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Unformatted text preview: PHYS 2212 Test 3
Spring 2011
Name(print)
Instructions
• Read all problems carefully before attempting to solve them.
• Your work must be legible, and the organization must be clear.
• You must show all work, including correct vector notation.
• Correct answers without adequate explanation will be counted wrong.
• Incorrect work or explanations mixed in with correct work will be counted wrong. Cross out anything
you don’t want us to read!
• Make explanations correct but brief. You do not need to write a lot of prose.
• Include diagrams!
• Show what goes into a calculation, not just the ﬁnal number, e.g.:
5 × 104 a·b
c·d = (8×10−3 )(5×106 )
(2×10−5 )(4×104 ) = • 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
Problem
Problem
Problem Score
1
2
3
4 (25
(25
(25
(25 pts)
pts)
pts)
pts) Grader 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 n1 = 6 × 1025 mobile electrons per cubic meter. The central wire is L2 =
5 cm long, and has n2 = 2.5 × 1025 mobile electrons per cubic meter. All wires have the crosssectional area 4 × 10−6 m2 and the electron mobility of each material is 5 × 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. (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. (c 5pts) How many electrons per second cross the junction between the two wires? Show your work. Problem 2 (25 Points) × A round light bulb is connected to two batteries, each with emf = 1.5 V, and a conventional current I0 = 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. A (a 5pts) Sketch a graph of conventional current vs. time in the new circuit (including the
capacitor): (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 Ef ringe , 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.
(c 8pts) Calculate the charge on the positive plate of the fully charged capacitor. Problem 3 (25 Points) A battery with an unknown emf is connected
in series to an ohmic resistor with resistance
R1 = 5 Ω and a nonohmic resistor. The resistance of the nonohmic resistor is given by the
formula Rn (T ) = R0 + r0 T , where R0 = 20 Ω,
r0 = 0.018 Ω/kelvin, and T is the temperature
of the nonohmic resistor in kelvin. (a 15pts) When the circuit is ﬁrst connected, the nonohmic 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 nonohmic
resistor is at a temperature of 400 K. What is the conventional current in the circuit after the 20 minutes? (b 10pts) Considering the microscopic view of resistance, what material or geometric property of the nonohmic resistor likely changed during the 20 minutes? As best you can, describe how temperature aﬀected
this property. Problem 4 (25 Points)
At a particular instant a proton is moving with velocity 5.5 × 107 , 0, 0 m/s, and an electron is moving with
velocity 0, −7 × 107 , 0 m/s. The electron is located 1.5 × 10−6 m below the proton (in the y direction). y 1.5e6 meters v
p x
z (out of page)
ve
(a 4pts) On the diagram draw an arrow representing the electric ﬁeld at the location of the electron, 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 Fel .
(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.
Show all work. This page is for extra work, if needed. Things you must know
Relationship between electric ﬁeld and electric force
Electric ﬁeld of a point charge
Relationship between magnetic ﬁeld and magnetic force
Magnetic ﬁeld of a moving point charge Conservation of charge
The Superposition Principle Other Fundamental Concepts
dp
dp
= Fnet
and
≈ ma if v << c
dt
dt
f
∆V = − i E • dl ≈ − (Ex ∆x + Ey ∆y + Ez ∆z )
Φmag = B • ndA
ˆ dv
dt
∆Uel = q ∆V
Φel = E • ndA
ˆ
qinside
E • ndA =
ˆ a= 0 emf = EN C • dl = B • ndA = 0
ˆ
dΦmag
dt Iinside path + B • dl = µ0 B • dl = µ0
0 d
dt Iinside path E • ndA
ˆ Speciﬁc Results
1 2qs
(on axis, r
s)
4π 0 r 3
Q
1
(r ⊥ from center)
Erod =
4π 0 r r 2 + (L/2)2
1 2Q/L
Erod ≈
(if r
L)
4π 0 r
z
Q/A
(z along axis)
1− 2
Edisk =
20
(z + R2 )1/2
Q/A
(+Q and −Q disks)
Ecapacitor ≈
Edipole,axis ≈ 0 µ0 I ∆ × r
ˆ
∆B =
(short wire)
2
4π
r
µ0
µ0 2I
LI
Bwire =
≈
(r
2 + (L/2)2
4π r r
4π r 1 −qa⊥
4π 0 c2 r
i = nAv
¯ Erad = σ = q  nu
Edielectric = Eapplied
K 1 qs
(on ⊥ axis, r
4π 0 r 3 s) electric dipole moment p = qs, p = α Eapplied
qz
1
(z along axis)
2 + R2 )3/2
4π 0 (z
Q/A
Q/A
z
Edisk ≈
≈
(if z
R)
1−
20
R
20
Q/A s
just outside capacitor
Ef ringe ≈
2R
0
Ering = ∆F = I ∆l × B
L) µ0 2IπR2
2IπR2
µ0
≈
(on axis, z
4π (z 2 + R2 )3/2
4π z 3
µ0 2µ
Bdipole,axis ≈
(on axis, r
s)
4π r 3
Bloop = Edipole,⊥ ≈ Bwire = Bearth tan θ
R) µ = IA = IπR2 Bdipole,⊥ ≈ µ0 µ
(on ⊥ axis, r
4π r 3 ˆ
ˆ
v = Erad × Brad
ˆ Brad = I = q  nAv
¯
I
J=
= σE
A
1
q
1
∆V =
−
4π 0 rf
ri v = uE
¯
L
R=
σA
due to a point charge s) Erad
c I=  ∆V 
for an ohmic resistor (R independent of ∆V );
R Q = C ∆V  power = I ∆V Power = I ∆V K ≈ 1 mv 2 if v
2 c I= circular motion: dp
v 
mv 2
=
p ≈
dt ⊥
R
R ∆V 
(ohmic resistor)
R Math Help a × b = ax , ay , az × bx , by , bz
= (ay bz − az by )ˆ − (ax bz − az bx )ˆ + (ax by − ay bx )ˆ
x
y
z dx
= ln (a + x) + c
x+a dx
1
=−
+c
2
(x + a)
a+x a dx = ax + c
Constant
Speed of light
Gravitational constant
Approx. grav ﬁeld near Earth’s surface
Electron mass
Proton mass
Neutron mass
Electric constant
Epsilonzero
Magnetic constant
Muzero
Proton charge
Electron volt
Avogadro’s number
Atomic radius
Proton radius
E to ionize air
BEarth (horizontal component) ax dx =
Symbol
c
G
g
me
mp
mn
1
4π 0
0 µ0
4π
µ0
e
1 eV
NA
Ra
Rp
Eionize
BEarth a2
x +c
2 dx
1
=−
+c
3
(a + x)
2(a + x)2
ax2 dx = a3
x +c
3 Approximate Value
3 × 108 m/s
6.7 × 10−11 N · m2 /kg2
9.8 N/kg
9 × 10−31 kg
1.7 × 10−27 kg
1.7 × 10−27 kg
9 × 109 N · m2 /C2
8.85 × 10−12 (N · m2 /C2 )−1
1 × 10−7 T · m/A
4π × 10−7 T · m/A
1.6 × 10−19 C
1.6 × 10−19 J
6.02 × 1023 molecules/mole
≈ 1 × 10−10 m
≈ 1 × 10−15 m
≈ 3 × 106 V/m
≈ 2 × 10−5 T ...
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 Spring '10
 SHATZ

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