Exam 3 coversLecture, 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 1Phy208 Lect. 25

Tues. Dec. 1, 2009 Phy208 Lect. 25 2

Current loops & magnetic dipoles Current loop produces magnetic dipole field. Magnetic dipole moment:

€

rμ

€

rμ =IA

currentArea of loop

€

rμ

magnitude direction

In a uniform magnetic field

Magnetic field exerts torqueTorque rotates loop to align with

€

rτ =

rμ ×

rB ,

€

rμ

€

rB

€

rτ =

rμ

rB sinθ

Works for any shape planar loop

Tues. Dec. 1, 2009 Phy208 Lect. 25 3

I

€

rμ =IA

€

rμ perpendicular to loop

Torque in uniform magnetic field

€

rτ =

rμ ×

rB ,

€

rτ =

rμ

rB sinθ

Potential energy of rotation:

€

U = −r μ ⋅

r B = −μBcosθ

Lowest energy aligned w/ magnetic fieldHighest energy perpendicular to magnetic field

4

Which of these loop orientations has the largest magnitude torque? Loops are identical apart from orientation.

(A) a (B) b (C) c

Question on torque

a bc

Tues. Dec. 1, 2009 Phy208 Lect. 25

Magnetic flux

€

ε =−d

dtΦB

Magnetic flux through surface bounded by path

EMF around loop

Time derivative

Faraday’s law

If path along conducting loop, induces current I=EMF/R

• Magnetic flux is defined exactly as electric flux

• (Component of B surface) x (Area element)

€

ΦB = B • dA∫

Tues. Dec. 1, 2009 5Phy208 Lect. 25

Quick quizWhich of these conducting loops will have

currents flowing in them?

Constant I I(t) increases

Constant IConstant v

Constant I

Constant v

A.

B.

C.

D.

Tues. Dec. 1, 2009 6Phy208 Lect. 25

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.

Tues. Dec. 1, 2009 7Phy208 Lect. 25

Quick QuizA 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)?

B=1T

=0 in orientation shown

0 45 90 135 180 225 270 315 360ANGLE ( DEGREES )

0 45 90 135 180 225 270 315 360-1.5

-1

-0.5

0

0.5

1

1.5

ANGLE ( DEGREES )

0 45 90 135 180 225 270 315 360-1.5

-1

-0.5

0

0.5

1

1.5

ANGLE ( DEGREES )

0 45 90 135 180 225 270 315 360-1.5

-1

-0.5

0

0.5

1

1.5

ANGLE ( DEGREES )

0 0

0 0

A.

B.

C.

D.

Tues. Dec. 1, 2009 8Phy208 Lect. 25

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.

Tues. Dec. 1, 2009 9Phy208 Lect. 25

Quick QuizA person moves a conducting loop with constant velocity away from a wire as shown. The wire has a constant currentWhat is the direction of force on the loop from the wire?

A. LeftB. RightC. UpD. DownE. Into pageF. Out of page

I

v

€

rF top−side =

r I induced Lloop ×

r B top = ILBˆ y

€

rF bottom−side =

r I induced Lloop ×

r B bottom = −ILBˆ y

€

Btop > B bottom⇒ Force is up

€

rF left−side +

r F left−side = 0 cancel

Tues. Dec. 1, 2009 10Phy208 Lect. 25

Inductors

• Flux = (Inductance) X (Current)

€

Φ=LI

• Change in Flux = (Inductance) X (Change in Current)

€

ΔΦ=LΔI

• Potential difference

€

V = −LdI

dt Constant current -> no potential diffTues. Dec. 1, 2009 11Phy208 Lect. 25

Energy stored in ideal inductor

• Constant current (uniform charge motion)– No work required

to move charge through inductor

• Increasing current:

– Work required

to move charge across induced EMF

–

– Total work €

ΔVLq = ΔVLIdt

€

dW = ΔVLIdt = LdI

dtIdt = LIdI

€

W = LIdI =1

20

I

∫ LI2 Energy stored in inductor

Tues. Dec. 1, 2009 12Phy208 Lect. 25

Tues. Dec. 1, 2009 Phy208 Lect. 25 13

€

VL t = 0( ) = −Vbattery = −LdIL

dt

⇒dIL

dt=

Vbattery

L

IL

IL(t)

Time ( t )0

0

Slope dI / dt = Vbattery / L

€

IL t = 0( ) = 0

IL instantaneously zero, but increasing in time

Inductors in circuits

Just a little later…

A short time later ( t=0+Δt ), the current is increasing …

Tues. Dec. 1, 2009 Phy208 Lect. 25 14

IL(t)

Time ( t )0

0

Slope dI / dt = Vbattery / L

A. More slowlyB. More quicklyC. At the same rate

IL>0, and IR=IL

VR≠0, so VL smallerVL= -LdI/dt, so dI/dt smaller

Switch closed at t=0

In empty space: sinusoidal wave propagating along x with velocity

E = Emax cos (kx – ωt)

B = Bmax cos (kx – ωt)

• E and B are perpendicular oscillating vectors

•The direction of propagation is perpendicular to E and B

λ

Electromagnetic waves

€

Bmax = E /c

Tues. Dec. 1, 2009 15Phy208 Lect. 25

Quick Quiz on EM waves

x

z

y

c

EB

Tues. Dec. 1, 2009 16Phy208 Lect. 25

EM wave transports energy at its propagation speed.

Intensity = Average power/area =

EM wave incident on surface exerts a radiation pressure prad (force/area) proportional to intensity I.

Perfectly absorbing (black) surface:

Perfectly reflecting (mirror) surface:

Resulting force = (radiation pressure) x (area) €

prad = I /c

€

prad = 2I /c

Radiation Pressure

Power and Intensity

€

I =EmaxBmax

2μo

=Emax

2

2μoc=

cBmax2

2μo

=cεoEmax

2

2

Spherical wave:

€

I = Psource /4πr2

Tues. Dec. 1, 2009 17Phy208 Lect. 25

Polarization Linear polarization:

E-field oscillates in fixed plane of polarization

Linear polarizer: Transmits component of E-field parallel to

transmission axis Absorbs component perpendicular to

transmission axis. Intensity

Circular Polarization E-field rotates at constant magnitude

€

Eaftermax = Ebefore

max cosθ

€

I ∝ Emax2 ⇒ Iafter = Ibefore cos2 θ

Tues. Dec. 1, 2009 18Phy208 Lect. 25

Quantum Mechanics

• Light comes in discrete units:– Photon energy

• 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

€

E photon = hf = hc /λ =1240 eV ⋅nm /λ

Electron ejected only if hf >

Minimum photon energy required

€

E photonmin = hf

Tues. Dec. 1, 2009 19Phy208 Lect. 25

Photon properties of light• Photon of frequency f has energy hf

–

–

–

• 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

€

E photon = hf = hc /λ

€

h = 6.626 ×10−34 J ⋅s = 4.14 ×10−15eV ⋅s

€

hc =1240eV ⋅nm

Tues. Dec. 1, 2009 20Phy208 Lect. 25

Photon energy

What is the energy of a photon of red light ( =635 nm)?

A. 0.5 eV

B. 1.0 eV

C. 2.0 eV

D. 3.0 eV

€

E =hc

λ=

1240 eV − nm

635 nm=1.95 eV

Tues. Dec. 1, 2009 21Phy208 Lect. 25

Bohr’s model of Hydrogen atom Planetary model:

Circular orbits of electrons around proton. Quantization

Discrete orbit radii allowed: Discrete electron energies: Each quantum state labeled by quantum # n

€

rn = n2ao

€

En = −13.6 /n2 eV

How did he get this?

Quantization of circular orbit angular mom.

€

rL =

r r ×

r p = mvr = nh

Tues. Dec. 1, 2009 22Phy208 Lect. 25

Consequences of Bohr model• Electron can make transitions

between quantum states.• Atom loses energy: photon emitted

• Photon absorbed: atom gains energy:€

E photon = Eatom ninitial( ) − Eatom n final( )

€

E photon = Eatom n final( ) − Eatom ninitial( )

€

Eatom n( ) = −13.6eV

n2

Tues. Dec. 1, 2009 23Phy208 Lect. 25

24

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.

Spectral Question

n=2n=3

n=1

A. λ31 < λ21

B. λ31 = λ21

C. λ31 > λ21

E31 > E21 so λ31 < λ21

Wavelength is smaller for

larger jump!

€

ΔE = hf =hc

λ

Tues. Dec. 1, 2009 Phy208 Lect. 25

QuestionThis quantum system (not a

hydrogen atom) has energy levels as shown. Which photon could possibly be absorbed by this system?

E1=1 eV

E2=3 eV

E3=5 eV

E3=7 eV

€

E photon =hc

λ=

1240 eV ⋅nm

λ

A. 1240 nm

B. 413 nm

C. 310 nm

D. 248 nm

Tues. Dec. 1, 2009 25Phy208 Lect. 25

Matter Waves• deBroglie postulated that matter has wavelike

properties.

• deBroglie wavelength

€

λ =h / p

Example: Wavelength of electron with 10 eV of energy:

Kinetic energy

€

EKE =p2

2m⇒ p = 2mEKE

λ =h

2mEKE

=hc

2mc 2EKE

=1240eV ⋅nm

2 0.511×106eV( ) 10eV( )= 0.39nm

Tues. Dec. 1, 2009 26Phy208 Lect. 25

Tues. Dec. 1, 2009 Phy208 Lect. 25 27

Heisenberg Uncertainty Principle

Using Δx = position uncertainty Δp = momentum uncertainty

Heisenberg showed that the product

( Δx ) ( Δp ) is always greater than ( h / 4 )

Often write this as

where is pronounced ‘h-bar’

Planck’sconstant

€

Δx( ) Δp( ) ~ h /2

€

h=h

2π