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Unformatted text preview: Version on] — Third Exam — Hofhnarm — {some} 1 This print—out should have 15 questions.
Miﬂtiple—choice questions may continue on
the next column or page — ﬁnd all choices
before answering. lUahie [15
[I131 11].!) points
The ﬁgure below shows a cylindrical coaxial
cable of radii as h, and c in which Equal, uni—
formly tiistl'ihutevtl1 but. antiparallel currents ii
exist in the two conductors.
{I Which expression gives the magnitude
3&3} at D of the magnetic ﬁeld in the re
gion ii <1 r3 4: a? in;
1. em} = iiim2 imilo2+r§ — 2s?)
2 airr3 [n2 — {12} i
3. em} = Jig—3 2. sea} = i1".2—h2 _ mi [a2  bi)
5' 3”“ — m 6.3{13}: ““1“ 2st:2
 2 2
7_ B = M
{r3} 2171'3 [a2 _ :52] correct
_ theirs
E. B{r3] — 2M?
9. Eli3] = U no. em.) = E
or T3
Explanation: Ampere’s Law states that the line inte
gral if E  (iii arouud any closed path equals pal, where I is the total steady current poser
ing through any surface hounded by the closed
path. Considering the symmetry of this problem,
we choose a circular path, so Ampere‘a Law
simpliﬁes to B{21TT3)=,UOI€", where 1"; is the radius of the circle and Im is the current enclosed.
Aer]. _ Tr {Ta — b2} Aeyﬁnder " all? —b2}’
r3 a: o for the cylinder. Since when b s: B: Mien
2am .airibii] ”'3' [12—irrfc2—h2} Earn; 1: G2 ' TE
#0 '32 _ 52 2TH"; _ imitatei)
_' 21rr3{a2—i12} I Rectangular Loop and Long Wire
["12 mo points Consider a long wire and a rectangular current
loop. '—I1‘—3" H——£’—HC Determine the magnitude and direction of
the net magnetic force exerted on the rectan
gular current loop due to the current I1 in the long straight wire shove the loop. 1I.I’ersion [lill — Third Exam — Hoffman]: — {:5de5] ‘2. 1.F=M( a ).down
251' n+1!)
2.s=i‘222;22{et),iee
—_nul'1l'2t‘ _ .
3'F_—2w {o b},rrght
—« #011125? tr
4_ =— —
F 21w [n{n+h] ,upcorrect
5,F=M{a+b]rup
2w
— mflfzd.
ELF: tu
2WE [ ajlup
T.F‘=‘uﬂflh£ob,down
‘lrr
3 F=lm1112(ﬂh),down
2w o+b
Explanation: To compute the net force on the loop. we
Editionsider—the forces on segments E.
BC, CD, and BA. The net force on the loop
is the vector sum of the forces on the pieces of
the loop. The magnetic force on E due to
the straight wire can be calculated by using ._ B _.
Fgﬁ=hf {$37K B.
A
In order to use this, we need to know the
magnitude and direction of the magnetic ﬁeld
at each point on the wire loop. We can apply
the BiotSavart Law. The result of this is that the magnitude of the magnetic ﬁeld due to the
straight wire is itch
B:
2n!" and the direction of the magnetic field is giwn
by the right hand rule; the ﬁeld curls around
the straight wire with the ﬁeld coming out of
the page above the wire and the ﬁeld going
into the page below the wire. We can not.r
ﬁnd the force on the segment E; applying
the right hand rule to ﬁnd the direction of the
cross product, ds'x B. we see that the force will
be in the up ﬂection. Since the wire along
the segment ALB is straight and always at a
right angle to B, the cross product simpliﬁes to Bds. Since the magnitude of the—magnetic
ﬁeld is constant along segment AB, it can
come out of the integral which simpliﬁes to
give us the result, FAB = I2 E 31
= 1'25 (so 11) .
2 a a.
Following the same argument1 we see that the
force on the segment CD is #1111 F =Ir —
22 2 l2nio+hi ’ and its direction is down. This is because
the direction of the current is new in in the
opposite direction along segment ﬁr We can i the nithe same procedure for
segments BC and DA. but because the mag—
netic ﬁeld decreases with distance from the
straight wire. Ev“ is changing along these seg—
ments. This means that the integrals are not
as simple. Using the right hand rule, we see
that the force on segment E is directed to
wards the right and the force on segment DA
is directed towards the left. Because the two
segments of wire are symmetrically placed,
their magnitudes will he equal. Since these
forces on the loop have equal magnitudes but
opposite directions, theyr will cancel. Look—
ing more closely at segments FE and m,
we see Lhat for each small portion of the seg
ment BC, we can ﬁnd a small portion on the
segment EYE such that the forces on these
two portions are the same magnitude {The
small portions are the same distance from the
straight wire. II; because the currents are in op
posite directions, the forces on them will be
in opposite directions. Their contributions to
the net force cancel. The net force on the loop is then Fioop=FrABTFCD
#nIr .U'oir
=1 E —I E’ —
2 (217a) 2 {seem}
_pnr1.r2e t
_ 2n n[n+h] I Version 881 — Third Exam — Hofi'mann — {55T45} 3 If B F
A 23" léiﬂ B
'2 n ‘2® C
I2 8 X .31 ‘8“ Fan
Here, a positive force is in the up direction.
and a negative force is in the down direction. So. the direction of the net force is up. Metal Ring
063 10.0 points
No units are shown for the answer choices.
The units are Amperes {A}.
The magnetic ﬂux threading a metal ring
varies with time t according to e3=3aﬁ—bﬁ. wM:i=EAs4nﬁE mob:
5 s'2 r1112 T. The resistance of the ring is
2.8 ﬂ. Determine the maximum current induced
in the ring during the interval from :1 = —2 s
to t; = 2 s. 1. 8.12344
8.838889
8.488891
D.2TT?T8
[1.258518
[1.488111
0.15581
5.155999
. 0.233245
18. 8.182153 weeeeewm Correct answer: 0.15501 A. Explanation: From Faraday’5 law, the induced emf should he dcpﬂ 2
g=__._.._.=_ t —25
in [9% ti. so the maximum 5 occurs when rif.‘
E ——l8o£+25=ﬂ r—b
“—8.: and the maximum emil.r is e 2a a _ﬁ _ Ezﬂo' Thus the maximum current is I _mm_rﬂ
m”_ R _8oR
[5 s'Em2 91‘]? :QmAs4mLTH28m Conducting Bar and Wire
554 10.1] points
No units are given for the answer choices. The
units are milli‘v’oits [m‘iﬂ
a conducting bar moves as shoum near a.
long wire carrying a constant I = 57 a cur— rent.
in" A e
l—H—L—e
a _l_._ I Ifu =15 mm1 L = T9 cm. ando = 25 mfs.
what is the potential diﬁemnce. av E V4 —
V3? . 25.48 . 18.3211
. 37.7143
88.1??4
57.2
183.3?6
414084
. 155.486
El. 18.8I 18. 82.3833 mummhmMn—t Correct answer: 155.455 mV. Version 1301  Third Exam — Hoﬁmann — {55745} Explanation:
Given; _dﬂ dt =~EEBEri dt
(it
_“BEE =—Bi‘p. 5: From Ampere’s law, the strength of the
magnetic ﬁeld created by the long current
carrying wire at a distance a from the wire IS I
3:”) 2 it a L
Hence the potential difference is raV=BLe
#01
= — L
(2wa)l a] AP EM 1993 MC 63 v2
[105 11].!) points
In the ﬁgure shown, the magnet is ﬁrst with
drawn upward from the loop of wire, then
moved downward toward the loop of wire. 11F
then
down
Counter—
Clockwise {7 ____ “\clockwme
induced I :5 I induced
eurren \ 5/, current As viewed from above, the induced current
in the loop is 1. ﬁrst counter—clockwise, then clockwiso. 2. for both cases clockwise with increasing
magnitude. 3. for both cases clockwise with decreasing
magnitude. 4. for both cases counterclockwise with in
creasing magnitude. 5. ﬁrst clockwise, then counterclockwise.
correct E. for both cases counterclockwise with de—
creasing magnitude. Explanation: When the north pole of the magnet is with—
drawn npward from the loop, there is a de—
crease in the downward magnetic flux at the
loop. This means that the direction of the
change in the magnetic ﬂux is upward. Lane‘s
law implies that the loop generates an induced
magnetic ﬁeld Em: to oppose this change, so
Bind is points downward. The right hand
rule implies that the indumd current is clock
wise as viewed ﬁ'om above. As the magnet
is moved downward toward the loop, follow— ing the same reasoning, the induoed current
should now ﬂow in the opposite direction. Falling Magnet [11 DDE 19.0 points
a magnetic dipole is falling in a conducting
metallic tube. Consider the induced current
in an imaginaryr current loop when the magnet
is moving advise,r from the upper loop and when
the magnet is moving toward the lower loop. Version Dill — Third Exam — Hoffmann — {513745} 5 Determine the directions of the induced
currents lame and Ibelom in an imaginary loop
shown in the ﬁgure1 as viewed from ebot'e1
when the loop is above the falling magnet and
when the loop is below the falling magnet. 1. lm : Wilmer—clockwise and
Ibﬂm = counter—clockwise 2. no current ﬂow 3. lame = clockwise and
Imam = counterclockwise correct 4. Lime = counterclockwise end
Imam = clockwise 5. fume = clockwise and
Emma] = ClEIIEkWiSE Explanation: 1When the felling magnet is below the upper
loop1 Hind must be down to attract the falling
magnet and slow it down: i.e., clockwise as
viewed from above. Before reaching the lower loop, Tilm must
be up to oppose the falling magnet; ten,
counterclockwise as viewed from above. Inner Coaxial Coil [It] Short
ﬂﬂ‘i’ 10.0 points
A long solenoid has a. tilted coil inside it made
of ﬁne wire1 as shown in the ﬁgure below. Inside coil has N turns
Outside solenoid has it turns per meter What is the mutual inductance M between
the solenoid and the coil? 1.M=manREcoslicorrect 1M =p43nNirrrzsinEl' 3M=#DTlNT 2
4..M =ngnN%ocsH ELM =hganR2sinﬁl 2
ﬁ.M=_uunN% Version 001 — Third Exam — Hoffmann  [55745) E 2 7.M=mﬂN%sinﬂ 3..M =pnnN1rr2coeﬂ
0.M=nnnNirr2 10.M=nunN1rrR2 Explanation:
Tho mutual inductance of jliﬁrc'gil satisﬁes MI— — N03. Thus M: 1 which
yields, I M=ngnN1rﬂzoosH. Inductance of an LC Circuit
003 10.0 points No units are given for answer choices. The
units are Henrg.r (H). Calculate the inductance of an LC circuit
that oscillates at 130 Hz when the capacitance
is 4.2 pF. 1. 0.425462 2. 1.15504 3. 0.151811 4. 0.11349? 5. 0.503137 5. 0.453459 7. 0.19480? 3. 0.159525 9. 0.312149 10. 0.215407 Correct answer: 0.312140 H. Explanation: Let: f=13lZIHz and
C=4.2nF. The frequency of osnillations of an LC circuit is l
f: 2 11th c“ ‘
Therefore, the inductance is
1 Linnea 1
_ [2u{139 H2312 (4.2 x 105 F} ELL{S Current 04 000 10.0 points
No units are given for answer choices. The
units are Amperes (A). An AC voltage of peak 1ralue 85.7 V and
ifetlljenltljgr 48.3 Hz is applied to a 24.5 HF
capacitor. What is the rms current? 1. 0.342335
0.450557
0.303053
0.3Tﬁl3123
. 0.483197
. 0.403751
. 0.3712041
. 0. 35503
“£1 318857
[1. 0. 413353 _goonqmcngap:5¢ Correct answer: 0.45050? 15L. Explanation: Let: V... = saw,
f = 483 Hz. and
c = 24.5 “F = 2.45 x 10—5 F. The capacitive ['eﬂﬂtaIICE is 1 1 XC=E=esfd so the rms current is = (E) 21:1!me (Via 211185.?” a) (43.3 Hz] x {2.45 x 1D'5 F} = 0.4505575. . ll Version Uﬂl — Third Exam _. Hoffmann — [5517145) "I rFransformer
ﬂlﬂ 10.0 points
a transformer has input 1roltage and current
of 14 V and 2 a respectively. and an output
current of ISLE a. If there are 1325 turns turns on the see
ondary side of the transformer. how many
turns are on the primary.r side? 1. 397.5
. 224.0
. 129.2
_ 1T3.ﬂ5
. 223.1] . 1T9.85
T. 144.114
3. 213.03
9. 184.2
10. 145.529 menarche Correct answer: 32?.5 turns.
Explanation: Let 1 at, = 1325 turns. IF = 2 A. and
I3 = 0.6 A.
Energyr is conserved, so
Pp: Pf
In V = I V;
V: __ I
V _ I
For the transformer
V cc n
a _ E a
m‘n‘a
I_ [LB a
n: itsﬁ— — {1325 turns] 2 A
= _ Maximum Displacement Current
1311 10.1] points
No units are given for answer choices. The units are Amperes {A}. A sinusoidal voltage is applied directl},r
acrom a 16.29 nF capacitor. The frequency
of the source is 2.85} kHz. and the voltage
amplitude is 59.4 V. Find the maximum 1u'alue of the displace—
ment current in the capacitor. 1. 15.223 2. 3.73459 3. IDJSSEI 4. 2].8364 5. 6.13293 5. 45.241
’4". 3ﬂ.33ﬁ4
8. 12.5705
9. 8.111555
1]. 13.1241 Correct answer: 1T.5Tﬂﬁ A. Explanation: Let: C = 15.22 11F = 1.329 X 13—5 F, f = 2.39 kHz = 2890 Hz, and
EM: = 59.4 V.
The angular frequencyr of the source is
w = 2 1r 1"
= 2 it {2890 Ha)
= 13153.4 3“. The voltage across the capacitor as a function
of t is V = V...” sinﬁwt}
= {59.4 if) sin [(131534 s’ljnt] . Now we can ﬁnd the charge on the capacitor,
Q = CV. and to calculate the displacement
current L11 by differentiating Q with respect to
t: :1 Q d d V
Id— 5— E[CV}— C— = to‘ C V7,“; oos[w t] . Thus. the maximum value of the displacement
current in the capacitor is Jrthermal: = ”J0 me
= {13153.1 s_1][1.ﬁ2‘3} x 111—5 F)[59.4 V] Version ﬂﬂl — Third Exam — Hoifrnann — [56T45j S Alternative eolut ion:
V = IXC
1 1
X = = —
‘3 211' for no
V
I = — = C V
XE: a: Id‘mz = we Vim: = {13153.4 s'ljtlﬁﬂﬁ >< 105 F]{59.4 in RC and LC Circuits [III
012 lﬂﬂ points
Consider the following circuit. After leaving
the S‘W'ltch at the position “a,” for a long time...
more the switch from ”c” to ”b". There will
be current oscillations. The maximum current Twill be given by lufmmﬁg'lﬂLC IIIL 2IIm1=£ E
3. Ln” = 81.9% correct 5 L 4. In“ = \/g E 5
511mm:— H E
ﬂImar=—R“VLC E {'0
7.1mmz—F — R L
at ~5H’i
'TTMII_ LC
Explanation: l l
2 2
ELImax Ecqrrmm 1 2
E05 C
Iﬂlﬂm‘ — EE Electron in Varying E Field v
313 10.1] points An electron under the inﬂuence of some
central force moves at speed tr; in a counter—
clockwise circular orbit of radius R. A uni—
form magnetic ﬁeld B perpendicular to the
plane of the orbit is turned on [see ﬁgure}. B increasing ._..':‘" circular path of electron Suppose that the magnitude of the ﬁeld d B
changes at a given rate E' What is the magnitude of the electric ﬁeld
induced at the radius of the electron orbit? 1.r«:=2«.rr3rrE
er
2.E=%%correct cs .13: 2—
3 R or
are
4 E: is?—
” alt
5E=nR2d—B
di RdB as 7E: a—
” dt
8E=rrERd—B Version {1131 — Third Exam — Hoﬁ‘mann — {551'45} 9 dB
‘3. E = R —
(it
1o. E = 211' R E
dt
Explanation: Basic concepts: Faraday's lattr states dtIi 5=—a Magnetic ﬂux, lJrhital magnetic moment of
an electron, and Newton‘s second law mag,r he
germane. Solution: To determine the induced elec—
tric ﬁeld apply.r Faraday‘s lattT .. dilig
Ee“=e~—.
jet 5 ext The line integral around the circular electric
path reduces to Efds=E2rrR_. and the magnetic ﬁlec through the surface
enclosed by the electron’s path is ee=eaﬂa Since we are interested in the magnitude of E .
we will drop the "‘—” sign in Faraday’s law. Thus
2
E2IrR= tittﬂ‘R B}!
dt
RdB
"3 e a The induced electric ﬁeld at the electron gen—
erates an induced magnetic ﬁeld which op—
poses the increase of the magnetic ﬂux. Thus
the induced magnetic ﬁeld points downward.
correspondingly the induced electronic ﬁeld
pointing clockwise as viewed from above. Induced EMF [IT
(114 10.0 points
No units are given for the answer choices The units are niilli‘tl'olts {miter}. The current in a solenoid is increasing at a
rate of 4.16 life. The crosssectional area of
the solenoid is e cm2, and there are 435 turns
on its 7.35 cm length. What is the induced emf oppodng the in.
creasing current“? The permeability of free
space is 4 it x 15‘? T . m/A. 1. 5.52341 2. 0.630823 3. 4.22235 4. 4.59288 5. 4.89291
5. 2.51539
7. 11.9202
3. 1.3325?
9. 1.15538
15'. 2.36141 Correct answer: 4.22235 mV. Explanation:
d I
Let : — = 4.16 A
[it is 1‘
A = 1r ern2 .
n — 435 , and L = 1'35 em = 5.5735 m and
on =4rr x ltl—TTamfA. The magnetic ﬁeld induced by the current is
B _ MN I the solenoid. , so applying Faraday‘s Law for n_ E__ @1324}
5‘ th_ N tit _ maria er _ L E _ (4e 1: inTT.mm}{435}2
_ notes m x [If cm2}{4.lﬁ Afsj( 1m )2 Hit] em
15GB mi."
1 V Inductor in a Circuit [12
ms 1o.o points 5 X Version [I'D] — Third Exam — Hoﬂ‘n'iann — {SENS} . No units are given for answer choices. The
units are Henr},r {H}
An inductor that has a resistance of 1m} H!
is connected to an ideal :Ir.I.’E.terj..r of 184 V.
E14 n15 seconds after the switch is thrown the
current in the circuit is 11112 not.
Calculate the inductance.
1. 25.9495
2. 193.453
3. 139.304
4. 443.303
5. 24LUE‘2
5. 145.117
7. 50ﬂ934
8. 133475
9. 91.3231
1]. 113123 Correct answer: 5313934 H. Explanation: Let: R=1neko=1x105n. . 8:184V. 11 = 11312 mi. = Dﬂﬂlﬂlﬂ a, and
t1 = 0.4 me = cairn s. The magnitude of the current in the circuit
as a function of time is in; = E. (1—33415) R
Rth=1_E
E 5
an
L: 1 LR
111(1— T)
—{c.ccees}{1 x 105 a}
1n (1 H [11130112112 A} {1 x 105 oi) 184i!r L: J;
CL; f? fig
Mm  ,1; 2 (3:1
9 t m ML
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 Fall '11
 HOFFMAN

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