Version 041 – QZ3 – ditmire – (58565)
This print-out should have 30 questions.
Multiple-choice questions may continue on
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before answering.
001
10.0 points
A conducting loop (octagonal) around a
magnetic field (circular) contains three lightbulbs (labeled A, B, and C). The wires
connecting the bulbs are ideal, with no resistance. The loop is ground at one point as
shown in the figure. The magnetic field is
increasing rapidly.
B
B
B
B
B
A
C
Rank order the brightness of the three
bulbs, from brightest to least bright.
1. A > B > C
2. A = C > B
3. None of these correct
1
002 (part 1 of 3) 10.0 points
An LC circuit is shown in the figure below.
The 49 pF capacitor is initially charged by the
10 V battery when S is at position a. Then S
is thrown to position b so that the capacitor
is shorted across the 10 mH inductor.
10 mH
49 pF
S b
a
10 V
What is the maximum value for the oscillating current assuming no resistance in the
circuit?
1. 0.00043293
2. 0.000252763
3. 0.000347297
4. 0.000313786
5. 0.0003047
6. 0.000413023
7. 0.0003371
8. 0.000209762
9. 0.0007
10. 0.000241523
Correct answer: 0.0007 A.
Explanation:
4. A > C > B
5. A > B = C
Let :
6. B > C > A
Explanation:
The current in the loop is the same throughout the loop. Consequently, all the lightbulbs
burn at the same brightness
C = 49 pF = 4.9 × 10−11 F ,
L = 10 mH = 0.01 H , and
E = 10 V .
The maximum charge on the capacitor is
Qmax = C V . The maximum current is
Imax = ω Qm = ω C V ,
A=B =C.
where ω is the oscillatory frequency given by
Since the ground maintains a relative potential, it provides nothing of consequence to
this argument.
ω=√
1
.
LC
Version 041 – QZ3 – ditmire – (58565)
Thus
2
Therefore
Imax =
r
C
V =
L
r
4.9 × 10−11 F
(10 V)
0.01 H
Umax,ind =
= 0.0007 A .
1
1
C V 2 = (4.9 × 10−11 F) (10 V)2
2
2
= 2.45 × 10−9 J .
003 (part 2 of 3) 10.0 points
What is the natural angular frequency of the
circuit?
1. 1889820.0
2. 1165630.0
3. 1725160.0
4. 1869890.0
5. 1581140.0
6. 1025980.0
7. 1282470.0
8. 1428570.0
9. 1315330.0
10. 1000000.0
Correct answer: 1.42857 × 106 rad/s.
Explanation:
005 10.0 points
Gauss’ law for magnetism tells us
1. the net charge in any given volume.
2. that the line integral of a magnetic field
around any closed loop vanishes.
3. that the magnetic monopoles do not exist.
correct
4. that charges must be moving to produce
magnetic fields.
5. the magnetic field of a current element.
Explanation:
Correct answer: 2.45 × 10−9 J.
Explanation:
Umax,ind = Umax,cap
Q2max
1
=
= CV2.
2C
2
2T
5 m/s
m≪1 g
004 (part 3 of 3) 10.0 points
What is the maximum energy stored in the
magnetic field of the inductor?
1. 1.3365e-09
2. 1.7e-09
3. 2.45e-09
4. 1.539e-09
5. 9.92e-10
6. 8.64e-10
7. 8.82e-10
8. 1.45e-09
9. 1.1025e-09
10. 1.296e-09
In the arrangement shown in the figure,
the resistor is 4 Ω and a 2 T magnetic field
is directed into the paper. The separation
between the rails is 8 m . An applied force
moves the bar to the left at a constant speed
of 5 m/s .
4Ω
= 1.42857 × 106 rad/s .
006 (part 1 of 2) 10.0 points
8m
I
1
1
ω=√
=p
LC
(0.01 H) (4.9 × 10−11 F)
2T
Calculate the applied force required to
move the bar to the left at a constant speed of
5 m/s. Assume the bar and rails have negligible resistance and friction. Neglect the mass
of the bar.
Version 041 – QZ3 – ditmire – (58565)
1. 1045.33
2. 72.0
3. 170.667
4. 196.0
5. 28.125
6. 54.0
7. 2592.0
8. 4374.0
9. 16.0
10. 320.0
3
Correct answer: 1600 W.
Explanation:
The power dissipated in the resistor is
P = I 2 R = (20 A)2 (4 Ω)
= 1600 W .
Correct answer: 320 N.
Explanation:
The motional emf induced in the circuit is
E = B ℓ v = (2 T) (8 m) (5 m/s)
= 80 V .
From Ohm’s law, the current flowing through
the resistor is
I=
E
80 V
=
= 20 A ,
R
4Ω
so the magnitude of the force exerted on the
bar due to the magnetic field is
FB = I ℓ B = (20 A)(8 m)(2 T)
= 320 N .
To maintain the motion of the bar, a force
must be applied on the bar to balance the
magnetic force
008 10.0 points
Two long parallel wires are separated by
7.1 cm. One of the wires carries a current
of 30 A and the other carries a current of
64 A.
Determine the magnitude of the magnetic
force on a 3.3 m length of the wire carrying
the greater current.
1. 2.13224
2. 4.9536
3. 12.4502
4. 11.214
5. 5.98154
6. 17.8479
7. 6.69388
8. 0.937143
9. 4.75655
10. 13.9208
Correct answer: 17.8479 mN.
Explanation:
Let : a = 7.1 cm = 0.071 m
I1 = 30 A
I2 = 64 A
ℓ = 3.3 m
F = FB = 320 N .
007 (part 2 of 2) 10.0 points
At what rate is energy dissipated in the resistor?
1. 600.0
2. 10368.0
3. 348.444
4. 33075.0
5. 1843.2
6. 522.667
7. 2508.8
8. 1600.0
9. 22050.0
10. 28.125
The magnetic field due to a long straight
µ0 I
wire is B =
.
2πr
Since the wires are parallel, the field is perpendicular to the wires and therefore, the
force on the wire is equal to F = I ℓ B. Combining these two equations, we get
µ0 I 1 I 2 ℓ
2πa
(1.25664 × 10−6 N/A2 )(30 A)(64 A)
=
2 π(0.071 m)
F =
Version 041 – QZ3 – ditmire – (58565)
× (3.3 m) (1000 mN/N)
= 17.8479 mN
4
The direction of the circulating eddy current in the plate is
1. counter-clockwise. correct
009 I10.0 points
~ = µ0 I , the inte~ · dl
In Ampere’s Law,
B
gration must be over
1. any surface.
2. any closed path that surrounds all the
current producing B.
3. any path.
4. any closed path. correct
5. any closed surface.
Explanation:
Ampere’s Law states :
~ around any closed
~ · dl
The line integral of B
path equals µ0 I, where I is the total steady
current passing through any surface bounded
by the closed path.
Notice that Ampere’s Law applies to any
closed path, not necessarily one surrounding
all the current producing B.
keywords:
010 (part 1 of 3) 10.0 points
A pendulum consists of a supporting rod and
a metal plate (see figure). The rod is pivoted
at O. The metal plate swings through a
region of magnetic field. Consider the case
where the pendulum is entering the magnetic
field region from the left.
O
2. clockwise.
Explanation:
Using the right-hand rule the circulating
eddy current in the plate is counter-clockwise.
011 (part 2 of 3) 10.0 points
The direction of the induced magnetic field at
the center of the circulating eddy current is
1. parallel to the direction of motion; i.e.,
the arrow in the sketch.
2. along the rod toward O.
3. out of the paper. correct
4. along the rod away from O.
5. into the paper.
Explanation:
Because the magnetic field is pointing into
the paper, the induced flux of the portion of
the plate entering the magnetic region should
be out of the paper. Hence, by the right hand
rule, the induced current is counter-clockwise.
O
FB
i
ent
er
012 (part 3 of 3) 10.0 points
The direction of the force which the magnetic
field exerts is
ent
er
1. into the paper.
2. along the direction of swing.
Version 041 – QZ3 – ditmire – (58565)
5
6. B(r3 ) = 0
3. out of the paper.
4. opposite to the direction of swing. correct
5. along the rod toward the pivot point.
Explanation:
Because the magnetic field only exerts a
force on the current segment already in the
magnetic field region, the net magnetic force
is opposite to the direction of swing, see the
figure in the explanation of the previous Part.
013 10.0 points
The figure below shows a cylindrical coaxial
cable of radii a, b, and c in which equal, uniformly distributed, but antiparallel currents i
exist in the two conductors.
a
b
iout ⊙
c
µ0 i r 3
2 π a2
µ0 i (a2 − b2 )
8. B(r3 ) =
2 π r3 (r32 − b2 )
µ0 i r 3
9. B(r3 ) =
2 π b2
µ0 i (a2 + r32 − 2 b2 )
10. B(r3 ) =
2 π r3 (a2 − b2 )
7. B(r3 ) =
Explanation:
Ampere’s
Law states that the line inteI
~ · d~ℓ around any closed path equals
B
gral
µ0 I, where I is the total steady current passing through any surface bounded by the closed
path.
Considering the symmetry of this problem,
we choose a circular path, so Ampere’s Law
simplifies to
B (2 π r3 ) = µ0 Ien ,
iin ⊗
O
F
E
D
C
r1
r2
r3
r4
Which expression gives the magnitude
B(r3) at D of the magnetic field in the region b < r3 < a?
1. B(r3) =
2. B(r3) =
3. B(r3) =
4. B(r3) =
5. B(r3) =
µ0 i (a2 − r32 )
correct
2 π r3 (a2 − b2 )
µ0 i
π r3
µ0 i
2 π r3
µ0 i r 3
2 π c2
µ0 i (r32 − b2 )
2 π r3 (a2 − b2 )
where r3 is the radius of the circle and Ien is
the current enclosed.
Aen
π (r32 − b2 )
Since
, when b <
=
Acylinder
π (a2 − b2 )
r3 < a for the cylinder,
µ0 Ien
2 π r3
π (r32 − b2 )
µ0 i − i
π (a2 − b2 )
=
2 π r3
2
a − r32
µ0 i
a 2 − b2
=
2 π r3
B=
=
µ0 i (a2 − r32 )
.
2 π r3 (a2 − b2 )
014 (part 1 of 3) 10.0 points
A long solenoid has a coil made of fine wire
inside of it and coaxial with it.
Version 041 – QZ3 – ditmire – (58565)
r
I
R
6
6. M = π r 2 µ0 n N
7. M = π R2 µ0 n N correct
8. M = 2 π R µ0 n N
Assuming the inner coil is absent, determine the self-inductance for the long solenoid
with length b. The inside coil has N turns
and the outside solenoid n turns per meter.
1. L = π µ0 n r 2 b2
2. L = π µ0 n r 2 b
3. L = π µ0 n b2
4. L = π µ0 n2 r 2 b correct
Explanation:
The magnetic field of a solenoid is
B = µ0 n I .
The magnetic flux is
ΦB = B · A = (µ0 n I) (πR2) ,
so by Faraday’s Law the emf is
E = −N
d ΦB
dI
= −π R2 µ0 n N
.
dt
dt
7. L = π µ0 n r b2
We are interested in the emf in the inner coil.
The area to use is the area of the smaller inner
coils rather than the larger solenoid area.
The emf created through a mutual inducdI
tance M is E = −M
, so we have
dt
8. L = π µ0 n2 r b
M = π R 2 µ0 n N .
2 2
5. L = π µ0 n r
6. L = π µ0 n r b
9. L = π µ0 n r 2
3. M = 2 π r µ0 n N
016 (part 3 of 3) 10.0 points
If the current in the solenoid is I0 (1 − e−t/τ )
with I0 = 30 A, τ = 0.63 s, n = 413 turns/m,
r = 25 cm, N = 260 turns, and R = 8 cm,
what is the magnitude of the induced emf at
time t = 0? The permeability of free space is
4 π × 10−7 T · m/A .
1. 117.185
2. 85.6091
3. 129.194
4. 132.766
5. 96.183
6. 147.373
7. 112.58
8. 77.5988
9. 134.665
10. 152.855
4. M = 2 π R µ0 n
Correct answer: 129.194 mV.
Explanation:
The self-inductance for unit length for a
single long solenoid is µ0 n2 (πr 2 ), for length
b,
L = π µ0 n 2 r 2 .
015 (part 2 of 3) 10.0 points
What is the mutual inductance between the
solenoid and the inner loop?
1. M = 2 π r µ0 n
2. M = π R2 µ0 n
2
5. M = π r µ0 n
Explanation:
Version 041 – QZ3 – ditmire – (58565)
I0
τ
π (0.08 m)2 (4 π × 10−7 T · m/A)
=
0.63 s
× (413 turns/m) (260 turns) (30 A)
|E0 | = π R2 µ0 n N
= 129.194 mV .
017 10.0 points
What current is required in the windings of
a long solenoid that has 483 turns uniformly
distributed over a length of 0.526 m in order
to produce inside the solenoid a magnetic field
of magnitude 0.0001274 T? The permeablity
of free space is 1.25664 × 10−6 T m/A.
1. 110.407
2. 8.49259
3. 34.3403
4. 77.7035
5. 19.7548
6. 36.2653
7. 16.3019
8. 39.2972
9. 14.6124
10. 7.32377
Correct answer: 110.407 mA.
Explanation:
Let : N = 483 ,
ℓ = 0.526 m , and
B = 0.0001274 T .
Magnetic field inside the solenoid is
µ0 N I
, then
ℓ
Bℓ
I=
µ0 N
(0.0001274 T) (0.526 m)
=
µ0 (483)
B=
= 110.407 mA .
keywords:
7
018 10.0 points
The electric field between the plates of an air
capacitor of plate area 0.2 m2 is changing at a
rate of 7200 V/m/s.
The permitivity of free space is 8.85419 ×
10−12 C2 /N/m2 .
What is Maxwell’s displacement current?
1. 1.2797e-07
2. 2.3862e-08
3. 7.98736e-08
4. 1.7062e-07
5. 1.12891e-07
6. 4.66244e-07
7. 1.48662e-08
8. 1.59216e-07
9. 1.275e-08
10. 1.87266e-08
Correct answer: 1.275 × 10−8 A.
Explanation:
Let : ǫ0 = 8.85419 × 10−12 C2 /N/m2 ,
A = 0.2 m2 , and
dE
= 7200 V/m/s ,
dt
dE
where A is the area and
is the changing
dt
rate of the electric field.
The displacement current is
dE
dt
= 8.85419 × 10−12 C2 /N/m2
× 0.2 m2 (7200 V/m/s)
ID = ǫ0 A
= 1.275 × 10−8 A .
019
10.0 points
A solenoid with circular cross section produces a steadily increasing magnetic flux
through its cross section. There is an octagonally shaped circuit surrounding the solenoid
as shown. The increasing magnetic flux gives
rise to a counterclockwise induced emf E.
Version 041 – QZ3 – ditmire – (58565)
Initial Case: The circuit consists of two
identical light bulbs of equal resistance R connected in series, leading to a loop equation
E − 2 i R = 0.
i
Figure 1:
B
B
X
Y
B
B
i
The corresponding electrical power consumed by bulb X and bulb Y are PX and
PY , respectively.
Primed′ Case: Now the loop is doubled
so that the wire circumscribes the increasing
magnetic flux twice, as in figure 2.
Figure 2:
B
B
X
Y
B
B
8
PX′
PY′
= 0 and
=0
PX
PY
PY′
P′
=1
5. X = 1 and
PX
PY
PX′
PY′
6.
= 2 and
=2
PX
PY
Explanation:
Let E and R be the induced emf and resistance of the light bulbs, respectively.
For the first case, since the two bulbs are
in series, the equivalent resistance is simply
Req = R + R = 2 R and the current through
the bulbs is
E
i=
,
2R
so the power consumed by bulb X is
2
E2
E
R=
.
PX =
2R
4R
4.
For the second (primed) case, since the number of loops is doubled, the emf is doubled to
2E :
E
2E
= ,
i′ =
2R
R
and the power consumed by bulb X is
2
E
E2
′
PX =
, so
R=
R
R
E2
= R2 = 4 .
PX
E
4R
For the second (primed) case, bulb Y has
the same current as bulb X, so
PX′
i′
The corresponding electrical power consumed by bulb X and bulb Y are PX′ and
PY′ , respectively.
PX′
PY′
What are the ratios
and
?
PX
PY
P′
PY′
1. X = 4 and
= 4 correct
PX
PY
PY′
P′
1
1
and
=
2. X =
PX
4
PY
4
′
′
PY
P
1
1
and
=
3. X =
PX
2
PY
2
PY′
PY
E2
= R2 = 4 .
E
4R
020 (part 1 of 2) 10.0 points
The two-loop wire circuit is 90.1845 cm
wide and 60.123 cm high. The wire circuit
in the figure is located in a magnetic field
whose magnitude varies with time according
Version 041 – QZ3 – ditmire – (58565)
to B = (0.001 T/s) t and its direction is out
of the page.
9
1 2
ℓ = 0.361477 m2 ,
2
AR = ℓ2 = 0.722955 m2 , and
dB
d
=
α t = α = 0.001 T/s .
dt
dt
AL =
P
B
60.123 cm
B
30.0615 cm
60.123 cm
The resistance for the wire is proportional
to the wire’s length. For a length of 60.123 cm,
the resistance is
R = δ ℓ = (0.0757 Ω/m) (0.60123 m) = 0.0455131 Ω
Q
When the magnetic field is 0.7 T , find the
magnitude of the current through the middle
leg P Q of the circuit. Assume the resistance
per length of the wire is 0.0757 Ω/m .
1. 417.125
2. 214.851
3. 1367.26
4. 909.349
5. 984.852
6. 382.361
7. 430.317
8. 360.788
9. 540.987
10. 661.856
Applying Ohm’s law V = I R for the right
and left perimeters of the loops in the circuit,
ER = 4 ℓ δ IR = 4 R IR and
E L = 3 ℓ δ IL = 3 R IL .
When the magnetic field changes with time,
there is an induced emf in both the right- and
the left-hand sides. Applying Faraday’s law
d ΦB
E =−
where ΦB = A B ,
dt
dB
dB
= ℓ2
and
dt
dt
dB
1
dB
EL = AL
= ℓ2
.
dt
2
dt
ER = AR
Explanation:
ℓ = 0.60123 m ,
δ = 0.0757 Ω/m ,
α = 0.001 T/s .
E R = 4 ℓ δ IR = ℓ 2
1
4
1
E L = 3 ℓ δ IL =
2
1
IL =
6
IR =
and
P
IR
B
B
ℓ
2
ℓ
ℓ
IP Q
Q
(4)
dB
dt
ℓ
α and
δ
dB
ℓ2
dt
ℓ
α.
δ
Using the junction rule, the current flowing
through the middle leg P Q of the circuit is
IP Q
IL
(3)
Equating Eqs. 1 and 3, and Eqs. 2 and 4,
we have
Correct answer: 661.856 µA.
Let :
(1)
(2)
1 1 ℓα
1 ℓα
= IR − IL =
−
=
4 6
δ
12 δ
1 (0.60123 m) (0.001 T/s)
=
12
(0.0757 Ω/m)
= 661.856 µA .
Version 041 – QZ3 – ditmire – (58565)
10
Explanation:
021 (part 2 of 2) 10.0 points
The direction of the current in the middle leg
of the circuit is from
Let : c = 4.28 cm ,
a = 2.2 cm ,
b = 35.9 cm , and
I = 0.0257 A .
1. Q to P . correct
2. P to Q.
dr
Explanation:
IP Q is positive, so the current flows up the
page, from Q to P , as indicated by the direction of the arrows in the figure.
The directions of the currents IL and IR in
the figure were chosen using Lenz’s law, which
indicates that the direction of the magnetic
field produced by the induced current will
oppose the increasing external magnetic field.
4.28 cm
35.9 cm
0.0257 A
022 10.0 points
A rectangular loop located a distance from a
long wire carrying a current is shown in the
figure. The wire is parallel to the longest side
of the loop.
2.2 cm
Find the total magnetic flux through the
loop.
1. 1.94708e-09
2. 1.88189e-09
3. 2.51436e-09
4. 1.59883e-09
5. 7.65354e-10
6. 5.91877e-10
7. 3.76245e-09
8. 7.74471e-10
9. 1.67408e-09
10. 6.7682e-10
Correct answer: 7.65354 × 10−10 Wb.
r
I
c
b
a
From Ampère’s law, the strength of the
magnetic field created by the current-carrying
wire at a distance r from the wire is (see
figure.)
µ0 I
,
2πr
so the field varies over the loop and is directed
~ is parallel
perpendicular to the page. Since B
~ the magnetic flux through an area
to dA,
element dA is
Z
Φ ≡ B dA
Z
µ0 I
=
dA .
2πr
B=
~ is not uniform but rather depends on
Note: B
r, so it cannot be removed from the integral.
In order to integrate, we express the area
element shaded in the figure as dA = b dr.
Since r is the only variable that now appears
in the integral, we obtain for the magnetic
flux
Z a+c
µ0 I
dr
ΦB =
b
2π
r
c
a+c
µ0 I b
ln r =
2π
c
a+c
µ0 I b
ln
=
2π
c
Version 041 – QZ3 – ditmire – (58565)
11
µ0 (0.0257 A)(0.359 m)
a+c
The loop in the figure carries a current 1.17 A.
=
ln
The semicircular portion has a radius 7.81 cm.
2π
c
The permeability of free space is
µ0 (0.0257 A)(0.359 m)
=
(0.414768)
1.25664 × 10−6 T · m/A .
2π
Correct answer: 2.74537 mT.
Explanation:
A
1.17 A
7.81 cm
7. 8
023 10.0 points
A rectangular coil of 40 turns, 0.25 m by
0.31 m, is rotated at 47 rad/s in a magnetic
field so that the axis of rotation is perpendicular to the direction of the field. The maximum
emf induced in the coil is 0.4 V.
What is the magnitude of the field?
1. 1.13543
2. 0.924139
3. 0.787638
4. 0.810153
5. 1.31898
6. 0.432821
7. 2.74537
8. 1.09266
9. 1.11913
10. 3.43209
1c
m
= 7.65354 × 10−10 Wb .
2 × 7.81 cm
Determine the magnitude of the magnetic
field at A.
1. 2.80942e-05
2. 6.84999e-05
3. 9.92292e-06
4. 8.94356e-06
5. 4.4549e-05
6. 4.20442e-05
7. 8.19106e-06
8. 2.16058e-05
9. 3.20735e-05
10. 3.45873e-05
Correct answer: 8.94356 × 10−6 T.
Explanation:
Let : N = 40 turns ,
ω = 47 rad/s ,
ǫmax = 0.4 V ,
x = 0.25 m , and
y = 0.31 m .
Let : µ0 = 1.25664 × 10−6 T · m/A
I = 1.17 A and
R = 7.81 cm .
From ǫmax = N A B ω, where the area A is
R
A = (0.25 m)(0.31 m) = 0.0775 m2 ,
so
A
R
ǫmax
B=
N Aω
I
0.4 V
=
(40 turns)(0.0775 m2 )(47 rad/s)
= 0.00274537 T = 2.74537 mT .
024 (part 1 of 2) 10.0 points
2R
The Biot-Savart Law is given by
~ =
dB
µ0 I d~ℓ × r̂
.
4π
r2
(1)
Version 041 – QZ3 – ditmire – (58565)
Note: The distance from a current element
on a circle to the center is a constant, namely
r, so we can pull this out of the integral. Also,
the current element, I d~l, is always perpendicular to r̂, so sin θ = 1.
Hence,
Z
µ0 I
Bf ull circle =
dℓ
4 π r2
µ I
= 0 2 2πr
4π r
µ0 I
=
,
2r
where a half-circular arc is one-half this value.
Assume: Out of the page to be positive.
The angle θ (between the vector ~ℓ and the
vector ~r ) is from the Biot-Savart Law, Eq.
~ a distance a
(1), for the magnetic field B
away from a wire segment
Z
µ0 I θ 2
sin θ dθ
B=
4 π a θ1
µ I
= 0 (cos θ1 − cos θ2 ) ,
4πa
where a = R for the sides of the semi-square
in figure.
12
3π
π
µ0 I
cos
− cos
+
4πR
4
4
µ0 I
3π
π
+
cos
− cos
4πR
4
2
"
√ #
µ I 1
2
+
=− 0
R
4 2π
bottom
lef t edge
(2)
(1.25664 × 10−6 T · m/A) (1.17 A)
=−
0.0781 m
× 0.475079
= −8.94356 × 10−6 T
~ = 8.94356 × 10−6 T ,
kBk
~ is pointed into the page.
where B
Alternate Solution: Consider a square loop
with each side a length 2 R.
θ1
R
θ2
For the line segment (one-fourth of a
square), the magnitude of the magnetic field
is
µ0 I
(cos θ1 − cos θ2 )
4πR µ0 I
1
−1
√ −√
=
4πR
2
2
√
µ I
2 .
= 0
4πR
B=
lb
π
4
π
2
ll
r
3π
4
ll
r
R
R
r
lr
π
2
3π
4
lr
Then the magnetic field at the center is
r
π
4
The magnetic field due to the semicircular
portion and the straight wire segments is
1
µ I
B=
− 0
semi−circle
2
2R
π
π
µ0 I cos − cos
right edge
+
4πR
2
4
Bsquare = 4 B
√
2 µ0 I
=
.
πR
The magnetic field at the center of a circular
loop is
Bcircle =
µ0 I
.
2R
Therefore the magnetic field at the center
of the “half-circular loop” plus “half-square
Version 041 – QZ3 – ditmire – (58565)
loop” is
13
3. 3.83409
4. 3.71794
5. 0.675966
6. 4.98226
7. 1.74241
8. 4.77643
9. 1.72089
10. 1.89157
1
[Bcircle + Bsquare ]
2"
#
√
1 µ0 I
2 µ0 I
=
+
2 2R
πR
"
√ #
µ I 1
2
= 0
,
+
R 4 2π
B=
Correct answer: 4.24841 ms.
which is the same as Eq. (2).
Explanation:
025 (part 2 of 2) 10.0 points
In which direction is the magnetic field pointing at A ?
1. into the page correct
Let : R = 2.56 Ω ,
L = 201 mH ,
E = 8.7 V .
and
2. out of the page
R
3. not enough information given
Explanation:
The current is circulating in the clockwise
direction. Using the right hand rule, with the
thumb pointing in the direction of the current,
our fingers projected unto point A is into the
page.
026 (part 1 of 3) 10.0 points
An inductor and a resistor are connected
with a double pole switch to a battery as
shown in the figure.
The switch has been in position b for a long
period of time.
2.56 Ω
201 mH
8.7 V
S b
a
If the switch is thrown from position b
to position a (connecting the battery), how
much time elapses before the current reaches
179 mA?
1. 4.24841
2. 1.8427
L
E
S b
a
The time constant of an RL circuit is
τ=
0.201 H
L
=
= 0.0785156 s .
R
2.56 Ω
The final current reached in the circuit is
E
8.7 V
=
= 3.39844 A .
R
2.56 Ω
The switch is in position a in an RL circuit
connected to a battery at t = 0 when I = 0;
then
I = I0 1 − e−t/τ
I1
t1 = −τ ln 1 −
I0
0.179 A
= −(0.0785156 s) ln 1 −
3.39844 A
I0 =
= 4.24841 ms .
027 (part 2 of 3) 10.0 points
Version 041 – QZ3 – ditmire – (58565)
What is the maximum current in the inductor
a long time after the switch is in position a?
1. 1.38508
2. 0.902873
3. 1.16928
4. 2.51337
5. 1.55844
6. 1.53722
7. 3.39844
8. 1.80272
9. 1.18196
10. 2.03704
Correct answer: 3.39844 A.
Explanation:
After a long time compared to τ , we have a
dc circuit with a battery supplying an emf E,
which is equal to the voltage drop I R across
the resistor and
I=
E
8.7 V
=
= 3.39844 A .
R
2.56 Ω
028 (part 3 of 3) 10.0 points
The switch has brushes within it so that the
switch can be thrown from a to b without
internal sparking. Now the switch is smoothly
thrown from a to b, shorting the inductor and
resistor.
How much time elapses before the current
falls to 123 mA?
1. 16.1511
2. 260.584
3. 44.95
4. 119.763
5. 73.0216
6. 90.6805
7. 100.1
8. 193.808
9. 82.4148
10. 155.255
Correct answer: 260.584 ms.
Explanation:
The current decay in an RL circuit when
there is no voltage source present, and the
initial current is I0 , is
I = I0 e−t /τ
14
I
t = −τ ln
.
I0
The time t3 that elapses for the current to fall
to I3 = 0.123 A is
I3
t3 = −τ ln
I0
0.123 A
= −(0.0785156 s) ln
3.39844 A
= 260.584 ms .
029 10.0 points
A wire in which there is a current of 3.09 A is
to be formed into a circular loop of one turn.
If the required value of the magnetic field
at the center of the loop is 12 µT, what is
the required radius? The permeability of free
space is 1.25664 × 10−6 T · m/A .
1. 120.637
2. 62.2285
3. 50.2501
4. 27.5314
5. 24.322
6. 16.1792
7. 32.3553
8. 41.2209
9. 48.1632
10. 15.4481
Correct answer: 16.1792 cm.
Explanation:
Let : B = 12 µT = 1.2 × 10−5 T ,
I = 3.09 A , and
µ0 = 1.25664 × 10−6 T · m/A .
The magnetic field at the center of circular
loop is
µ0 I
B=
2R
µ0 I
R=
2B
(1.25664 × 10−6 T · m/A) (3.09 A)
=
2 (1.2 × 10−5 T)
100 cm
×
1m
= 16.1792 cm .
Version 041 – QZ3 – ditmire – (58565)
030 10.0 points
An inductor of 460 turns has a radius of 3 cm
and a length of 29 cm.
The permeability of free space is
1.25664 × 10−6 N/A2 .
Find the energy stored in it when the current is 0.7 A.
1. 0.000635164
2. 0.00153556
3. 0.000462638
4. 4.3525e-06
5. 0.00423369
6. 1.26618e-05
7. 0.00253236
8. 0.000440598
9. 9.12109e-05
10. 0.000571027
Correct answer: 0.000635164 J.
Explanation:
Let : r = 3 cm = 0.03 m ,
ℓ = 29 cm = 0.29 m ,
N = 460 turns ,
I = 0.7 A , and
µ0 = 1.25664 × 10−6 N/A2 .
The inductance is
A
π r2
= µ0 N 2
ℓ
ℓ
= (1.25664 × 10−6 N/A2)
π (0.03 m)2
× (460 turns)2
0.29 m
= 0.00259251 H .
L = µ0 N 2
The energy stored in an inductor is
1
L I2
2
1
= (0.00259251 H)(0.7 A)2
2
= 0.000635164 J .
U=
15