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I Problem 6 Assume that the circular Ioop from Probiem 4 has a radius 0f 10 cm and a
resistance of 5 51, Given an external. magnetic ﬁeid E : 0.2sa:n(103t)atz + may T , determine the vatlue and direction of the current that is induced in the 10011 g; ﬂm?‘ I: g "M”; “A mm},
{a Z§Lﬂdfﬁﬁg + 2m?)  5&3 :f’ée
j K J g f
"r m MW) {50% th r: “ fig w a; al picture frame of side length 5,
a height H above a region
wnwar
d that, while ent S) H I Consider a square met dropped from rest from frame page. The frame accelerates do
d. it is observe D the magnetic fiel
moves with constant velocity.
(a) What is the frame’s spee
(b) What is the strength 0
physical quantities (s, M,
(c) After the frame has comp
sides are fully "immersed" in the your answer. 0
(:1 under the influence of gravi
ering the magnetic field, the frame e magnetic field?
B in terms of the given stants you need. d when it begins to enter th R, and H), and any con
letely entered the m
field), what is agnetic field (i.e., after all four
the frame’s acceleration? justify use kinematics. As We
acquires final speed the positive direction. 80, v0 = 0, and a 2 g. Solve for v to get gy conservation. 1’11 fashioned kinematics, or ener
through distance Ay (a) You can solve using oid~
ect accelerating at rate a saw last semestex, an obj
v2 = via + Edy. 1’11 take downward as v = 42313}; 3 42311. The same answer pops out of energy conservation, according to which gravi— tational potential energy mgH converts into kinetic energy, émvz. TI: ms
obiem is ﬁguring out where to begin. We can gain X x x (b) The hardest part of this pr
gnetic field, the frame moves
X X g X IX insight by thinldng physically. As it enters the ma
acceleration goes to zero. Therefore,
2% X X at constant velocity. In other words, its
according to Newton’s 2nd law, the net force goes to zero. 80, an upward mag— a
tional force. The magnetic
Since the frame has no ance
Since the niagnetic force acceleration, these
d by settmg Fmag == Mg. forces must cancel. 1 the downWard gravita ake sure this happens.
an soive for that ﬁel netic force, Fmag, must (1
ﬁeid must be strong enough to In depends on the magnetic ﬁeld, We c 438 Portable T. A. gnetic force acting on the frame. Well, the mag—
ding to Faraday’s law, a voltage and curw
ers the field. So first, we need an expression for the ma
netic force on each side depends on the current. And accor
rent get induced because the magnetic flux through the frame Changes as it cut Let me tie all this together, in the form of a strategy. 1) Using Faraday’s law, calculate the induced EMF—and the corresponding induced currentm—in the frame, as it enters the magnetic field. 2) Given that current, use magnetic force reasoning to find the f
Add them up to obtain the resulting total magnetic force, Fmag.
B, the magnetic field strength. 3) Finally, set Fmg : Mg, and solve for B.
the frame has no acceleration (constant velocity) while enterin area on each side of the frame.
That total force will depend on This gives you the field strength needed to ensure that
g the field. We’ve completed the tricklest part of the problem, formulating an overall strategy. Before reading the rest of this answer, see if you can solve everything on your own. Subprobiem 1: Using Faraday's Iaw,ﬁnd the induced EMF and current as the frame enters theﬁeld.
at which the flux through the frame changes To apply Faraday's law, we must find the rate
for the flux through the frame. Then, I’ll dif as it falls into the field. So first, I’ll write an expression ferentiate with respect to time, to obtain dog/dz.
Let y denote the ”height” of the frame that’s inside the magnetic 3 field at an arbitrary moment. 50, y changes with time. The area of the frame
made the field 15 Ah, ﬁeld = ys.
y I 3< >< K Since the magnetic field is constant, we can pull it outside the flux
3: x x x In: integral: o, a jam a BAmﬁeld = Bys. efore (51),; increases. To find the rate of change of As the frame falls into the field, 3; increases, and that
the differentiation only ”hits" y: (by, differentiate it with respect to time. Since B and s are constant,
43% m #3315) _ it _
dt " dt ‘ 3351:“ BS”
ocity is the rate at which position changes: 1; 2 dy/ dt. By Faraday’s law, Einduced == 44%}; = mst. Therefore, by Ohm’s law, the induced current in In the last step, I remembered that vel the frame is . _ gmduced BSU :— 2w— R R ’ where v ”2 .IZgH, as we found in part (a). The minus sign indicates that the induced current ”fights” _ the change in flux. Here, as the frame falls into the field, the intentheupage flux through it increases Chapter '55 ” to generate an outof—the—page field, at points inside the
lockwise. I figured ﬁns out using the point your right thumb with
pro current “wants
the induced must ﬂow counterc
square, and
d fingertips represent the field lines quare, they point out of the
ofthe—page field, at points Therefore, the induced frame. To generate such a field,
right—hand rule shortcut, as follows. Pick any side of the the current, assuming a counterclockwise flow. Your curle
duced by that current. When you curl them towards the center of the 5 page. This goes to show that a coun creates an out—
inside the square. em 2: Write an expression for the net magnetic force on thefmme Subprobl
Era; "the magnetic field can exert a force only on Sides 1, 2 and 4, as
3 labeled here. Starting from the basic magnetic force iaw Fmg x qv x 3, you
y [ x 4mg): x can derive the infinitesimal force acting on an infinitesimal length of wrre
K R 1% 1 Y.
x x n X y dFmag W 153le
W S i—e—i
F2 and F4 must cancel. Since those two sides carry the same current, and F
since the same length of those sides (namely, length y} is immersed in 1:4,. 3‘ 2::
the field, F1 and F4 are equaily strong. But along side 2, d1 points down— 3; I x x x x ward; while along side 4, all points upward. Therefore, F2 points in the x
‘K, )4
opposite direction from F4. Since those two forces have the same F1 x x strength, but point in opposite di For this reason, the total
1, £11 and B are perpendicula
2» inl. Since that whole si rections, they cancel. magnetic force felt by
r. 50, the
de is immersed in the field, the frame is simply F1. Let's figure out that magnitude ot the cross product reduces to a force. Along side
regular product: HIP] P1=JdF1 = has ester : iBs, because side 1 has length 5.
To tind the direction of that force, apply the righthand rule to the cross product in 151 —r 1‘le B. Point your fingers rightward with oil. Then curl your fingertips into the page with B.
Your thumb now points upward. That’s the direction of F1. Indeed, we already knew that F1
points upward, since the magnetic force must cancel the downward gravitational force. Without
this cancellation, the frame would accelerate downward, instead of drifting downward at steady speed.
Before continuing, let me express P1 in terms of the given physical quantities Fr
B 2 H
problem 1, 1' = E? 2 RE; 5, since the frame enters the field with speed '0 = «lZgH . Therefore, the ward magnetic force has strength B 2 2 FM”—
P1=iBs:[ ‘VggHsleB S “23H. up R R 440 Portable T. A. Subproblem 3: By balancing the forces, solvefor the magnetic ﬁeld, B. Since the frame enters the field at constant velocity (zero acceleration), the net force on it
must vanish. In other words, the upward magnetic force cancels the downward gravitational force: Raﬁ/lg
22
BS— «errI'm, R Solve for B to get B .. MgR
52 ZgH’
If B is weaker than this, then gravity overcomes the upward magnetic force, and the frame acceler ates downward.
(c) A magnetic force acts 0n the frame because current ﬂows around it. And current ﬂows because the magnetic flux through the frame changes, thereby inducing an EMF. But once the frame becomes
totally immersed in the magnetic ﬁeld, the ﬂux through it no longer increases. At that stage, (135 stays
constant, and hence, dth/dt = 0. Therefore, no EMF gets induced. Therefore, the current around the frame stops flowing. As a resuit, no magnetic force acts on the frame.
Once the magnetic force vanishes, only gravity acts on the frame. It accelerates downward, with acceleration g. Problem‘ 4 (8 points) A cyiindrical conducmr along the z axis has a spatiajiy varying conductivity of
J m if) . emp(w150p) kS/m. An electric ﬁeld sf 3062 V/m is present.
Make use of Ampere’s circuital iaw to ﬁnd H as a function of ,0. an!” I' . . 47T€OR\
where R is th“ ‘*  _ . ‘ _.=ch incremental 1301113, ”F the ring and the point i = a}? m 458mp{~150p}52 kA/m2 Totai current cressing the circular surface 220 with radius p: ’ 271}? I
Ixffd§mjf45exp{~:50p)pdpd¢: 22.6{1 — (1+150p)e$'p(E50p}] A
0 0 I  . / Symmetry suggests that I? will be qb—directed only, and so we consider a c1rcu1ar
path of integration, centered on and perpendicuier he the z—axis. ZWqub {2 Janet {/Cf‘: %[! (Hf WEUJQ7<$K {ft/j] dd)” («yd/Lilia” hath“? Yazéxwg QQ 7 026 gm) Two conducting spheres, one wi’dx radius 5 2 0.10 m, the 0mm mu. m
any charge Q m 1.0 X 10‘11 coulombs 1) mﬁar apart. Both spheres c H é“: a) What is the potential right at the surface of the sma‘der sphere? The bigger sphere contributes
.x tial, since it’s so far away. Please set the ”zero" of potential at infinity: negﬁgibly to this poten
V=Owhererr~oq (b) A particle of charge .3 2 . 2 3.0 X 10’7 kg is released from rest
from the surface of the smaller sphere. H ' ' ' when it reaches a distance
35 = 0.30 m from the center of the smal‘ier sphere? Negiect gravity, and neglect the iarger sphere.
of resistance R = 2.0 9. Immediately after (c) Now the two spheres
the wire gets comecte (d) As time passes, does this curren
(6} When current stops ﬂowing, What is the charge on are connected by along wire, d, what current ﬂows through it?
t? Explain. t increase, decrease, or stay constan
the smaller sphere?
1 ' 1 ..I , . ' e _ _. , ' 34.“. La...“ W». At points outside a spherically symmetric char potential created by a point charge: V r: 47:5 1'
a
sphere. At the surface of the sphere, r = s. formula also applies to
two objects produce the
we’ve seen charged spheres.
same electric field, then they create ”—9” the same ﬁeld produced by ap .
the same electric field, they also create the and hence, E = 4 2 ,
agar
and a spherically symmetric charge distribution create
same potential (at points outside the sphere).
Many students figure out the force on the particle, calculat
rmulas apply 0 kinematic equation. But those old kinematic to
arther from the charged sphere, it feels a decreasing force, e its acceleration, and then use an old nly to constant accelerations. Here, as (b)
and hence, a decreasing the particle getsf
forces; Here, as the particle rushes acceleration.
Fortunately, conserVation laws allo
' ' ' " nergy gets pro the charged sphere,
the answer. no heat or other away from
conservation gets us to 444 Portabfa T. A. The particle starts at rest, with no kinetic energy. Using U = 3V, we get
K0 + U0 2 Kf + U; 0+qu : éMrﬁ +qu. To solve for v, the ”final” velocity when the particle reaches a distance 35 from the center, we just
need to know V0 and V;, the initial and final potential. (Actually, '0 depends only on the change in
potential. But here, we can ﬁnd that change most easily by calculating Vf and V0 separately, and then subtracting.) From part (a), we already know the potential created by the small sphere. It’s
V = . Since the particle starts at r = s, and reaches its ”final” position at r = 35, 4:71:30!“ V‘Qand Q w 4158.33 Vi = 42reo(35)' Solve the above energy conservation equation for v, and substitute in these potentials, to get 2, Q _, Q l
a 2q(Vo—Vf)_ tineGs 47.55085)
M "“ M
m 462
BaeosM (1.0 x we one x 1011 C)
W
3:43.85 x 1017 F / m)(0.10 m)(3.0 x 10" kg; 2 2.0 m/s. (c) A common mistake is to think that no current ﬂows, because both spheres carry the same charge. I
But the spheres have different potentials. Remember, current flows through a wire because of the
potential difference (voltage) across the wire. Usually, a battery generates the voltage. But here,
the relevant voltage is simply the potential difference between the two spheres. Current flows
from the higher«potential sphere to the lower—potential sphere. Indeed, according to Ohm’s law
(V a IR), the current is proportional to this potential difference. To figure out the potential difference, I’ll just calculate the potential of each separate sphere, and subtract. In part (a), we already found the potential on the surface of the Small sphere. It’s
_, Q V ..
small sphere 47580 5 . Since the big sphere also carries charge Q, but has radius r = 25, the potential on 3 its surface is Vbigsphere = 47r€9Q(ZS)' Notice that the smaller sphere has higher potential. That‘s because the charges on the
small sphere are more concentrated, as drawn here.
Consequently, charges "want” to move from the small sphere
to the big one. Current flows as if driven by a battery of voltage as . wire 't’ =_Q.__._Q_._2_Q_ V = Vsmall sphere "' Vbig sphere 473805 4.7580 (25) 875805. 445 Chapter 56 wire initially carries current Therefore, according to Ohm's law, the ifference between the two spheres. But
tial difference decreases. Here’s al goes down. And as the big sphere gains charge,
gets smaller. As a (6.) As just explained, the current gets " driven" by the potential at as charge flows from the small sphere to the big why. As the small sphere loses charge, its potenti
its potential goes up. Therefore, the potential diﬁerence, V : Vsmn sphm — result, the current gets smaller.
A numerical example cl
coulombs. So, the spheres initially generate p arifies this reasoning. Initially, both spheres carry Q = 1.0 X 10‘” otentials :2 $339.12....”—
= x a. 0.90 it
Vsmauspherems snags arr(8.85><10‘12F/m)(010ml W S Q Winnie...” 045 1
. . . z W 2 n 3:: , l: .
Vlarge sphere, initial 43150 (25) 471.1835 X 10—12 F/mXOQO m) V0 S Therefore, the initial potential difference across the wire is V = anspm  Vbigsphm r. 0.90 V — 0.45 v = 0.45 v. after 0.1 x 10’“ coulomb But at a later time,
” coulombs. 80, small sphere carries only 0.9 x 10' 0.9 x 10“ C
.7; W 2: 081 015:5
Vsmall sphere, later 4K(8.85 X 10—12 F/m)(0_10 m) v 0”” coulombs, which generates a potential And now the large sphere carries 1.1 x 1 1.1 x 10'“ C
__—,, W 2 050 V it .
Vlarge sphere, later 47!:(885 X 10—12 F /m)(0_20 m) 0 S tween the two spheres is only 0.81 V  0.50 V = 0.31 volts, less Therefore, the potential difference be
f 0.45 volts. Since the potential difference goes down, so does than the original potential difference 0 the current.
(e) Eventually, the current otential difference drops to zero~— pens when die p
anishes A common mistake is to say the current v
easons too subtle to explain here, have slightly different potentials.
ou’re interested. stops entirely. This hap i.e., when both spheres have the same potential. when both spheres carry the same surface charge density. But for r
the same surface charge density, they Ask your instructor about this, if y
B When Vsmall sphere = Vbig sphere And potentials are what matters.
and big sphere. Do we know So, this problem wan
TO ﬁgure It out, let Qsmall and Qbig ts to know the charge on the small spher
denote the final charge on the small 446 Portable T. A. the total charge, Qsmau ~§~ ng? Yes. Intuitively, every bit of charge that ﬂows off the small Sphere
flows onto the large sphere. Therefore, the total charge on the two spheres stays constant. Initially,
the two spheres carry total charge 2Q. So, that’s the total final charge, too: Qsmall + Qbig = 2Q. Given this insight, we can set the final potential of the small sphere equal to the ﬁnal poten
tial of the large sphere: VanI211} sphere = Vbig sphere Qsmail _. Qbig 4.75805 .. 4mu(25)’ and hence, (21315 a 2Qsmaﬂ. In words, the two spheres generate the same potential when the larger one
carries twice as much charge. Since the total charge is 2Q, 2Q = Qsmai! + Qbig
= Qsmall + 2Qsmal!
= 3Qsmall:
and hence
gm” = §Q = get) x 1011 C) = 5.7 x10‘12 c. g Problem 4 (5/25) A long straight cylindrical wire carries a uniformly distributed current ef 3 A into the paper plane
and has a radius of R22 cm. The ciosed loop shown in the sketch is a semiwcircie of radius ,0 = 4cm
that passes through the center of the Wire. [,9 (a) Use HAmpere s iaw to ﬁnd the magnetic ﬁeid along the circuiar are (5 points). m H4! C [9) 67:; glue +6 “the Asymmetry G) (if) . 1,J 17¢
I HOW}; H AL Heme; Felt he ‘ Hate”) [M mt: H44?) 27:va
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(b) BONUS problem: Calculate the line integrai of I? aiong the arc and the straight section
and explain why the result is different from the answer in (a) (4 extra points). ' “J a.) _ ‘ ...__.3 _J I "a“, {yr
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C(Dﬁeié Loaf) Problem 2 (5/25) Use j 2 (IE to prove that two resisters R1 and R2 in parallel behave like one resistor with a total
resistance of RM 2 (R1R2)/{R1 + R2). 3m; :AwI + 4120
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