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Physics 2113
Jonathan Dowling
Lecture 32: MON 09 NOV
Review Session A : Midterm 3
EXAM 03: 6PM WED 11 NOV in Cox Auditorium
The exam will cover: Ch.27.4 through Ch.30
The exam will be based on: HW08–11
The formula sheet and practice exams are here:
http://www.phys.lsu.edu/~jdowling/PHYS21133-FA15/lectures/index.html
Note this link also includes tutorial videos on the various right hand rules.
You can see examples of even older exam IIIs here:
http://www.phys.lsu.edu/faculty/gonzalez/Teaching/Phys2102/Phys2102OldTests/
Problem: 27.P.046. [406629]
In an RC series circuit, E = 17.0 V, R = 1.50 M, and C =
1.80 µF.
(a) Calculate the time constant.
(b) Find the maximum charge that will appear on the
capacitor during charging.
(c) How long does it take for the charge to build up to 16.0
µC?
Magnetic Forces and Torques
v
r=
F
L
mv
qB
Top view
t net = iAB sin q
Side view
CFnet = 0
C
(28-13)
U = mB
U = -m B
(28-14)
Ch 28: Checkpoints and Questions
28.3: Finding the Magnetic Force on a Particle:
Always assume particle is POSITIVELY charged to work
Out direction then flip your thumb over if it is NEGATIVE.
(a) + z
(b) - x
(c) sin180° = 0
F1 = 0 B + ⊙
F3 = 0 B + ⊙
E
F2 = ⊙ B + ⊙
E
(a) F2 > F1 = F3
(b) F4 = 0
F4 = - ⊙ B + ⊙
E
E
ICPP
The left face is at
a lower electric potential (minus charges) and
the right face is at a higher
electric potential (plus charges).
Circular Motion:
v
F
r
B into blackboard.
Since magnetic force is perpendicular to motion,
the movement of charges is circular.
2
v
out
Fcentrifugal
= ma = mrw 2 = m
r
in
Fmagnetic
= qvB
FB = FC
mv
® qv B =
r
2
mv
Solve : r =
qB
In general, path is
a helix (component of v parallel to
field is unchanged).
.
electron
C
.
r
mv
r=
qB
v qB
w= =
r m
2pr 2pmv 2pm
Tº
=
=
v
qBv
qB
1 qB
f º =
T 2pm
Radius of Circlcular Orbit
Angular Frequency:
Independent of v
Period of Orbit:
Independent of v
Orbital Frequency:
Independent of v
+
⊙
⊙
(a) The electron because q = e for both particles
but m p >> me Þ rp >> re
(b) Proton counter-clockwise and electron
clockwise by right-hand-rule.
v
F
⊙
Å
r
Which has the longer period T?
2p r
B into blackboard.
T=
v
Since v is the same and rp >> re the proton has the longer period T. It
mv
r=
has to travel around a bigger circle but at the same speed.
qB
ICPP
Two charged ions A and B traveling with a constant velocity
v enter a box in which there is a uniform magnetic field
directed out of the page. The subsequent paths are as shown.
What can you conclude?
A
v
B
v
(a) Both ions are negatively charged.
(b) Ion A has a larger mass than B.
(c) Ion A has a larger charge than B.
(d) None of the above.
mv
r=
qB
RHR says (a) is false. Same charge q, speed v, and same B for both masses. So: ion with larger
mass/charge ratio (m/q) moves in circle of larger radius. But that’s all we know! Don’t know m or q
separately.
Problem: 28.P.024. [566302]
In the figure below, a charged particle moves into a region of uniform
magnetic field , goes through half a circle, and then exits that region. The
particle is either a proton or an electron (you must decide which). It
spends 160 ns in the region.
(a) What is the magnitude of B?
(b) If the particle is sent back through the magnetic field (along the same
initial path) but with 3.00 times its previous kinetic energy, how much
time does it spend in the field during this trip?
Magnetic Force on a Wire.
L
28.8.2. A portion of a loop of wire passes between the poles of a magnet as shown.
We are viewing the circuit from above. When the switch is closed and a current
passes through the circuit, what is the movement, if any, of the wire between the
poles of the magnet?
a) The wire moves toward
the north pole of the magnet.
b) The wire moves toward the
south pole of the magnet.
c) The wire moves upward
(toward us).
d) The wire moves downward (away from us into board).
e) The wire doesn’t move.
i
Example
Wire with current i.
Magnetic field out of page.
What is net force on wire?
F1 = F3 = iLB
dF = iBdL = iBRdq
By symmetry, F2 will only
have a vertical component,
p
p
0
0
F2 = ò sin(q )dF =iBR ò sin(q )dq =2iBR
Ftotal = F1 + F2 + F3 = iLB + 2iRB + iLB = 2iB( L + R)
Notice that the force is the same as that for a straight wire of length R,
and this would be true no
matter what the shape of
L
R
R
L
the central segment!.
-y direction ¯
i
SP28-06
Fgrav = mg Fmag = iLB
-3
m / L = 46.6 ´10 kg/m
To balance:
Fgrav = Fmag
Þ mg = iLB
mg æ m ö g
ÞB=
=ç ÷
iL è L ø i
ICPP:
Find Direction of B
i
B=
46.6 ´ 10 –3 kg
m
=
1.6 ´ 10 –2 kg
s×C
|
9.8 m
s2
| 28 Cs
= 16 ´ 10 T = 16 mT
–3
Example 4: The Rail Gun
• Conducting projectile of length
2cm, mass 10g carries constant
current 100A between two rails.
• Magnetic field B = 100T points
outward.
• Assuming the projectile starts
from rest at t = 0, what is its
speed after a time t = 1s?
• Force on projectile: F= iLB
• Acceleration: a = F/m = iLB/m
• v = at = iLBt/m
rails
B
I
L
projectile
(from F = iL x B)
(from F = ma)
(from v = v0 + at)
= (100A)(0.02m)(100T)(1s)/(0.01kg) = 2000m/s
= 4,473mph = MACH 8!
= - m Bcosq
B
Highest Torque:
 = ±90° sin = ±1
Lowest Torque:
 = 0° & 180° sin = 0
 = 180°
–cos = +1
 = 0°
–cos = –1
Principle behind electric motors.
Torque on a Current Loop:
Rectangular coil: A=ab, current = i
Net force on current loop = 0
But: Net torque is NOT zero!
F1 = F3 = iaB
F^ = F1 sin(q )
Torque = t = F^b = iabB sin(q )
For a coil with N turns,
τ = N I A B sin ,
where A is the area of coil
Magnetic Dipole Moment
We just showed: τ = NiABsinθ
N = number of turns in coil
A = area of coil.
Define: magnetic dipole moment m
Right hand rule:
curl fingers in direction
of current;
thumb points along 
As in the case of electric dipoles, magnetic dipoles tend to align
with the magnetic field.
Electric vs. Magnetic Dipoles
+Q
p=Qa
-Q
q
QE
QE
1 and 3 are “downhill”.
2 and 4 are “uphill”.
U1 = U4 > U2 = U3
t1 = t 2 = t 3 = t 4
τ is biggest when B is at
right angles to μ
Right Hand Rule: Given Current i Find Magnetic Field B
Checkpoints/Questions
Magnetic field?
Force on each wire due to
currents in the other wires?
Ampere’s Law: Find Magnitude of ∫B∙ds?
Superposition: ICPP
I-OUT
• Magnetic fields (like electric
⊙
fields) can be
“superimposed” -- just do a
vector sum of B from different
sources
• The figure shows four wires
located at the 4 corners of a
square. They carry equal
Ä
I-IN
currents in directions
indicated
• What is the direction of B at
the center of the square?
I-OUT
⊙
Ä
I-IN
B
Right Hand Rule: Given Current i Find Magnetic Field B
The current in wires A,B,D is out of the
page, current in C is into the page. Each wire
produces a circular field line going through
P, and the direction of the magnetic field for
each is given by the right hand rule. So, the
circles centers in A,B,D are
counterclockwise, the circle centered at C is
clockwise. When you draw the arrows at the
point P, the fields from B and C are pointing
in the same direction (up and left).