Version 121 – Exam 3 – hoffmann – (57505)
This print-out should have 23 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
A capacitor of capacitance C has a charge Q
at t = 0. At that time, a resistor of resistance
R is connected to the plates of the charged
capacitor. Find the magnitude of the displacement current between the plates of the
capacitor as a function of time.
Q −t/Q
e
RC
RC t/(RC)
2.
e
Q
Q t/(RC)
3.
e
RC
RC −t/(RC)
4.
e
Q
Q −t/(RC)
5.
e
correct
RC
Explanation:
Basic Concept
RC circuits. Displacement Current.
The displacement current is defined to be
1.
Id = ǫ0
d ΦE
.
dt
The electric field inside a capacitor is essenq
tially uniform and E =
. Since the charge
ǫ0 A
on a capacitor in a discharging RC circuit is
given by q(t) = Q e−t/RC , the displacement
current is found by
d ΦE
dt
d
q
= ǫ0
A
dt ǫ0 A
Id = ǫ0
=
dq
Q −t/(RC)
= −
e
.
dt
RC
Note that the displacement current equals the
actual current in the wires to the capacitor.
Thus, the Ampere-Maxwell law tells us that
~ will be the same regardless of which current
B
we evaluate.
1
002 (part 2 of 2) 10.0 points
Given C = 2 µF, Q = 28 µC, R = 229 kΩ,
and ǫ0 = 8.85419 × 10−12 C2 /N · m2 , at what
rate is the electric flux between the plates
changing at time t = 0.1 s?
1. -4084300.0
2. -2947880.0
3. -5550340.0
4. -3881590.0
5. -4027370.0
6. -2231700.0
7. -5845500.0
8. -4826340.0
9. -4899350.0
10. -3442650.0
Correct answer: −5.55034 × 106 Vm/s.
Explanation:
Let :
ǫ0 = 8.85419 × 10−12 C2 /N · m2 ,
t = 0.1 s ,
C = 2 µF ,
Q = 28 µC = 2.8 × 10−5 C , and
R = 229 kΩ = 2.29 × 105 Ω .
From the discussion in the previous part, we
know that
d ΦE
Id
=
dt
ǫ0
Q
e−t/(RC)
ǫ0 RC
2.8 × 10−5 C
=−
ǫ0 (2.29 × 105 Ω) (2 × 10−6 F)
−0.1 s
5
−6
× e (2.29 × 10 Ω)(2 × 10 F)
=−
= −5.55034 × 106 Vm/s .
003 10.0 points
A long straight wire carries a current 20 A. A
rectangular loop with two sides parallel to the
straight wire has sides 2.5 cm and 10 cm, with
its near side a distance 1 cm from the straight
wire, as shown in the figure.
Version 121 – Exam 3 – hoffmann – (57505)
2
1 cm
20 A
10 cm
dx
x
I
Find the magnetic flux through the rectangular loop.
The permeability of free space is 4 π ×
10−7 T · m/A.
1. 2.31127e-06
2. 1.50332e-06
3. 7.16704e-07
4. 3.89247e-07
5. 2.05446e-06
6. 1.00221e-06
7. 9.80829e-07
8. 5.01105e-07
9. 9.41596e-07
10. 7.51658e-07
Correct answer: 5.01105 × 10−7 Wb.
Explanation:
The magnetic flux through the strip of area
dA is
dΦ = B dA
µ0 I
b dx
=
2π x
µ0 2 b I dx
,
=
4π
x
so the total magnetic flux through the rectangular loop is
Z d+a
Φtotal =
dΦ
d
Z d+a
µ0
dx
=
(2 b I)
4π
x
d
µ0
d+a
=
(2 b I) ln
4π
d
= (1 × 10−7 N/A2 ) 2 (0.1 m) (20 A)
0.01 m + 0.025 m
× ln
0.01 m
= 5.01105 × 10−7 Wb .
004
P
Let :
I = 20 A ,
a = 2.5 cm = 0.025 m ,
b = 10 cm = 0.1 m ,
d = 1 cm = 0.01 m , and
µ0
= 1 × 10−7 N/A2 .
4π
a
d
2.5 cm
b
10.0 points
a
2a
X
Y
Two long parallel wires are a distance 2a
apart, as shown above. Point P is in the
Version 121 – Exam 3 – hoffmann – (57505)
plane of the wires and a distance a from the
wire X. When there is a current I in wire X
and no current in wire Y , the magnitude of
the magnetic field at P is B0 . when there are
equal currents I in the same direction in both
wires, the magnitude of the magnetic field at
P is
1. B0
2.
10
B0
9
3. 2B0
2
4. B0
3
4
5. B0 correct
3
Explanation:
The superposition principle is applicable to
the magnetic field. The magnetic field at P
is the sum of two components magnetic fields
due to wires X and Y respectively. Furthermore, we know the fields due to X and Y are
in the same direction. So we can consider the
magnitudes only.
BX is known to be B0 . Apply Ampere’s
law to magnetic field due to a long straight
µ0 I
1
wire. We have B =
∝ . Therefore,
2πr
r
1
1
BY = BX = B0 . Finally, we obtain
3
3
1
4
B = BX + BY = 1 +
B0 = B0 .
3
3
005 I10.0 points
~ = µ0 I , the inte~ · dl
In Ampere’s Law,
B
gration must be over
1. any closed surface.
2. any closed path that surrounds all the
current producing B.
3. any closed path. correct
4. any surface.
5. any path.
3
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:
006 10.0 points
A plane loop of wire of N turns, each of
area A, is perpendicular to a magnetic field
whose magnitude changes in time according
to B(t) = B0 sin(2 π f t).
What is the induced emf in the loop as a
function of time?
1. −N 2 A B0 f π cos(2 π f t) correct
2. −N A B0 f π cos(π f t)
3. N 2 A B0 f π cos(2 π f t)
4. N A B0 f π sin(π f t)
5. −N A B0 f π cos(2 π f t)
6. N 2 A B0 f π sin(2 π f t)
7. −N A B0 f π sin(π f t)
8. N 2 A B0 f cos(2 f t)
9. −N 2 A B0 f π sin(2 π f t)
Explanation:
From Faraday’s Law of induction
E =−
dΦ
d (N B A)
=−
dt
dt
where N is the number of coils, A is the area
of the coils and B is the magnetic field, so
d B0 sin(2 π f t)
dt
= −N 2 A B0 f π cos(2 π f t)
E = −N A
Version 121 – Exam 3 – hoffmann – (57505)
007 10.0 points
The segment of wire in the figure carries a
current of 3 A, where the radius of the circular
arc is 5 cm.
The permeability of free space is
1.25664 × 10−6 T · m/A .
m
5c
3A
O
Determine the magnitude of the magnetic
field at point O , the origin of the arc.
1. 31.4159
2. 17.952
3. 23.5619
4. 47.1239
5. 13.7445
6. 10.472
7. 15.708
8. 9.42478
9. 6.28319
10. 20.944
Correct answer: 9.42478 µT.
Explanation:
Let : I = 3 A and
R = 5 cm = 0.05 m .
For the straight sections d~s × r̂ = 0. The
quarter circle makes one-fourth the field of a
µ0 I
full loop B =
into the paper. Or, you
8R
can use the equation
µo I
θ,
B=
4π
π
where θ = . Thus the magnetic field is
2
µ0 I
B=
,
8R
(1.25664 × 10−6 T · m/A) (3 A)
=
8 (0.05 m)
= 9.42478 µT , into the paper .
4
008 (part 1 of 3) 10.0 points
At t = 0, S is closed and C begins charging.
C
R
E
S
i
Determine the initial plate charge q0 (at
t = 0) stored in C immediately after S is
closed and a final charge q∞ (at t = ∞), i.e.,
at asymptotic time.
1.
2.
E
,
C
E
q0 = ,
C
q0 =
q∞ = 0
E
C
E
=
C
q∞ =
3.
q0 = 0,
q∞
4.
q0 = 0,
q∞ = 0
5.
q0 = 0,
q∞ = E C correct
6.
q0 = EC,
q∞ = E C
Explanation:
Characteristic time of a RC-circuit is τ =
R C. Current in a charging RC-circuit is
i(t) =
E −t/τ
e
.
R
Before S is closed, there is no charge on
the capacitor so immediately after S is closed,
(i.e., at t = 0),
q0 = 0 .
When t = ∞, the current in the circuit will
diminish to nothing, so the voltage across the
capacitor is the same as the emf, therefore
q∞ = V C = E C .
Version 121 – Exam 3 – hoffmann – (57505)
009 (part 2 of 3) 10.0 points
Consider 2 RC-circuits. Both have E = 20 V.
1044 µF
27 Ω
20 V
S
87 µF
324 Ω
20 V
i
S
i
Correct answer: 12.
E −t/τ
e
i1
R1
=
E −t/τ
i2
e
R2
=
324 Ω
R2
=
= 12 ,
R1
27 Ω
i1
is the same for all time;
i2
i.e., t = 12 τ is a meaningless criterion.
and the ratio
1. τ1 > τ2
2. τ1 = τ2 correct
011 (part 1 of 3) 10.0 points
3. τ1 < τ2
Explanation:
R1
R2
C1
C2
E
3. 21.0
4. 11.0
5. 19.0
6. 15.0
7. 12.0
8. 9.0
9. 8.0
10. 20.0
Explanation:
τ1 = τ2 = τ , so
Compare the characteristic times τ1 and τ2
of the circuits. The subscript 1 applies to the
top circuit.
Let :
5
= 27 Ω ,
= 324 Ω ,
= 1044 µF = 0.001044 F ,
= 87 µF = 8.7 × 10−5 F , and
= 20 V .
τ1 = R1 C1 = (27 Ω) (0.001044 F)
= 0.028188 s , and
τ2 = R2 C2 (2) = (324 Ω) (8.7 × 10−5 F)
= 0.028188 s .
Thus τ1 = τ2 .
010 (part 3 of 3) 10.0 points
Consider the same setup as the previous part.
i1
Find the ratio
at t = 12 τ .
i2
1. 17.0
2. 18.0
In the LC circuit the capacitor is charged
to its maximum 27.3 µC while the switch S is
open. Then the switch is closed at t = 0.
1 µF
1H
Q
S
Find the energy stored in the inductor at
the quarter of a period of oscillations in the
circuit.
1. 340.402
2. 414.72
3. 253.5
4. 372.645
5. 297.562
6. 18.0267
7. 35.5267
8. 88.935
9. 128.0
10. 23.2067
Correct answer: 372.645 µJ.
Version 121 – Exam 3 – hoffmann – (57505)
Explanation:
Let :
T
,
4
C = 1 µF = 1 × 10−6 F ,
L = 1 H , and
q0 = 27.3 µC = 2.73 × 10−5 C .
t=
C
L
Q
S
Oscillation in LC circuit:
q = qmax cos(ω t + δ)
dq
= −ω qmax sin(ω t + δ)
I=
dt
2π
1
with ω =
, where T is the period
=√
T
LC
of oscillation.
When the switch closes, initially there is
zero current in the circuit. Thus the solution
has the correct phase at t = 0
dQ
I=
= −ω qmax sin(ω t) .
dt
1
2π
=√
ω=
T
LC
1
=p
(1 H)(1 × 10−6 F)
= 1000 1/C .
T
Now at t = ,
4 2π T
×
I = −q0 ω sin
T
4
= −q0 ω = −(27.3 µC)(1000 1/C)
= −0.0273 A , so
1
1
L I 2 = L (−q0 ω)2 = ULmax
2
2
1 2
= UCmax =
q
2C 0
(2.73 × 10−5 C)2 106 µJ
=
2 (1 × 10−6 F)
1J
6
T
Observe that at t = , the magnitude of
4
the current flow through the inductor is at a
maximum, and the charge q = qmax cos ω t is
T
π
zero at t =
=
. Thus, at this moment
4
2ω
there is no energy stored in the capacitor and
all the energy is stored in the inductor. At
any other time, one can show that for an
ideal LC circuit where there is no dissipation
from resistance, the total energy stored is the
sum of the energy in the capacitor and the
energy in the inductor, and this energy is
constant in time (although the energy in each
one oscillates).
012 (part 2 of 3) 10.0 points
T
Find total energy at t = .
12
1. 81.9025
2. 372.645
3. 217.202
4. 85.5625
5. 48.735
6. 796.005
7. 222.042
8. 549.903
9. 1230.08
10. 174.845
Correct answer: 372.645 µJ.
Explanation:
The total energy does not change with respect
to time, so
Utotal = ULmax = UCmax = 372.645 µJ .
UL =
= 372.645 µJ .
013 (part 3 of 3) 10.0 points
Consider a new initial condition at t = 0:
q0 = 0, I0 = +1.0 A. Let q = qmax cos(ω t+δ),
dq
I=
where qmax is assumed to be positive.
dt
Version 121 – Exam 3 – hoffmann – (57505)
Find the new phase angle δ for this new
situation.
1. δ = 45◦
2. δ = 0◦
3. δ = 135◦
4. δ = 90◦
5. δ = 225◦
Find the frequency of the oscillations.
1. 1239020.0
2. 1553190.0
3. 1186270.0
4. 1299490.0
5. 1139730.0
6. 1452880.0
7. 1677640.0
8. 1061030.0
9. 1369790.0
10. 1098270.0
Correct answer: 1.29949 × 106 Hz.
6. δ = 270 correct
◦
7. δ = 180
7
Explanation:
◦
8. δ = 315◦
Explanation:
We find the correct δ by fitting the choices
into the formulas below:
Let : C = 10 pF = 1 × 10−11 F ,
L = 1.5 mH = 0.0015 H , and
E = 9 V.
q = qmax cos(ω t + δ)
dq
= −ω qmax sin(ω t + δ)
I=
dt
At t = 0, q should be 0, while I is positive.
Remember qmax is assumed to be positive.
Then only δ = 270◦ fits all the requirement;
i.e., at t = 0
q = qmax cos 270◦ = 0
L
C
E
and
I = −ω qmax sin 270◦ = ω qmax > 0 .
014 (part 1 of 3) 10.0 points
An LC circuit is shown in the figure below.
The 10 pF capacitor is initially charged by the
9 V battery when S is at position a. Then S
is thrown to position b so that the capacitor
is shorted across the 1.5 mH inductor.
9V
a
1
, the frequency is
LC
1
ω
√
=
2π
2π LC
1
p
=
2 π (0.0015 H) (1 × 10−11 F)
f=
= 1.29949 × 106 Hz .
1.5 mH
10 pF
Since ω = √
S b
S b
a
015 (part 2 of 3) 10.0 points
What is the maximum value of charge on the
capacitor?
1. 9.6e-11
2. 6.6e-11
3. 1.92e-10
Version 121 – Exam 3 – hoffmann – (57505)
4. 1.12e-10
5. 1.2e-10
6. 5.4e-11
7. 2.4e-10
8. 9.0e-11
9. 1.08e-10
10. 1.95e-10
Correct answer: 9 × 10−11 C.
Explanation:
The initial charge on the capacitor equals
the maximum charge, so
Qmax = C E = (1 × 10−11 F)(9 V)
= 9 × 10−11 C .
8
where a1 = 0.0189 T/s, a2 = 0.0348 T/s2 are
constants, time t is in seconds and field B is
in Tesla.
Find the magnitude of the induced emf in
the coil at t = 5.68 s.
1. 0.64452
2. 0.125402
3. 0.0804163
4. 0.0101629
5. 1.15521
6. 1.70915
7. 0.0737051
8. 0.187688
9. 0.290736
10. 0.822529
Correct answer: 1.15521 V.
016 (part 3 of 3) 10.0 points
What is the maximum value of current in the
circuit?
1. 0.000816497
2. 0.000774597
3. 0.00102762
4. 0.000695701
5. 0.00109301
6. 0.00100698
7. 0.000734847
8. 0.0013
9. 0.000751443
10. 0.000885438
Explanation:
Let :
The area of the circular coil is
Correct answer: 0.000734847 A.
A = π r2
= π (0.0977 m)2
= 0.0299874 m2 ,
Explanation:
I = −ω Qmax sin ω t = −Imax sin ω t
Imax = ω Qmax = 2 π f Qmax
= 2 π(1.29949 × 106 Hz)(9 × 10−11 C)
= 0.000734847 A .
017 10.0 points
A 93 turns circular coil with radius 9.77 cm
and resistance 5.68 Ω is placed in a magnetic
field directed perpendicular to the plane of
the coil. The magnitude of the magnetic field
varies in time according to the expression
B = a 1 t + a 2 t2 ,
n = 93 turns ,
r = 9.77 cm = 0.0977 m ,
R = 5.68 Ω ,
a1 = 0.0189 T/s ,
a2 = 0.0348 T/s2 , and
t = 5.68 s .
so from Faraday’s law,
d ΦB
dt
dAB
= −n
dt
dB
= −n A
dt
= −n A (a1 + 2 a2 t)
= −(93 turns) (0.0299874 m2 )
× [0.0189 T/s + 2 (0.0348 T/s2 ) (5.68 s)]
= −1.15521 V ,
E = −n
Version 121 – Exam 3 – hoffmann – (57505)
which has a magnitude of 1.15521 V .
018
10.0 points
An ideal transformer has a primary with
40 turns and secondary with 16 turns. The
load resistor is 69 Ω and the source voltage is
90 Vrms .
9
radius 4.66 cm consists of N = 560 turns of
wire that carries a current I = I0 sin ω t, with
I0 = 43.3 A and a frequency f = 70.5 Hz. A
loop that consists of Nℓ = 19 turns of wire
links the toroid, as in the figure.
69 Ω
16 turns
90 Vrms
40 turns
N
R
a
What is the rms electric potential across
the 69 Ω load resistor?
1. 68.0
2. 76.0
3. 38.0
4. 32.0
5. 84.0
6. 52.0
7. 36.0
8. 56.0
9. 78.0
10. 48.0
Correct answer: 36 Vrms .
b
Determine the maximum E induced in the
loop by the changing current I.
1. 0.102028
2. 0.260595
3. 0.843929
4. 0.300991
5. 0.41881
6. 0.683905
7. 0.274187
8. 0.161627
9. 0.187319
10. 0.256164
Correct answer: 0.274187 V.
Explanation:
Let : N1 = 40 turns ,
N2 = 16 turns ,
V1 = 90 Vrms .
Nl
Explanation:
Basic Concept: Faraday’s Law
and
The rms voltage across the transformer’s
secondary is
N2
16 turns
V1 =
(90 Vrms )
N1
40 turns
= 36 Vrms ,
V2 =
which is the same as the electric potential
across the load resistor.
019 10.0 points
A toroid having a rectangular cross section
(a = 1.08 cm by b = 4.02 cm) and inner
E =−
d ΦB
.
dt
Magnetic field in a toroid
B=
µ0 N I
.
2πr
Solution: In a toroid, all the flux is confined
to the inside of the toroid
B=
µ0 N I
.
2πr
So, the flux through the loop of wire is
Z
ΦB1 =
B dA
Version 121 – Exam 3 – hoffmann – (57505)
10
Z b+R
µ0 N I 0
a dr
=
sin(ω t)
2π
r
R Let : ω = 588 rad/s ,
b+R
µ0 N I 0
.
a sin(ω t) ln
=
Vmax = 166 V , and
2π
R
C = 7.65 µF = 7.65 × 10−6 F .
Applying Faraday’s law, the induced emf can
The capacitive reactance is
be calculated as follows
d ΦB1
dt
µ0 N I 0
b+R
= −Nℓ
cos(ω t)
ω a ln
2π
R
= −E0 cos(ω t)
E = −Nℓ
where ω = 2πf was used.
The maximum magnitude of the induced
emf, E0 , is the coefficient in front of cos(ω t).
d ΦB1
E0 = −Nℓ
dt
ω
b+R
a ln
2π
R
= −(19 turns) µ0 (560 turns)
× (43.3 A) (70.5 Hz) (1.08 cm)
(4.02 cm) + (4.66 cm)
× ln
(4.66 cm)
= −0.274187 V
|E| = 0.274187 V .
= −Nℓ µ0 N I0
020 10.0 points
The generator in a purely capacitive AC circuit has an angular frequency of 588 rad/s.
If Vmax = 166 V and the capacitance is
7.65 µF, what is the rms current in the circuit?
1. 0.527997
2. 0.0537123
3. 0.159865
4. 0.135589
5. 0.424616
6. 0.457576
7. 0.250212
8. 0.295326
9. 0.106228
10. 0.323942
XC =
1
ωC
and the maximum voltage is
√
Vmax = 2 Vrms
Vmax
Vrms = √ .
2
Thus, the rms current in the circuit is
Vrms
ω C Vmax
√
=
XC
2
(588 rad/s) (7.65 × 10−6 F) (166 V)
√
=
2
= 0.527997 A .
Irms =
keywords:
021 (part 1 of 2) 10.0 points
A small-diameter solenoidal inductor with
self inductance 147 mH and a large-diameter
solenoidal inductor with self inductance
90 mH are connected in parallel as shown.
Consider the two inductors to be far apart so
that the mutual inductance between them is
0 mH.
Correct answer: 0.527997 A.
Explanation:
a
b
Version 121 – Exam 3 – hoffmann – (57505)
Determine the equivalent self-inductance
for the system. Assume the current enters
terminal a and exits terminal b.
1. 42.9524
2. 49.064
3. 40.1616
4. 45.5
5. 36.7725
6. 55.8228
7. 52.7788
8. 50.8704
9. 43.9362
10. 47.7273
Correct answer: 55.8228 mH.
Explanation:
Let : Ls = 147 mH and
Lℓ = 90 mH .
The source of the current through the smallerdiameter inductor Is and the larger-diameter
inductor Iℓ is from the same terminal. That
is, the total current I in this parallel inductor
complex is the sum of the currents in the two
inductors I = Is + Iℓ .
The emf E is the same across both inductors because they are connected in parallel.
The emf is related to the total current I by
E = −Leq
Leq = −
In general the total induced emf E across
an inductor is due to self inductance with
itself and mutual inductance with the other
inductor. But with M = 0 mH, we only have
self-inductance, so
E
d Is
=−
dt
Ls
d Iℓ
dt
E
d Iℓ
=− .
dt
Lℓ
I = Is + Iℓ
d Is d Iℓ
E
E
dI
=
+
=−
−
dt
dt
dt
Ls Lℓ
E (Ls + Lℓ )
=−
, so
Ls Lℓ
E
Ls Lℓ
=
dI
Ls + Lℓ
dt
(147 mH) (90 mH)
= 55.8228 mH .
=
147 mH + 90 mH
Leq = −
Note that parallel inductors combine like
parallel resistors (if no mutual inductance is
present).
022 (part 2 of 2) 10.0 points
The same two inductors are reconfigured so
that the smaller-diameter inductor is inside
the larger-diameter inductor; the inductors
overlap as shown below. Now, these two
solenoidal inductors have a mutual inductance of magnitude 38 mH .
dI
dt
E
.
dI
dt
E = −Ls
E = −Lℓ
11
d Is
dt
and
a
b
Find the equivalent self-inductance for the
system. As before, assume the current enters
terminal a and exits terminal b. The direction
of the magnetic field in the inductors must be
considered.
1. 33.7993
2. 35.7467
3. 69.5223
4. 23.3412
5. 60.4688
6. 29.1852
7. 46.4672
Version 121 – Exam 3 – hoffmann – (57505)
8. 37.655
9. 49.9692
10. 32.6866
To find
Correct answer: 37.655 mH.
Explanation:
The emf E across the parallel combination
is related to the total current I = Is + Iℓ by
E = −Leq
dI
dt
E
E
=−
.
dI
d Is d Iℓ
+
dt
dt
dt
From the figure, currents in both inductors
will produce magnetic fields (on the axis) that
are in opposite directions. As a consequence,
the partially induced emf in one of the inductors due to changing current in the other
inductor will have the opposite sign as the
partially induced emf from self inductance.
Since M =
6 0 mH, the total emf induced
across the smaller-diameter inductor is due to
d Is
the partial emf of the self inductance −L
dt
and the partial emf from mutual inductance
dI
+M ℓ , so
dt
Leq = −
d Is
d Iℓ
+M
dt
dt
d Iℓ
d Is
1
E + Ls
.
=+
dt
M
dt
Similarly, the emf across the largerdiameter inductor is
E = −Ls
d Iℓ
d Is
+M
dt
dt
d Iℓ
1
d Is
.
=−
E −M
dt
Lℓ
dt
Equating these,
E = −Lℓ
1
d Is
1
d Is
+
E + Ls
=−
E −M
M
dt
Lℓ
dt
d Is
d Is
= Lℓ E + Ls Lℓ
−M E + M2
dt
dt
E (Lℓ + M)
d Is
=−
.
dt
Ls Lℓ − M 2
12
d Iℓ
,
dt
d Is
d Iℓ
+M
dt
dt
d Is
d Iℓ
1
E −M
=−
dt
Ls
dt
E = −Ls
and
d Is
d Iℓ
+M
dt dt
d Is
d Iℓ
1
E + Lℓ
.
=+
dt
M
dt
E = −Lℓ
Equating these,
d Iℓ
1
d Iℓ
1
E + Lℓ
=−
E −M
+
M
dt
Ls
dt
d Iℓ
d Iℓ
= Ls E + Ls Lℓ
−M E + M2
dt
dt
E (Ls + M)
d Iℓ
=−
, so
dt
Ls Lℓ − M 2
d Is d Iℓ
dI
=
+
dt
dt
dt
E (Ls + Lℓ + 2 M)
=−
Ls Lℓ − M 2
and
Ls Lℓ − M 2
E
=
dI
Ls + Lℓ + 2 M
dt
(147 mH) (90 mH) − (38 mH)2
=
147 mH + 90 mH + 2 (38 mH)
Leq = −
= 37.655 mH .
Note: If the two inductors are wound like
this
a
b
Version 121 – Exam 3 – hoffmann – (57505)
the magnetic field they produce (on the axis)
are in the same direction, so the partial
d Iℓ
emf from mutual inductance is −M
, and
dt
d Iℓ
d Is
−M
and
E = −Ls
dt
dt
E = −Lℓ
13
iout ⊙
iin ⊗
O
d Iℓ
d Is
−M
.
dt
dt
F
E
Therefore, by either repeating (or simply
using the fact that +M ⇒ −M) in the previous calculations,
Ls Lℓ − M 2
Ls + Lℓ − 2 M
(147 mH) (90 mH) − (38 mH)2
=
147 mH + 90 mH − 2 (38 mH)
= 73.205 mH .
Leq =
Note: In the special case when the two inductors merge together (have the same radius,
same diameter, same length, same number of
windings, and occupy the same space) then
Ls = Lℓ = L = M and
Leq
L2 − M 2
(L − M) (L + M)
=
=
2 (L ∓ M)
2 (L ∓ M)
1
1
= L ± M = L or 0 ,
2
2
with the upper/lower sign indicating
same/opposite magnetic field directions from
each solenoid inside the smaller inductor. If
we have the same number of windings carrying a total current I, it is understandable that
Leq = L .
If Leq = 0, the currents are such that Is =
1
Iℓ = I, but the directions of the current in
2
each solenoid are opposite each other, so the
total magnetic field they both produce is zero,
as would be the case for zero inductance.
023 10.0 points
The figure below shows a straight 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
c
D
C
r1
r2
r3
r4
Which expression gives the magnitude
B(r2 ) E of the magnetic field in the region
c < r2 < b?
1. B(r2 ) =
2. B(r2 ) =
3. B(r2 ) =
4. B(r2 ) =
5. B(r2 ) =
6. B(r2 ) =
7. B(r2 ) =
8. B(r2 ) =
µ0 i r 2
2 π a2
µ0 i
correct
2 π r2
µ0 i (a2 + r22 − 2 b2 )
2 π r2 (a2 − b2 )
µ0 i (a2 − b2 )
2 π r2 (r22 − b2 )
µ0 i (a2 − r22 )
2 π r2 (a2 − b2 )
µ0 i r 2
2 π c2
µ0 i
π r2
µ0 i (r22 − b2 )
2 π r2 (a2 − b2 )
9. B(r2 ) = 0
10. B(r2 ) =
µ0 i r 2
2 π b2
Explanation:
Ampere’s
Law states that the line inteI
~ · d~ℓ around any closed path equals
gral
B
µ0 I, where I is the total steady current passing through any surface bounded by the closed
path.
Considering the symmetry of this problem,
Version 121 – Exam 3 – hoffmann – (57505)
we choose a circular path, so Ampere’s Law
simplifies to
B (2 π r1 ) = µ0 Ien ,
where r1 is the radius of the circle and Ien is
the current enclosed.
For c < r2 < b,
µ0 Ien
2 π r2
µ0 (i)
=
2 π r2
µ0 i
=
.
2 π r2
B=
14