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Unformatted text preview: Induced Voltages and Inductance 1'79 Problem Solutions 20.1 The magnetic flux through the area enclosed by the 100p is
<1), = BAcosa = B(7rr2)cosO° = (0.30 T)[n‘(0.25 mf] = 5.9x10‘2 T m2 20.2 The magnetic ﬂux through the loop is given by (DB = BA cost) where B is the magnitude of the magnetic field, A is the area enclosed by the loop, and (9 is the angle the magnetic
field makes with the normal to the plane of the loop. Thus, 10'2 m
1 cm 2
q), =BACO36=(5.00><10‘5 T)[20.0 cm2[ J ]c056=(1.00x10'7 Tm2)cosl9 (a) When i is perpendicular to the plane of the loop, 19 = 0" and «1),: 1.00x10'7 Tmz
(b) If 9:300; then «b, =(1.OO><10‘7 Tm2)cos.30.0°= 3.66x10's Tm2 (c) If 6:900", then o,=(1.00x107 Tm2)cosgo.0°='§ ' 20.3 The magnetic ﬂux through the loop is given by (113 = BA c056 where B is the magnitude of the magnetic field, A is the area enclosed by the loop, and n9 is the angle the magnetic
field makes with the normal to the plane of the loop. Thus, a, = sAcosa = (0.300 T)(2.00 rn)2 cosSOD" = 7.71x 101 T  m2 20.4 The magnetic field lines are tangent to the surface of the cylinder, so that no magnetic
field lines penetrate the cylindrical surface. The total flux through the cylinder is l
i 20.5 (a) Every field line that comes up through the area A on one side of the wire goes back
down through area A on the other side of the wire. Thus, the net ﬂux through the coil is (b) The magnetic field is parallel to the plane of the coil, so 6 = 90.0°. Therefore,
(1),, = BACOSB = BAC0590.0° =IEI 130 CHAPTER 20 20.6 The magnetic field generated by the current in the solenoid is 250 B =,uonI «1 (47rx10'7 Tm/A)[W
. m
(15.0 A)=2.36><10'2 T and the flux through each turn on the solenoid is (1),, =BAcosl9 ”(4.00x10‘2 m)2 4 cosO°= 2.96x10'5 Tm‘z =(2.3cs><102 T) 20.7 (a) The magnetic flux through an area A may be written as (133 =(BcosB)A (component of B perpendicular to A) ~ A Thus, the flux through the shaded side of the cube is o, = B, A =(5.0 T)(2.5><10"1 m)2 = (b) Unlike electric field lines, magnetic field lines always form closed loops, Without
beginning or end. Therefore, no magnetic ﬁeld lines originate or terminate within
the cube and any line entering the cube at one point must emerge frOm the cube at some other point. The net flux through the cube, and indeed through any closed
surface, is zero . A¢IB (AB)ACO56 _ (1'5 T"0)[ﬂ(16X10'3 m)2]c050° 20.3 8 = =—— m=10 10"1 V: 0.1 V
I I At At 120x10”3s X A¢E _ BAAcosB
At At 20.9 From IE I = , we find that 8 18x10'3 V '
B=%:———= .1 T “
(AA/At)c056 (0.10 mZ/s)cosO° I : Induced Voltages and Inductance 181 ; AKDB _ B(AA)C056
l 20.10 £ =
At At W (0.15 T)[m(0.12 m)2 —0:c050° ———————~——:3.4x10'2 v: 020 s A<I>B _ A(Bcos6l)A 20.11 The magnitude of the induced emf is [5 = At H At If the normal to the plane of the loop is considered to point in the original direction of
the magnetic field, then 61 = 0° and 9f : 180°. Thus, we find . T 1 °— .301" 00 0.30 2
5_£'_20__122880_w_225__e(_m)9_4m2 v: 1.5 S
20,12 g=A¢B =M§£§Bﬂ so B=M ;
At At ’ NA[A(cosB)] i .1 . '3 5
or B ____(£)(277:1fo)mnu“=5._20>.;10’5 T: = 500[7r(0.150 m)2/4][COSD° —c0390°] : 20.13 The required induced emf is 8 = IR = (0.10 A)(8.0 Q) = 0.80 V. From '5': AER =[i—ijAcosd A_B_ IEI _ 0.80V : At _ NACOSE) _ (75)[(0.050 m)(0.080 m)]c050° 20.14 The initial magnetic field inside the solenoid is 100
0.200 m B =,uanI =(4nx10’7 Tm/A)[ )(300 A) =1.88><10‘a T (a) £135 =BACOSB=(1.88X10'3 T)(1.00><10’2 m)2c050° = 1.88X10’7 T  In2 l 182 CHAPTER 20 (b) When the current is zero, the ﬂux through the loop is (1),, = 0 and the average
induced emf has been 47 . 2_
e= A3,, = 1.33m: 00Tsm 0 : 6.28x10'3 V 20.15 If the solenoid has current I, the magnetic field inside it is 300
0.200 m B=yunI=(47r><10‘7 Tm/A)[ ]1=(6.00:rr><10‘4 T/A)I
(a) Afbﬂ =(AB)Ac056
:[(6.007rx10‘4 T/A)(5.0 A—2.0 A)][:~;(1.5><102 m)2:lcosO° =— N Ad) 4 4.0x10'6 Tm2
(b) €l= ( “)=——_( )=1.8x10‘5V= 18.;1V
At 0.905 20.16 The magnitude of the average emf is N(AI§DB) 2 NBA[A(C056)] '8' 2 At At 2000.1 T)(100x10“‘ m2)(coso°—c05180°)
_ 0.10 s =44V Therefore, the average induced current is I = I? = $6 = 20.17 If the magnetic field makes an angle of 280" with the plane of the coil, the angle it makes
with the normal to the plane of the coil is 19 = 62.0°. Thus, N(A<I)B) m NB(AA)C058
At _ At _ 200(50.0><10*‘ T)[(39.0 cm2)( 1 1112/10" cm2)]cossz.0° 5: _ =1.02><10"5 V: 10.2 v
3.; 1.805 1
i
l
IEI=
l
I
l Induced Voltages and Inductance 183 20.18 From a = 350, the required speed is 20.19 8 = B iEv , where B i is the component of the magnetic field perpendicular to the velocity
‘ 6. Thus, a =[(50.0x10"’ T)sm53.0°](60.0 m)(300 m/s) = 20.20 The speed of the beam after falling freely for 9.00 m, starting from “35* (on = 0) I is = v§y+2ay(Ay) = 0+2(~—9.so m/szxngno m) =13,3 m/S Since the induced emf is 8 = B LEI) , where BL is the component of the magnetic field
perpendicular to the velocity i’r , we find .9 =(18.0x10*‘ T)(12.0 m)(13.3 m/s) 2 2.87x10“3 v = "1 2021 (a) Observe that only the horizontal component, 3,, , of Earth’s magnetic field is effective in exerting a vertical force on charged particles in the antenna. For the
magnetic force, Fm = quh sinB , on positive charges in the antenna to be directed upward and have maximum magnitude (when 6:90”), the car should move toward the east through the northward horizontal component of the magnetic ﬁeld. (b) 6 = kav , where 8,, is the horimntal component of the magnetic field. h :3'=[(50.0><106 T)cosss.oo](1.20 m)[[65_ 0 k_m][9m_m/SH 1 km/h = 4.58><1(J“‘1 V 184 20.22 20.23 20.24 CHAPTER 20 During each revolution, one of the rotor blades sweeps out a horizontal circular area of
radius f: A = 7:62 . The number of magnetic ﬁeld lines cut per revolution is
NDB = B J_A = B A . The induced emf is then vertical 8— At _ 1/f __ 0.505 “2.8x10 Vum (a) To oppose the motion of the magnet, the magnetic field generated by the induced
current should be directed to the right along the axis of the coil. The current must then be left to right through the resistor. (b) The magnetic field produced by the current should be directed to the left along the axis of the coil, so the current must be right to left through the resistor. (a) As the bottom conductor of the loop falls, it cuts across l“_w—"
the magnetic field lines coming out of the page. This    ‘
induces an emf of magnitude 8 = Bwv in this conductor, IT
with the left end at the higher potential. As a result, an
induced current of magnitude I"§_Bwv
R R flows clockwise around the loop. The ﬁeld then exerts
an upward force of magnitude 2 2
Pm =BIw—B[m]wa w v
R R on this currentcarrying conductor forming the bottom of the loop. If the loop is
falling at terminal speed, the magnitude of this force must equal the downward
gravitational force acting on the loop. That is, when U = 11,, we must have 321020, _ MR (b) A larger resistance would make the current smaller, so the loop must reach higher
speed before the magnitude of the magnetic force will equal the gravitational force. (c) The magnetic force is proportional to the product of the field and the current, while .
the current itself is proportional to the field. If B is cut in half, the speed must
become four times larger to compensate and yield a magnetic force with magnitude
equal to the that of the gravitational force. Induced Voltages and Inductance 185  20.25 (a) After the right end of the coil has
entered the field, but the left end
has not, the ﬂux through the area _V;.
enclosed by the coil is directed into T 
the page and is increasing in w 
magnitude. This increasing flux _i_ H ii induces an emf of magnitude I j? l E = & =———NB(AA) = NEW:
At At in the loop. Note that in the above equation, AA is the area enclosed by the coil that
enters the field in time At . This emf produces a counterclockwise current in the
loop to oppose the increasing inward flux. The magnitude of this current is I = S/R = NBwv/R . The right end of the loop is now a conductor, of length Nw, carrying a current toward the top of the page through a field directed into the page.
The' field exerts a magnetic force of magnitude 2 2 2
P : BI(Nw) = B[Niwv](Nw) = directed toward the left on this conductor, and hence, on the loop. When the loop is entirely within the magnetic field, the ﬂux through the area
enclosed by the loop is constant. Hence, there is no induced emf or current in the loop, and the field exerts force on the loop. After the right end of the 1001:) emerges from the field, and before the left end
emerges, the flux through the loop is directed into the page and decreasing. This
decreasing ﬂux induces an emf of magnitude ]£ = NBwv in the loop, which produces an induced current directed clockwise around the loop so as to oppose the
decreasing flux. The current has magnitude I = B/R = NBwv/R . This current ﬂowing upward, through conductors of total length Nw, in the left end of the loop,
experiences a magnetic force given by 2 2 2
F = BI (Nw) = B[Niwv](Nw) : directed toward the left . 20.26 When the switch is closed, the magnetic field due to the current frorn the battery will be
directed to the left along the axis of the cylinder. To oppose this increasing leftward flux,
the induced current in the other loop must produCe a field directed to the right through the area it encloses. Thus, the induced current is left to right through the resistor. 186 CHAPTER 20 20.27 Since the magnetic force, PM = qu sin6 , on a positive charge is directed toward the top
of the bar when the velocity is to the right, the right hand rule says that the magnetic field is directed . 20.28 When the switch is closed, the current from the battery produces a magnetic ﬁeld
directed toward the right along the axis of both coils. (a) As the battery current is gmwing in magnitude, the induced current in the
rightmost coil opposes the increasing rightward directed field by generating a field toward to the left along the axis. Thus, the induced current must be left to right
through the resistor. (b) Once the battery current, and the field it produces, have stabilized, the ﬂux through the rightmost coil is constant and there is . (c) As the switch is Opened, the battery current and the field it produces rapidly
decrease in magnitude. To oppoSe this decrease in the rightward directed field, the
induced current must produce a field toward the right along the axis, so the induced current is right to left through the resistor. 20.29 When the switch is closed, the current from the battery produces a magnetic ﬁeld
directed toward the left along the axis of both coils. (a) As the current from the battery, and the leftward field it produces, increase in
magnitude, the induced current in the leftmost coil opposes the increased leftward field by ﬂowing right to left through R and producing a field directed toward the
right along the axis. (b) As the variable resistance is decreased, the battery current and the leftward field
generated by it increase in magnitude. To oppose this, the induced current is right to left through R, producing a field directed toward the right along the axis. (c) Moving the circuit containing R to the left decreases the leftward field (due to the
battery current) along its axis. To oppose this decrease, the induced current is left to right through R, producing an additional field directed toward the left
along the axis. (d) As the switch is opened, the battery current and the leftward field it produces
decrease rapidly in magnitude. To oppose this decrease, the induced current is left to right through R, generating additional magnetic field directed toward the
left along the axis. Induced Voltages and Inductance 187 : 20.30 gm“ : NBhorizuntaIAa) = 100(20X 105 T) (0.20 m): [1500 re'v ][ 2n. rad ][1 mm]
: mm 1 rev 60 S =1sx102 V: S 20.31 Note the similarity between the situation in this problem and a generator. In a generator,
one normally has a loop rotating in a constant magnetic ﬁeld so the flux through the
loop varies sinusoidally in time. In this problem, we have a stationary loop in an
oscillating magnetic field, and the ﬂux through the loop varies sinusoidally in time. In
both cases, a sinusoidal emf 8 = 8W sinwt where 8m : NBAaJ is induced in the loop. The loop in this case consists of a single band (N = 1) around the perimeter of a red blood cell with diameter d = 8.0x10'ﬁ m . The angular frequency of the oscillating flux
through the area of this loop is a) : 2% f = 23(60 Hz) 2 120:: rad/s. The maximum induced emf is then 3 —s 2 1
8W:NBM=B[§:_2JM(L0><10 T)2r(8.0:10 m) (120“, )=_ 20.32 (a) Immediately after the switch is closed, the motor coils are still stationary and the E 240V
13 k mf'sz .Thus,I=—=—m= 8.0A
ac e 1 ero R 309 (b) At maximum speed, Shack = 145 V and seem 240v—145v = r =__+———= .2A I R 30..
(c) em. = E—IR2240 v—(6.0 A)(30 o) = 188 CHAPTER 20 20.33 (a) When a coil having N turns and enclming area A rotates at angular frequency a) in
a constant magnetic field, the emf induced in the coil is 8 =5”): sinwt where 8",“ = NBlAa) Here, Bl is the magnitude of the magnetic field perpendicular to the rotation axis of
the coil. In the given case, B i = 55.0 pl"; A = 7m!) where a = (10.0 crn)/2 and
b = (4.00 cm)/2; and a) : 27: f = 24100 “a," M1 mm ] = 105 rad/s
mm 60.0 5 Thus, 8m: (10.0)(55.0><10r6 T)[Z:(O.100 m)(0.040 0 111)](105 rad/s) or am =1.81><10'5 v = (b) When the rotation axis is parallel to the field, then B l = 0 giving 8 = [El max It is easily understood that the induced emf is always zero in this case if you recognize that the magnetic field lines are always parallel to the plane of the coil,
and the flux thrOugh the coil has a constant value of zero. 20.34 (a) Using me=NBAw, 2.7 rad
a =1000 0.20T 0.10 t 60 BI J =7.5><103= 75 kV
max ( l( m l[[ s 1 W (b) Em“ ocmrs when the flux through the loop is changing the most rapidly. This is when the plane of the 100p is parallel to the magnetic field . 20.35 w=[120 113v J£1mm)(27zrad]=4ﬂ ﬂ min 60 s 1 rev 5
and the period is T = 2;}; = 0.50 s (a) em = NBAa): 500(0.60 T)[(0.080 m)(0.20 m)](4rr rad/s) = (b) 8 = Em“ sin(a)t) =(60 V)sin[(47r rad/s)[3—7; 5]] : Induced Voltages and Inductance 189 (c) The emf is first maximum at t = I = 0'50 S = 0.36 Treating the coiled telephone cord as a solenoid, , 2 73' _ 2
”ONZA =(4nx10’ T.m/A)(70.0) [Z(1.30x10 2) ]
e 0.600m =1.36x10"6 H: L: AI , 1.50 A“0.20 A
=L—= 3.00x103II m =2.0 1 “2 = 20 v
1M1 )[ 0203 ) xov 83V @37 :' 2
.33 The units of N‘DB are T 'm
A
From the force on a moving charged particle, P = qu , the magnetic field is B a i and
q’U
we find that
1 T = 1 N = 1 N ‘ S ThUS, T4112 =(éqS jmz =(—N;(I:1—).—Sl:[%]~s=v.s I 2 r
and T m = D which is the same as the units of E
A A AI/At #DNZA (4:7)(10‘7 Tm/A)(400)2[2r(2.5x10'2 Inf]
8 : 0.20m :2.O><10‘3 H: ‘ 9 (a) L: A_I:]E = 75x10'3 V F = — —u——:
(b) mmlgl MAI/M’At L 2.0x10’3H l'
E
E
190 CHAPTER 20 E 20.40 From s = L(AI/At) , the self—inductance is s . ‘3
L=______=El_4_[_).)_(10_v=2.40>~<10'6 H AI/At 10.0 A/s Then, from L = NEDB /I , the magnetic ﬂux through each turn is
. ZZJIUXIO'3 H 4.00 A
(DB =H=# = 1.92).:10‘5 Tm2
. N 500 20.41 The inductive time constant is z“ = UK . From € = MAI/At) , the selfinductance is : JEL with units of ——V— = [X]  s = $2  5. Thus, the units of the he constant are
AI/At A/s A 20.42 (a) The time constant of the RL circuit is 'r = L/R , and that of the RC circuit is r = RC . If
the two time constants have the same value, then L {L I 3.00H
RC=—, R: —= —=1.00x103§2= 1.00 kg
R or C 3.00x10“ F (b) The common value of the two time constants is
L 3.00 H a t:=—+—+——=3.00x10 s= 3.001115 R 1.00x103 g 20.43 The maximum current in a RL circuit I max = 8 / R , so the resistance is R=i 6.0V :ZOQ I _ 0.300 A max The inductive time constant is r : L/R , so i
i
} L=r«R=(600><10‘6 s)(20 Q)=1.2><10‘2 H:
i Induced Voltages and Inductance 191 . 20.44 The current in the RL circuit at time t is I 2%(1— 2"") . The potential difference across the resistor is AVR = R1 = E (1 — 8"” ) , and from Kirchhoff’s loop rule, the potential difference across the inductor is AVL = 8 ~ AVR = £[1 —(1 —e*f*)] = ger/r
(a) At t=0, AVR =£(1—e'°)=5(1_1)[email protected]
(b) Att=r, AVR=5(1—E'1):(6.0V)(1—0.368)= (c) At t=0, AVL=8€_0 :8:
(d) At t=r, AVL =85"‘ =(6.o V)(U.368)= 20.45 From I = 1m (1  e4“) , 5‘” = 1 71—
If —I—=0.900 at t= 3.00 s , then
e—W/r : 0.100 or r =—i§:m—5=1.30 s
ln(0.100) Since the time constant of an RL circuit is 1' = L/ R , the resistance is (d) 1 = IN (16”) yields a”? = 1 4/1,“, , and t=—2'ln(1dI/1mx)=—(2.00 ms)ln(1—0.800) = 192 CHAPTER 20 1 20.47 PEL =§L12 =§(70.O><10'3 H)(2.00 A)2 = 20.48 (a) The inductance of a solenoid is given by L = ,uoNzA/ﬂ , where N is the number of turns on the solenoid, A is its cross«5ectional area, and ﬂ is its length. For the given
Solenoid, L #0N2(71'r2) (Ammo7 Tm/A)(300)2zr(5.00x10'2 mf L %= = 4.44x10'3 H
I, 0.200... _ (b) When the solenoid described above carries a current of I = 0.500 A , the stored energy is
_ 1 2 1 1 3 2 _ —4
PEL “EU —E(4.44><10 H)(0.500 A) — 5.55x10 I
20.49 The current in the cirCuit at time t is I 2%(1— 34” ) , and the energy stored in the
inductor is PEL szI2
(a) As t—>oo, balmx =£:ﬂ=3.0 A, and
R 8.0 Q
1 1 2
PE eLiz :— 4.0 H 3.0 A = 18
L LP) 2 max 2( )( ) u (b) At i=1, I=Imax(1ee'1)=(3.0 A)(1—0.368):l.9 A and PE, minis H)(1.9 A): = 7.2 I 20.50 (a) When the two resistors are in series, the total resistance is "a R, = R + R = 2R , and the time constant of the circuit is T = i =
" Rm 2R (b) With the resistors now connected in parallel, the total resistance is 31’:
R R
( )(R) = B , and the time constant is 2' = —LM
2 R... R “'2 R+R 20.51 ._' 20.52 20.53 Induced Voltages and Inductance 193 According to Lenz’s law, a current will be induced in the coil to oppose the change in
magnetic flux due to the magnet. Therefore, current must be directed from b to a through the resistor, and V2 —Vb will be —. While the coil is between the poles of the magnet, the component of the field
perpendicular to the plane of the coil is B,. = 0.10 T. After the coil is pulled out of the field, B), = 0. The magnitude of the average induced emf as the coil is moved is A1133 :N A(BA) =N (AB)A lgl =N
At At At and the average induced current in the galvanorneter is gl _ N(AB)A = 10(0.10 T—0)[¢r(0.020 Inf] R R(At) (2.0 Q)(0.20 s)
: 3.1><10'3 A = 3 100 ,uA
This means the galvanometer will definitely show the induced current and even be overloaded. (a) The current in the solenoid reaches I = 0.6321,,“ in a time of t= r = L/R , Where ,uONZA _ (4nx10’7 Trn/A)(12 500)2(1.00><10*L m2)
1? " 7.00x10'2 m 0.280H_ 4 _
Thus, t= 14.09 —2.00x10 5— (b) The change in the solenoid current during this time is L: = 0.280 H AI = 0.632% — 0 = 0.53293) = 0.632(ﬂ9X] = 2.71 A
R 14.0 u so the average back emf is M 2.7M
s 2L m = 0.280H _— = 37.9V
“K“ (At) ( )[ZOOXIU'ZS] 194 CHAPTER 20 (c) &= (ABM AM %N_._(.A__I_)A ‘
At At At 212 .(At)
(42mm7 Tm/A)(12 500)(2.71 A)(1.00><10*1 m2)
=W= 1.52 *3 v .~
2(7.00>~<10'2 m)(2.00><10‘2 s) _
8m,., N¢M(A¢B/At) (820)(1.52><10‘3 v) ».
d = =m=W:O. : 1,
() I R R 2409 0519A mil call 20.54 2‘:
10 mV (3)
Original Curve .5
5 mV “ '15
0 .
0.5 0.75 1.0 2.0
—5 mV v
—10 mV (a) Doubling the number of turns doubles the amplitude but does not alter the period. (b) Doubling the angular velocity doubles the amplitude and also cuts the period in
half. (c) Doubling the angular velocity while reducing the number of turns to one half the
original value leaves the amplitude unchanged but does cut the period in half. 20‘55 Q :Iav (At): Ea. _1 00:3 _B(M)
.. w .. JimT 15.0 10“6T 0.200 2—0
or QL—XTTEEJTElwl=LZOXIO'° C: Induced Voltages and Inductance 195 PEL :%L12 =%(50.0 H)(50.0><103 A)2 =— ﬂUIIIZ (422x104 Tm/A)(50.0x103 A):
2nd _ 2;:(0250 m) =2.00x103 ﬂW
m m To move the bar at uniform speed, the magnitude of the applied force must equal
that of the magnetic force retarding the motion of the bar. Therefore, Fm = B I t7 . The 5
z magnitude of the induced current is I = Lei = (NBS/At) EMA/At) = EL” R R R R so the field strength is B = "255’ giving Fm,p =1 R /v Thus, the current is (Fa iv (1.00 N)(2.00 m/s) = PF = ——«———u : . 0 I R 8.00 Q
(h) 9’ = 12R: (0.500 A): (8.00 9) = (c) 9°11:qu PW v= (1.00 N)(2.00 111/5): 196 CHAPTER 20 20.58 When A and B are 3.00 rn apart, the area enclosed A
by the loop consists of four triangular sections, 300 m 3.00 m
each having hypotenuse of 3.00 m, altitude of 1.50 m, and base of (3.00 m)2 —(1.50 m)2 = 2.60 m
The decrease in the enclosed area has been 3.00 m 3.00 m e” AA :A, —A, =(3.00 m)2 —4[«:—(1.50 m)(2.60 111)] = 1.21 m2 The average induced current has been I = ea, =(some) B(AA/At) (0.100 T)(1.21 1112/0100 5) As the enclosed area decreases, the flux (directed into the page) through this area also
decreases. Thus, the induced current will be directed around the loop to create additional ﬂux directed into the page through the enclosed area. 20.59 If :1 is the distance from the lightning bolt to the center of the coil, then N(A(I>B) N(AB)A = NM. (AI)/2mi]A = N;10(AI)A lgavl=—_—_= At At At 2mm) 100(42zx10'? Tm/A)(6.02><106 A—0)[ar(0.800 mf]
_ 2::(200 m)(10.5x10*' s) :1.15><105 V: 115 kV 20.60 The flux through the surface area of the tent is the same as that through the tent base.
Thus, as the tent is ﬂattened, the change is flux is so, =B(AA,,M,) = B[L(2L)—L(2Lcosa)] =2LzB(1~cosﬁ) The magnitude of the average induced emf is then AtIJB 2LaB(1—c05t9) 2(15m)2(0.30T)(1—cos60°)
= =___r_,__.—W= 6.8V 8 3V At At 0.10 S Induced Voltages and Inductance 197 no IBM W At At At 8 6V E 20.61 (a) _ (25.0 mT)7r(2i00><10‘2 m)2 4(50.0><10’3 s) =— As the inward directed flux through the loop decreases, the induced current goes
clockwise around the loop in an attempt to create additional inward ﬂux through
the enclosed area. With positive charges accumulating at B, point B is at a higher potential than A ,5ch _ (AB) A _ [(100—250) mT]7r(2.00><10‘2 mf At At _ 4(400x10'3 s) : (b) 18av= As the inward directed ﬂux through the enclosed area increases, the induced
current goes counterclockwise around the loop in an attempt to create ﬂux directed
outward through the enclosed area‘ With positive charges now accumulating at A, point A is at a higher potential than B 20.62 The induced emf in the ring is 8 EV __ A‘I’B _ (AB ) Asoienoia' Y (AB solenoid / 2) Asoleuoid _ 1 Mgolgnoid A
_. uM———"‘ —_ ”on At solenoid All At At 2
=%[(4”><10’7 T'm/Al(1000)(270 I‘l/S)(%[300><10'2 mm] =4.80x10‘4 v Thus, the induced current in the ring is .4
I TIEavLMl/z ”"3 ' R ‘ 3.00x104 9 20.63 (a) As the rolling axle (of length 6 = 1.50 m) moves perpendicularly to the uniform
magnetic field, an induced emf of magnitude 8 = BED will exist between its ends. The current produced in the closed—loop circuit by this induced emf has magnitude [8 :(A¢B/At) EMA/At) BEE) (0.800 T)(1.50 m)(3.00 m/s) av ____ _..._...._.._ m :————H—————= . A
R R R R 0.4009 198 CHAPTER 20 (b) The induced current through the axle will cause the magnetic field to exert a
retarding force of magnitude Pr = BI E on the axle. The direction of this force will be opposite to that of the velocity if so as to oppose the motion of the axle. If the axle ‘_
is to continue moving at constant speed, an applied force in the direction of a and ' ' having magnitude I—jw = F, must be exerted on the axle. PW =BI£ = (0.800 T)(9.00 A)(1.50 m) = (c) Using the righthand rule, observe that positive charges within the moving axle
experience a magnetic force toward the rail containing point b, and negative charges
experience a force directed toward the rail containing point n. Thus, the rail
containing b will be positive relative to the other rail. Point I) is then at a higher potential than a , and the current goes from b to a through the resistor R. (d) . Both the velocity v of the rolling axle and the magnetic field E are unchanged. Thus, the polarity of the induced emf in the moving axle is unchanged,
and the current continues to be directed from b to a through the resistor R. 20.64 (a) The ﬂowing water is a conductor moving through Earth’s magnetic field. A
motional emf given by [8 = B(w) U will exist in the water between the plates and the induced current in the load resistor is I = lgl/Rlatal = 3(w) U/Rmmr where s....=R..,..+R=p£+R=p3+R
A ab Thus, I: B(w)v = abvB
p(w/ab)+R p+abR/w (b) IfR=0, (100 m)(5.00 m)(3.00 rn/s)(50.0x10“" r)
100 o.m+0 I: =7.50><10'4 A: Induced Voltages and Inductance 199 20.65 Consider the closed conducting path made up by
the horizontal wire, the vertical rails and the path
containing the resistance R at the bottom of the
figure. As the wire slides down the rails, the ‘
outward directed ﬂux through the area enclosed
by this path is decreasing as the area decreases.
This decreasing flux produces an induced current
which flows counterclockwise around the
conducting path (and hence, right to left through
the horizontal wire) to oppose the decrease in ﬂux. The wire is now carrying a current toward the left
thrOugh a magnetic field directed out of the page.
The ﬁeld then exerts an upward magnetic force on
the wire of magnitude ‘F=BIﬂsin90°=B[H]ﬂ=— _____.____
R R R Observe that this upward directed magnetic force opposes the weight of the wire, and
its magnitude is proportional to the speed of the falling wire. The wire will be at its terminal speed (I) = 0,) when the magnitude of the magnetic force equals the weight of
the wire. That is, when 2 2
B :2), =mg which gives 0, = :5: The ﬂux through the area enclosed by the coil is
given by (D3 2 BA , where B is the magnetic field perpendicular to the plane of the coil, and A is the
enclosed area. Compare Figures (a) and (b) at the
right and observe that when B > 0 , d1 B is directed into the page through the interior of the coil. The
induced current in the coil will have a magnitude
of I=l8= l
R R M=[M]£ AtR where R is the resistance of the coil. 200 CHAPTER 20 (a) From the above result, note that the induced current has greatest magnitude whenAB/At is the greatest (that is, when the graph of B vs. 1 has the steepest slope). From Figure (a) above, this is seen to be in the interval between t : 0 and t: 2.0 s . I = 0 when [ABI/At = 0 (that is, when B is constant). This is true in the interval between t = 2.0 s and in 4.0 s . No. The direction of the induced current will always be such as to oppose the
change that is occurring in the flux through the coil. When B is decreasing the
change in the flux (and hence the induced current) is directed opposite to what it is
when B is increasing. Between 1‘: 0 and t = 2.0 s , the flux through the coil is directed into the page and
decreasing in magnitude. The induced current must flow around the coil so that the flux it generates through the interior of the coil is into the page opposing
the change that is oCCurring in the primary ﬂux. The magnitude is AB A [0.60 1710.20 m2 
[ At R 2.0 s 0.25 o Between tn 2.0 s and t = 4.0 s , the induced because the flux is constant. Between t= 4.0 and t: 6.0 s , the flux through the coil is directed into the page and increasing in magnitude. The induced current must flow counterclockwise around the coil so that the ﬂux it generates through the interior of the coil is out of the page
opposing the change that is occurring in the primary ﬂux. The current magnitude is 2
h£ﬂJ§=£oso r]o.20 m : At R 2.03 0.259 ...
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