29.18: a) With current passing from
ba
and is increasing the magnetic, field
becomes stronger to the left, so the induced field points right, and the induced current
must flow from right to left through the resistor.
b) If the current passes from
ab
, and is decreasing, then there is less magnetic
field pointing right, so the induced field points right, and the induced current must flow
from right to left through the resistor.
c) If the current passes from
,ab
and is increasing, then there is more magnetic
field pointing right, so the induced field points left, and the induced current must flow
from left to right through the resistor.
29.19: a)
B
is ⊙ and increasing so the flux
ind
of the induced current is clockwise.
b) The current reaches a constant value so
B
is constant.
0 dtd
B
and there is
no induced current.
c)
B
is ⊙ and decreasing, so
ind
is ⊙ and current is counterclockwise.
29.20: a)
)m50.1)(T750.0)(sm0.5(
vBl
V6.5
b) (i)
Let q be a positive charge in the moving bar. The
magnetic force on this charge
,BvF
q
which
points upward. This force pushes the current in a
counterclockwise direction through the circuit.
(ii) The flux through the circuit is increasing, so the induced current must cause a
magnetic field out of the paper to oppose this increase. Hence this current must flow in a
counterclockwise sense.
c)
Ri
A22.0
25
V6.5
R
i
29.21:
.V
C
J
C
mN
m
mC
sN
s
m
Tm
s
m
vBL
29.22: a)
.V675.0)m300.0)(T450.0)(sm00.5(
vBL
b) The potential difference between the ends of the rod is just the motional emf
.V675.0
V
c) The positive charges are moved to end b, so b is at the higher potential.
d)
.
m
V
25.2
m300.0
V675.0
L
V
E
e) b
29.23: a)
.sm858.0
)m850.0)(T850.0(
V620.0
BL
vvBL
b)
.A827.0
750.0
V620.0
R
I
c)
N598.0)T850.0)(m850.0)(A827.0(
ILBF
, to the left, since you must
pull it to get the current to flow.
29.24: a)
.V00.3)m500.0)(T800.0)(sm50.7(
vBL
b) The current flows counterclockwise since its magnetic field must oppose the
increasing flux through the loop.
c)
,N800.0
50.1
)T800.0)(m500.0)(V00.3(
R
LB
ILBF
to the right.
d)
.W00.6)sm50.7)(N800.0(
mech
FvP
00.6
50.1
)V00.3(
2
2
elec
R
P
W. So both rates are equal.
29.25:
For the loop pulled through the region of magnetic field,
a)
b)
Where
.and
22
00
R
LvB
ILBF
R
vBL
IIRvBL
29.26: a) Using Equation (29.6):
.T833.0
)m120.0)(sm50.4(
V450.0
vL
BvBL
b) Point a is at a higher potential than point b, because there are more positive
charges there.
29.27:
ldE
dt
dI
nAnIA
dt
d
BA
dt
d
dt
d
B
and)()(
00
.
2
2
2
00
dt
dI
nr
dt
dI
r
nA
r
E
a)
.mV1070.1)sA60(
2
)m0050.0)(m900(
cm50.0
4
1
0
Er
b)
m.V1039.3cm00.1
4
Er
29.28: a)
.
2
1
dt
dB
r
dt
dB
A
dt
d
B
b)
.
222
1
1
1
2
1
1
dt
dB
r
dt
dB
r
r
dt
d
r
E
B
c) All the flux is within r < R, so outside the solenoid
.
222
1
2
2
2
2
2
dt
dB
r
R
dt
dB
r
R
dt
d
r
E
B
29.29:
a) The induced electric field lines are concentric circles since they cause the
current to flow in circles.
b)
)sT0350.0(
2
m100.0
2
2
1
2
1
2
1
dt
dBr
dt
dB
A
r
dt
d
r
r
E
B
,mV1075.1
3
E
in the clockwise direction, since the induced magnetic field
must reinforce the decreasing external magnetic field.
c)
.A1075.2)sT0350.0(
00.4
)m100.0(
4
2
2
dt
dB
R
r
R
I
d)
V.1050.5
2
)00.4)(A1075.2(
2
4
4
TOT
IRIR
e) If the ring was cut and the ends separated slightly, then there would be a potential
difference between the ends equal to the induced emf:
V.1010.1)sT0350.0()m100.0(
322
dt
dB
r
29.30:
nA
rE
dt
dI
dt
dI
nAnIA
dt
d
BA
dt
d
dt
d
0
00
B
2
)()(
.A21.9
)m0110.0()m400(
)0350.0(2)mV1000.8(
21
0
6
s
dt
dI
29.31:
a)
.J1014.1)m0350.0(2)mV1000.8)(C1050.6(2
1166
RqEdW lF
(b) For a conservative field, the work done for a closed path would be zero.
(c)
.
dt
di
BAEL
dt
d
d
B
lE
A is the area of the solenoid.
For a circular path:
dt
di
BArE
2
constant for all circular paths that enclose the solenoid.
So
rqEW
2
constant for all paths outside the solenoid.
cm.00.7ifJ1014.1
11
rW
29.32:
t
nINA
t
BBNA
t
N
o
if
B
)(
.V1050.9
s0400.0
)A350.0)(m9000)(m1000.8)(12(
4
124
o
29.33:
23311
)smV100.24)(mF105.3( t
dt
d
i
E
D
s0.5givesA1021
6
ti
D
29.34: According to Eqn.29.14
3343
12
)s101.26)(smV1076.8(4
A109.12
dt
d
i
E
D
.mF1007.2
11
Thus, the dielectric constant is
.34.2
0
K
29.35: a)
.mA7.55
)m0400.0(
A280.0
2
2
0
00
A
i
A
i
dt
dE
j
cc
D
b)
.smV1029.6
mA7.55
12
0
2
0
D
j
dt
dE
c) Using Ampere’s Law
.T100.7)A280.0(
)m0400.0(
m0200.0
22
:
7
2
0
2
0
D
i
R
r
BRr
d) Using Ampere’s Law
.T105.3)280.0(
)m0400.0(
)m0100.0(
2
2
:
7
2
0
2
0
D
i
R
r
BRr
29.36: a)
.C1099.5
m1050.2
)V120)(m1000.3()70.4(
10
3
24
0
V
d
A
CVQ
b)
.A1000.6
3
c
i
dt
dQ
c)
.A1000.6
3
0
0
cDc
cc
D
iij
A
i
AK
i
K
dt
dE
j
29.37:a)
C100.900s)10(0.500A)10(1.80
963
tiq
c
V.406m)102.00()mV10(2.03
.mV102.03
)m10(5.00
C100.900
A
35
5
0
24
9
00
EdV
q
E
b)
s,m/V1007.4
)m10(5.00
A1080.1
11
0
24-
3
0
c
A
i
dt
dE
and is constant in time.
c)
2
11
00
mA3.60s)mV1007.4(
dt
dE
j
D
A,1080.1)m10(5.00)m/A60.3(
32-42
Aji
DD
which is the
same as
.
c
i
29.38: a)
m./V15.0
m102.1
A)m)(16100.2(
26-
8
A
I
JE
b)
s.m38)4000(
m101.2
m100.2
26
8
VsA
dt
dI
A
ρ
A
ρI
dt
d
dt
dE
c)
.m104.3s)sV38(
2
10
00
A
dt
dE
j
D
d)
A1014.7)m101.2()A/m104.3(
1626210
Aji
DD
,T1038.2
)m060.0(2
)A1014.7(
2
21
16
00
r
I
B
D
D
and this is a
negligible contribution.
.T1033.5
)m060.0(
)A16(
22
5
00
r
I
B
c
c
29.39:
In a superconductor there is no internal magnetic field, and so there is no changing
flux and no induced emf, and no induced electric field.
,0)(0
00encl0
material
Inside
ccDc
IIIIId
lB
and so there is no current inside the material. Therefore, it must all be at the surf
ace of the
cylinder.
29.40:
Unless some of the regions with resistance completely fill a cross-sectional area of
a long type-II superconducting wire, there will still be no total resistance. The regions of
no resistance provide the path for the current. Indeed, it will be like two resistors in
parellel, where one has zero resistance and the other is non-zero. The equivalent
resistance is still zero.
29.41:
a) For magnetic fields less than the critical field, there is no internal magnetic
field, so:
Inside the superconductor:
.
ˆ
)mA1003.1(
ˆ
)130.0(
,0
5
00
0
i
i
MB
μ
T
μ
B
Outside the superconductor:
.0,
ˆ
)130.0(
0
MiBB T
b) For magnetic fields greater than the critical field,
00 M
both inside
and outside the superconductor, and
,
ˆ
)T260.0(
0
iBB
both inside and outside the
superconductor.
29.42: a) Just under
1c
B
(threshold of superconducting phase), the magnetic field in the
material must be zero, and
.
ˆ
)mA1038.4(
ˆ
T1055
4
0
3
0
1
i
iB
M
c
b) Just over
2c
B
(threshold of normal phase), there is zero magnetization, and
.
ˆ
)T0.15(
2
iBB
c
29.43:a) The angle
between the normal to the coil and the direction of
.30.0is B
.||and)(||
2
RIdtdBrN
dt
d
B
For
0and0||,0s,00.1and0 IdtdBtt
For
πtπdtdBt sinT)120.0(s,00.10
πtπtπrNπ sinV)(0.9475T)sin120.0()(||
2
m100150.0,m1072.1;:wirefor
38
2
w
r
r
L
A
L
RR
m125.7m)0400.0()2()500(2
rNNcL
3058
w
R
and the total resistance of the circuit is
36586003058R
tRI
sinmA)259.0(/||
b)
B increasing so
is
B
⊙ and increasing
Isois
ind
is clockwise
29.44: a) The large circuit is an RC circuit with a time constant of
s.200F)1020()10(
6
RC Thus, the current as a function of time is
s200
10
V100
t
ei
At
s,200
t
we obtain
A.7.3)(A)10(
1
ei
b) Assuming that only the long wire nearest the small loop produces an appreciable
magnetic flux through the small loop and referring to the solution of Problem 29.54 we
obtain
ac
c
B
c
a
ib
dr
r
ib
)(1ln
2
2
00
So the emf induced in the small loop at
iss200
t
)
s10200
A3.7
((3.0)ln
2
m)200.0()104(
1ln
2
6
mA
Wb
7
0
2
π
π
dt
di
c
a
π
bμ
dt
d
ε
Thus, the induced current in the small loop is
A.54
)m(1.0m)25(0.600
mV81.0
R
i
c) The induced current will act to oppose the decrease in flux from the large loop.
Thus, the induced current flows counterclockwise.
d) Three of the wires in the large loop are too far away to make a significant
contribution to the flux in the small loop–as can be seen by comparing the distance
c
to
the dimensions of the large loop.
29.45:
a)
b)
c)
.mV4.0
s50.0
T80.0
2
)m50.0(
22
1
2
1
2
2
max
dt
dBr
dt
dB
r
rdt
dB
NA
rN
E
29.46: a)
.
sin)cos(
11
R
tBA
dt
tBAd
R
dt
d
R
R
I
B
b)
.
sin
2222
2
R
tAB
RIP
c)
.
R
sin
2
tBA
IA
d)
.
sinB
Bsinsin
222
R
tA
tB
e)
,
sin
2222
R
tAB
P
which is the same as part (b).
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