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Unformatted text preview: PHYS 2212 Test 4
Dec 1st 2010 Name (print) 6/ Instructions 0 Read all problems carefully before attempting to solve them. 0 Your work must be legible, and the organization must be clear. a You must show all work, including correct vector notation. 0 Correct answers without adequate explanation will be counted wrong. 0 Incorrect work or explanations mixed in with correct work will be counted wrong. Cross out anything
you don’t want us to read! 0 Make explanations correct but brief. You do not need to write a lot of prose. 0 Include diagrams! 0 Show what goes into a calculation, not just the ﬁnal number, e.g.: 3—17 = éﬁgfﬁiﬁg‘ig =
5 X 104 0 Give standard SI units with your results. Unless speciﬁcally asked to derive a result, you may start from the formulas given on the
formula sheet, including equations corresponding to the fundamental concepts. If a formula
you need is not given, you must derive it. If you cannot do some portion of a problem, invent a symbol for the quantity you can’t
calculate (explain that you are doing this), and use it to do the rest of the problem. Honor Pledge “In accordance with the Georgia Tech Honor Code, I have neither given
nor received unauthorized aid on this test.” Sign your name on the line above PHYS 2212
Do not write on this page! Problem 1 (25 lots) — A Problem 2 25 pts)
Problem 3 25 pts)
Problem 4 25 pts) Problem 1 (25 Points) to power supply A bar made out of a conducting ma
terial is 0.21 m high with a rectan
gular cross section 0.05 m Wide and
0.004 m deep. The bar is part of a cir—
cuit and carries a steady current (the
diagram shows only part of the cir—
cuit). A uniform magnetic ﬁeld of 0.95
tesla is applied perpendicular to the
bar, going into the page (using some
coils that are not shown). Two volt
meters are connected along and across
the bar and read steady voltages as
shown (mV 2 millivolt = 10“3 volt).
The connections across the bar were
carefully placed directly across from
each other to eliminate false readings
corresponding to the much larger volt—
age along the bar. Assume that there
is only one kind of mobile charge in
this material. to power supply (a 5pts) What is the sign of the mobile charges, and which way do they move? To receive full credit, explain your answer with a supporting diagram.
. 4 1
PostHM Char?” “(CI/n4!va on. “HAb W7“; S‘J‘: "f 3%" baa (val Wk“ )
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(VJ Pond—s down 9%") If ,Chﬂ¢~ CM“! e3
A QCW 0W Max/h!» [>675 7, ;
5./\£V\, «not ﬂan/L13 7‘!) M 5:5 “ﬂu MW“th 0k“/'65 4/4 Pasnhw—e and! MdvL Jan/916m,“ (b 5pts) What is the drift speed 1; of the mobile charges? Explain your reasoning. :‘L 551641752Lq¢o 7%, Mob/c char/g) mge oil/o 3.4 7€fc¢ W 04.111506 dc’MD‘I/I‘Dﬂ/ (c 5pts) What is the mobility u of the mobile charges? \/: ME“ 2: VLAV“/9~ Vs “55"(0'4 “4% 0‘2"“) ’ / /_
\A’ AV“ ’ (2‘1V) ,5 M
m: 50840 V/m (d 5pts) The current running through the bar was measured to be 0.55 ampere. If each mobile charge is
singly charged = 6), how many mobile charges are there in 1 m3 of this material? ’5 jigv’lAV :17 m: “J; i/iv’ 0.5571
AM ._ __ «4
Vi : 0‘9‘(a‘iqc)<o.05m)(0.004m>(("5 3'“) "4/5) 25' ’3
': 21(34'0 W‘ (e 5pts) What is the resistance in ohms of this length of bar? D
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H)? Problem 2 (25 Points) A neutral rectangular copper bar with length L = 0.55 m, height = depth = w = 0.02 m, is pulled to the
right by a rope (which is not shown), at constant speed 11 = 0.3 m/s., through a region in which there is a
uniform magnetic ﬁeld of 0.43 tesla7 directed into the page. The copper bar slides along metal rails, with
negligible friction. (You may not need to use all the information given in this problem.) In the following questions, if any quantity is zero, state this explicitly. :3 wka . A 3 J) (Adah/‘05 (F:% J )(8 )
(a 5pts) On the diagram, draw ”+’s” and ”—’s” indicating the charge distribution n and/ or on the bar.
Explain your answer. Full credit will only be awarded to a complete explanation. “NH/u— t/f‘Uw’of (hitch cm
vita, Cltlotféle SCEOJuclts _
(b 5pts) On the diagram, draw an arrow indicating the direction of conventional current through t e
resistor, and label it ”conventional current”. (c 10pts) What is the absolute value of the potential difference between the ends of the bar? To receive full
credit you must start from fundamental principles (do not start with a formula that is not on the formula
sheet). Show all steps in your work. If vii/w, know Is (JUKJJ a} (oquwi’ Speedl/ 44m. CUW‘LVL‘i’ :5 akawi“ Ami ‘Hu, Wolm‘i’fc. "Q/CL 330m “HA4 altarxlés iv.
“HAL bow is )Uéi’ («my Waqui/u, éieo‘l’hc f‘ofci Aida 41. “We 9~€J>AKKM oil/WWTC. lfmi'l' iii :7 $1132 glib” Dr {W 2 «B :> WWI: EWL :VBL 3(0‘3m8)(0‘%7)(m‘m)
(Moat: 0.07)\/ (d 5pts) Assuming a 109 resistor, calculate the power that is necessary to move the bar at the constant
speed 2)? CM». ofo “HMVJ +000 WW Si 1"“ W» «Li
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D P J when, j P: Q [(911 {.0 /0 Problem 3 (25 Points) / \
\
/
\ \ \
Two point charges, one positive and \
one negative, are brought close to an /
uncharged metallic bar magnet. As I shown in the Figure, the bar mag
net becomes polarized. Consider an
arbitrarily shaped Gaussian surface
around the bar magnet. The two point
charges remain outside the Gaussian /
surface. ' / K \ \ sw) (a 6pts) Determine the net Electric ﬂux through this Gaussian surface? Be sure to show your work to
receive full credit. (b 6pts) Determine the net Magnetic ﬂux through this Gaussian surface? Be sure to show your work to
receive full credit. (PM : @ﬁMA : 0 Via Mir MKYMHC ﬂux“ (c 6pts) A new Gaussian surface is shown in the Figure on
the right [This surface includes one of the point charges].
In this situation, does. your answer to part (a) or (b)
change (circle one)? If so, you must explain why? Electric Flux is now: @@ Same
Magnetic Flux is now: Different éucloseoa malt Cl/totvje
220m): ' a $13: 16:: ﬁgwﬁ Wt M 3 0 film? ((1 7 pts) A second uncharged bar magnet is brought inside
the Gaussian surface shown in Figure 1 (see Figure 3). In this situation, does your answer to part (a) or (b) change? / \ \
. . / (Cerle one) If so, you must explaln wh? \ \
Electric Flux is now: Different \
Magnetic Flux is now: Different @ I I , We enclosecﬂ M'lV’CQ/L‘C/‘ﬁﬁ
so ﬂame<0 ' / ¢Z§E.RJLA:Q l \
E \ @E’M/P :0 @430 Problem 4 (25 Points) A coil of wire is connected to a power
supply, and a current runs in the coil.
A single loop of wire is located near
the coil, with its axis on the same line
as the axis of the coil. The radius of
the loop is 4 cm. At time t1 the mag
netic ﬁeld at the center of the loop,
due to the coil, is 0.5 T, in the direc
tion shown in the diagram; the current
in the coil is constant. trail (a 3pts) What is the absolute value of the magnetic ﬂux through the loop at time t1? ®M=\%'hd¥l 2: BA
3) 2: 313,: {osﬂﬁXaollMZ : 2.;i(o'3TmZ
M (b 3pts) What approximations or assumptions did you make in calculating your answer to part (a)?
VW Mac‘vxxln‘z. e\ «l (s: appmkamkb M
‘ Sam WWJW swan ‘HA—L «AER we” bog / m :JdM O‘HOW M; L ls (Ammas WWqu “(but £5“ £332 (0 3pts) When Viewed from the coil, What is the direction of the ” curly” electric ﬁeld inside the wire of the
loop at time t1? (the current in the coil is constant.) EQCRUDC “HA1 OWN—(4+ l‘s (Kassie/wt} “liar £2162
£065 woi' Cl/MVLZCwm" ‘kmb that“) °Huzm :5 Mo Co’s/[L7 eléolm‘c «Mal a} t,. (d 3pts) At a later time t2, the current in the coil begins to decrease. When viewed from the coil, what is the direction of the ”curly” electric ﬁeld in the loop at this time? Indicate this on the diagram with
arrows. see/litM “L” “likl‘ CM 9 imp b§~ i; 4?
15g mm: “M (allot 5.53.); l 2 ‘
30 Mi ’45 Powds ﬁg: H~UA€¢ girl/UL (WM/V! élfﬁlVl‘C/
ﬁeld Pm‘vtts {lean/41V clockwis€
WW“ view/(LO “(l/ow» % Carl. (e 9pts) At time t2 the rate of change of the magnetic ﬁeld at the center of the loop, due to the coil, is
—0.25 T/s. At this time, what is the magnitude of the electric ﬁeld at location P, which is inside the wire? (f 4pts) Now the Wire loop is removed. Everything else remains as it was at time t2; the magnetic ﬁeld is still changing at the same rate. What is the magnitude of the electric ﬁeld at location P? Explain how you
know this to receive full credit. . I" «311/
Saw. £5145 / tug :53“ /C w 4M {5 CNcULCA? wit” “Hams, 4/00/ M Vlo C'Ulfv‘ev‘t‘i 14/09“ [7/6 {/Lo (90?. This page is for extra work, if needed. Things you must know Relationship between electric ﬁeld and electric force Conservation of charge
Electric ﬁeld of a point charge The Superposition Principle
Relationship between magnetic ﬁeld and magnetic force Magnetic ﬁeld of a moving point charge Other Fundamental Concepts d" d" _, d“
6:67: d—f: net and—d—fzmé’ifv<<c
AUel = qnv AV 2 — Lilli" o dfw — 2 (Ean: + EyAy + EzAz)
(EelszoﬁdA @mangBoﬁdA
fﬁoﬁdAzw féoﬁdA=0
0
lemfl = fENC' . dl : ’ dygag fB . Z #0 ZIimside path —' —’ d —a A
f B . : :u'O Iinside path + 60% f E . Speciﬁc Results —» 1 2 s . a 1 8 . Edipole,awis % 47T€0 "‘53— (On 3X15, 7' >> 3) Edipole,i‘ % 47r60 % (on J— aXISa T >> 8) —; 1 —;
Emd 2 47m) :3ng (r J. from center) electric dipole moment p = qs, 16’ = a Eapphed .. 1 262/ L . ’ 1 qz .
Erod ~ 47160 r (If T < E1"ng 2 (Z along aXlS) a Q/A z _ a N Q/A z Q/A ,
‘Edisk = 260 1 — m (2 along aX18) ‘Edisk ~ 260 [1 — ~ 260 (1f 2 < R)
Emmato," % Q/ (+Q and —Q disks) ‘Efmnge x Q/ just outside capacitor 60 60 2R
6 [AZ A a a 1
AB 2 £9 2X T (short wire) AF = [Al x B
7r 7" r 0 LI 2] a _,
'Bwire = W 7' < L) Bwire = ‘Bearth tang _. _ ,uO 2I7rR2 N no 2]7rR2 , _ _ 2 Bloop —EW~E Z3 (on ax15,z>>R) ,u—IA—I7rR ~ #0 2 . a . Bdipole,ams % (on aXISa T >> 8) ‘Bdipole,.l_‘ % (on J— aXlSa T >> 5) 4 1 —q[i_L A A A _. Erad
Erad : 47T60 Czr '0 : rad X Brad Brad : c
i=nA17 I=IqnA17 17=uE I L
or Iql nu A a 0A
E  1 1
Edielectric = algzw AV 2 4 q [— — —] due to a point charge
71' 60 T‘f Ti IAVl I = R for an ohmic resistor (R independent of AV); power 2 I AV
[AVI . .
Q = 0 [AV] Power = I AV I = R (ohmic res1stor)
dﬂ ' 2
K m @7202 if 1) << 0 circular motion: d—IZL = 11R, m %
Math Help
(ix 5: (az,ay,az) >< (bx,by,bz)
= (ay bz — a2 byﬁ — (am bz — az bx)? + (am by — ay (7&2
/d:c 1(+m)+ / dx 1 +c/ dx 1 +0
2 n a c —— = — —— = ———
x—l—a (510+a)2 (1+9: (a+a:)3 2(a—l—ac)2
a 2 2 a 3
/adm=ax+c faxdmzix +c /aa: dxzgac +0
Constant Symbol Approximate Value
Speed of light 0 3 X 108 m/s
Gravitational constant G 6.7 x 10—11 N . m2/kg2
Approx. grav ﬁeld near Earth’s surface 9 9.8 N / kg
Electron mass me 9 x 10—31 kg
Proton mass mp 1.7 X 10—27 kg
Neutron mass mn 1.7 x 10“27 kg
1
Electric constant 4% 9 x 109 N  mZ/C2
0
Epsilon—zero 60 8.85 X 10—12 (N  InZ/C2)_1
Magnetic constant 5—0 1 x 10—7 T  m/ A
7r
Mu—Zero M0 47r x 10’7 T  m/A
Proton charge 6 1.6 X 10“19 C
Electron volt 1 eV 1.6 X 10—19 J
Avogadro’s number N A 6.02 x 1023 molecules/mole
Atomic radius R, w 1 x 10—10 m
Proton radius R1, m 1 x 10‘15 m
E to ionize air Eiom'ze rd 3 x 106 V/m BEarth (horizontal component) BEarth % 2 X 10—5 T ...
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