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30.1: a)
V,0.270/s)A830(H)1025.3()/(
4
12
dtdiM
and is constant.
b) If the second coil has the same changing current, then the induced voltage is the
same and
V.270.0
1
30.2: For a toroidal solenoid,
.2/and,/
110B12
22
rAiNiNM
B
So,
.2/
210
rNANM
30.3: a)
H.1.96A)(6.52/Wb)0320.0()400(/
12
2
iNM
B
b) When
Wb.107.11(700)/H)(1.96A)54.2(/A,54.2
3
12B2
1
NMii
30.4: a)
H.106.82s)/A0.242(/V1065.1)/(/
33
2
dtdiM
b)
A,20.1,25
12
iN
Wb.103.27
25/H)10(6.82A)20.1(/
4
3
21
2
NMi
B
c)
mV.2.45s)/A(0.360H)1082.6(/ands/A360.0/
3
212
dtMdidtdi
30.5:
Ωs.1A)s/(V1AC)s/(J1A/J1A/Nm1A/Tm1A/Wb1H1
222
30.6: For a toroidal solenoid,
)./(// dtdiiNL
B
So solving for
N
we have:
turns.238
s)/A(0.0260Wb)(0.00285
A)(1.40V)106.12(
)/(/
3
dtdiiN
B
30.7: a)
V.104.68s)/A(0.0180H)260.0()/(
3
1
dtdiL
b) Terminal
a
is at a higher potential since the coil pushes current through from
b
to
a
and if replaced by a battery it would have the
terminal at
.
a
30.8: a)
H.130.0
m)120.0(2
)m1080.4()1800()500(
2/
252
0
2
0m
m
π
μ
πr
ANμK
K
L
b) Without the material,
H.102.60H)130.0(
500
11
4
m
m
K
L
K
L
30.9:
For a long, straight solenoid:
.//and/
2
00
lANμLlNiAμiNL
BB
30.10: a) Note that points
a
and
b
are reversed from that of figure 30.6. Thus, according
to Equation 30.8,
s./A00.4
H0.260
V04.1
L
VV
dt
di
ab
Thus, the current is decreasing.
b) From above we have that
.)s/A00.4( dtdi
After integrating both sides of this
expression with respect to
,
t
we obtain
A.4.00s)(2.00A/s)(4.00A)0.12(s)/A00.4(
iti
30.11: a)
H.0.250A/s)(0.0640/V)0160.0()/(/
dtdiL
b)
Wb.104.50(400)/H)(0.250A)720.0(/
4
NiL
B
30.12: a)
J.540.02/A)(0.300H)0.12(
2
1
22
LIU
b)
W.2.16)180(A)300.0(
22
RIP
c) No. Magnetic energy and thermal energy are independent. As long as the current is
constant,
constant.
U
30.13:
πr
AlNμ
LIU
4
2
1
22
0
2
turns.2850
A)0.12()m1000.5(
J)(0.390m)150.0(44
224
0
2
0
μ
π
AIμ
πrU
N
30.14: a)
J.101.73h)/s3600h/day(24W)200(
7
PtU
b)
H.5406
A)(80.0
J)1073.1(2
2
2
1
2
7
2
2
I
U
LLIU
30.15:
Starting with Eq. (30.9), follow exactly the same steps as in the text except that
the magnetic permeability
is used in place of
.
0
30.16: a) free space:
J.3619)m0290.0(
2
T)(0.560
V
2
V
3
0
2
0
2
B
uU
b) material with
J.04.8)m0290.0(
)450(2
T)(0.560
V
2
V450
3
0
2
0m
2
m
K
B
uUK
30.17: a)
3
2
6
0
2
0
0
2
m1.25
T)(0.600
J)1060.3(22
Volume
2
B
U
B
Vol
U
u
.
b)
T.9.11T4.141
m)(0.400
J)1060.3(22
2
3
6
00
2
B
Vol
U
B
30.18: a)
.mT35.4
)m0690.0(2
)A50.2()600(
2
00
r
NI
B
b) From Eq. (30.10),
.m/J53.7
2
)T1035.4(
2
3
0
23
0
2
B
u
c) Volume
.m1052.1)m1050.3()m0690.0(22V
3626
rA
d)
.J1014.1)m1052.1()m/J53.7(
5363
uVU
e)
.H1065.3
)m0690.0(2
)m1050.3()600(
2
6
262
0
2
0
r
AN
L
J1014.1)A50.2()H1065.3(
2
1
2
1
5262
LIU
same as (d).
30.19: a)
.s/A40.2
H50.2
V00.6
0When.
dt
di
i
L
iR
dt
di
b) When
.A/s800.0
H50.2
)00.8()A500.0(V00.6
A00.1
dt
di
i
c) At
A.413.0)1(
8.00
V00.6
)1(s200.0
s)(0.250H)50.2/00.8()/(
ee
R
ε
it
tLR
d) As
A.750.0
8.00
V00.6
R
it
30.20: (a)
mA,30A030.0
1000
V30
max
i
long after closing the switch.
)b
V4.0V26V30ε
V26A)(0.0259Ω)(1000
A0.0259
e1A0.030)e(1
RBatteryL
R
)(
max
10
20
VV
RiV
ii
μs
μs
R/L/t
(or, could use
s)20at
tLV
dt
di
L
c)
30.21: a)
RLeRi
t
/),1(/
/
2
1
2
1
maxmax
and,)(1when2/so/
τ/tτ/t
eeiiRεi
μs
L
t
τ/t 3.17
50.0
H)10(1.252)ln(
R
ln2
and)(ln
3
2
1
b)
max
2
2
1
max
2
2
1
; LiULiU
2/when
maxmax
2
1
iiUU
2929.02/11eso211
//
τtτt
e
μsLt 30.7R/ln(0.2929)
30.22: a)
A13.2
H0.115
J)260.0(2
2
2
1
2
L
U
ILIU
V.256)(120A)13.2(
IR
b)
2
0
)/(222)/(
2
1
2
1
2
1
2
1
2
1
and
LIUeLiLiUIei
tLRtLR
s.1032.3
2
1
ln
)2(120
H115.0
2
1
ln
2
2
1
4
)/(2
R
L
t
e
tLR
30.23: a)
A.250.0
240
V60
0
R
I
b)
A.137.0A)250.0(
)s10(4.00H)0.160/(240)/(
0
4
eeIi
tLR
c)
,V9.32)240()A137.0( iRVV
abcb
and c is at the higher potential.
d)
.s1062.4
2
1
ln
)240(
)H160.0(
2
1
ln
2
1
4
2/1
)/(
0
2/1
R
L
te
I
i
tLR
30.24: a)
.V60and00At
bcab
vvt
b)
.0andV60As
bcab
vvt
c)
.V0.24V0.36V0.60andV0.36A150.0When
bcab
viRvi
30.25: a)
)1(
00.8
)V00.6(
)1()1(
)H50.2/00.8(
2
)/(
2
)/(
0
ttLRtLR
ee
R
eIiP
).1()W50.4(
)s20.3(
1
t
eP
b)
2)H50.2/00.8(
2
2)/(
2
2
)1(
00.8
)V00.6(
)1(
ttLR
R
ee
R
ε
RiP
.)1()W50.4(
2)s20.3(
1
t
R
eP
c)
)()1(
)/(2)/(
2
)/()/(
tLRtLRtLRtLR
L
ee
R
ε
e
L
ε
Le
R
ε
dt
di
iLP
).()W50.4(
)s40.6()s20.3(
11
tt
L
eeP
d) Note that if we expand the exponential in part (b), then parts (b) and (c) add to
give part (a), and the total power delivered is dissipated in the resistor and inductor.
30.26:
When switch 1 is closed and switch 2 is open:
.)/(ln
0
)/(
00
0
0
LRt
i
I
t
eIit
L
R
Ii
td
L
R
i
id
L
R
i
dt
di
iR
dt
di
L
30.27: Units of
s/)s(/H/ RL
units of time.
30.28: a)
πf
LC
ω 2
1
.H1037.2
)F1018.4()106.1(4
1
4
1
3
1226222
Cf
L
b)
F.1067.3
)H1037.2()1040.5(4
1
4
1
11
3252
min
22
max
Lf
C
30.29: a)
)F1000.6()H50.1(22
2
5
πLCπ
ω
π
T
s.rad105,s0596.0 ω
b)
.C1020.7)V0.12)(F1000.6(
45
CVQ
c)
.J1032.4)V0.12)(F1000.6(
2
1
2
1
3252
0
CVU
d)
.0)cos(,0At
ωtQQqt
)F1000.6)(H50.1(
s0230.0
cos)C1020.7()cos(s,0230.0
5
4
ωtQqt
C.1043.5
4
Signs on plates are opposite to those at
.0
t
e)
)sin(,s0230.0 ωtωQ
dt
dq
it
A.0.0499
H)10(6.00H)(1.50
s0.0230
sin
H)10H)(6.00(1.50
C107.20
55
4
i
Positive charge flowing away from plate which had positive charge at
.0
t
f) Capacitor:
.J102.46
F)102(6.00
C)10(5.43
2
3
5
242
C
q
U
C
Inductor:
.J101.87A)(0.0499H)50.1(
2
1
2
1
322
LiU
L
30.30: (a) Energy conservation says (max)=(max)
CL
UU
A0.871
H1012
F1018
V)22.5(
CV
2
1
2
1
3
6
max
22
max
LCVi
Li
The charge on the capacitor is zero because all the energy is in the inductor.
(b)
LCT
2
2
at
41
period:
F)10(18H)10(12
2
)2(
4
1
4
1
63
LCT
s1030.7
4
at 43 period:
s1019.2)s1030.7(3
4
3
34
T
(c) CCVq
405)V(22.5F)18(
0
30.31:
F0.30
V
10
29
.
4
C10150
3
9
V
Q
C
For an L-C circuit,
LCω 1
and
LCωπT
22
mH601.0
)2(
2
C
T
L
30.32:
rad/s1917
)F1020.3()H0850.0(
1
6
ω
a)
C1043.4
srad1917
A1050.8
7
4
max
maxmaxmax
ω
i
Q
ωQi
b) From Eq. 31.26
2
1
4
2722
s1917
A1000.5
)C1043.4(
LCiQq
.C1058.3
7
30.33: a)
)sA80.2()F1060.3()H640.0(0
1
6
2
2
dt
di
LCqq
LC
dt
qd
C.1045.6
6
b)
.V36.2
F
10
60
.
3
C1050.8
6
6
C
q
30.34: a)
.
max
max
maxmaxmax
LCi
ω
i
Q
ωQi
J450.0
)F1050.2(2
)C1050.1(
2
.C1050.1)F1050.2()H400.0()A50.1(
10
25
max
2
max
510
max
C
Q
U
Q
b)
14
10
s1018.3
)F1050.2()H400.0(
11
2
2
2
LC
f
(must double the frequency since it takes the required value twice per period).
30.35:
.ssA
A
1
s
s
C
V
Ω
V
C
s
Ω
V
C
HFH][
222
sLCLC
30.36: Equation (30.20) is
.0
1
2
2
q
LC
dt
qd
We will solve the equation using:
.
11
0)cos()cos(
1
).cos()sin()(cos
22
2
2
2
2
2
LC
ω
LC
ωt
LC
Q
ωtQωq
LCdt
qd
ωtQω
dt
qd
ωtωQ
dt
dq
ωtQq
30.37: a)
.
)(cos
2
1
2
1
222
C
ωtQ
C
q
U
C
.
1
since,
)(sin
2
1
)(sin
2
1
2
1
2
22
2222
LC
ω
C
ωtQ
ωtQLωLiU
L
b)
)(sin
2
1
)(cos
2
1
2222
2
Total
ωtQLωωt
C
Q
UUU
LC
)(sin
1
2
1
)(cos
2
1
222
2
ωtQ
LC
L
ωt
C
Q
))(sin)((cos
2
1
22
2
ωtωt
C
Q
Total
2
2
1
U
C
Q
is a constant.
30.38: a)
)cos(
)2/(
tωAeq
tLR
)sin(
2
2)cos(
2
).sin()cos(
2
)2/()2/(
2
2
2
)2/()2/(
tωe
L
R
A
ωtωe
L
R
A
dt
qd
t
ωAeωtωe
L
R
A
dt
dq
tLRtLR
tLRtLR
).cos(
)2/(2
tωAeω
tLR
2
2
2
2
2
2
2
2
2
4
1
0
1
22
L
R
LC
ω
LCL
R
L
R
q
LC
q
dt
dq
L
R
dt
qd
b)
:0,,0
dt
dq
iQqtAt
.
4//12
2
tan
2
and
cos
0sincos
2
andcos
22
LRLCL
R
ωL
R
Q
ω
L
QRQ
A
A
ωA
L
R
dt
dq
QAq
[...]... kx d 2 q R dq q 0 Eq. (30. 27) : 2 0 a) Eq (13.41): 2 dt m dt m dt L dt LC 30. 39: Subbing x q, m L, b R, k k 1 b2 R2 Eq. (30. 28) : ω 2 m 4m 2 LC 4 L ( b / 2 m )t c) Eq (13.42): x Ae cos(ωt ) Eq. (30. 28) : q Ae ( R / 2 L )t cos(ωt ) b) Eq (13.43): ω L L H s V 2 30. 40: A C F C V C 30. 41: 2 1 1 R2 1 1 1 ... a 2π a 30. 45: a) 2 30. 46: a) M b) ε 2 N 2 μ NN A μ N N πr N2 N A N A μ N IA B2 2 2 B1 2 2 0 1 1 0 1 2 2 0 1 2 2 I I A1 IA1 l1 l1 l1 d B2 dt 2 N2 μ0 N1 A2 di1 μ0 N1 N 2 π r2 di1 l1 l1 dt dt 2 μ N N π r di2 di di c) ε1 M 12 2 M 2 0 1 2 2 dt l1 dt dt 30. 47: a) ε L di L ε /(di / dt ) (30. 0 V) /( 4.00 A / s ) 7.5 H dt d f i εt f (30. 0 V)(12.0... the inductance for any height L L0 1 m D Height of Fluid Inductance of Liquid Oxygen Inductance of Mercury dD 4 0. 6302 4 H 0. 6300 0 H dD 2 0. 6304 8 H 0.62999 H d 3D 4 0. 6307 2 H 0.62999 H dD 0. 6309 6 H 0.62998 H Where are used the values m (O 2 ) 1.52 10 3 and m (Hg) 2.9 10 5 d) The volume gauge is much better... μ 1 2 2U Li L 2 l 0 ln (b / a), which is the same as in Problem 30. 50 2 i 2π 30. 52: a) L1 L2 N1 B1 i1 N 2 B2 i2 2 N1 A μ0 N1i1 μ0 N1 A , 2π r i1 2π r 2 N 2 A μ 0 N 2 i2 μ 0 N 2 A i2 2π r 2π r 2 2 2 μ N N A μ N A μ0 N 2 A b) M 0 1 2 0 1 L1 L2 2π r 2π r 2π r 2 30. 53: u B u E ε0 E 2 B2 B ε 0 μ0 E 2 2 2 μ0 ε0 μ0 (650... i f (1 e ( R / L )t ) ln (1 i / i f ) L L ln (1 i / i f ) 30. 54: a) R L (1860 )(7.25 10 4 s) 0.963 H ln (1 (4.86 / 6.45)) 30. 55: a) After one time constant has passed: 6.00 V (1 e 1 ) 0.474 A i (1 e 1 ) 8.00 R 1 2 1 U Li (2.50 H)(0.474 A) 2 0.281 J 2 2 Or, using Problem (30. 25(c)): 3/ 7 U PL dt (4.50 W ) (e (3.20)t e ( 6.40)t )dt 0... the same as part (a) 30. 57: Multiplying Eq (30. 27) by i, yields: i 2 R Li 2 di q di q dq d 1 d 1 q i i 2 R Li i 2 R Li 2 dt C dt C dt dt 2 dt 2 C PR PL PC 0 That is, the rate of energy dissipation throughout the circuit must balance over all of the circuit elements 3T 2 3T 3 q Q cos( t ) Q cos Q cos 8 T 8 4 30. 58: a) If t 1... steady state is reached, so the inductor has no potential drop across it Simplified circuit becomes i ε /R 50 V 0.385 A 130 V1 (100 )(0.385 A) 38.5 V ; V2 0 V3 V4 50 V 38.5 V 11.5 V i1 0.385 A, i2 i3 11.5 V 0.153 A 75 11.5 V 0. 230 A 50 30. 65: a) Just after the switch is closed the voltage V5 across the capacitor is zero and there is also no current through the inductor,... reads zero c) Use the results of problem 30. 49 to combine the inductor network into its equivalent: R 75.0 is the equivalent resistance Eq. (30. 14) says i (ε R )(1 e t τ ), with τ L / R (10.8 mH) (75.0 ) 0.144 ms ε 20.0 V , R 75.0 , t 0.115 ms, so i 0.147 A VR iR (0.147 A)(75.0 ) 11.0 V 20.0 V VR VL 0 so VL 20.0 V VR 9.0 V 30. 74: (a) Steady state: i 75.0 V... (0.299 V)sin ((125.6 s 1 )(0.0180 s)) (t ) 0. 230 V 30. 48: a) L di1 di di L2 2 Leq , dt dt dt di di di but i1 i 2 i for series components so 1 2 , thus L1 L2 Leq dt dt dt di1 di2 di L2 Leq , where i i1 i 2 b) Parallel: Now L1 dt dt dt Leq di Leq di di di di di1 di2 So But 1 and 2 dt dt dt dt L2 dt dt L2 dt 30. 49: a) Series: L1 1 1 di Leq di Leq di 1 ... J U C U max 2 4 4 3 Q 4 2 q 2C 2C 30. 61: The energy density in the sunspot is u B B 2 / 2 μ0 6.366 10 4 J /m 3 The total energy stored in the sunspot is U B u BV The mass of the material in the sunspot is m ρV K U B so 1 2 mv U B ; 2 1 ρVv 2 u BV 2 The volume divides out, and v 2u B / ρ 2 10 4 m /s 30. 62: (a) The voltage behaves the same as the current .
A.750.0
8.00
V00.6
R
it
30. 20: (a)
mA,30A 030. 0
1000
V30
max
i
long after closing the switch.
)b
V4.0V26V30ε
V26A)(0.0259Ω)(1000
A0.0259
e1A0. 030) e(1
RBatteryL
R
)(
max
10
20
VV
RiV
ii
μs
μs
R/L/t
(or,.
LCT
s1 030. 7
4
at 43 period:
s1019.2)s1 030. 7(3
4
3
34
T
(c) CCVq
405)V(22.5F)18(
0
30. 31:
F0 .30
V
10
29
.
4
C10150
3
9
V
Q
C
For