Unformatted text preview: PHYS 2212 Test 3
Nov 10th 2010
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 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 FINAL EXAM CONFLICT RESOLUTION
The ﬁnal exam period has been set by the Registrar’s Oﬃce. The Final Exam for all sections of 2212 will
be held during Period Seven, on Wedesday December 15, from 8:00 AM to 10:50 AM. If you think you
have a conﬂict please read the full announcement posted on WebAssign and email Dr. Greco the
requested information.
The departmental deadline for notifying us about a potential conﬂict is Monday, November 15! Problem 1 (25 Points)
A compass is located at the center of a square
coil of wire consisting of 3 turns. Each side of
the square is 6 cm long. Originally, the compass
points North, in the −x direction. Then the wire
is connected to a battery (which is not shown),
and a conventional current of 0.45 A runs in the
coil, in the direction indicated on the diagram.
(a 5pts) What direction does the compass needle
point while the current is running? Draw the direction clearly on the diagram below (which is a
top view of the compass). N (b 8pts) Determine the magnitude of the compass needle’s angular deﬂection while the current is running.
Clearly show each step in your work. (c 7pts) Determine the magnetic dipole moment of the square coil. Show all steps in your work. (d 5pts) Compute the magnitude of the magnetic ﬁeld at location 0, 1.5, 0 m due only to the current in
the coil. Show all steps in your work. Problem 2 (25 Points)
A horizontal pipe contains a dilute solution of sulfuric acid (H2 SO4 ). There are 4 × 1025 sulfate ions (SO2− )
4
per m3 (charge 2e) and 8 × 1025 hydrogen ions (H+ ) per m3 (charge +e). An electric ﬁeld is applied to
the right along the pipe, and the hydrogen ions acquire a drift speed of 9 × 10−5 m/s while the sulfate ions
acquire a drift speed of 2 × 10−5 m/s. The pipe has a crosssectional area 2.4 × 10−4 m2 . Determine the
magnitude and direction of the conventional current. Problem 3 (25 Points)
A capacitor consists of two circular plates, each 1.5 m in radius, separated by a gap of 2 mm. The capacitor
is charged and is connected to a resistive wire 5 m long, as shown in the diagram. A current runs through
the wire. At a particular instant, the magnitude of the electric ﬁeld at the location inside the wire marked
by the “x” is 3.5 V/m. At this instant, determine how much charge is on the positive plate of the capacitor.
Show your work.
2 mm R = 1.5 m L =5 m × Problem 4 (25 Points)
A Ushaped piece of wire hangs from supports and is connected to a battery, as shown. A steady current
of 4.5 A runs though the Ushaped wire. The bottom piece of the wire is 25 mm long, and is oriented
along the z axis. The vertical sides of the U are 4 cm long. y 4 cm A coil of wire is placed on the −x axis,
with its axis oriented along the x axis.
When the coil is connected to a power
supply that is not shown in the diagram, the 25 mm section of the Ushaped wire is acted on by a force in
the +y direction 0, 6 × 10−3 , 0 N. 25 m m x z
(a 10pts) If you stand on the +x axis and look toward the coil, which way is conventional current running
in the coil? In addition to describing the direction in words, show it on the diagram. To receive full credit,
explain brieﬂy using diagrams that include vectors. (c 15pts) Determine the magnetic ﬁeld at the location of the 25 mm section of the wire, due to the coil.
Your answer should be a vector and be sure to show all steps in your 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 ...
View
Full Document
 Spring '10
 SHATZ
 Electron, Magnetic Field, Electric charge, Erad

Click to edit the document details