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Unformatted text preview: PHYS 2212, Test 3, October 28th, 2009 Name (print) ___________________ “gig: ______________________________________________________________ __ 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. a 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—2 = W =
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
Problem 1 (15 pts)
Problem 2 (30 pts) Problem 3 (30 pts) —_ Problem 4 (25 pts) Problem 1 (15 Points) The following VPython program is intended to calculate and display the magnetic ﬁeld due to a single
moving charged particle, at a single location. from visual import *
from __future__ import division
##constants
mu_over_4pi = 1e7
L 2.
scalefactor = 7e9
deltat = 9e20
#objects & initial conditions
particle = sphere(pos=vector(0,0,4e—10), radius=1e11, color=color.red)
velocity = vector(0,0,4e4)
barrow=arrow(pos=vector(8e11,0,0), axis=(0,0,0), color=color.cyan)
scene.autoscale = O
#calculations
while particle.y > 5e10:
particle.pos = particle.pos + velocity*deltat
r = barrow.pos  particle.pos
rmag = sqrt(r.x**2+r.y**2+r.z**2)
rhat = r/rmag
B = mu_over_4pi*q*cross(velocity,rhat)/rmag**2
barrow.axis = B*scalefactor (a 5pts) In what direction does the magnetic ﬁeld point?
’5 a” .8?!» (Jiltch 0i: Xckvmivqﬁ. ,b/Dbﬁr
E, {5 in ,5 Aimvch’ou . ( r {ski Rama WIch +1 (b 5pts) Where is the particle located when the magnetic ﬁeld is largest? Give the position as a vector and explain how you determined this. 4\ ‘9‘ Mo WV"? where, {s smut/54' lg act JAM WiUivL , (DID/a) (c 2pts) In order to change the particle to an alpha particle (a helium nucleus, containing two protons and two neutrons), what line or lines of code would you need to change? Circle the line(s) in the program, and
write the changed line(s) here: 2: 2" Lee—l0! (d 3pts) After changing the particle to an alpha particle, you run the program. In What way will the
display look different in this version? Give as precise and clear a description as you can. m 0W (bud mesew‘kwl “RN, «443th \MC‘kN v36“ be Lou/01¢ V V/(, 7) VS
ahead—Cr but 5H“ T05A’i M ' Problem 2 (30 Points) A long wire is formed into a vertical rectangular
coil in the my plane, as shown in the diagram, and
connected to a battery (not shown). The height of
the rectangle is h = 6 cm, and the width is w = 3
cm. A compass is placed a distance 6 = 4 mm
beneath the bottom wires. When a current runs in
the coil, the compass needle is deflected 20° East
when viewed from the origin; Magnetic North is in
the +m direction. (a 15pts) Determine the magnitude of the con
ventional current in the loop. Indicate the direction
of this current on the diagram. The compass is close
enough to the bottom wire that the magnetic ﬁeld
from the other three wires can be neglected when
solving for the current in the loop. Bk) : Baal/£6 Ma L '
3.. 4; , we I: 0.;th (b 15pts) Determine the approximate magnetic ﬁeld expressed as a vector at location (0,0, ~—0.7) In, due
only to the current in the coil. Show every step in your work. If you make any approximations or
assumptions, state them clearly. Appmxn'mqsk /00/ M Mﬁ7m/‘7‘c o/ipa/e,
g: i“: 231 M:I/4=Izw
‘HT r3 / 2(0/‘4(0H)(é/013)(3/o',:) M, 2:11“) ‘ ' 47”, ______/—————
5: If"? r3 ' (In) 7*.) (0«7’M)3
’IO
 , T ’7 ,, . ~’° *
 /.§3/o 53(0/0/ Irilo 7/ Problem 3 (30 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 ma
terials. Two of the wires are identical:
they are each L1 = 7 cm long, and are
made of material having n1 = 6 x 1025
mobile electrons per cubic meter. The
central wire is L2 = 5 cm long, and
has 112 = 2.5 x 1025 mobile electrons
per cubic meter. Both wires have the crosssectional area 4 X 10’6 In2 and
the electron mobility of each material
is 5 X 10"5 (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 5pts) Determine the magnitude of the electric ﬁeld in wire 2. Show all steps in your work. Av: .. 3 343 E’ ,3 “New m4.»
Av 1.33\/ 
I?” a: t ’ “WM (b 7pts) In the steady state, calculate the magnitude of the electric ﬁeld in the left wire, whose length is
L1. Show all steps in your work. Moi: rule 3 AV 0 (00,9 t too r em; ~ 5": ' FLLI ' EIL’ :0
P V‘gfz AV 25L! : crvtc ’ AV; £Av aﬁv' 1.33V V
“A 2 v , = //.2 /
El: T ' l(¥~/ozm) m (c lOpts) 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. ’y _q
’ ‘ r ‘/ : . ' m
\Il 1 ME. — (5/0 Wmﬂna. IM) 5 bl (0 l5
(5 ' h ,3
v1.4a}: (9.10 ﬁwa/M)  1.33/0 m/s
Ad:va+ )
L! L; A In k LZ
T: 7 + / +,— 1 «— _._
l VL V. V\ VZ
1(7 (0 W‘ (ilo’zm
T'I +. _#——’— =231'N 5
(Cu (o'qW/s) (/3340 3.445) (d 8pts) How many electrons per second cross the junction between the two wires? Show your work. blzbl /
.. §N/5
;,,m,AAE.=(O'/°”7§a)(4'/0 may” m)(”'2“/”) Silt/tall error b/L mumﬂivta_
‘ a
1/1: “ZﬂMEL : '/0l e/S Problem 4 (25 Points) A metal slab of width 3 / 2 and area A is centered between the plates of a capacitor with a gap width 3
and area A. The plates are connected to a battery with emf by Nichrome wires and allowed to charge
completely. Then the metal slab is removed quickly. (a 4pts) On the above diagrams above, show the electric ﬁeld at location A inside the Nichrome wire, just
before and just after removing the metal slab. If the electric ﬁeld is zero state this explicitly. (b 9pts) Calculate the total charge on the right plate before the metal slab is removed?
look) rule 3 Avloo‘) 7' (c 7pts) Just after removing the metal slab, will an electron current ﬂow? If so, in what direction will the
electron current ﬂow at A? To receive full credit you must provide a clear explanation. T55 I Cow/Mr Clockwise lo Macc Achf W; W W H Um“ WWWQ W mum access ‘H’LL Cox‘oacﬂvr {s new l4?ij *fl/(avt {MC . (d 5pts) After the metal slab is removed and the circuit has reached equilibrium, calculate the total charge
on the right plate. Avlm‘o : gym9’ Q[_AVM(’ ’ g: 0 0 o
_ an
AVMP'vew/Li' AVMe : E3 ' éo 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‘dﬁ’ d5_ 4 dﬁN r,
a—dt EE— net dandarvmalfv<<c
AUel=quV AV:—ﬁ:Eodlm—Z(E$Ax+EyAy+EzAz)
<I>el=onﬁdA <I>mag=fBofidA
fﬁmczAzilgmﬂ fémdAzo
0
demag f E o elf: #0 Z Iinside path lemfl = fE—lNC Odi=‘ dt
—a —a d —» A
f B . = M0 [inside path ‘l' 503% f E 0 Speciﬁc Results —» 1 2qs , —' 1 (18 .
‘Edipoleﬁxis % (on 3X13, 7' >> 3) Edipoledl % 471.60 “7'3” (on J a‘XIS: T >> 3)
a 1 Q a
E = ————— _L f t l t ’ d' 1 t = 4: E 
rod 47m) r T2 + (L/2)2 (7“ rom cen er) e ec I'lC 1pc e momen p qs, p or applied
~ 1 ZQ/L , .4 1 qz .
Emd N R T (if 1“ < L) Ting = mm (2 along ax15)
~ Q/A Z . a Q/A Z Q/A .
Edisk = 260 1 — m (2 along 3X18) Edisk ~ 260 — ~ 260 (1f 2 << A a A
IEmpaaW’ % Q/ (+62 and —Q disks) {ElmWe z 62/ just outside capacitor
50 60 2R
» [AZ ‘ a a a
AB = g X T (short Wire) AF = I Al x B
47r r2
—‘ _ _ [1,0 LI N [1,0 2] ¢  _ .i
lezre —— Em N E? (7‘ < L) iszre — ‘Bearth‘ tang
—» _ [L0 ZIﬂRg N [1,0 21WR2 _ _ _ 2
.Bloop —EW AJE 23 (on ax1s, z>>R) ,u—IA—I7rR
a 2 , a _
depolemis % ﬁll—73%: (on ems, 7' >> 3) leipoleaL‘ % ﬁll—73% (on J. ems, 7" >> 8)
~ 1 —qc'i A A A _. Erad
Erad : 4 2 J— ’U = rad X Brad Brad =
7m) 0 r c
i=nA17 I=qlnA17 77=uE
I L
= = — : E 2 —
0 [611% J A a R 0A
E  1 1
Edielectm'c = applied AV = —q— — — —— due to a point charge
K 47reo rf n IAVI I = R for an ohmic resistor (R independent of AV); power 2 I AV
[AV] . .
Q = C IAV[ Power = I AV I = R (ohmic res1stor)
dt * 2
K m %mv2 if’u << 0 circular motion: d—i = liRl m LL]:
Math Help
a x '6: (am,ay,az) x (bx,by,bz)
= (ay bz — a2 by)§: — (ax b2 — a2 bag + (am by — ay bm)2
/ dm _1n(a+m)+c/ dry __ 1 +C/ d3: _ 1 +6
m+a_ (m+a)2— a+x (a+m)3_ 2(a+ar)2
‘1 2 2 a 3
/adm=am+c /aacdm=§x +c /aa: dm=§$ +c
Constant Symbol Approximate Value
Speed of light 0 3 X 108 m/s
Gravitational constant G 6.7 x 10—11 N  mZ/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 ~ mg/C2
0
Epsilonzero 60 8.85 X 10—12 (N  m2/C2)_1
Magnetic constant 2%? 1 x 10—7 T  m/A
Mu—zero ,uo 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 m 1 X 10—10 m
Proton radius Rf, % 1 X 10—15 m
E to ionize air Eiomze % 3 X 106 V/m
BEmnth (horizontal component) BEarth % 2 x 10—5 T ...
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This note was uploaded on 04/10/2011 for the course PHYS 2212 taught by Professor Kindermann during the Spring '09 term at Georgia Institute of Technology.
 Spring '09
 Kindermann

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