This preview shows pages 1–12. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHYS 2212 Test 4
Spring 2011 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. 0 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.: 7—3 = W = C
5x104 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 The ﬁnal exam will be held during exam period seven
Wednesday May 4th from 8:00 to 10:50am Locations TBD. The conﬂict ﬁnal exam will be held during exam period ten
Thursday May 5th from 8:00 to 10:50am in Howey N208. PHYS 2212
Do not write on this page! Problem 1 (25 Points) 0.05 m
+L.—t+ To power $11 A bar made out of a conducting ma
terial is 0.21 m high with a rectangu
lar cross section 0.05 m wide and 0.004
m deep. The bar is part of a circuit
and carries a steady current (the dia
gram shows only part of the circuit). 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). 0.21 m _Y_
To power B = 0.95 T supply Two voltmeters are connected along and across the bar and read steady voltages as shown (mV 2 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 voltage along the bar. There is only one kind of mobile
charge in this material. (a 5pts) What is the sign of the mobile charges, and which way do they move? Explain in detail using
diagrams to support your explanation. W “dot/{J 09‘ VDl'l'mbl—fzrz 5".“5 us (REGHN pt
Chm/vies have, callioiM CM HM. V‘f‘ll’d' side 09— +04. bar (364 jl M Hula/i3 Rom wll—welw 1 gives us wcgl— 4km COW/Wi‘latl cdvmvi+,f, ﬁlms Mus—ch bur. Ii: we.  . r‘ kl
“55VWe. + dag(Qt Caulk/s] “Ha; Waowkcgfct 15 ’lb 'H": "I ’5 2x3
(b 5pts) What is the drift speed 1) of the mobile charges? Explain your reasoning. F 1 Z V ‘ .
Alzlor skaJTS‘imk, Mars loam €Sﬁbligk¢rﬂj . So meile
m viomcavdak ( Crass wt“) ant on Cup/{la 05246.
 '  dds/‘9. . a '
CW7“ '5 aw i “My “M '“ 5"“"7 ‘h‘ {’36 MW _
“us 00w“. 94w. 54/. l "V" m
4 ... / qwf‘tls
f ' "—5" F"??” gr“; 6 ’mj
W’ “W” I 5* amoeba) W“ '4"
EL [ML/w (0.03140 “We, a; I) (a 5 Law—£ rﬁ‘i m/s‘] amt r; M (c 5pts) What is the mobility u of the mobile charges? E V V
V:,d( “ —7 MsE“ : Mu/L (d 5pts) The current running through the bar was measured to be 0.55 ampere. If each mobile charge is
singly charged = e), What is the mobile charge density n for this material? I "" I
“W” ’7 = 137?? (e 5pts) What is the resistance in ohms of this length of bar? VSTK : Au — R I gal” :y X
K: 0.53% 4.71.}7, / Problem 2 (25 Points) A neutral copper bar of length L and resistance R is free to slide along perfectly conducting metal rails in
a region in which there is a uniform magnetic ﬁeld B directed into the page, as shown in the diagram. A
power supply is connected across the metal rails and supplies a constant voltage emf. ® ® ®
® ® ®
® ® ®
® ® ® (a 5pts) As soon as the power supply is connected current run through the length of the bar. What is the
magnitude and direction of the force on the bar due to this current. Label this force on the diagram. A‘+ vHM. inslavvi’ "H'u. (owesl1)“, {5 Mode / m
CWNVVl’ in “ll/w (00M {S Sim‘aia em§/@_ 1‘4.
‘l’lom 44M lop Mil '40 “HM— lOO‘H‘OWL (ail (Snail. (b 5pts) As the bar slides down the rails, the motion of the bar causes it to polarize. Indicate this
polarization of the bar by drawing pluses and minuses on the diagram. Note, we are not asking for the polarization created by the power supply. ‘ m
Clmqes in “We bar an with 7L0 44w 0714+. a ’4’ ‘
M~1Mm apprcc on Wc char?“ {5 6 ‘ [56C (C 5pts) As the bar moves along the rails, the current decreases. Explain Why. "ML low academics 41, ~th New}. As {HS 8M {VICNVlw/S ‘l’W0 “Hiltvbﬁ “ﬁr/mm: (’) Chap/yes ,‘h “71% 5 and ‘AI’Cfol 7‘6 ‘ﬂe (MG dual [2) am e/(ch'c
ar I —>)
.50 4L1M‘5 QM FE ((41.1. 744/1 .‘n W 17v 7mm L4 Marin/(j. [
7L“. fad/1L ' 1's #4 «#511le (Sew—er mzemzij'Cs It I‘ 1,; 7'0; éﬂr age ,4: 54% 0.9
p a f 4’ [’2 AM (‘5’! Off Jt’p. )' 7&4/5 34“ (Witt/UP 49“? Mp4 NM CC is Nod/cc 9?. (d 10pts) The bar quickly reaches a steady state (constant) velocity 1). Determine the magnitude and
direction of this velocity. You must start only from fundamental equations or equations on the formula
sheet to receive full credit. Hint: Do not attempt to use Faraday’s Law for this problem. [.F M bar is modi‘t] aL—l Cowslaud sececll I Problem 3 (25 Points) The electric ﬁeld is measured at locations on every surface of a box—shaped surface oriented with its long
side on the x—axis, as shown. The electric ﬁeld at all locations on each surface of the box is found to have
the value E = (—100, 500, O) V/m. For clarity, the electric ﬁeld vector has been drawn only on four faces
of the box in the diagram, but the electric ﬁeld has the same value on the other faces (front and back). [New V rbc“
E f; a augiaué vealor so 22:42:» (a 20pts) Calculate the net electric flux through the surface of the box. Be sure to show all of your work
to receive full credit. I, a c .  ,  : » $00 07 <—I 0,07 (003)(o043
alrﬁ' I ¢‘ EA! <loO/ I I — 4>l:+. l1. ‘1‘“
(0.0?)(°«°”‘3 : 0 VM o.oH) :—.IZVM
2 (’(00) 503/07 .<+')ol 07(0'03)( .7 (0.0?)(00'4) v: 0"“ : <_loo FOP/07 ‘ 4010/,
I (0.0?)(0'03) :ﬂ'ogvm 45 :CHZVM)+(O) +(112Kn)+(o)+(1.0§vm)+(—(.oS‘\/.a) : 0v,k
+ , (b 5pts) Calculate the amount and sign of the charge enclosed inside this box—shaped surface. (FET: ZZCVK 6° :0 L5 Zwlctm / / Problem 4 (25 Points)
Circular ring Ewe} VHW Power supply radius r2
6’ In the ﬁgure a very long solenoid of
length d and radius 7'1 is tightly wound
uniformly with N turns of Wire. A vari—
able power supply forces current to run
in the solenoid of amount I1 = p — kt,
where p and k are positive constants. A
circular metal ring of radius 7‘2 and re— resistance R sistance R is placed around the middle Very long solenoid: B
0f the SOIenmd' radius r1 N turns along length d 0...; an «Era
? mk NKe, . (a lOpts) Calculate the magnitude and direction of the constant magnetic ﬁeld inside the solenoid. Re—
member that the ﬁeld outside of a solenoid is zero. Consult.” “HM. aMp4n.M 1006) I __, —’) —>;_ Z]:
igimim law— “‘0 u. e . ‘HM; SARVlou‘oi Ponds ‘l'b ‘HM rchlr (56¢ «ML 4.4"a jam was up «w.
is (wole camsined (sce «Ci«‘3. {Wu/(Whine, else. ii— i”; 124w. 3 E, L+0(w)+(o)L+(o)u.> =33sz
C M L.
14011;“ = MON: 3,; W
rheaHo” c4: cunncf' pguiwﬁ‘nj *“M Sdf‘cuLe . (b 5pts) Determine the ﬂux through the circular ring of radius r2 as a function of time? .a ? n JloNrt ‘
42M: BHA ’13— d : Thukip)
A: TT‘QZ l7/C. 3:0 outside—
_ 4L“ galeuoicil
J4 NTW‘I’L
7520 (c 10pts) What is the magnitude of the induced current in the circular ring of radius 7’2? Nike). / go /5
N1‘4?! Makes Jtﬂse l7/c, ‘H/u. CVVNVLf‘th dean(“5037 / (d 5pts) Extra Credit: On the diagram, indicate the direction of the current you calculated in part Brieﬂy explain how you determined this. f (see a l7d fe m4 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 _, _ d1? d1? _ —o d}? N _, ,
a—dt dt— net ﬁanddt~ma1fv<<c
AUel = qAV AV = — ff E . dl % — Z (Exm + EyAy + EzAz)
@elzfﬁoﬁdA <I>mag=f§oﬁdA
fEﬁdA=Z—q:”sﬂi find/1:0
0
r 4 do a r r
lemfl : fENC . = ' 6;: g ° : M0 ZIinside path
r r d r A
f3 . = M0 Iinsidepath + 60% f E . Speciﬁc Results
4 1 2qs , » N 1 qs _
Edipole,azzis R: arm—7:3— (On 3X13; 7‘ >> 3) IEdipoleA.’ mg (on J— a‘XlS) 7“ >> 5)
_. 1 Q _ _ _.
ET = —————————— 'r' .L from center electric d1 ole moment = s, “’2 oz Ea 16
0d 4mm 7. T2 + (L/2)2 ( ) p p q 17 ppl d
a 1 2Q/L _ a _ 1 qz .
Emd N F60 T (1f 1' < L) Em”g —— m (Z2 + R2)3/2 (2 along axrs)
_. A Q/A z _ ‘ a Q/A Z Q/A .
‘ = — —— s z — — m f R
’Edlsk 260 1 (Z2 +R2)1/2 (2 along axrs) Ed1 k 260 [1 R] 260 (1 z < )
~ Q/A . e Q/A s . . .
Emmator % 60 (+Q and ——Q dlsks) ‘Efmnge z 60 Just outsrde capamtor
AB; = ﬂIAE X T (short Wire) AF 2 IAfX E
471' r2
" #0 LI #0 2I _. ..
Bwire = —————— % ——— 7‘ < L }Bwire = B a, tan6
l 47M1 /7~2 + (L/2)2 47r 7" ( ) 6 M
—o _ ,LLo 2I7TR2 N no 2.771'R2 , _ __ 2
‘Bloop — E (22 + R2)3/2 ~ 4” Z3 (on axrs, 2 > R) p, —— IA — I7rR
~ N #0 2M . ~ N #0 H 
leipole,aIis ~ E773 (on mm, 7" >> 3) Edipole; ~ E773 (on 1. 3X18, 7" >> 3)
a E
+ 1 —an_ A A A _. rad
= —— = Ta. Bra B'Nl :
Erad 4WD 027, v d X d d c
i=nAﬁ I=qnA17 27=uE
I L
__—_ = — = R : —“
a q nu J A (TE 0A
Eapplied q 1 1 .
. . = = _ _ _ t
Edwlecmc K AV 4WD Tf n due to a pom charge _ [AVI I R for an ohmic resistor (R independent of AV); power = I AV
IAV , _
Q = C AVI Power = I AV I = R (ohmic reastor)
di —' 2
K m 5—va if 1) << 0 circular motion: ﬁi : % x L]:
Math Help
a x 13': (amayﬂz) x (bm,by,bz)
= (ay bz — az byﬁ: — (asc bZ — a2 bag? + (ax by — ay bmﬁ
/ dac __1n(a+m)+c/ dm _ 1 +0/ dm _ 1 +
x+a— (:c—l—a)?‘ (1+3: (a+a:)3_ 2(a+:1:)2 c
a 2 2 a 3
/adm=ax+c /axdx=—2x +0 fact dnga: +c
Constant Symbol Approximate Value
Speed of light 0 3 X 108 m/s
Gravitational constant G 6.7 X 10—11 N  n12/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 EE— 9 X 109 N  m2/C2
0
Epsilon—zero 60 8.85 X 10‘12 (N  m2/C2)_1
Magnetic constant 52 1 X 10‘7 T v m/A
Mu—zero no 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 Ra % 1 X 10—10 m
Proton radius RP % 1 X 10—15 m
E to ionize air Eiom'ze % 3 X 106 V/m B Earth (horizontal component) BEarth % 2 X 10—5 T ...
View
Full
Document
This note was uploaded on 05/23/2011 for the course PHYISCS 2212 taught by Professor Shatz during the Spring '10 term at Georgia Institute of Technology.
 Spring '10
 SHATZ

Click to edit the document details