 6B
5.
11
B
3
6. 3B correct
B ## = 3B