Transcript Document 7412351

Exam 3 covers
Lecture, Readings, Discussion, HW, Lab
Exam 3 is Thurs. Dec. 3, 5:30-7 pm, 145 Birge
Magnetic dipoles, dipole moments, and torque
Magnetic flux, Faraday effect, Lenz’ law
Inductors, inductor circuits
Electromagnetic waves:
Wavelength, freq, speed
E&B fields, intensity, power, radiation pressure
Polarization
Modern Physics (quantum mechanics)
Photons & photoelectric effect
Bohr atom: Energy levels, absorbing & emitting photons
Uncertainty principle
Tues. Dec. 1, 2009
Phy208 Lect. 25
1
Current loops & magnetic dipoles


Current loop produces magnetic dipole field.
Magnetic dipole moment: 

  IA
current
Area of
loop


direction
magnitude
In a uniform magnetic field
Magnetic field exerts torque    B,    B sin 
Torque rotates loop to align  with B
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2
Works for any shape planar loop
r
  IA
r
 perpendicular to loop
I
Torque in uniform magnetic field
   B,    B sin 
r r
Potential energy of rotation: U   B  Bcos

Lowest energy aligned w/ magnetic field

Highest energy perpendicular to magnetic field

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3
Question on torque
Which of these loop orientations has the largest magnitude
torque? Loops are identical apart from orientation.
(A) a (B) b (C) c
a
b
c
4
Tues.
Dec. 1, 2009
Phy208 Lect. 25
Magnetic flux

• Magnetic flux is defined


B
dA
B
exactly as electric flux
• (Component of B  surface) x (Area element)
Faraday’s law


Time
derivative
d
  B
dt
EMF around loop
Magnetic flux through
surface bounded by path
If path along conducting loop,
 induces current I=EMF/R
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Phy208 Lect. 25
5
Quick quiz
Which of these conducting loops will have
currents flowing in them?
A.
C.
I(t) increases
Constant I
B.
Constant v
Constant I
Tues. Dec. 1, 2009
D.
Constant v
Constant I
Phy208 Lect. 25
6
Lenz’s law & forces
• Induced current produces a magnetic field.
– Interacts with bar magnet just as another bar magnet
• Lenz’s law
– Induced current generates a magnetic field
that tries to cancel the change in the flux.
– Here flux through loop due to bar magnet is increasing.
Induced current produces flux to left.
– Force on bar magnet is to left.
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7
Quick Quiz
A square loop rotates at frequency f in a 1T uniform
magnetic field as shown. Which graph best
represents the induced current (CW current is
positive)?
A.
B=1T
C.
1.5
1
0.5
0
0
0
-0.5
-1
0
45 90 135 180 225 270 315 360
-1.5
0
ANGLE ( DEGREES )
45 90 135 180 225 270 315 360
ANGLE ( DEGREES )
1.5
1.5
B.
D.
1
1
0.5
0.5
0
0
0
0
-0.5
-0.5
-1
-1
-1.5
-1.5
0
45 90 135 180 225 270 315 360
ANGLE ( DEGREES )
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0
45 90 135 180 225 270 315 360
ANGLE ( DEGREES )
Phy208 Lect. 25
=0 in
orientation
shown
8
Lenz’s law & forces
• Induced current produces a magnetic field.
– Interacts with bar magnet just as another bar magnet
• Lenz’s law
– Induced current generates a magnetic field
that tries to cancel the change in the flux.
– Here flux through loop due to bar magnet is increasing.
Induced current produces flux to left.
– Force on bar magnet is to left.
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Phy208 Lect. 25
9
Quick Quiz
A person moves a conducting loop with constant
velocity away from a wire as shown.
The wire has a constant current
What is the direction of force on the loop from the wire?
A.
B.
C.
D.
E.
F.
I
Left
Ftopside  Iinduced Lloop  Btop  ILByˆ
Right
Up
Down
v

Fbottomside  Iinduced Lloop  Bbottom  ILByˆ
Into page
Out of page
F
F
 0 cancel
leftside
Tues. Dec. 1, 2009

leftside
Btop  B bottom Force is up
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Inductors
• Flux = (Inductance) X (Current)
  LI
• Change in Flux
= (Inductance) X (Change in Current)
  LI
• Potential difference
dI
V  L
Constant current -> no potential diff

dt
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
Energy stored in ideal
inductor
• Constant current (uniform charge motion)
– No work required
to move charge through inductor
• Increasing current:
– Work VLq  VL Idt
required
to move charge across induced EMF
dI
– dW  VL Idt  L Idt  LIdI
dt

– Total work
W 
I

0
Tues. Dec. 1, 2009
1 2 Energy stored
LIdI  LI
in inductor
2
Phy208 Lect. 25
12
Inductors in circuits
IL t  0  0
VL t  0  Vbattery  L
dIL
dt
IL
dIL Vbattery


dt
L
IL instantaneously zero, but increasing in time
IL(t)
Slope dI / dt = Vbattery / L
0
0
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Time ( t )
Phy208 Lect. 25
13
Just a little later…
Switch closed at t=0
IL(t)
Slope dI / dt = Vbattery / L
0
0
Time ( t )
A short time later ( t=0+Δt ), the current is increasing
…
A. More slowly
B. More quickly
C. At the same rate
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Phy208 Lect. 25
IL>0, and IR=IL
VR≠0, so VL smaller
VL= -LdI/dt, so dI/dt smaller
14

Electromagnetic waves
In empty space:
sinusoidal wave propagating along x with velocity

E = Emax cos (kx –  t)
B = Bmax cos (kx –  t)
Bmax  E /c
• E and B are perpendicular oscillating vectors
•The direction of propagation is
perpendicular to E and B
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Quick Quiz on EM waves
z
c
E
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x
B
y
16
Power and Intensity
EM wave transports energy at its propagation speed.
Intensity = Average power/area =
2
2
2
E max Bmax E max
cBmax
co E max
I



2o
2oc
2o
2
Spherical wave: I  Psource / 4 r 2

Radiation Pressure
EM wave incident on surface exerts a radiation pressure

prad (force/area)
proportional to intensity I.
Perfectly absorbing (black) surface: prad  I /c
Perfectly reflecting (mirror) surface: prad  2I /c
Resulting force = (radiation pressure) x (area)
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Polarization

Linear polarization:


Linear polarizer:




E-field oscillates in
fixed plane of polarization
Transmits component of E-field parallel to
max
max
 E before
cos
transmission axis E after
Absorbs component perpendicular to
transmission axis.
2
Intensity I 
 E max
 Iafter  Ibefore cos2 
Circular Polarization

E-field rotates at constant magnitude

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Quantum Mechanics
• Light comes in discrete units:
– Photon energy E photon  hf  hc /   1240 eV  nm / 
• Demonstrated by Photoelectric Effect
– Photon of energy hf collides with electron in metal
– Transferssome or all of hf to electron
– If hf >  (= workfunction) electron escapes
Electron ejected only if hf > 
min
Minimum photon energy E photon
 hf required
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Photon properties of light
• Photon of frequency f has energy hf
– E photon  hf  hc / 
– h  6.626 1034 J  s  4.14 1015 eV  s
– hc 1240eV  nm
 • Red light made of ONLY red photons

– The intensity of the beam can be increased by
increasing the number of photons/second.
– (#Photons/second)(Energy/photon) =
energy/second = power
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Photon energy
What is the energy of a photon of red light
(=635 nm)?
A. 0.5 eV
B. 1.0 eV
1240 eV  nm
E

1.95 eV

635 nm
hc
C. 2.0 eV
D. 3.0 eV

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21
Bohr’s model of Hydrogen atom


Planetary model:
Circular orbits of electrons around proton.
Quantization



2
r

n
ao
Discrete orbit radii allowed: n
2
Discrete electron energies: E n  13.6 /n eV
Each quantum state labeled by quantum # n
How did he get this?


Quantization of circular orbit angular mom. L  r  p  mvr  n
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Consequences of Bohr model
• Electron can make transitions
between quantum states.
• Atom loses energy: photon emitted
E photon  E atomninitial  E atomn final
• Photon absorbed: atom gains energy:

E photon  E atomn final E atomninitial
E atom n  

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Phy208 Lect. 25
13.6eV
n2
23
Spectral Question
Compare the wavelength of a photon produced from
a transition from n=3 to n=1 with that of a photon
produced from a transition n=2 to n=1.
A.  31 <  21
n=3
n=2
B.  31 =  21
C.  31 >  21
E31 > E21
so
E  hf 
hc

 31 <  21

Wavelength is smaller for
larger
jump!
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1, 2009
Phy208 Lect. 25
n=1
24
Question
This quantum system (not a
hydrogen atom) has energy
levels as shown. Which photon
could possibly be absorbed by
this system?

A. 1240 nm
B. 413 nm
E photon 
hc


1240 eV  nm

E3=7 eV
E3=5 eV
E2=3 eV
C. 310 nm
E1=1 eV
D. 248 nm
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Phy208 Lect. 25
25
Matter Waves
• deBroglie postulated that matter has wavelike
properties.
• deBroglie wavelength   h / p
Example:
Wavelength of electron with 10 eV of energy:

Kinetic energy
p2
E KE 
 p  2mE KE
2m
h
hc
1240eV  nm



 0.39nm
2
6
2mE KE
2mc E KE
20.51110 eV 10eV 
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26
Heisenberg Uncertainty Principle

Using



x = position uncertainty
p = momentum uncertainty
Planck’s
constant
Heisenberg showed that the product
( x )  ( p ) is always greater than ( h / 4 )
Often write this as
x p ~
/2
h

is pronounced ‘h-bar’
2
where

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