Law
CHAPTER 31
Faraday's OUTLINE
31.1 31.2 31.3 31.4 31.5 31.6 31.7 Faraday's Law of Induction Motional emf Lenz's Law Induced emf and Electric Fields Generators and Motors Eddy Currents Maxwell's Equations
ANSWERS TO QUESTIONS
Q31.1 Magnetic flux measures the "flow" of the magnetic field through a given area of a loop--even though the field does not actually flow. By changing the size of the loop, or the orientation of the loop and the field, one can change the magnetic flux through the loop, but the magnetic field will not change. The magnetic flux is B = BA cos . Therefore the flux is maximum when B is perpendicular to the loop of wire and zero when there is no component of magnetic field perpendicular to the loop. The flux is zero when the loop is turned so that the field lies in the plane of its area. The force on positive charges in the bar is F = q v B . If the bar is moving to the left, positive charge will move downward and accumulate at the bottom end of the bar, so that an electric field will be established upward.
Q31.2
Q31.3
a
f
Q31.4 Q31.5
No. The magnetic force acts within the bar, but has no influence on the forward motion of the bar. By the magnetic force law F = q v B : the positive charges in the moving bar will flow downward and therefore clockwise in the circuit. If the bar is moving to the left, the positive charge in the bar will flow upward and therefore counterclockwise in the circuit. We ignore mechanical friction between the bar and the rails. Moving the conducting bar through the magnetic field will force charges to move around the circuit to constitute clockwise current. The downward current in the bar feels a magnetic force to the left. Then a counterbalancing applied force to the right is required to maintain the motion. A current could be set up in the bracelet by moving the bracelet through the magnetic field, or if the field rapidly changed. Moving a magnet inside the hole of the doughnut-shaped toroid will not change the magnetic flux through any turn of wire in the toroid, and thus not induce any current.
a
f
Q31.6
Q31.7 Q31.8
213
214 Q31.9
Faraday's Law
As water falls, it gains speed and kinetic energy. It then pushes against turbine blades, transferring its energy to the rotor coils of a large AC generator. The rotor of the generator turns within a strong magnetic field. Because the rotor is spinning, the magnetic flux through its turns changes in time as -Nd B . This induced emf is the B = BA cos t . Generated in the rotor is an induced emf of = dt voltage driving the current in our electric power lines. Yes. Eddy currents will be induced around the circumference of the copper tube so as to fight the changing magnetic flux by the falling magnet. If a bar magnet is dropped with its north pole downwards, a ring of counterclockwise current will surround its approaching bottom end and a ring of clockwise current will surround the receding south pole at its top end. The magnetic fields created by these loops of current will exert forces on the magnet to slow the fall of the magnet quite significantly. Some of the original gravitational energy of the magnet will appear as internal energy in the walls of the tube. Yes. The induced eddy currents on the surface of the aluminum will slow the descent of the aluminum. It may fall very slowly. The maximum induced emf will increase, increasing the terminal voltage of the generator. The increasing counterclockwise current in the solenoid coil produces an upward magnetic field that increases rapidly. The increasing upward flux of this field through the ring induces an emf to produce clockwise current in the ring. The magnetic field of the solenoid has a radially outward component at each point on the ring. This field component exerts upward force on the current in the ring there. The whole ring feels a total upward force larger than its weight.
Q31.10
Q31.11 Q31.12 Q31.13
FIG. Q31.13 Q31.14 Oscillating current in the solenoid produces an always-changing magnetic field. Vertical flux through the ring, alternately increasing and decreasing, produces current in it with a direction that is alternately clockwise and counterclockwise. The current through the ring's resistance produces internal energy at the rate I 2 R . (a) The south pole of the magnet produces an upward magnetic field that increases as the magnet approaches. The loop opposes change by making its own downward magnetic field; it carries current clockwise, which goes to the left through the resistor. The north pole of the magnet produces an upward magnetic field. The loop sees decreasing upward flux as the magnet falls away, and tries to make an upward magnetic field of its own by carrying current counterclockwise, to the right in the resistor.
Q31.15
(b)
Chapter 31
215
Q31.16
(a)
The battery makes counterclockwise current I 1 in the primary coil, so its magnetic field B1 is to the right and increasing just after the switch is closed. The secondary coil will oppose the change with a leftward field B 2 , which comes from an induced clockwise current I 2 that goes to the right in the resistor. At steady state the primary magnetic field is unchanging, so no emf is induced in the secondary. The primary's field is to the right and decreasing as the switch is opened. The secondary coil opposes this decrease by making its own field to the right, carrying counterclockwise current to the left in the resistor.
(b)
(c)
FIG. Q31.16
Q31.17
The motional emf between the wingtips cannot be used to run a light bulb. To connect the light, an extra insulated wire would have to be run out along each wing, making contact with the wing tip. The wings with the extra wires and the bulb constitute a single-loop circuit. As the plane flies through a uniform magnetic field, the magnetic flux through this loop is constant and zero emf is generated. On the other hand, if the magnetic field is not uniform, a large loop towed through it will generate pulses of positive and negative emf. This phenomenon has been demonstrated with a cable unreeled from the Space Shuttle. No, they do not. Specifically, Gauss's law in magnetism prohibits magnetic monopoles. If magnetic monopoles existed, then the magnetic field lines would not have to be closed loops, but could begin or terminate on a magnetic monopole, as they can in Gauss's law in electrostatics. (a) (b) A current is induced by the changing magnetic flux through the a ring of the tube, produced by the high frequency alternating current in the coil. The higher frequency implies a greater rate of change in the magnetic field, for a larger induced voltage. The resistance of one cubic centimeter in the bulk sheet metal is low, so the I 2 R rate of production of internal energy is low. At the seam, the current starts out crowded into a small area with high resistance, so the temperature rises rapidly, and the edges melt together. The edges must be in contact to allow the induced emf to create an electric current around the circumference of the tube. Additionally, (duh) the two edges must be in contact to be welded at all, just as you can't glue two pieces of paper together without putting them in contact with each other.
Q31.18
Q31.19
(c)
(d)
216
Faraday's Law
SOLUTIONS TO PROBLEMS
Section 31.1 Section 31.3 P31.1 Faraday's Law of Induction Lenz's Law B NBA = = 500 mV t t
=
a
f
P31.2
2.50 T - 0.500 T 8.00 10 -4 m 2 B B A = = = 1.00 s t t
a f a
fe
j FG 1 N s IJ FG 1 V C IJ H 1 T C mK H 1 N mK
= 1.60 mV and I loop =
1.60 mV = = 0.800 mA R 2.00
P31.3
= -N
cos f - cos i BA cos = - NB r 2 = -25.0 50.0 10 -6 T 0.500 m t t
FG H
IJ K
e
j a
f FGH cos 180- cos 0 IJK 0.200 s
2
= +9.82 mV
P31.4 (a)
=-
d B ABmax - t dB e = -A = dt dt
2 -4.00 2.00
(b) (c) P31.5
e0.160 m ja0.350 Tf e =
2.00 s At t = 0 = 28.0 mV
= 3.79 mV
Noting unit conversions from F = qv B and U = qV , the induced voltage is +200 1.60 T 0.200 m 2 cos 0 1 N s d BA 0 - Bi A cos = -N = -N = dt t 1 T C m 20.0 10 -3 s 3 200 V = 160 A I= = R 20.0
a f
FG H
IJ K
a
fe
j
FG H
IJ FG 1 V C IJ = 3 200 V KH N m K
P31.6
= -N
N BA - 0 d B =- dt t = NB r 2
a
f
-2 2
t =
NBA
e j = 500a0.200f e5.00 10 j
10.0 10 3
= 7.85 10 -5 s
Chapter 31
217
P31.7
=
d BA dI = 0.500 0 nA = 0.480 10 -3 V dt dt I ring =
a f
(a)
4.80 10 -4 = = 1.60 A R 3.00 10 -4 0I = 20.1 T 2rring
(b) (c)
Bring =
Coil's field points downward, and is increasing, so Bring points upward . FIG. P31.7
P31.8
=
d BA dI I = 0.500 0 nA = 0.500 0 n r22 dt dt t I ring =
a f
(a)
0n r22 I = R 2 R t
(b) (c)
B=
2 0 n r22 I 0I = 2r1 4r1 R t
The coil's field points downward, and is increasing, so Bring points upward . FIG. P31.8
P31.9
(a)
d B = B dA =
h+ w 0I 0 IL dx 0 IL h+ w Ldx : B = = ln 2 x h 2 x 2 h
z
FG H
IJ K
(b)
=- =-
d B d 0 IL h+w =- ln dt dt 2 h
e4 10
-7
IJ OP = - LM L lnFG h + w IJ OP dI K Q N 2 H h K Q dt T m A ja1.00 mf F 1.00 + 10.0 I lnG H 1.00 JK b10.0 A sg = 2 FG H
0
LM N
-4.80 V
The long wire produces magnetic flux into the page through the rectangle, shown by the first hand in the figure to the right. As the magnetic flux increases, the rectangle produces its own magnetic field out of the page, which it does by carrying counterclockwise current (second hand in the figure). FIG. P31.9
218 P31.10
Faraday's Law
B = 0 nI Asolenoid
b
g
= -N
d B dI 2 = - N 0 n rsolenoid dt dt
e
j
= -15.0 4 10 -7 T m A 1.00 10 3 m -1 0.020 0 m = -14.2 cos 120 t mV
P31.11
e
a f
je
jb
g b600 A sg cosa120tf
2
For a counterclockwise trip around the left-hand loop, with B = At d At 2 a 2 cos 0 - I 1 5 R - I PQ R = 0 dt
e j
a f
and for the right-hand loop, d Ata 2 + I PQ R - I 2 3 R = 0 dt where I PQ = I 1 - I 2 is the upward current in QP. Thus, and 2 Aa 2 - 5 R I PQ + I 2 - I PQ R = 0
2 PQ 2
a f d
i Aa + I R = I a3 Rf 5 2 Aa - 6 RI - e Aa 3
2 PQ
FIG. P31.11
2
+ I PQ R = 0
j
I PQ = I PQ
Aa 2 upward, and since R = 0.100 m 0.650 m = 0.065 0 23 R
b
ga
f
e1.00 10 = FG IJ H K
-3
23 0.065 0
b
T s 0.650 m
ja
g
f
2
= 283 A upward .
P31.12
=
B dB A = N 0.010 0 + 0.080 0t A =N dt t
b
g
At t = 5.00 s , = 30.0 0.410 T s 0.040 0 m P31.13 B = 0 nI = 0 n 30.0 A 1 - e -1.60 t
b
g b
j
g
2
= 61.8 mV
a
fe
a fe jz dA = na30.0 A fe1 - e j R d = -N = - N na30.0 A f R a1.60fe dt
B = BdA = 0 n 30.0 A 1 - e -1.60 t
B 0 -1.60 t 2 B 0 2
z
-1.60 t
= - 250 4 10
=
a fe a68.2 mVfe
-7
N A
2
je400 m ja30.0 Af b0.060 0 mg
-1
FIG. P31.13
2
1.60 s e
-1 -1.60 t
-1.60 t
counterclockwise
Chapter 31
219
*P31.14
(a)
Each coil has a pulse of voltage tending to produce counterclockwise current as the projectile approaches, and then a pulse of clockwise voltage as the projectile recedes. v= d 1.50 m = = 625 m s t 2.40 10 -3 s
V 0 V1 V2 FIG. P31.14 t
(b)
P31.15
=
d NB dt
e
2
cos =
j
N 2 B cos t
-3
e80.0 10 Vja0.400 sf a50fe600 10 T - 200 10 Tj cosa30.0f = 1.36 m Length = 4 N = 4a1.36 mfa50f = 272 m
t = = NB cos
-6 -6
*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 Ampere's Law.
z
B ds = 0 I : B =
0 I max sin t 2 R
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 BA =
0 I max A sin t . 2 R
The toroid has 2 Rn turns. As the magnetic field varies, the emf induced in it is
= -N
I A d d sin t = - 0 I max nA cos t . B A = -2 Rn 0 max dt 2 R dt
This is an alternating voltage with amplitude max = 0 nA I max . Measuring the amplitude determines the size I max 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. (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. Ampere'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.
220 P31.17
Faraday's Law
In a toroid, all the flux is confined to the inside of the toroid. B=
0 NI 500 0 I = 2 r 2 r
B = BdA = B =
z
500 0 I max adr sin t r 2
500 0 I max b+R a sin t ln R 2
= N =
= *P31.18
d B 500 0 I max b+R = 20 cos t a ln R dt 2
FG H
FG H
z
IJ K
IJ K
FG H
IJ K
FIG. P31.17
3.00 + 4.00 cm 10 4 4 10 -7 N A 2 50.0 A 377 rad s 0.020 0 m ln cos t 2 4.00 cm
e
ja
fb
gb
a0.422 Vf cos t
g FGH a
f IJ K
The upper loop has area 0.05 m
a
f
2
= 7.85 10 -3 m 2 . The induced emf in it is
= -N
d dB BA cos = -1 A cos 0 = -7.85 10 -3 m 2 2 T s = -1.57 10 -2 V . dt dt
b
g
The minus sign indicates that it tends to produce counterclockwise current, to make its own magnetic field out of the page. Similarly, the induced emf in the lower loop is
= - NA cos
dB 2 = - 0.09 m 2 T s = -5.09 10 -2 V = +5.09 10 -2 V to produce dt
a
f
counterclockwise current in the lower loop, which becomes clockwise current in the upper loop . The net emf for current in this sense around the figure 8 is 5.09 10 -2 V - 1.57 10 -2 V = 3.52 10 -2 V . It pushes current in this sense through series resistance 2 0.05 m + 2 0.09 m 3 m = 2.64 .
3.52 10 -2 V = 13.3 mA . The current is I = = R 2.64
a
f
a
f
Section 31.2 Section 31.3 P31.19 (a)
Motional emf Lenz's Law For maximum induced emf, with positive charge at the top of the antenna, F+ = q + v B , so the auto must move east .
a
f
(b)
= B v = 5.00 10 -5 T 1.20 m
e
ja
0 fFGH 65.3 10 s m IJK cos 65.0 = 600
3
4.58 10 -4 V
Chapter 31
221
P31.20
I=
R
=
B v R
v = 1.00 m s
FIG. P31.20 P31.21 (a)
FB = I B = I B
When and we get
R =B v
I= FB = B v R
a Bf = B R v = a2.50f a1..20f a2.00f = 3.00 N . 6 00
2 2 2 2
The applied force is 3.00 N to the right . (b) P31.22
FIG. P31.21
P = I2R =
B2
2 2
v
R
= 6.00 W or P = Fv = 6.00 W
FB = I B and = B v IR B v I= = so B = R R v (a) (b) (c) FB = I2 R and I = v FB v = 0.500 A R
I 2 R = 2.00 W
For constant force, P = F v = 1.00 N 2.00 m s = 2.00 W .
a
fb
g
*P31.23
Model the magnetic flux inside the metallic tube as constant as it shrinks form radius R to radius r: 2.50 T R 2 = B f r 2
2
Bf
e j F RI = 2.50 TG J HrK
= 2.50 T 12
a f
2
= 360 T
222 *P31.24
Faraday's Law
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 = dt . The outer brush slides over distance rd . The radial line to the outer brush sweeps over area 1 1 dA = rrd = r 2dt . 2 2 d = -N B A The emf generated is dt dA 1 = - 1 B cos 0 = - B r 2 dt 2 (We could think of this as following from the result of Example 31.4.) The magnitude of the emf is
af
FG H
IJ K
FIG. P31.24
=B
FG 1 r IJ = b0.9 N s C mgLM 1 a0.4 mf b3 200 rev mingOPFG 2 rad rev IJ H2 K N2 QH 60 s min K
2 2
= 24.1 V
A free positive charge q shown, turning with the disk, feels a magnetic force qv B outward. Thus the outer contact is positive . *P31.25 The speed of waves on the wire is v= T radially
=
267 N m 3 10 -3 kg
= 298 m s .
In the simplest standing-wave vibration state, d NN = 0.64 m = and f = (a) v =
2
= 1.28 m
298 m s = 233 Hz . 1.28 m
The changing flux of magnetic field through the circuit containing the wire will drive current to the left in the wire as it moves up and to the right as it moves down. The emf will have this same frequency of 233 Hz . The vertical coordinate of the center of the wire is described by
(b)
f b g dx = -a1.5 cmfb 2 233 sg sinb 2 233 t sg . Its velocity is v = dt Its maximum speed is 1.5 cma 2 f 233 s = 22.0 m s .
x = A cos t = 1.5 cm cos 2 233 t s . The induced emf is = -B v , with amplitude
a
max = B v max = 4.50 10 -3 T 0.02 m 22 m s = 1.98 10 -3 V .
a
f
Chapter 31
223
P31.26
= -N
d A BA cos = - NB cos dt t
FG IJ H K
2
= -1 0.100 T cos 0
I= 1.21 V = 0.121 A 10.0
a
f
a3.00 m 3.00 m sin 60.0f - a3.00 mf
0.100 s
= 1. 21 V FIG. P31.26
The flux is into the page and decreasing. The loop makes its own magnetic field into the page by carrying clockwise current. P31.27
= 2.00 rev s 2 rad rev = 4.00 rad s
=
1 B 2
2
b
gb
g
= 2.83 mV
P31.28
(a)
B ext = B ext i and Bext decreases; therefore, the induced field is B0 = B0 i (to the right) and the current in the resistor is directed to the right .
(b)
B ext = B ext - i increases; therefore, the induced field B0
0
e j = B e+ i j is to the right, and the current in e j
the resistor is directed to the right . (c) B ext = B ext - k into the paper and Bext decreases; therefore, the induced field is B0 = B0 - k into the paper, and the current in the resistor is directed to the right . (d) FIG. P31.28
e j
f
By the magnetic force law, FB = q v B . Therefore, a positive charge will move to the top of the bar if B is into the paper .
a
224 P31.29
Faraday's Law
(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
a f a f
=N
so the current is I =
d Bwx d B =N = NBwv . dt dt
a f
NBwv = counterclockwise. R R
The force on the leading side of the coil is then: F=N (b)
FG NBwv IJ wB = H R K
N 2B2w 2 v to the left . R
Once the coil is entirely inside the field, B = NBA = constant , so
=0, I =0,
and
F= 0 .
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: F= N 2B2 w 2 v to the left again . R
P31.30
Look in the direction of ba. The bar magnet creates a field into the page, and the field increases. The loop will create a field out of the page by carrying a counterclockwise current. Therefore, current must flow from b to a through the resistor. Hence, Va - Vb will be negative . Name the currents as shown in the diagram: Left loop: Right loop: At the junction: Then, I3 = So, I 1 = Bd I1
2 3 3
P31.31
+ Bdv 2 - I 2 R 2 - I 1 R1 = 0 + Bdv 3 - I 3 R3 + I 1 R1 = 0 I 2 = I1 + I 3 Bdv 2 - I 1 R 2 - I 3 R 2 - I 1 R1 = 0 FIG. P31.31 Bdv 2 - I 1 R1 + R 2 -
2
Bdv 3 I 1 R1 + . R3 R3
b
g
FG v R - v R IJ upward HR R +R R +R R K L b4.00 m sga15.0 f - b2.00 m sga10.0 f OP = = b0.010 0 Tga0.100 mfM MN a5.00 fa10.0 f + a5.00 fa15.0 f + a10.0 fa15.0 f PQ
1 2 1 3 2 3
Bdv 3 R 2 I 1 R1 R 2 - =0 R3 R3
145 A upward.
Chapter 31
225
Section 31.4 P31.32 (a)
Induced emf and Electric Fields dB = 6.00t 2 - 8.00t dt At t = 2.00 s ,
=
E=
d B dt
R 2 dB dt
2 r2
b
g = 8.00 b0.025 0g 2 b0.050 0g
2
F = qE = 8.00 10 -21 N FIG. P31.32 (b) P31.33 When 6.00t - 8.00t = 0 ,
2
t = 1.33 s
d B dB = r12 = 2 r1 E dt dt
dB = 0.060 0t dt At t = 3.00 s , E=
2 1
=
F r I dB = 0.020 0 m e0.060 0 T s jb3.00 sgFG 1 N s IJ GH 2 r JK dt H 1 T C mK 2
2 1
E = 1.80 10 -3 N C perpendicular to r1 and counterclockwise P31.34 (a)
FIG. P31.33
z
Ed =
d B dt so E=
2 rE = r 2 (b)
e j dB dt
b9.87 mV mg cosb100 tg
The E field is always opposite to increasing clockwise B. .
Section 31.5 P31.35 (a) (b)
Generators and Motors
max = NAB = 1 000 0.100 0.200 120 = 7.54 kV
t = NBA sin t = NBA sin is maximal when sin = 1
or =
b
ga
fa
fa
f
af
2
FIG. P31.35
so the plane of coil is parallel to B .
226 P31.36
Faraday's Law
For the alternator, = 3 000 rev min
b
= -N
(a) (b) P31.37
d B d = -250 2.50 10 -4 dt dt
e
rad min gFGH 21rev IJK FGH 160 s IJK = 314 rad s T m j cosb314t sg = +250e 2.50 10
2
-4
T m 2 314 s sin 314t
jb
g a f
= 19.6 V sin 314t max = 19.6 V
a
f a f je ja f
B = 0 nI = 4 10 -7 T m A 200 m -1 15.0 A = 3.77 10 -3 T For the small coil, B = NB A = NBA cos t = NB r 2 cos t . Thus,
e
e j
2
=-
d B = NB r 2 sin t dt
= 30.0 3.77 10 -3 T 0.080 0 m
P31.38 As the magnet rotates, the flux through the coil varies sinusoidally in time with B = 0 at t = 0 . Choosing the flux as positive when the field passes from left to right through the area of the coil, the flux at any time may be written as B = - max sin t so the induced emf is given by
a fe
jb
g e4.00 s j sinb4.00 tg = a28.6 mVf sinb4.00 tg
-1
.
1
0.5
I/I max
0 0 0.5 0.5 1 1.5 2
1
t/T = ( t/2 )
=-
d B = max cos t . dt
FIG. P31.38
The current in the coil is then I = 850 mA *P31.39 120 V M
max = cos t = I max cos t . R R
850 mA
To analyze the actual circuit, we model it as 120 V
11.8
back
.
(a) (b)
The loop rule gives +120 V - 0.85 A 11.8 - back = 0
a
f
back = 110 V .
The resistor is the device changing electrical work input into internal energy: 2 P = I 2 R = 0.85 A 11.8 = 8.53 W .
a
fa
2
f
(c)
With no motion, the motor does not function as a generator, and back = 0 . Then 120 V - I c 11.8 = 0
c 2 c
I = 10.2 A a f P = I R = a10.2 A f a11.8 f = 1.22 kW
c
Chapter 31
227
P31.40
(a)
max = BA = B max
FG 1 R IJ H2 K = a1.30 Tf a0. 250 mf b 4.00 rad sg 2
2 2
t
max = 1.60 V
(b) (c) (d) (e) P31.41 (a)
Figure 1
2 0
=
2 0
z
BA d = 2 2
z
sin d = 0
t
The maximum and average would remain unchanged. See Figure 1 at the right. See Figure 2 at the right. B = BA cos = BA cos t = 0.800 T 0.010 0 m 2 cos 2 60.0 t =
Figure 2
FIG. P31.40
a
fe
j
a f e8.00 mT m j cosa377tf
2
(b)
=-
I=
d B = dt
a3.02 Vf sina377tf
(c) (d) (e)
= R
a3.02 Af sina377tf
a9.10 Wf sin a377tf
2
P = I2R =
P = Fv = so =
P =
a24.1 mN mf sin a377tf
2
Section 31.6 P31.42
Eddy Currents upward magnetic field, so the N and S poles on the
The current in the magnet creates an
solenoid core are shown correctly. On the rail in front of the brake, the upward flux of B increases as the coil approaches, so a current is induced here to create a downward magnetic field. This is clockwise current, so the S pole on the rail is shown correctly. On the rail behind the brake, the upward magnetic flux is decreasing. The induced current in the rail will produce upward magnetic field by being counterclockwise as the picture correctly shows.
228 P31.43
Faraday's Law
(a)
At terminal speed, Mg = FB = IwB = or
FG IJ wB = FG Bwv IJ wB = B w v H RK H R K R
T 2 2
T
vT =
MgR B 2 2
.
(b)
The emf is directly proportional to vT , but the current is inversely proportional to R. A large R means a small current at a given speed, so the loop must travel faster to get FB = mg .
FIG. P31.43
(c)
At a given speed, the current is directly proportional to the magnetic field. But the force is proportional to the product of the current and the field. For a small B, the speed must increase to compensate for both the small B and also the current, so vT B 2 .
Section 31.7
Maxwell's Equations i j k -e E + v B where v B = 10.0 0 0 = -4.00 j m 0 0 0.400
11
P31.44
F = ma = qE + qv B so a =
-19
e-1.60 10 j 2.50 i + 5.00 j - 4.00 j = e-1.76 10 j 2.50 i + 1.00 j 9.11 10 a = e -4.39 10 i - 1.76 10 jj m s
a=
-31 11 11 2
P31.45
F = ma = qE + qv B
a= i j k e E + v B where v B = 200 0 0 = -200 0.400 j + 200 0.300 k m 0.200 0.300 0.400
a
f
a
f
a=
1.60 10 -19 50.0 j - 80.0 j + 60.0k = 9.58 10 7 -30.0 j + 60.0k 1.67 10 -27
a = 2.87 10 9 - j + 2k m s 2 =
e-2.87 10 IJ K
9
j +5.75 10 9 k m s 2
j
Additional Problems P31.46
a f e j FGH a fLMN e j OPQa f = -a30.0fL e 2.70 10 mj Oe3.20 10 MN PQ = -e7.22 10 V j cos 2 e523t s j
= -N
-3 2 -3 -1
d dB BA cos = - N r 2 cos 0 dt dt 2 d = - 30.0 2.70 10 -3 m 1 50.0 mT + 3.20 mT sin 2 523 t s -1 dt
a
-3
T 2 523 s -1
j e
f e j j cose2 523t s j
-1
Chapter 31
229
P31.47
(a)
Doubling the number of turns.
Amplitude doubles: period unchanged
(b) Doubling the angular velocity.
doubles the amplitude: cuts the period in half
(c) Doubling the angular velocity while reducing the number of turns to one half the original value.
FIG. P31.47
Amplitude unchanged: cuts the period in half
P31.48
= -N
B 1.50 T - 5.00 T BA cos = - N r 2 cos 0 = -1 0.005 00 m 2 1 = 0.875 V t t 20.0 10 -3 s
a
f
e j
e
ja fFGH
IJ K
(a)
I=
R
=
0.875 V = 43.8 A 0.020 0
(b) P31.49
P = I = 0.875 V 43.8 A = 38.3 W
a
fa
f
In the loop on the left, the induced emf is
=
d B dB =A = 0.100 m dt dt
a
f b100 T sg = V
2
and it attempts to produce a counterclockwise current in this loop. In the loop on the right, the induced emf is
=
d B = 0.150 m dt
a
f b100 T sg = 2.25 V
2
FIG. P31.49
and it attempts to produce a clockwise current. Assume that I 1 flows down through the 6.00- resistor, I 2 flows down through the 5.00- resistor, and that I 3 flows up through the 3.00- resistor. From Kirchhoff's junction rule: Using the loop rule on the left loop: Using the loop rule on the right loop: Solving these three equations simultaneously, I3 = I1 + I 2 6.00 I 1 + 3.00 I 3 = 5.00 I 2 + 3.00 I 3 = 2.25 (1) (2) (3)
I1 = 0.062 3 A , I 2 = 0.860 A , and I 3 = 0.923 A .
230 P31.50
Faraday's Law
The emf induced between the ends of the moving bar is
= B v = 2.50 T 0.350 m 8.00 m s = 7.00 V .
The left-hand loop contains decreasing flux away from you, so the induced current in it will be clockwise, to produce its own field directed away from you. Let I 1 represent the current flowing upward through the 2.00- resistor. The right-hand loop will carry counterclockwise current. Let I 3 be the upward current in the 5.00- resistor. (a) Kirchhoff's loop rule then gives: and (b)
a
fa
fb
g
a f +7.00 V - I a5.00 f = 0
+7.00 V - I 1 2.00 = 0
3
I1 = 3.50 A I 3 = 1.40 A .
The total power dissipated in the resistors of the circuit is
P = I1 + I 3 = I 1 + I 3 = 7.00 V 3.50 A + 1.40 A = 34.3 W .
(c) Method 1: The current in the sliding conductor is downward with value I 2 = 3.50 A + 1.40 A = 4.90 A . The magnetic field exerts a force of Fm = I B = 4.90 A 0.350 m 2.50 T = 4.29 N directed moving. Method 2: The agent moving the bar must supply the power according to P = F v = Fv cos 0 . The force required is then: F= P31.51
b
g a
fa
f
a
fa
fa
f
toward the right on this to the left to keep the bar
conductor. An outside agent must then exert a force of 4.29 N
P 34.3 W = = 4.29 N . v 8.00 m s
Suppose we wrap twenty turns of wire into a flat compact circular coil of diameter 3 cm. Suppose we use a bar magnet to produce field 10 -3 T through the coil in one direction along its axis. Suppose we then flip the magnet to reverse the flux in 10 -1 s . The average induced emf is then BA cos B cos 180- cos 0 = -N = - NB r 2 t t t -2 2 = - 20 10 -3 T 0.015 0 m ~ 10 -4 V 10 -1 s
= -N
a fe
jb
g FGH
e jFGH IJ K
IJ K
Chapter 31
231
P31.52
I=
+ induced R
and
induced = -
F=m
d BA dt
a f b g
FIG. P31.52
To solve the differential equation, let
dv = IBd dt dv IBd Bd = = + induced dt m mR dv Bd = - Bvd dt mR u = - Bvd du dv = - Bd dt dt 1 du Bd - = u Bd dt mR
a
f
so Integrating from t = 0 to t = t , or Since v = 0 when t = 0 , and
u0
z
u
t Bd du dt . =- u mR 0
za f
a f
2
Bd u t =- ln u0 mR 2 2 u = e - B d t mR . u0 u0 = u = - Bvd
2
- Bvd = e - B
Therefore, *P31.53 The enclosed flux is The particle moves according to v=
2 2
d t mR
2 2
.
Bd
e1 - e
- B d t mR
j
.
B = BA = B r 2 .
F = ma :
qvB sin 90 = r= mv . qB
mv 2 r
Then
B =
B m 2 v 2 q2B2
2
.
(a)
v=
Bq 2B
m2
=
e15 10
-6
T m 2 30 10 -9 C
je
2 10
e
-16
kg
j
2
j a0.6 Tf =
2.54 10 5 m s
(b)
Energy for the particle-electric field system is conserved in the firing process:
Ui = K f :
qV =
1 mv 2 2
2 10 -16 kg 2.54 10 5 m s mv 2 V = = 2q 2 30 10 -9 C
e
e
je
j
j
2
= 215 V .
232 *P31.54
Faraday's Law
(a)
Consider an annulus of radius r, width dr, height b, and resistivity . Around its circumference, a voltage is induced according to
= -N
d d B A = -1 Bmax cos t r 2 = + Bmax r 2 sin t . dt dt
b
g
The resistance around the loop is The eddy current in the ring is dI = The instantaneous power is
2 r = . Ax bdr
b g
Bmax r 2 sin t bdr Bmax rbdr sin t . = = 2 resistance 2 r
b g
b
g
dPi = dI =
2 Bmax r 3 b 2 dr sin 2 t . 2
The time average of the function
sin 2 t =
1 1 1 1 - cos 2 t is - 0 = 2 2 2 2
so the time-averaged power delivered to the annulus is dP =
2 Bmax r 3 b 2 dr . 4
The power delivered to the disk is
P = dP =
z z
R
P=
(b) (c) (d) I=
2 2 Bmax b 2 R 4 Bmax R 4 b 2 -0 = . 4 4 16 2 Bmax and P get 4 times larger.
F GH
I JK
2 Bmaxb 2 3 r dr 4 0
When Bmax gets two times larger,
When f and = 2 f double, 2 and P get 4 times larger. When R doubles, R 4 and P become 2 4 = 16 times larger.
P31.55
B A = R R t
so q = It =
b15.0 Tga0.200 mf
0.500
2
= 1.20 C
P31.56
(a)
I=
dq d B = where = -N so dt dt R
z
dq =
N R
2 1
z
d B N 2 - 1 . R
and the charge through the circuit will be Q =
b
g
(b)
IJ OP = BAN KQ R RQ a 200 fe5.00 10 C j = = so B = NA a100fe 40.0 10 m j
Q= N BA cos 0 - BA cos R 2
LM N
FG H
-4
-4
2
0.250 T .
Chapter 31
233
P31.57
(a) (b) (c)
= B v = 0.360 V
FB = I B = 0.108 N
I=
= 0.900 A R
Since the magnetic flux B A is in effect decreasing, the induced current flow through R is from b to a. Point b is at higher potential.
FIG. P31.57
(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.
P31.58
= B v at a distance r from wire =
F II v GH 2 r JK
0
v
FIG. P31.58 P31.59 (a) At time t, the flux through the loop is B = BA cos = a + bt r 2 cos 0 = a + bt r 2 . At t = 0 , B = ar 2 . (b)
a
fe j
a
f
=-
d a + bt d B = - r 2 = - br 2 dt dt
a
f
(c)
I=
br 2 = - R R
(d)
P =I= -
F GH
br 2 R
I e- br j = JK
2
2b 2r 4 R
P31.60
=-
d dB NBA = -1 a2 = a2K dt dt Q = C = C a 2 K B into the paper is decreasing; therefore, current will attempt to counteract this. Positive charge will go to upper plate . The changing magnetic field through the enclosed area induces an electric field , surrounding the B-field, and this pushes on charges in the wire.
a
f
FG IJ H K
(a) (b)
(c)
234 P31.61
Faraday's Law
The flux through the coil is B = B A = BA cos = BA cos t . The induced emf is
= -N
(a) (b) (c)
d cos t d B = - NBA = NBA sin t . dt dt
b
g
max = NBA = 60.0 1.00 T 0.100 0.200 m 2 30.0 rad s = 36.0 V
d B d B = , thus dt N dt =
max
a
fe
jb
g
max 36.0 V = = 0.600 V = 0.600 Wb s N 60.0
At t = 0.050 0 s , t = 1.50 rad and = max sin 1.50 rad = 36.0 V sin 1.50 rad = 35.9 V .
a
f a a
f a
f
(d)
The torque on the coil at any time is
= B = NIA B = NAB I sin t =
When = max
FG IJ FG IJ sin t . H KH RK a36.0 V f = = , sin t = 1.00 and = R b30.0 rad sga10.0 f f
max 2 max 2
4.32 N m .
P31.62
(a)
We use = -N
B , with N = 1 . t Taking a = 5.00 10 -3 m to be the radius of the washer, and h = 0.500 m , B = B2 A - B1 A = A B2 - B1 = a 2
b
g
F I - I I = a I FG 1 - 1 IJ = - ahI . GH 2 ah + af 2 a JK 2 H h + a a K 2ah + af
0 0 2 0 0
The time for the washer to drop a distance h (from rest) is: Therefore, =
t =
2h . g
and (b)
a f 2 h = 2a h + a f 2 e4 10 T m Aje5.00 10 mja10.0 Af e9.80 m s ja0.500 mf = = 2 2b0.500 m + 0.005 00 mg a f
-7 -3 2
0 ahI ahI = 0 2 h + a t 2 h + a
g
0 aI
gh
97.4 nV .
Since the magnetic flux going through the washer (into the plane of the paper) is decreasing in time, a current will form in the washer so as to oppose that decrease. Therefore, the current will flow in a clockwise direction .
P31.63
Find an expression for the flux through a rectangular area "swept out" I by the bar in time t. The magnetic field at a distance x from wire is B=
0I and B = BdA . Therefore, 2 x 0 Ivt r + dx where vt is the distance the bar has moved in time t. 2 r x
d B 0 Iv = ln 1 + dt r 2
z
r
v
vt
B =
z
FIG. P31.63
Then, =
FG H
IJ K
.
Chapter 31
235
P31.64
The magnetic field at a distance x from a long wire is B = through the loop. d B =
0I . Find an expression for the flux 2 x
0I dx so 2 x
a f
B =
0I 2
r+w r
z
w dx 0 I = ln 1 + r x 2
FG H
IJ K a f
.
Therefore,
d B 0 I v w 0I v w =- = and I = = dt 2 r r + w R 2 Rr r + w
a
f
P31.65
We are given and Maximum E occurs when which gives Therefore, the maximum current (at t = 1.00 s ) is
B = 6.00t 3 - 18.0t 2 T m 2
e
j
=-
d B = -18.0t 2 + 36.0t . dt
d = -36.0t + 36.0 = 0 dt t = 1.00 s . I=
-18.0 + 36.0 V = = 6.00 A . R 3.00
a
f
P31.66
For the suspended mass, M: For the sliding bar, m: Mg -
F = Mg - T = Ma .
B = ma , where
F =T -I
B2 2 v = m + M a or R
a
f
B v = R R Mg dv B2 2v a= = - dt m + M R M + m
I=
a
f f
2
zb
v 0
t dv = dt where - v 0
g
z
=
v=
Mg B2 2 . and = R M+m M+m
a
Therefore, the velocity varies with time as
2 MgR 1 - e- t = 1 - e-B 2 2 B
e
j
t R M +m
a
f
.
P31.67
(a)
= -N
d B dB d = - NA = - NA 0nI where A = area of coil dt dt dt
b
g
N = number of turns in coil and Therefore, n = number of turns per unit length in solenoid. d 4 sin 120 t = N 0 An 480 cos 120 t dt 2 = 40 4 10 -7 0.050 0 m 2.00 10 3 480 cos 120 t
= N 0 An
=
V R
a f b g b g g e e j b ja f b a1.19 Vf cosb120 tg
P = VI =
cos 2 =
2
g
(b)
I=
and
a1.19 V f
f
2
2
cos 2 120 t
b
g
8.00
From 1 the average value of cos is , so 2
1 1.19 V P = = 88.5 mW . 2 8.00
a a
1 1 + cos 2 2 2
f
236 *P31.68
Faraday's Law
(a)
= -N
d a2 Ba 2 d 1 1 d 2 BA cos = -1 B cos 0 = - = - Ba 2 = - 0.5 T 0.5 m 2 rad s dt dt 2 2 dt 2 2 = -0.125 V = 0.125 V clockwise
a
fa
f
The sign indicates that the induced emf produces clockwise current, to make its own magnetic field into the page. (b) At this instant = t = 2 rad s 0.25 s = 0.5 rad . The arc PQ has length r == 0.5 rad 0.5 m = 0.25 m . The length of the circuit is 0.5 m + 0.5 m + 0.25 m = 1.25 m its 0.125 V resistance is 1.25 m 5 m = 6.25 . The current is = 0.020 0 A clockwise . 6.25
a
fa
b
f
a
f
g
*P31.69
Suppose the field is vertically down. When an electron is moving away from you the force on it is in the direction given by qv B c as - away down = -
b
g
= - left = right .
Therefore, the electrons circulate clockwise. FIG. P31.69 (a) As the downward field increases, an emf is induced to produce some current that in turn produces an upward field. This current is directed counterclockwise, carried by
negative electrons moving clockwise. Therefore the original electron motion speeds up. (b) At the circumference, we have
Fc = ma c :
q vBc sin 90 =
mv 2 r
mv = q rBc .
The increasing magnetic field Bav in the area enclosed by the orbit produces a tangential electric field according to dB d r dBav E 2 r = r 2 av E= E ds = - B av A . dt dt 2 dt dv qE=m . An electron feels a tangential force according to Ft = mat : dt r dBav r dv q q B av = mv = q rBc =m Then dt 2 dt 2 Bav = 2Bc . and
z
b g
P31.70
The induced emf is = B v where B = y f = yi -
0I , v f = vi + gt = 9.80 m s 2 t , and 2 y
e
j
f a0.300 mf 9.80 m s t = e1.18 10 jt e j 0.800 - 4.90t 2 0.800 m - e 4.90 m s jt e1.18 10 ja0.300f V = 98.3 V . At t = 0.300 s , = 0.800 - 4.90a0.300 f
=
e4 10
1 2 gt = 0.800 m - 4.90 m s 2 t 2 . 2
e
j
-7
T m A 200 A
2
ja
-4
2
2
2
V
-4
2
Chapter 31
237
P31.71
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 I magnitude of the field is B = 0 . Thus, the flux linkage is 2 r N B =
0 NIL h+ w dr 0 NI max L h+ w = ln sin t + . h r 2 2 h
z
FG H
IJ b K
g
Finally, the induced emf is
FIG. P31.71
FG IJ b g H K e4 10 ja100fa50.0fa0.200 mfe200 s j lnFG 1 + 5.00 cm IJ cosb t + g =- H 5.00 cm K 2 = -a87.1 mV f cosb 200 t + g The term sinb t + g in the expression for the current in the straight wire does not change
=- 0 NI max L w ln 1 + cos t + h 2
-7 -1
appreciably when t changes by 0.10 rad or less. Thus, the current does not change appreciably during a time interval t <
e200 s j
-1
0.10
= 1.6 10 -4 s .
We define a critical length, ct = 3.00 10 8 m s 1.6 10 -4 s = 4.8 10 4 m equal to the distance to which field changes could be propagated during an interval of 1.6 10 -4 s . 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. If the frequency w were much larger, say, 200 10 5 s -1 , the corresponding critical length would be only 48 cm. In this situation propagation effects would be important and the above expression for would require modification. As a "rule of thumb" we can consider field propagation effects for circuits of laboratory size to be negligible for frequencies, f = d B = -BA sin ; dt
e
je
j
2
, that are less than about 10 6 Hz.
P31.72
B = BA cos I - sin
IB sin - sin 2
FIG. P31.72
238
Faraday's Law
ANSWERS TO EVEN PROBLEMS
P31.2 P31.4 P31.6 P31.8 0.800 mA (a) see the solution; (b) 3.79 mV ; (c) 28.0 mV 78.5 s P31.40 P31.42 P31.44 I counterclockwise; t I ; (c) upward t P31.46 P31.48 P31.50 P31.52 P31.54 (a) 1.60 V ; (b) 0; (c) no change; (d) and (e) see the solution both are correct; see the solution
0 n r22 2R 2 0 n r22 (b) 4r1 R
(a)
e-4.39i - 1.76 jj10 m s -a7.22 mV f cosb 2 523 t sg
11 2
(a) 43.8 A ; (b) 38.3 W (a) 3.50 A up in 2 and 1.40 A up in 5 ; (b) 34.3 W ; (c) 4.29 N see the solution (a)
2 Bmax R 4 b 2 ; (b) 4 times larger; 16 (c) 4 times larger; (d) 16 times larger
P31.10 P31.12 P31.14 P31.16 P31.18 P31.20 P31.22 P31.24 P31.26 P31.28 P31.30 P31.32
-14.2 mV cos 120 t
61.8 mV
a f
(a) see the solution; (b) 625 m/s see the solution 13.3 mA counterclockwise in the lower loop and clockwise in the upper loop. 1.00 m s (a) 500 mA; (b) 2.00 W ; (c) 2.00 W 24.1 V with the outer contact positive 121 mA clockwise (a) to the right; (b) to the right; (c) to the right; (d) into the paper negative; see the solution (a) 8.00 10 -21 N downward perpendicular to r1 ; (b) 1.33 s (a) 9.87 mV m cos 100 t ; (b) clockwise
P31.56 P31.58 P31.60
(a) see the solution; (b) 0.250 T see the solution (a) C a 2 K ; (b) the upper plate; (c) see the solution (a) 97.4 nV ; (b) clockwise
P31.62 P31.64
0I v w 2 Rr r + w
a
f
2 2
P31.66 P31.68
MgR B
2 2
1 - e-B
t R M +m
a
f
(a) 0.125 V to produce clockwise current; (b) 20.0 mA clockwise 1.18 10 -4 ; 98.3 V 0.800 - 4.90t 2 see the solution
P31.34 P31.36 P31.38
b g b g (a) a19.6 V f sina314tf ; (b) 19.6 V
see the solution
P31.70 P31.72