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Unformatted text preview: PHYS 2212 Test 3 Spring 2011
Name(print) K6 Instructions o 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! a Show what goes into a calculation, not just the ﬁnal number, e.g.: a—b — M51193) 2 ad — (2x10— )(4x10 )
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 The Final Exam will be held during Exam Period Seven,
Wednesday, May 4, from 8:00 to 10:50am. If you have a
conﬂict, please contact Dr. Greco by Monday April 4th. PHYS 2212
Do not write on this page! Problem
Problem 1 25 pts)
 Problem 2 25 pts) Problem 3 25 pts)
Problem 4 25 pts) AAAA Problem 1 (25 Points) A circuit contains one battery and three wires in series as shown in the diagram. The wires have different
lengths and are made of different materials. Two of the wires are identical: they are each L1 = 7 cm long, and are made of material having 111 = 6 X 1025 mobile electrons per cubic meter. The central Wire is L2 =
5 cm long, and has n2 2 2.5 x 1025 mobile electrons per cubic meter. All wires have the cross—sectional area 4 X 10"6 m2 and the electron mobility of each material is 5 x 10—4 (m/s)/(V/m) The emf of the battery is 2.9 volts. In the steady state, the potential drop across the length
L2 of wire 2 is 1.33 V. (a 15pts) In the steady state, what is the magnitude of the electric ﬁeld in the left wire, of length L1?
Show all steps in your work. / , ‘(0 M (C; —2
.. 7. .10
Ex , 2L, + a—‘LLZ (mm W 7’7» (b 5pts) How long would it take a single electron to travel from the negative end of the battery all the way
around the circuit to the positive end of the battery? Show all steps in your work. ¢:‘tl2_+t +t\ :2t'+t7_
Ml
L L l
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l V; / (05 M/
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vz , l 33 /o "4/5 [ ,‘+ shat/3! 1m
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7 10"") (M :
T: l (ﬂirt/5) + (l«%3/o"ZM/s) 28 015 Problem 2 (25 Points) A round light bulb is connected to two bat—
teries, each with emf = 1.5 V, and a conven
tional current IO = 25mA runs in the circuit.
An uncharged capacitor (C = 1 F) exactly
like the one you used in class and lab is connected in series to the battery and the
light bulb. 25 mA (a 5pts) Sketch a graph of conventional cur—
rent vs. time in the new circuit (including the
capacitor): 305 (b 12pts) The circuit with the capacitor is allowed to fully charge. On the diagram, at location A inside
the wire, draw and label three arrows representing the: (1) net electric ﬁeld Enet, (2) the electric ﬁeld due
to the capacitor plates E fringe, and (3) the electric ﬁeld due to charge on the surface of the wires and in and on the batteries Esurf If any ﬁelds are zero, explicitly state this. (0 8pts) Calculate the charge on the positive plate of the fully charged capacitor. [00? NILZ AVWO‘VL O 3],”? : 2em¥ «I but“ '5 ” Problem 3 (25 Points) A battery with an unknown emf is connected
in series to an ohmic resistor with resistance
R1 = 5 Q and a non—ohmic resistor. The resis
tance of the non—ohmic resistor is given by the
formula Rn(T) = R0 +r0T, where R0 = 20 (2;
7’0 2 0.018 Q/kelvin, and T is the temperature
of thenon—ohmic resistor in kelvin. non—ohmic resistor (a 15pts) When the circuit is ﬁrst connected; the non—ohmic resistor is at a temperature of 273 kelvin,
and the conventional current in the circuit is measured to be 3.00 A. Twenty minutes later, the non—ohmic
resistor is at a temperature of 400 K. What is the conventional current in the circuit after the 20 minutes? Vise [004’ Mt Mtge:
T:7_?5K : who? — emhtﬁ. 40mm) 50/ 571+ ZofL—k o‘Olfér—E (£43K)
afﬂ/ It: Io KtaKnHOO) : 34 SJUZo/LPOD‘SMJ‘EHW‘)
l": 2.23:4 < Io (b 10pts) Considering the microscopic view of resistance, what material or geometric property of the non
ohmic resistor likely changed during the 20 minutes? As best you can7 describe how temperature affected
this property. m MENU \/Lo.s decmqsecgi 24$ vH/tL resielzsr 6 MS Vank; Wowe,
With U? [m whim/18c o \ (ollc’J/KD MS Problem 4 (25 Points)
At a particular instant a proton is moving with velocity (5.5 x 107, 0, 0) m/s, and an electron is moving with
velocity <0, —7 X 107, 0) m/s. The electron is located 1.5 x 10’“6 m below the proton (in the y direction). V
p
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y {7‘3 ‘ ‘.
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9! ’3 (I’__»
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"‘ : EMSLT ei
"‘ I 2<" .
(9 x _l_ 59$ _:.)
z (out of page) (a 4pts) On the diagram draw an arrow representing the electric ﬁeld at the location of the electron7 due
to the proton. Label it Ep. (b 4pts) On the diagram draw an arrow representing the magnetic ﬁeld at the location of the electron, due
to the proton. Label it Bp. (c 4pts) On the diagram draw an arrow representing the electric force on the electron at this instant. Label
it Fez (d 4pts) On the diagram draw an arrow representing the magnetic force on the electron at this instant.
Label it Fmag. (e 9pts) Calculate the magnetic force on the electron at this instant. Your answer should be a vector. W wr. v? A
Sho all 0; J a __ 7g : £9 gp VP xr
(Ma /0"'7C) {gr/07,0,» W5 x<o, «i, 0) ——>> ~¥TM //'T——_—’—
BP:(/'/0 7( (LgIO'6M) 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 1 d6 er a d"
azg V £23161; Handﬁwmaifv<<c
Ayelzqév AV:—f1/:Eodl~—Z(EmAa:+EyAy+EzAz)
<I>el=onﬁdA (PmangBoﬁdA
fE.ﬁdA:;M fﬁoﬁdAzO
60
demag emf = fENC 0 df: dt
f g . dz: #0 [2 [inside path + 60d dtf E . ndA] Speciﬁc Results 7» 1 2 s . —» 1 s .
Edipole,aa:is 7“) 47T€0 TQ3_ (on 3X13: 7’ >> 3) 'Edipole,_L[ z 47r€0 (:13 (on J aXIS, T >> 3)
a 1 Q a
E = —— J. f t l t ‘ d' 1 t = ”2 E ~
Tod 47T€0 7‘ r2 + (L/2)2 (7“ rom cen er) e ec r1c 1p0 e momen p qs, p a applied
—' 1 2Q/L _ —. 1 qz _
Erod N 47"60 7" (1f 7' << L) ring :04’IT6 W (2 along axrs)
a Q/A z . Q/A Q/A .
[Edisk = —2 60 1 — ———(z2 + R2)1/2 (z along ax13) 'Edwkz 260 [1 — El ~ 260 (if z << R)
Ecapacitor ~ Q—e/ (+62 and —Q disks) [Efringe N Q/ (2—S‘R) just outside capacitor
0 6
~ I AZ _. _. _,
AB: ”—9 J T (short Wire) AF = [Al x B
47r 7'
‘4 M0 [11%de 2—1 { —¢ _.
B ~ = ———————— L B . = I t a
”LU’LTE 471’ 7” 7'2 + (L/2)2 WEI—7T 7’ (7' << ) wwe Bearth an
no 217TR2 N [10 2171'R2 . 2
ZZWWNE Z3 (on ax15,z>>R) [.LZIAZI’ITR
#0 2M . » ,uo ,
m E? (on ax1s, 7‘ >> 8) Bdipole,.l.‘ ’5 Eff), (on —L 3X15: 7" >> 3)
~ E
—. 1 — a A A A _. Tad
Erad : Q2 J— 7} : rad X Brad Brad
47T€0 c 7’ c
iznAr“) I=[qnA?7 v—uE
I L
0 IQ W A a 0A
E ~ 1 1
Edielecm'c = applied AV 2 q —— — — due to a point charge
K 47reo rf 7‘2 A
121w for an ohmic resistor (R independent of AV); power = I AV R
AV ‘ _
Q = C IAV Power = I AV 1 = — (ohmic I‘eSlStOI')
d” ' 2
K % %mv2 if 11 << 0 circular motion: d_1:1 = [—17%] I]?! % mRi
Math Help
a x 5: (am, ay,az) >< (bmby, bz)
= (ay bz — az by)a": — (ax bz  az bag + (am by — ay b$)2
/ dm 1 ( + ) + / dm 1 + c / da: ' 1 + c
= n a a: c ———— = — = ____
x+a (96+a)2 a+m (a+a:)3 2(a+a:)2
a 2 2 a 3
/adx:a$+c /amdm=§x +c fax dngm +c
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 e 9 X 109 N  rri2/C2
7F 0
Epsilon—zero 50 8.85 X 10“12 (N  m2/C2)“1
Magnetic constant g9 1 X 10“7 T  In/ A
7T
Mu—zero H0 47r x 10“7 T  m/A
Proton charge e 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 "NV 1 x 10—10 In
Proton radius R1, m 1 X 10“15 In
E to ionize air Emma z 3 x 106 V/m BEarth (horizontal component) 7 BEaTth m 2 x 10—5 T ...
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