Solucionário Cap 31 Lei de Faraday

Prévia do material em texto

Chapter 31 - Faraday’s Law 
 
*P31.4 (a) We evaluate the average emf: 
( )22cos 0 0.11 Tcos 12 0.0210 m cos0 0.0102 V
0.180 s
f iB BBAN N r
t t
θ π θ πε
−⎛ ⎞Δ −⎛ ⎞⎡ ⎤= − = − = − ° =⎜ ⎟ ⎜ ⎟⎣ ⎦Δ Δ ⎝ ⎠⎝ ⎠
 
The average induced current will then be 0.0102 V/2.3 Ω = 4.42 mA. If the meter has a 
sufficiently short response time, it will register the current. The average current may 
even run the meter offscale by a factor of 4.42, so you might wish to slow down the 
motion of the coil. 
 
(b) Positive. The coil sees decreasing external magnetic flux toward you, so it makes 
some flux of its own in this direction by carrying counterclockwise current, that enters 
the red terminal of the ammeter. 
P31.5 (a) max tB ABd dBA e
dt dt
τ
τ
ε −Φ= − = − = 
 
(b) 
( )( )2 4.00 2.000.160 m 0.350 T 3.79 mV
2.00 s
eε −= = 
 
(c) At 0t = 28.0 mVε = 
 
P31.7 Noting unit conversions from q= ×F v B
r rr and U qV= , the induced voltage is 
 
( ) ( )( )2
3
200 1.60 T 0.200 m cos00 cos 1 N s 1 V C 3 200 V
20.0 10 s 1 T C m N m
3 200 V
160 A
20.0 
i
d B AN N
dt t
I
R
θε
ε
−
⋅ + °− ⋅ ⋅⎛ ⎞ ⎛ ⎞⎛ ⎞= − = − = =⎜ ⎟⎜ ⎟⎜ ⎟Δ × ⋅ ⋅ ⋅⎝ ⎠⎝ ⎠⎝ ⎠
= = =
Ω
B A
r r
 
P31.14 ( )0.010 0 0.080 0B dBN A N t
t dt
ε ΔΦ ⎛ ⎞= = = +⎜ ⎟Δ ⎝ ⎠
A 
 
 At , 5.00 st = ( ) ( )230.0 0.410 T s 0.040 0 m 61.8 mVπε ⎡ ⎤= =⎣ ⎦ 
 
P31.15 ( )( )1.600 0 30.0 A 1 tB nI n eμ μ −= = −
 
 ( )( )1.600 30.0 A 1 tB BdA n e dAμ −Φ = = −∫ ∫ 
 
 ( )( )1.60 20 30.0 A 1 tB n eμ π−Φ = − R
 
 ( ) ( )2 10 30.0 A 1.60 tB
dN N n R e
dt
μ πε −Φ= − = − .60 
 
 
FIG. P31.15 
 ( )( )( )( ) ( )27 2 1 1 1.60250 4 10 N A 400 m 30.0 A 0.060 0 m 1.60 s teπ πε − − ⎡ ⎤= − × ⎣ ⎦
− − 
 
 ( ) 1.6068.2 mV counterclockwiseteε −= 
P31.16 (a) Suppose, first, that the central wire is long and straight. The enclosed current of 
unknown amplitude creates a circular magnetic field around it, with the 
magnitude of the field given by Ampère’s law. 
 
 0d Iμ⋅ =∫B s
r r : 0 max
sin
2
I t
B
R
μ ω
π
= 
 
 at the location of the Rogowski coil, which we assume is centered on the wire. 
This field passes perpendicularly through each turn of the toroid, producing 
flux 
 
 0 max
sin
2
I A t
R
μ ω
π
⋅ =B A
r r
 
 
 The toroid has 2 Rnπ turns. As the magnetic field varies, the emf induced in it 
is 
 
 0 max 0 max2 sin2
I Ad dN Rn t I nA
dt R dt
cos tμπ ω μ ω ω
π
ε = − ⋅ = − = −B A
r r
 
 
 This is an alternating voltage with amplitude max 0 maxnA Iμ ωε = . Measuring the 
amplitude determines the size of the central current. Our assumptions 
that the central wire is long and straight and passes perpendicularly through 
the center of the Rogowski coil are all unnecessary. 
maxI
 
(b) If the wire is not centered, the coil will respond to stronger magnetic fields on 
one side, but to correspondingly weaker fields on the opposite side. The emf 
induced in the coil is proportional to the line integral of the magnetic field 
around the circular axis of the toroid. Ampère’s law says that this line integral 
depends only on the amount of current the coil encloses. It does not depend on 
the shape or location of the current within the coil, or on any currents outside 
the coil. 
 
P31.17 ( )
2
2 coscosd NNB
dt t
B θθε Δ= =
Δ
l
l 
 
 
( )( )
( )( ) ( )
3
6 6
80.0 10 V 0.400 s
1.36 m
cos 50 600 10 T 200 10 T cos 30.0
 t
N B θ
ε −
− −
×Δ
= = =
Δ × − × °
l 
 
 Length ( )( )4 4 1.36 m 50 272 mN= = =l 
 
P31.20 (a) For maximum induced emf, with positive charge at the top of the antenna, 
 
 , so the auto must move(q+ += ×F v
r rr )B east . 
 
(b) ( )( )
3
5 465.0 10 m5.00 10 T 1.20 m cos65.0 4.58 10 V
3 600 s
B vε − −⎛ ⎞×= = × ° = ×⎜ ⎟
⎝ ⎠
l 
 
P31.23 (a) B I I= × =F B
r r r
ll B 
 When I
R
ε
= 
 and B vε = l 
 we get ( ) ( ) ( ) ( )
2 22 2 2.50 1.20 2.00
3.00 N
6.00B
B v B vF B
R R
= = = =
l l
l . 
 
 The applied force is 3.00 N to the right . 
 
(b) 
2 2 2
2 6.00 WB vI R
R
= = =P l or 6.00 WFv= =P 
 
 
FIG. P31.23 
 
*P31.24 and BF I B= l B vε = l 
 B vI
R R
ε
= =
l so IRB
v
=
l
 
 
(a) 
2
B
I RF
v
=
l
l
and 0.500 ABF vI
R
= = 
 
(b) 2 2.00 WI R = 
 
(c) For constant force, ( )( )1.00 N 2.00 m s 2.00 W= ⋅ = =F vP
r r . 
 
(d) The powers computed in parts (b) and (c) are mathematically equal. More 
profoundly, they are physically identical. Each bit of energy delivered to the 
circuit mechanically immediately goes through being electrically transmitted to 
the resistor and there becomes additional internal energy. Counting the 2 W 
twice is like counting your lunch money twice. 
 
P31.25 Observe that the homopolar generator has no commutator and 
produces a voltage constant in time: DC with no ripple. In time dt, the 
disk turns by angle d dtθ ω= . The outer brush slides over distance rdθ . 
 The radial line to the outer brush sweeps over area 
 21 1
2 2
dA rrd r dtθ ω= = 
 The emf generated is dN
dt
ε = − ⋅B A
r r
 
 ( ) 211 cos0
2
dAB B
dt
r ωε ⎛ ⎞= − ° = − ⎜ ⎟
⎝ ⎠
 
 (We could think of this as following from the result of the example in 
the chapter text about the helicopter blade.) 
 
 The magnitude of the emf is 
 
 
FIG. P31.25 
 
( ) ( ) ( )22 2 rad rev1 10.9 N s C m 0.4 m 3 200 rev min
2 2
24.1 V
B r
π
ωε
ε
⎛ ⎞⎛ ⎞ ⎡ ⎤= = ⋅ ⋅ ⎜ ⎟⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦ ⎝ ⎠
=
60 s min 
 
 A free positive charge q shown, turning with the disk, feels a magnetic force q ×v B
rr 
 radially outward. Thus the outer contact is positive . 
 
P31.27 ( )( )2.00 rev s 2 rad rev 4.00 rad sω π= = π 
 21 2.83 mV
2
Bωε = =l 
 
P31.28 (a) and decreases; therefore, the 
induced field is (to the right) and the 
current in the resistor is directed
ext ext
ˆB=B i
r
extB
0 0
ˆB=B
r
i
to the right . 
 
(b) increases; therefore, the induced 
field is to the right, and the current in 
the resistor is directed out of the plane in the 
( )ext ext ˆB= −Br i
i( )0 0 ˆB= +Br
textbook picture and to the right in the diagram 
here. 
 
(c) into the paper and decreases; 
therefore, the induced field is 
(ext ext ˆB= −Br )k extB
( )0 0 ˆB= −B kr into the 
paper, and the current in the resistor is directed 
to the right . 
 
 
FIG. P31.28 
 (d) By the magnetic force law, ( )B q= ×F v B
r rr . Therefore, a positive charge will move 
to the top of the bar if B
r
 is into the paper . 
 
P31.29 (a) The force on the side of the coil entering the field 
(consisting of N wires) is 
 
 ( ) ( )F N ILB N IwB= = 
 
 The induced emf in the coil is 
 
 ( )B d BwxdN N NB
dt dt
ε Φ= = = wv 
 
 so the current is NBwvI
R R
ε
= = counterclockwise. 
 
 The force on the leading side of the coil is then: 
 
 
2 2 2
to the leftNBwv N B w vF N wB
R R
⎛ ⎞= =⎜ ⎟
⎝ ⎠
 
 
(b) Once the coil is entirely inside the field, 
, constantB NBAΦ = =
 
 so 0ε = , 0I = , and 0F = 
 
 
FIG. P31.29 
 
(c) As the coil starts to leave the field, the flux decreases at the rate Bwv, so the 
magnitude of the current is the same as in part (a), but now the current is 
clockwise. Thus, the force exerted on the trailing side of the coil is: 
 
 
2 2 2
to the left againN B w vF
R
= 
 
P31.33 0.060 0dB t
dt
= 21 12B
d dBr r
dt dt
π πε EΦ= = = 
 
 At , 3.00 st =
 
 ( )( )
2
21
1
0.020 0 m 1 N s0.060 0 T s 3.00 s
2 2 1 T
r dBE
r dt
π
π
⎛ ⎞ ⋅⎛ ⎞= =⎜ ⎟ ⎜ ⎟⋅ ⋅⎝ ⎠⎝ ⎠ C m
 
 
 3 11.80 10 N C perpendicular to and counterclockwiser
−= ×E
r
 
 
 
FIG. P31.33 
 
P31.37 ( )( )( )7 10 4 10 T m A 200 m 15.0 A 3.77 10 TB nIμ π − −= = × ⋅ = × 3− 
 
 For the small coil, ( )2cos cosB N NBA t NB r tω π ωΦ = ⋅ = =B A
r r
. 
 
 Thus, 2 sinBd NB r t
dt
π ω ωε Φ= − = 
 
 ( )( ) ( ) ( ) ( ) ( ) ( )23 130.0 3.77 10 T 0.080 0 m 4.00 s sin 4.00 28.6 mV sin 4.00t tπ π πε − −= × = π 
 
P31.55 (a) 0.360 VB vε = =l 0.900 AI
R
ε
= = 
 
(b) 0.108 NBF I B= =l 
 
(c) Since the magnetic flux ⋅B A
r r
 is in effect decreasing, the 
induced current flow through R is from b to a. Point b is 
at higher potential. 
 
 
FIG. P31.55 
 
(d) No. Magnetic flux will increase through a loop to the left of ab. Here 
counterclockwise current will flow to produce upward magnetic field. The 
current in R is still from b to a. 
 
P31.56 B vε = l at a distance r from wire 
 
 0
2
I v
r
μ
π
ε ⎛ ⎞= ⎜ ⎟
⎝ ⎠
l I
 
v
 
 
FIG. P31.56 
 
 
P31.64 For the suspended mass, M: F Mg T Ma= − =∑ . 
 For the sliding bar, m: , where F T I B ma= − =∑ l B vI R R
ε
= =
l 
 
 (
2 2B vMg m M a
R
− = +
l ) or 
( )
2 2Mgdv B va
dt m M R M m
= = −
+ +
l 
 
 
( )0 0
v tdv dt
vα β
=
−∫ ∫ where 
Mg
M m
α =
+
 and 
( )
2 2B
R M m
β =
+
l 
 Therefore, the velocity varies with time as
 
( ) ( )2 22 21 1 B t R M mt
MgR
v e e
B
βα
β
− +− ⎡ ⎤= − = −⎣ ⎦
l
l
 
 
*P31.66 (a) The induced emf is B vε = l where 0
2
IB
y
μ
π
= , ( )29.80 m sf iv v gt= + = t , and 
 
 ( )2 21 0.800 m 4.90 m s
2f i
y y gt t= − = − 2 
 
 
( )( )
( )
( )( ) ( )
7 4
2
22 2
4 10 T m A 200 A 1.18 10
0.300 m 9.80 m s V
0.800 4.902 0.800 m 4.90 m s
t
t
tt
π
π
ε
− −× ⋅ ×
= =
⎡ ⎤ ⎡ −− ⎣ ⎦⎣ ⎦ ⎤
 
 
 (b) The emf is zero at t = 0 
 
(c) The emf diverges to infinity at 0.404 s 
 
 At ,0.300 st =
( )( )
( )
4
2
1.18 10 0.300
V 98.3 V
0.800 4.90 0.300
με
−×
= =
⎡ ⎤−⎣ ⎦
. 
 
P31.67 The magnetic field produced by the current in the straight wire is 
perpendicular to the plane of the coil at all points within the coil. The 
magnitude of the field is 0
2
IB
r
μ
π
= . Thus, the flux linkage is 
 
 ( )0 0 max ln sin
2 2
h w
B
h
NIL NI Ldr h wN t
r h
μ μ
ω φ
π π
+ +⎛ ⎞Φ = = +⎜ ⎟
⎝ ⎠∫ 
 
 Finally, the induced emf is 
 
 
FIG. P31.67 
 
 
 
( )
( )( ) ( ) ( )( ) ( )
( ) ( )
0 max
7 1
ln 1 cos
2
4 10 100 50.0 0.200 m 200 s 5.00 cmln 1 cos
2 5.00 cm
87.1 mV cos 200
NI L w t
h
t
t
μ ω
ω φ
π
π π
ω φ
π
π φ
ε
ε
ε
− −
⎛ ⎞= − + +⎜ ⎟
⎝ ⎠
× ⎛ ⎞= − + +⎜ ⎟
⎝ ⎠
= − +
 
 
 The term (si )n tω φ+ in the expression for the current in the straight wire does not 
change appreciably when tω changes by 0.10 rad or less. Thus, the current does not 
change appreciably during a time interval 
 
 
( )
4
1
0.10 1.6 10 s
200 s
t
π
−
−
Δ < = × 
 
 We define a critical length, ( )( )8 43.00 10 m s 1.6 10 s 4.8 10 mc t −Δ = × × = × 4
1
equal to the 
distance to which field changes could be propagated during an interval of . 
This length is so much larger than any dimension of the coil or its distance from the 
wire that, although we consider the straight wire to be infinitely long, we can also 
safely ignore the field propagation effects in the vicinity of the coil. Moreover, the 
phase angle can be considered to be constant along the wire in the vicinity of the coil. 
41.6 10 s−×
 
 If the angular frequency ω were much larger, say, 5200 10 sπ −× , the corresponding 
critical length would be only 48 cm. In this situation propagation effects would be 
important and the above expression for ε would require modification. As a general 
rule we can consider field propagation effects for circuits of laboratory size to be 
negligible for frequencies, 
2
f ω
π
= , that are less than about Hz. 610