Version 078 – Test #3 – 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.
001 (part 1 of 2) 5.0 points
An electron circles at a speed of 8960 m/s in a
radius of 3.01 cm in a solenoid. The magnetic
field of the solenoid is perpendicular to the
plane of the electron’s path.
The charge on an electron is 1.60218 ×
10−19 C and its mass is 9.10939 × 10−31 kg.
Find the strength of the magnetic field inside the solenoid.
1. 3.65724e-06
2. 1.69247e-06
3. 2.39029e-06
4. 2.04832e-06
5. 3.25976e-06
6. 1.52343e-06
7. 3.1187e-06
8. 3.00526e-06
9. 4.2442e-06
10. 3.32391e-06
1
4. 2.03578
5. 0.703589
6. 0.978086
7. 0.727917
8. 0.687154
9. 0.954931
10. 2.10441
Correct answer: 0.687154 mA.
Explanation:
Let :
µ0 = 1.25664 × 10−6 N/A2 , and
n = 19.6 turns/cm = 1960 turns/m .
The current in the solenoid is from B = µ0 n I,
we get
I=
=
B
µ0 n
1.69247 × 10−6 T
(1.25664 × 10−6 N/A2 ) (1960 turns/m)
= 0.000687154 A = 0.687154 mA .
Correct answer: 1.69247 × 10−6 T.
Explanation:
003 (part 1 of 2) 5.0 points
Let : r = 3.01 cm = 0.0301 m ,
v = 8960 m/s ,
q = 1.60218 × 10−19 C , and
m = 9.10939 × 10−31 kg .
The strength of the magnetic field is
mv
qr
(9.10939 × 10−31 kg) (8960 m/s)
=
(1.60218 × 10−19 C) (0.0301 m)
B=
= 1.69247 × 10−6 T .
002 (part 2 of 2) 5.0 points
Find the current in the solenoid if it has
19.6 turns/cm.
1. 1.47825
2. 2.15805
3. 1.04011
In the figure shown above, the north pole
of a bar magnet points toward a thin circular
coil of wire containing 40 turns.
The magnet is moved away from the coil,
so that the flux through one turn inside the
coil decreases by 0.2 T · m2 in 0.3 s.
What is the magnitude of average emf in
the coil during this time interval?
1. 125.0
2. 100.0
3. 28.0
4. 90.0
Version 078 – Test #3 – Antoniewicz – (57030)
5. 26.6667
6. 80.0
7. 37.5
8. 41.6667
9. 14.0
10. 20.0
Correct answer: 26.6667 V.
Explanation:
The magnitude of the average emf around
the coil is
!
!
! ∆Φ !
!
|emf| = N !!
∆t !
0.2 T · m2
= (40)
0.3 s
= 26.6667 V .
004 (part 2 of 2) 5.0 points
As the bar-magnet is moving to the left,
please make the correct choice among each
pair of statements.
I. The direction of the magnetic flux
through the coils is pointing:
a) to the right
b) to the left.
II. The magnitude of the flux within the
coils is
a) increasing
b) decreasing
III. The current induced in the coils as
viewed from the right is:
a) clockwise
b) counterclockwise.
2
8. Ib, IIb, IIIa
Explanation:
! = B
!f − B
! i is to the left, toward the
∆B
! ind , i.e −∆B
! is tomagnet. The direction of B
ward the right, away from the magnet. Hence,
the answer is Ia.
The magnet is moved away from the coil,
so that the flux through one turn inside the
coil decreases. Hence, the answer is IIb.
! is toAs we have indicated earlier, −∆B
ward the right, away from the magnet. With
your thumb pointed to the right, your fingers curl counterclockwise around the loop,
as viewed from the right side of the coil, so
current will flow counterclockwise as viewed
from the right side of the coil. Hence, the
answer is IIIb.
005 (part 1 of 2) 5.0 points
Given: B = b t, where b = 0.3 T/s (lower
figure).
Consider a changing magnetic field (from
B " to B, upper figure) within a circle of radius
7 m.
The magnetic field is pointing out of the
page # and its magnitude is increasing.
7m
B"
B
3. Ia, IIb, IIIb correct
4. Ia, IIa, IIIa
7m
B
B
⇒
"
B
"
1. Ib, IIb, IIIb
2. Ib, IIa, IIIa
B
B
!
B
0
B"
!t
Find the direction of the induced emf Eind .
5. Ib, IIa, IIIb
1. into the page
6. Ia, IIa, IIIb
2. counter-clockwise
7. Ia, IIb, IIIa
3. out of the page
Version 078 – Test #3 – Antoniewicz – (57030)
3
From
4. clockwise correct
E=
Explanation:
Let : a = 7 m and
b = 0.3 T/s .
Faraday’s Law of Induction:
"
! ind · d!s = − d ΦB
Eind = E
dt
Lenz’s Law – Induced emf’s always oppose
magnetic flux changes.
Use a “bracket”, “[ ]”, to denote the directions. The direction content of Faraday’s Law
then reads,
#
$
dΦ
[Eind ] = −
= −# = ⊗ .
dt
The right hand rule then implies that the direction of Eind is clockwise. This is consistent
with Lenz’s Law: the magnetic field is increasing out of the page, so the induced emf
Eind creates a magnetic field into the page to
counter this change.
"
! ind · d!s ,
E
we have Eind = Eind 2 π a, so
Eind
2πa
Eind =
dB
dt
2πa
a dB
2 dt
(7 m)
(0.3 T/s)
2
1.05 V/m .
π a2
=
=
=
=
007 (part 1 of 3) 4.0 points
Consider a battery with an internal resistance
of 4.0 ohms connected to a 16-ohm and a 20ohm resistor in series. The current in the
20-ohm resistor is 0.3 amperes.
16 Ω
X
006 (part 2 of 2) 5.0 points
Find the magnitude of the induced electric
field Eind on the perimeter of the circular
magnetic field.
1. 1.5
2. 4.05
3. 1.35
4. 1.05
5. 0.45
6. 1.25
7. 2.1
8. 3.2
9. 0.6
10. 2.0
4. 10.8 V
Correct answer: 1.05 V/m.
5. 6.0 V
Explanation:
I = 0.3 A
ε
4Ω
Internal
Resistance
20 Ω
Y
What is the emf of the battery?
1. 1.2 V
2. 13.2 V
3. 12.0 V correct
Explanation:
d
(B A)
dt
dB
π a2
=
dt
= (0.3 T/s) π (7 m)2
= 46.1814 V .
|Eind | =
Let :
R1
R2
R3
I
= 4 Ω,
= 16 Ω ,
= 10 Ω , and
= 0.3 A .
Version 078 – Test #3 – Antoniewicz – (57030)
= 12.0 V .
010 10.0 points
A square loop of wire carries a current and is
located in a uniform magnetic field.
The left side of the loop is aligned and
attached to a fixed axis (dashed line in figure).
008 (part 2 of 3) 3.0 points
What is the potential difference across the
terminals X and Y of the battery?
→ 4.5 A →
← axis of rotation
! = 0.56 T
B
1. 6.0 V
2. 1.2 V
y
x
0.57 m
E = IRtotal
= (0.3 A)(16 Ω + 20 Ω + 4.0 Ω)
4
3. 12.0 V
0.57 m
! = 0.56 T
B
4. 13.2 V
1. 3.6 W
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. 0.539392
2. 0.0825552
3. 0.818748
4. 0.22743
5. 0.614656
6. 0.39204
7. 0.184512
8. 0.848016
9. 0.09438
10. 0.1344
2. 4.8 W
Correct answer: 0.818748 N m.
5. 10.8 V correct
Explanation:
The potential difference across the terminals of the battery is
VXY = E − Ir
= 12.0 V − (0.3 A)(4.0 Ω)
= 10.8 V .
009 (part 3 of 3) 3.0 points
What power is dissipated by the 4-ohm internal resistance of the battery?
Explanation:
3. 1.2 W
← axis of rotation
!
B
Explanation:
The power dissipated by the internal resistance is
Pinternal = I 2 r
= (0.3 A)2 (4.0 Ω)
= 0.36 W .
→ I →
5. 0.36 W correct
y
x
d
!
B
#
4. 3.2 W
Version 078 – Test #3 – Antoniewicz – (57030)
5
9. 2.11e-06
10. 1.4e-06
Let : d = 0.57 m ,
# = 0.57 m ,
B = 0.56 T ,
I = 4.5 A .
Correct answer: 2.69 × 10−6 N.
and
Only the right side of the loop contributes to
the torque. By definition, the torque is
!(
τ = (!d × F
= d×I #B
= (0.57 m) (4.5 A) (0.57 m) (0.56 T)
= 0.818748 N m .
011 10.0 points
A long wire carries a current I1 = 1 A upward,
and a rectangular loop of height h = 0.2 m
and width w = 0.1 m carries a current I2 =
8 A 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.
Find the magnitude of the net magnetic
force exerted by the long wire on the rectangular loop.
1. 1.92e-05
2. 0.000145
3. 1.87e-05
4. 8.73e-05
5. 6.55e-05
6. 2.69e-06
7. 3.08e-05
8. 8.07e-06
Explanation:
let :
I1 = 1 A ,
I2 = 8 A ,
d = 7 cm = 0.07 m ,
h = 0.2 m , and
w = 0.1 m .
The forces on the top and bottom wires
cancel each other. The force on the left
µ0 I 1
wire is FB = I2 h B = I2 h
to the
2πd
left. The force on the right wire is similarly
µ0 I 1
to the right.
FB = I2 h
2 π (d + w)
The net force exerted by the long wire on
the loop is
µ0 I 1
µ0 I 1
− I2 h
2πd
2 π (d + w)
&
%
1
1
−
= I 2 h µ0 I 1
2 π d 2 π (d + w)
= (1 A)(8 A)(0.2 m)(2 × 10−7 N · A−2 )
%
&
1
1
−
0.07 m 0.07 m + 0.1 m
= 2.69 × 10−6 N .
Fnet = I2 h
012 (part 1 of 2) 5.0 points
As in the figure below, a a thin spherical metal
shell of radius r1 has a charge Q (on its outer
surface) and is surrounded by a concentric
thin spherical metal shell of radius r2 which
has a charge −Q (on its inner surface).
−
−
+
+
+
r1 + − r2
− +
+
−
−
+
−
+
−
Version 078 – Test #3 – Antoniewicz – (57030)
Use the definition of capacitance
Solve for Q to get
Q = C |∆V |
Q= %
to find the capacitance of this spherical capacitor.
8π%0
&
1
1
−
r1 % r2
&
1
1
1
=
−
4π%0 r1 r2
%
&
1
1
1
−
=
8π%0 r1 r2
4π%0
& correct
=%
1
1
−
r1 % r2
&
1
1
= 4π%0
−
r1 r2
3. C
4. C
5. C
Explanation:
The electric potential outside a uniformly
charged spherical shell is that of a point particle at the center of the spherical shell:
1 Q
V =
.
4π%0 r
Inside the spherical shell, the electric field is
zero. Thus, at r = r1 ,
V =
1 Q
.
4π%0 r1
C=%
4π%0
&.
1
1
−
r1 r2
013 (part 2 of 2) 5.0 points
Assuming that the radii of the spherical shells
r1 and r2 are large and nearly equal to each
other, write C in terms of the surface area
A = 4 π r 2 , and s, which is the small gap
distance between the shells (r2 = r1 + s).
%0 s
A
%0 A
2. C ≈
correct
s
1. C ≈
3. C ≈ s A
4. C ≈ s A %0
sA
%0
A
6. C ≈
s
5. C ≈
Explanation:
We can write
At r ≥ r2 ,
V =
4π%0
& ∆V.
1
1
−
r1 r2
Comparing to Q = C ∆V gives
1. C = %
2. C
6
1 Q
1 −Q
+
= 0.
4π%0 r
4π%0 r
Just inside r2 ,
V =
1 Q
.
4π%0 r2
1
1
r1 − r2
s
−
=
=
r1 r2
r1 r2
r1 r2
%
&
r1 r2
C = 4π%0
.
s
And since r1 ≈ r2 = R, we know that
r1 r2 ≈ R2 , and
Thus,
|∆V | = V1 − V2
1 Q
1 Q
−
=
4π%0 r1 4π%0 r2
&
%
1
1
1
Q
−
=
4π%0
r1 r2
%0 4π R2
s
%0 A
.
≈
s
C≈
014 (part 1 of 2) 5.0 points
Version 078 – Test #3 – Antoniewicz – (57030)
7
n = 8.49 × 1028 electrons/m3 ,
q = 1.6021 × 10−19 C ,
I = 8 A , and
L = 2.2 m .
Assume: The mobile charge carriers are either electrons or holes.
The holes have
the same magnitude of charge as the electrons. The number of mobile charge carriers for this particular material is n =
8.49 × 1028 electrons/m3 .
Note: In the figure, the point at the upper
edge P1 and at the lower edge P2 have the
same x coordinate.
A constant magnetic field of magnitude
points out of the paper. There is a steady
flow of a horizontal current flowing from left
to right in the x direction.
y
!
B
a
x
P1
!
B
b
y
I
B = 1.2 T
8
cm
x
P1
L
8A
2.4 cm
!
B
For the Hall effect the magnetic force balances the electric force which means
q vd B = q E ,
V
P2
2.2 m
The charge on the electron is 1.6021 ×
10−19 C.
What is the magnitude of the electric field
between the upper and lower surfaces?
1. 4.84684e-07
2. 2.30549e-07
3. 5.16137e-07
4. 2.58369e-07
5. 2.73986e-07
6. 3.67598e-07
7. 3.47992e-07
8. 4.27113e-07
9. 3.53813e-07
10. 2.70739e-07
Correct answer: 3.67598 × 10−7 N/C.
Explanation:
Let :
V
P2
a = 8 cm = 0.08 m ,
b = 2.4 cm = 0.024 m ,
B = 1.2 T ,
or
E = vd B .
Also we know
I = n q vd A
I
vd =
,
nqA
or
so that the magnitude of the electric field is
E=
=
IB
nqA
(8 A) (1.2 T)
n (1.6021 × 10−19 C) (0.00192 m2 )
= 3.67598 × 10−7 N/C ,
where the area
A= a·b
= (0.08 m) (0.024 m)
= 0.00192 m2 .
015 (part 2 of 2) 5.0 points
Version 078 – Test #3 – Antoniewicz – (57030)
For this part of the problem a voltmeter (with
an internal resistance less than infinity) is
connected to the system, where the contact
points are on the upper and lower surfaces
and are in the same vertical plane.
Choose the correct answer for the case
where the sign of the charge of current carriers are either negative (electrons) or positive
(holes).
1. The direction of the current through the
voltmeter is upward for either positive or negative charge carriers.
8
and then allowed to discharge through the
inductor?
1. 91.4804
2. 61.5301
3. 56.9308
4. 84.7986
5. 89.8736
6. 66.3693
7. 83.0556
8. 55.3871
9. 68.7804
10. 101.49
Correct answer: 61.5301 Hz.
2. The direction of the current through the
voltmeter is downward for either positive or
negative charge carriers.
3. The direction of the current through the
voltmeter is downward for positive charge carriers and upward for negative charge carriers.
4. The current through the voltmeter is zero
for either positive or negative charge carriers.
5. The direction of the current through the
voltmeter is upward for positive charge carriers and downward for negative charge carriers.
correct
Explanation:
First consider the negative charge carrier
case. Applying the right hand rule, the force
on electrons moving to the left is downward.
This generates a counterclockwise electron
flow, or a clockwise current. In other words at
the voltmeter, the current flows downward.
For the positive charge carrier case, the
force on the positive charged carrier is again
downward. This generates a counterclockwise
current. At the voltmeter now the current
flows upward.
Explanation:
Let : L = 189 mH = 0.189 H , and
C = 35.4 µF = 3.54 × 10−5 F .
The natural frequency is given by
'
1
1
f=
2π LC
(
1
1
=
2 π (0.189 H) (3.54 × 10−5 F)
= 61.5301 Hz .
017 (part 2 of 2) 5.0 points
Find the current in the circuit when the capacitor has lost 61.8% of its initial energy.
1. 0.126357
2. 0.085402
3. 0.133793
4. 0.196949
5. 0.103575
6. 0.174158
7. 0.117938
8. 0.0924005
9. 0.141317
10. 0.0828094
Correct answer: 0.085402 A.
016 (part 1 of 2) 5.0 points
A circuit consists of an idealized 189 mH inductor and an idealized 35.4 µF capacitor.
What is the natural frequency of this circuit
if the capacitor is first charged to 281 µC
Explanation:
Let : Q = 281 µC = 0.000281 C ,
Elost = 61.8% = 0.618 .
and
Version 078 – Test #3 – Antoniewicz – (57030)
Q is the initial charge, q the final charge, and
I is the final current. The total energy is
conserved over time, so
Q2
2C
=
q2
2C
+
1
L I2 .
2
According to the problem,
% 2&
Q2
q2
Q
1
−
= 0.618
= L I2 .
2C 2C
2C
2
Thus the current in the circuit is
'
0.618
I=Q
LC
(
0.618
= (0.000281 C)
(0.189 H) (3.54 × 10−5 F)
= 0.085402 A .
018 (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 #.
Note: a and b are the contact points where
the rod touches the rails, and d > # .
a
B
d
m
R
v
B
#
b
Consider the relationship between the potentials Vb and Va and the direction of the
induced magnetic field.
Which is the correct pair?
9
4. Vb = Va
and
! ind is up
B
5. Va > Vb
and
! ind is out of the page.
B
6. Vb = Va
and
! ind is into the page.
B
7. Vb > Va
and
! ind is down
B
8. Va > Vb
and
! ind is into the page.
B
9. Vb > Va
correct
10. Va > Vb
and
and
! ind is into the page.
B
! ind is down
B
Explanation:
Note: 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. Hence, if you look at the potential
drop across the resistor, then you can see that
the potential at b is greater than the potential
at a.
019 (part 2 of 3) 4.0 points
What is the magnitude and the direction of
the induced current around the loop?
1. |Iind | =
2. |Iind | =
3. |Iind | =
1. Va > Vb
and
! ind is up
B
2. Vb = Va
and
! ind is out of the page.
B
4. |Iind | =
3. Vb > Va
and
! ind is out of the page.
B
5. |Iind | =
Bdv
, clockwise
R
B # v2
, clockwise
R
B#v
, clockwise
R2
B #2 v
, counter-clockwise
R
B#v
, counter-clockwise
R
Version 078 – Test #3 – Antoniewicz – (57030)
B # v2
, counter-clockwise
R
B#v
, counter-clockwise
7. |Iind | =
R2
B#v
8. |Iind | =
, clockwise correct
R
Bdv
9. |Iind | =
, counter-clockwise
R
B #2 v
10. |Iind | =
, clockwise
R
Explanation:
The rate of change of the area of the rectangular loop is
6. |Iind | =
dA
dx
=#
.
dt
dt
Then from Faraday’s law, the magnitude of
the induced emf is given by
E = B#v.
From Ohm’s law, E = I R so the magnitude
of the induced current is
Iind =
B#v
E
=
.
R
R
The direction of the induced current is
clockwise, see the explanation in part 1.
020 (part 3 of 3) 3.0 points
The magnitude of the force exerted on the
metal rod by the magnetic field through the
metal bar is given by
2
2
!( = B #v
1. (F
R
2
!( = B dv
2. (F
R
2
!( = B #v
3. (F
R2
2
!( = B dv
4. (F
R2
2
2
!( = B dv
5. (F
R2
B 2 #2 v
!
6. (F ( =
R2
10
2 2
! ( = B # v correct
7. (F
R
2
!( = B #v
8. (F
R
2
2
!( = B #v
9. (F
R2
2
2
!( = B dv
10. (F
R
Explanation:
The magnitude of the magnetic force exerted on the metal rod is given by
%
&
B#v
B 2 #2 v
F = I #B =
#B =
R
R
where I is the induced current found in Part
2.
! = 0 outside the
Note: # was used since B
rectangle.
Substituting in the expression for the induced current yields
F =
B 2 #2 v
.
R
021 10.0 points
A 38 µF capacitor is charged to an unknown
potential V0 and then connected across an
initially uncharged 5 µF capacitor.
If the final potential difference across the
38 µF capacitor is 51 V, determine V0 .
1. 40.5405
2. 29.0
3. 54.6667
4. 67.2727
5. 49.8667
6. 15.0
7. 82.7273
8. 28.0
9. 57.7105
10. 100.0
Correct answer: 57.7105 V.
Explanation:
Use conservation of charge as the basis for
this problem. Initially, before C2 is connected
across C1 , the total charge in the system is
Qi = C1 V0 .
Version 078 – Test #3 – Antoniewicz – (57030)
When C2 is connected, the capacitors are now
in parallel. This implies that the voltage
Vf across C1 is also across C2 . Further, we
can replace the parallel capacitor combination
with Ceq = C1 +C2 . Hence, using the fact that
charge is conserved in the system
Qi = Qf
C1 V0 = (C1 + C2 ) Vf .
(C1 + C2 ) Vf
C1
(38 µF + 5 µF) (51 V)
=
38 µF
V0 =
= 57.7105 V .
keywords:
11