Prof. Anchordoqui
Problems set # 10
Physics 169
April 28, 2015
1. A 12 V battery is connected into a series circuit containing a 10 Ω resistor and a 2 H inductor.
In what time interval will the current reach (i) 50% and (ii) 90% of its final value?
Solution This circuit has only one loop. From Kirchhoffs rule we get only one (differential)
dI
dt
equation ε − L dI
dt − IR = 0. We can separate the variables, ε−IR = L , and integrate both sides of
the equation (within appropriate limits). With the reference for time at the instant of closing the
R I(t) dI
Rt
circuit 0 ε−IR
= 0 dt
L . (Note that we used one symbol t for two different quantities!). From
the fundamental
theorem
of calculus ln ε−I(t)R
= −R
ε
L t. The solution is a time dependent function
Rt
−L
ε
I(t) = R 1 − e
. (i) The current reaches its final value after an infinite amount of time:
I(∞) = limt→inf ty Rε 1 − e−Rt/L = Rε . At instant t, the current is a fraction of its final value
I(t1 ) = I(∞) 1 − e−Rt/L . From which I(t)/I(∞) = 1 − e−Rt/L . Solving this equation for
L
2 H
t we obtain t = − R
ln [1 − I(t)/I(∞)]. Hence (i) t50% = − 10
Ω ln(1 − 0.5) = 0.14 s; (ii) t90% =
2 H
− 10 Ω ln(1 − 0.9) = 0.46 s.
2. In the circuit shown in Fig. 1, let L = 7.00 H, R = 9.00 Ω, and ε = 120 V. What is the
self-induced emf 0.200 s after the switch is closed?
Solution Duplicating the procedure of problem 1 we obtain I = Rε (1 − e−t/τ ). Since τ =
120 V
−0.26 ) = 3.05 A. Now ∆V
L/R = 0.78 s, we have I = 9.00
R = IR = 27.4 V and
Ω (1 − e
∆VL = ε − ∆VR = 92.6 V.
3. The switch in Fig. 2 is open for t < 0 and then closed at time t = 0. Find the current in the
inductor and the current in the switch as functions of time thereafter.
Solution Name the currents as shown in Fig. 2. Using Kirchhoff’s laws we obtain I1 = I2 + I3 ,
3
10.0 V − 4.00I1 − 4.00I2 = 0, and 10.0 V − 4.00I1 − 8.00I3 − 1.00 dI
dt = 0. From the first two
equations it follows that 10.0 V + 4.00I3 − 8.00I1 = 0 and I1 = 0.50I+ 3 + 1.25 A. Then the
3
last equation can be rewritten as 10.0 V − 4.00(0.500I3 + 1.25 A) − 8.00I3 − 1.00 H dI
dt = 0,
dI3
yielding 1 H dt + 10.0 ΩI3 = 5.00 V. We solve the differential equation to obtain I3 (t) =
5.00V
−10.0 Ωt/1.00 H = 0.50 A 1 − e−10t/s . Then I = 1.25+0.50I = 1.50 A−0.25 Ae−10t/s .
1
3
10.0 Ω 1 − e
4. Assume that the magnitude of the magnetic field outside a sphere of radius R is B =
B0 (R/r)2 , where B0 is a constant. Determine the total energy stored in the magnetic field outside
the sphere and evaluate your result for B0 = 5.00×10−5 T and R = 6.00×106 m, values appropriate
for the Earth’s magnetic field.
Solution The total magnetic energy is the volume integral of the energy density, u =
cause B changes with position, u is not constant.For B = B0 (R/r)2 , we have u =
B2
2µ0 .
B02
2µ0
Be-
(R/r)4 .
Next, we set up an expression for the magnetic energy in a spherical shell of radius r and thickness
2πB02 R4 dr
dr. Such a shell has a volume 4πr2 dr, so the energy stored in it is dU = u4πr2 dr =
.
µ0
r2
We integrate this expression for r = R to r = ∞ to obtain the total magnetic energy outside the
2πB 2 R3
sphere. This gives U = µ00 = 2.70 × 1018 J.
5. A large coil of radius R1 and having N1 turns is coaxial with a small coil of radius R2 and
having N2 turns. The centers of the coils are separated by a distance x that is much larger than R1
and R2 . What is the mutual inductance of the coils? Suggestion: John von Neumann proved that
the same answer must result from considering the flux through the first coil of the magnetic field
produced by the second coil, or from considering the flux through the second coil of the magnetic
field produced by the first coil. In this problem it is easy to calculate the flux through the small
coil, but it is difficult to calculate the flux through the large coil, because to do so you would have
to know the magnetic field away from the axis.
N µ I R2
0 1 1
Solution The large coil produces this field at the center of the small coil: 2(x12 +R
2 3/2 . The field
1)
is normal to the area of the small coil and nearly uniform over this area, so it produces a flux
N µ0 I1 R12
2
Φ12 = 2(x12 +R
2 )3/2 πR2 through the face area of the small coil. When the current I1 varies, this is
1
N µ R2 πR2
d 1 0 1
2
the emf induced in the small coil: ε2 = −N2 dt
I =−
2(x2 +R2 )3/2 1
M=
1
N1 N2 µ0 R12 πR22
.
2(x2 +R12 )3/2
N1 N2 µ0 R12 πR22 dI1
2(x2 +R12 )3/2 dt
1
= −M dI
dt , hence
6. Two inductors having self-inductances L1 and L2 are connected in parallel as shown in
Fig. 3(a). The mutual inductance between the two inductors is M . Determine the equivalent selfinductance Leq for the system shown in Fig. 3(b).
dI2
dI2
1
Solution With I = I1 + I2 , the voltage across the pair is: ∆V = −L1 dI
dt − M dt = −L2 dt −
2
dI1
dI2
dI2
dI
∆V
M dI2
M ∆V
1
+M
M dI
dt = −Leq dt . Hence, − dt = L1 + L1 dt and −L2 dt + L1
L1 dt = ∆V , yiedling
(−L1 L2 + M 2 )
2
By substitution, − dI
dt =
∆V
L2
+
M dI1
L2 dt
dI2
= ∆V (L1 − M ).
dt
(1)
dI1
= ∆V (L2 − M ).
dt
(2)
leads to
(−L1 L2 + M 2 )
∆V
Adding (1) to (2), (−L1 L2 +M 2 ) dI
dt = ∆V (L1 +L2 −2M ), and therefore Leq = − dI/dt =
L1 L2 −M 2
L1 +L2 −2M .
7. A long solenoid, which has n = 400 turns per meter, carries a current given by I =
30 A(1 − e−1.6t ). Inside the solenoid and coaxial with is a coil that has a radius of 6 cm and
consists of a total of N = 250 turns of fine wire. (i) What emf is induced in the coil by the changing
current? (ii) As indicated in Fig. 4, the current in the solenoid flows in the clockwise direction,
what is the direction of the current induced in the coil, clockwise or anticlockwise? Explain how
you arrived at your answer (iii) Consider now the situation in which an inductor is discharging, i.e.,
I = 30 Ae−1.6t , the direction of the current is still clockwise. What is the direction of the current
induced in the coil now (clockwise or anticlockwise)? Explain how you arrived at your answer.
Solution A long solenoid produces a uniform magnetic field. The strength B of the field depends
on the number of B turns per unit length n and the current in the solenoid I, B = µ0 nI, where
µ0 is the permeability of free space. Therefore the flux through the surface spanned by the coil
(in the direction of the field vector) is equal to the product of the magnetic field strength and the
R
R
◦
2
~ · dA
~ =
area of the coil ΦB = surface B
surface BdA cos 0 = BA = µ0 nIπR . From the Faraday’s
2 dI
B
law of induction, the electromotive force induced in the coil is ε = −N dΦ
dt = −N µ0 nπR dt =
1 −1.6t
N
2
2
−αt
−7
= −0.068 V · e−1.6t .
−µ0 N nπr I0 αe
= −4π10 A2 · 250 · 400 · π(0.06 m) · 30 A1.6 s e
(ii) The magnetic field through the coil is to the left and the magnetic flux is increasing, we need
to induce a current that oposes to the increase of the flux (Lenz law). The induced current is then
counterclockwise producing a magnetic field to the right. (iii) The magnetic field on the coil is to
the left as before, but the magnetic flux is now decreasing. The induced emf in the coil is now
0.068 V and the current is clockwise producing a magnetic field to the left.
8. In the circuit of Fig. 5, the battery emf is 50.0 V, the resistance is 250 Ω, and the capacitance
is 0.500 µF. The switch S is closed for a long time and no voltage is measured across the capacitor.
After the switch is opened, the potential difference across the capacitor reaches a maximum value
of 150 V. What is the value of the inductance?
Solution When the switch has been closed for a long time, battery, resistor, and coil carry
constant current Imax = ε/R. When the switch is opened, the current in battery and resistor
drops to zero, but the coil carries this same current for a moment as oscillations begin in the LC
loop. We interpret the problem to mean that the voltage amplitude of these oscillations is ∆V ,
−6 F(150 V2 )(250 Ω)2
2 2
C(∆V )2
1
1
2
2
= C(∆Vε2) R = 0.5×10 (50.0
= 0.281 H.
2 C(∆V ) = 2 LImax . Thus, L = I 2
V)2
max
9. An inductor consists of two very thin conducting cylindrical shells, one of radius a and one of
radius b, both of length h. Assume that the inner shell carries current I out of the page, and that
the outer shell carries current I into the page, distributed uniformly around the circumference in
both cases. The z-axis is out of the page along the common axis of the cylinders and the r-axis is
the radial cylindrical axis perpendicular to the z-axis. (i) Use Ampere’s law to find the magnetic
field between the cylindrical shells. Indicate the direction of the magnetic field on the sketch. What
is the magnetic energy density as a function of r for a < r < b? (ii) Calculate the inductance of this
long inductor recalling that U = 12 LI 2 and using your results for the magnetic energy density in (i).
R
~ · dA
~ and
(ii) Calculate the inductance of this long inductor by using the formula Φ = LI = open B
surface
your results for the magnetic field in (i). To do this you must choose an appropriate open surface
over which to evaluate the magnetic flux. Does your result calculated in this way agree with your
result in (ii)?
Solution (i) The enclosed current Ienc in the Ampere’s loop with radius r is given by


r<a
 0
Ienc =
I a<r<b .

 0
r>b
Applying Ampere’s law,
~ =
B
H
~ · d~s = B(2πr) = µ0 Ienc , we obtain
B





0
µ0 I
2πr φ̂
0
r<a
a < r < b (counterclockwise in the figure) .
r>b
The magnetic energy density for a < r < b is uB =
B2
2µ0
=
1
2µ0
µ0 I
2πr
2
=
µ0 I 2
.
8π 2 r2
It is zero elsewhere.
t
(ii) The volume element in this case is 2πrhdr. The magnetic energy is : UB =
V uB dV =
R b µ0 I 2
µ0 I 2 h
µ0 h
1
2
4π ln(b/a). Since UB = 2 LI , the inductance is L = 2π ln(b/a). (iii) The
a 8π 2 r2 2πhrdr =
magnetic field is perpendicular to a rectangular surface shown Fig. 6. The magnetic flux through
0I
0 Ih
a thin strip of area dA = hdr is dΦ = BdA = µ2πr
hdr = µ2πr
dr. Thus, the total magnetic flux is
R
R b µ0 Ih
µ0 Ih
0h
ΦB = dΦB = a 2πr dr = 2π ln(b/a). Finally, the inductance is L = ΦIB = µ2π
ln(b/a), which
agrees with that obtained in (ii).
10. The energy of an RLC circuit decreases by 1.00% during each oscillation when R = 2.00 Ω.
If this resistance is removed, the resulting LC circuit oscillates at a frequency of 1.00 kHz. Find
the values of the inductance and the capacitance.
Solution The period of damped oscillation is T =
RT
2π
ωd .
turning to the capacitor is Q = Qmax e− 2L = Qmax e
charge squared,so after one oscillation it is U = U0 e
2π2Ω
Lωd
After one oscillation the charge re-
2πR
− 2Lω
d
2π
− Lω
d
= ln(1.0101) = 0.001005. It follows that Lωd =
. The energy is proportional to the
2πR
. Then e Lωd =
2π2Ω
0.001005
1
0.99 ,
which leads to
1/2
R2
1
,
= 1250Ω = L LC
− 4L
2
2
L
L
yielding 1.563 × 106 Ω2 = C
− (2Ω)
or equivalently C
= 1.563 × 106 Ω2 . We are also given
4
1
1
3
ω = 2π × 10 /s = √LC , which leads to LC = (2π×103 /s)2 = 2.533 × 10−8 s2 . Solving simulta-
neously, C = 2.533 × 10−8 s2 /L, yields
C=
2.533×10−8
0.199 H
s2
= 127 nF.
L2
2.533×10−8 s2
= 1.563 × 106 Ω2 ; therefore L = 0.199 H and
perconducting
solenoid
42. Onshown
a printed
circuit
board,leta relatively
long straight conFor theisRL circuit
in Figure
P32.17,
the
19.
be 3.00 H,ductor
the resistance
8.00 #, and
the
r diameter ofinductance
6.20 cm and
and a conducting
rectangular
loop lie in the same
S
a
battery energy
emf 36.0 V. (a) plane,
Calculate
ratioinofFigure
the potential
(a) the magnetic
as the
shown
P31.9. Taking h ! 0.400 mm,
difference
across the resistor
to mm,
that across
w ! 1.30
and the
L !inductor
2.70 mm, find their mutual
he energy stored
in the
Chapter 32
245b
when the current is 1026
2.00inductance.
A. (b)
Calculate
the
voltage
d.
C H A P T E R 3 2 • Inductance across
the inductor when the current is 4.50 A.
ns is 8.00 cm long3 and has
43. Two solenoids A and B, spaced close to each other and
320.
.00 A1012.0-V battery is connected in series with a resistor and
2.0
1 energy
e
uch
is stored in its
sharing the same cylindrical axis, have 400 12.0
andV700 turns,
I (t )
an inductor.
The circuitrespectively.
has a time constant
of of
500
!Ss, Aand
1 200 Ω
A current
3.50
in coil A produces an averurrent of 0.770
A?
10.0 A
the maximum current age
is 200
mA.
thethrough
value ofeach
the turn of A and a flux of
flux
of What
300 &isWb
3
plied to
a coil
that
has an
3.00
10
200 µ s
inductance?
90.0 &Wb through each turn of B. (a) Calculate the
e of 30.0 %. (a) Find
istance
21. An inductor that has an
inductance
of 15.0of
H the
and two
a resismutual
inductance
solenoids. (b) What is the
tic field when
the current
3 tance
of
30.0
#
is
connected
across
a
100-V
battery.
What
is is induced in B when
12.0 Ω I (t )
self-inductance
of
A?
(c)
What
emf
00 (b)10Afterthethe
ue.
emf is
4 of the current ε(a) at t " 0 and (b) at
L
rate
of
increase
Figure
P32.26
.67 this10 s the current in A increases at the rate of 0.500 A/s?
e the current to 7
reach
100 Ω
n 0.020 0 t " 1.50 s?
44. A large coil of radius R 1 and having N 1 turns is coaxial
22. When the switch in Figure
P32.17
is closed,
theradius
currentRtakes
with
a small
coil of
2 and having N 2 turns. The
location, a 100-V/m
vertiA 140-mH
inductor and a 4.90-# resistor a
27.
3.00 ms to reach 98.0% centers
of its final
value.
If
R
"
10.0
#,
what
FIG.
P32.22
of the coils are separated by a distance
x that
is
e Earth’s surface.
At
the
4
nected
with
a
switch
to a 6.00-V battery as shownFiin
is the inductance?
7.67
10of
.0
7.much
67 mH
larger
R 1 Problems
and R 2. What
the
mutual
inducFigurethan
P32.17
17, 18,is19,
and
22.
c field
has a 10
magnitude
P32.27. (a) If the switch is thrown to the left (con
23. The switch in Figure P32.23
is the
opencoils?
forFigure
tSuggestion:
% 0 1:and
then
Problem
tance of
John 2.
von Neumann
proved
nergy densities of the two
the battery),
how much
before
26.time
Oneelapses
application
of the
an
closed at time t " 0. Find
in the must
inductor
and
thatthe
thecurrent
same answer
result
from considering
the
flux (b) What
reaches 220 mA?
is the current in the i
the current in the switch
as functions
of time
thereafter.
high voltage fr
the250
first
theswitch
magnetic
field
produced
byswitch is varying
10.0
s What
after is
the
closed? (c) Now the s
the circuit
!s coil
afterofthe
is closed.
(c)
the
mple
by proving that
n.
By32.4
Kirchhoff’s
laws: through
Figure
P32.26.
(a) Wh
the
second
orsteady-state
from considering
the(d)
flux
through
the
quickly
thrown
from a to b. How much time elapse
value
of thecoil,
final
current?
How
long
does
second
coilcurrent
of the to
magnetic
field produced
by thevalue?
firstto 160 mA?time after the switch h
current
falls
L
it take the
reach 80.0%
of its the
maximum
switch is thrown quick
4.00 Ωcoil. In this
8.00problem
Ω
!
(1)it is easy to calculate the flux through
2R
18. the
In the
circuit
shown
in
Figure
P32.17,
let
L
"
7.00
H,
voltage across each
small coil, but it is difficult to calculate the flux
R
"
9.00
#,
and
$
"
120
V.
What
is
the
self-induced
emf
(c) How much time e
H and R ! 5.00 % is conthrough the large coil, because to do so you would have to
a
b
0.200
s
after
the
switch
is
closed?
10.0
V
1.00
H
(2)
4.00
Ω
inductor
drops to 12.0
! 0. (a) What energy is
know the magnetic field away from the axis.
For
the
RL
circuit
shown
in
Figure
P32.17,
let
the
19.
current is 0.500 A? (b) At
S
S45. Two inductors having self-inductances L 1 and L 2 are coninductance
be
3.00
H,
the
resistance
8.00
#,
and
the
ed in dI
the inductor when
nected in parallel as shown in Figure P32.45a. The mutual
battery
emfbetween
36.0 V.
the ratio
the potential
s.00
being 3delivered
Figure
P32.23
0 to the
(3)(a)
inductance
the Calculate
two inductors
is M. of
Determine
the
L
ε
difference
across
the
resistor
to
that
across
the
inductor
.500 A?dt
equivalent self-inductance Leq for the system (Figure
when the current is 2.00 A. (b) Calculate the voltage across
and a 10.0-H inductor are
P32.45b).
Figure
2: Problem
3. A.
when
the current
is 4.50
A series
circuit
with
L "inductor
a series
RC circuit
I 1the
.00RL
I 1has
I 3the
0current
4.00in24.
4circuit
4.00
03.00 H and
FIG. P32.23
R
C "being
3.00 !F have
constants.
If the two
ulate (a) thewith
power
20. Aequal
12.0-Vtime
battery
is connected
in series with a resistor and
12.0 V
circuits to
contain
resistance
(a) what
the constant
the the same
an inductor.
hasisa time
I (tThe
) R , circuit
I (t ) of 500 !s, and
0ower
.vered
500being
Ito3thedelivered
1value
.
25
A
.
of
R
and
(b)
what
is
the
time
constant?
inductor, and
the maximum
is 200
L1 current
L2 mA. What is
Leqthe value of the Figure P32.27
M
25.ofAthe
current
pulse is fed toinductance?
the partial circuit shown in Figure
netic field
inductor.
dI 310.0 A of 15.0 H and a resisP32.25.
The current21.
begins
at zero, that
then has
becomes
nitude
throughAn 8
inductor
an inductance
V 4680
.00kV/m
0between
.500
I 3t " 01and
.25t "
A200
.
00
I
1
.
00
0 28. Consider two ideal inductors L 1 and L 2 that have z
!s, and 3then is zero once again.
n a total energy of 3.40 &J.
tance of 30.0
# is connected
across a 100-V battery. What is
dt
nal resistance and are far apart, so that their magne
Determine the current in the inductor
of
(a) as aoffunction
me region stores the same
the rate of increase
the current (a) at(b)
t " 0 and (b) at
do not influence each other. (a) Assuming theseFiin
time.
Figure P32.45
t
"
1.50
s?
Problems
993
dI 3
Figure
3: P32.17
Problem
6.
H
10.0 I 3 22.5When
.00 Vthe. switch in
Figure
is closed,
the current takes
dt
A 140-mH induc
27.
3.00 ms to reach 98.0% of its final value. If R "I 10.0 #, what
I
n turns/m
nected with a switch to
is the inductance?
P32.27. (a) If the switc
h
uation using Equations
and in
32.7:
× Figure
× × × × × ×P32.23
× × × × × is
× ×open
×××××
× × t× % 0 and then
23. 32.6
The switch
for
the battery), how mu
closed at time t " 0. Find the current
in the inductor and
R
reaches 220 mA? (b)
5.00 V w
the current
in the switch as functions
10.0
t 1.00
H
10 t of
s time thereafter.
10.0 s after the switch
1 e
0.500 A 1 e
quickly thrown from a
10.0
the current falls to 160
L
8.00 Ω
104.00
ts Ω
1Problems
.25 09.and
50071.I 3
1.50 A 0.250 A e N turns
b
g
ja f
e
f
b
fFGH IJK a
FG
H
IJ
K
a f
g
f
a
f
a
a
f
f
Figure P31.13
Figure
4: Problem
7.
10.0
V
H used to
14. An
instrument
based
been
4.00 on
Ω induced emf has1.00
adius 10.0 cm surrounds a long
measure
projectile
speeds
up
to
6
km/s.
A
small
magnet is
3
cm and 1.00 ! 103 L
turns/meter 3.00 H
S
we
get
1
.
00
10
1
.
00
k
.
R
imbedded in the projectile, as shown in Figure P31.14.
ent in the solenoid changes as
6
C
The F
projectile passes through two coils separated by a
3.00
10
ind the induced emf in the
P32.23
distance d. As theFigure
projectile
passes through each coil a
of time.
ε
ndrical
gth h.
page,
h
n axis
axis
etween
agnetic
sity as
calling that U B
the resistance is 250 ", and It
theiscapacitance
is 0.500 !F.
zero elsewhere.
what is the
The switch S is closed for a long time and no voltage is
is the perce
measured across the capacitor. After the switch is opened,
(b) the capacitor reaches
the potential difference across
56. Show that
rhd
volume
this caserule
is 2as app
a maximum value of 150 V.The
What
is the element
value of in
the
inductance?
57. The energ
each boscil
UB
uB dVol
removed, t
R
8
V
a kH
of 1.00
I 2l
b
1 capacitanc
Since U B
ε
L
0
C
4
ln
2
LI Electrical
, the indu
58.
o
2 taining a c
(a) If R %
elapsesLbef
off to 50.0%
the energy
a
S
Figure P32.48
Figure 5: Problem 8.
49. A fixed inductance L # 1.05 !H is used in series with a
variable capacitor in the tuning
(c) section of a radiotelephone on a ship. What capacitance tunes the circuit to the
signal from a transmitter broadcasting at 6.30 MHz?
50. Calculate the inductance of an LC circuit that oscillates at
120 Hz when the capacitance is 8.00 !F.
51. An LC circuit like the one in Figure 32.16 contains an
82.0-mH inductor and a 17.0-!F capacitor that initially carries a 180-!C charge. The switch is open for t % 0 and
then closed at t # 0. (a) Find the frequency (in hertz) of
the resulting oscillations. At t # 1.00 ms, find (b) the
charge on the capacitor and (c) the current in the circuit.
52. The switch in Figure P32.52 is connected to point a for
a long time. After the switch is thrown to point b, what
are (a) the frequency of oscillation of the LC circuit,
(b) the maximum charge that appears on the capacitor,
(c) the maximum current in the inductor, and (d) the
total energy the circuit possesses at t # 3.00 s?
10.0 Ω
a
b
0.100 H
S
12.0 V
1.00 µ
µF
which
Figure 6: Problem
9.
Figure P32.52
1 2
LI and using
2
An LC circuit
53.
agrees with that
Additional Pro
mag
59.The
Review
pro
surface
sh
Section 26
32.3.strip
(a) C
thin
o
large, clos
Show that
of
dthinking
Bd
B
a “negative
electric fie
ing electric
Thus,
the
ear curren
on one sh
other shee
the sheets
B
them. (d)
field betwe
sheet can
Thus,
the
field betwe
equal to
pressure ap
sheets of cu
60. Initially, th
obtained switch
in (b). is
discharge,
is one four
is known.
like that in Figure 32.16 consists of a
61. A 1.00-mH