Chapters 21-29
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(a) 6.8 Ω
Chapter 21:45,63
Chapter 22:25,49
Chapter 23:35,38,53,55,58,59
Chapter 24:17,18,20,42,43,44,50,52,53.59,63
Chapter 26:27,33,34,39,54
Chapter 27:17,18,34,43,50,51,53,56
Chapter 28: 10,11,28,47,52
Chapter 29: 9,15,17,24,25,26,27,33
Magnetic Force
Magnetic forces arise from the interactions of
moving charges in magnetic fields.
fields (The electrical
force is between two charges at rest)
rest
F=BQvsinΘ for a moving charge
F=BIlsinΘ
for a current
2r
•••••
•••••
FE=qE, FM=qvB
Stage 1: qE=qvB thus v=E/B
Stage 2: qvB’=mv2/r so m=qB’r/v=qBB’r/v
2. A charged particle is moving in a uniform, constant magnetic field. Which one of
the following statements concerning the magnetic force exerted on the particle is
false?
(a) It does no work on the particle.
X (b) It increases the speed of the particle.
(c) It changes the velocity of the particle.
(d) It can act only on a particle in motion.
(e) It does not change the kinetic energy of the particle.
A beam consisting of five types of ions labeled A, B, C, D, and E enters a region that
contains a uniform magnetic field as shown in the figure below. The field is perpendicular
to the plane of the paper, but its precise direction is not given. All ions in the beam travel
with the same speed.The table below gives the masses and charges of the ions. Note: 1
mass unit = 1.67 × 10–27 kg and e = 1.6 × 10–19 C.
Ion Mass Charge
A 2 units +e
B 4 units +e
C 6 units +e
D 2 units –e
E 4 units –e
Which one of the following statements best explains why a constant magnetic field can do
no work on a moving charged particle?
(a) The magnetic field is conservative.
(b) The magnetic force is a velocity dependent force.
(c) The magnetic field is a vector and work is a scalar quantity.
X(d) The magnetic force is always perpendicular to the velocity of the particle.
(e) The electric field associated with the particle cancels the effect of the magnetic field
on the particle.
Magnetic field at the center of a
current loop
B=μoI/2πr
I
B
r
1. Which ion falls at position 2?
(a) A (c) C (e) E
X(b) B (d) D
2. What is the direction of the magnetic field?
(a) toward the right X(c) into the page (e) toward the bottom
(b) toward the left
(d) out of the page of the page
B=μoI/2r
>
1. Complete the following statement: The magnitude of the magnetic force that acts
on a charged particle in a magnetic field is independent of
X (a) the sign of the charge.
(b) the magnitude of the charge.
(c) the magnitude of the magnetic field.
(d) the direction of motion of the particle.
(e) the velocity components of the particle.
×
B
<
<
r
I
>
>
F21.31.jpg
EMF Induced in Moving Conductor
Faraday’s Law
EMF= - ΔΦB/Δt
Motional EMF=
EMF=ΔΦB/Δt=BΔ
=BΔA/Δ
A/Δt=BlvΔ
lvΔt/Δ
t/Δt=Blv
Optics and Modern Physics
Transverse electromagnetic (EM) Waves
c=λf
The Lens Equation
Snell’s Law
1/do+1/di=1/f
f
ho
2F’
F’
O
do
F
di
hi
n1sin i =n2sin r
Or
sin i /sin r =v1/v2
d
Θ
Θ
Extra distance mλ
sinΘ=mλ/d or dsinΘ=mλ
m=0,1,2,3, . . . Constructive inference
m=1/2,3/2,5/2, . . . Destructive inference
A prism also disperses light
n depends on λ
The shorter λ, the greater n
Two Principles of Relativity:
• The laws of physics are the same for all uniformly
moving observers.
• The speed of light is the same for all observers.
Consequences:
• Different observers measure different times, lengths,
and masses.
• Only spacetime is observer-independent.
Polarization
I=Iocos2Θ
Time Dilation
Moving clocks runs slowly (as compared to
clock at rest)
t = T[1 - (v/c)2]1/2
(v has to be reasonably close to c)
Length Contraction
Moving objects are shorter in the direction of
motion than when at rest
l=L [1 - (v/c)2]1/2
Mass Increase and KE
The mass of a body is measured to increase
with speed
m = mo/[1 - (v/c)2]1/2
Kinetic Energy
KE=(m-mo)c2
Example: Find the speed for which the length of a
meterstick is 0.5 m.
Solution: Given: l=0.5 m L=1 m
v=?
Since l=L[1 - (v/c)2]1/2
v=c[1-(L/Lo)2]1/2=c[1-(0.5/1)2]1/2=0.866c
Example: An observer watching a high-speed
spaceship pass by notices that a clock on board
runs slow by a factor of 1.5. If the rest mass of the
clock is 0.32 kg, what is its kinetic energy?
Solution: Since t = T[1 - (v/c)2]1/2
v=c[1 - (T/t)2]1/2 =c[1 - (1/1.5)2]1/2 =0.745 c
KE=(m-mo)c2=moc2{1/[1 - (v/c)2]1/2-1}
=1.44x1016 J
Photoelectric Effect
h f = KEmax + Wo or KEmax = h f – Wo
Wo is the work function, minimum
energy required to get an electron out
through the surface.
Particle Nature of Light
h f = KEmax + Wo
Wave-Particle Duality
λ=h/mv
Bohr Model
Electron’s angular momentum
L=Iω=mvrn=nh/2π, n=1,2,3
n is called quantum number of the orbit
Radius of a circular orbit
rn=(n2/Z)r1
where r1=5.29x10-11 m (n=1)
r1 is called Bohr radius, the smallest orbit in H
Total energy for an electron in the nth orbit:
En=(Z2/n2)E1
where E1=-13.6 eV (n=1)
E1 is called Ground State of the hydrogen
1/λ=R(1/22-1/n2), n=3,4,… for Balmer series
where Rydberg constant R=1.097x107 m-1
Both orbits and energies depend on n, the quantum number
Example: A beam of white light containing frequencies
between 4.0x1014 Hz and 7.9x1014 Hz is incident on
sodium surface, which has a work function of 2.28 eV.
(a) What is the range of frequencies in this beam of light
for which electrons are ejected from the sodium surface?
(b) Find the maximum kinetic energy of photoelectrons
ejected from this surface.
Solution:
(a) fmin=Wo/h=2.28eV(1.6x10-19 J)/6.63x10-34 J•s=5.5x1014
Hz
Frequencies that eject electrons:
5.5x1014-7.9x1014 Hz
(b) Kmax=hf-Wo
=(6.63x10-34 J•s)(7.9x1014 Hz)- 2.28eV(1.6x10-19 J)
=1.59x10-19 J
Bohr Model
Early Quantum Theory
¾ Photoelectric effect: hf=KEmax+Wo
¾ De Broglie wavelength: λ=h/mv
¾ Wave-particle duality
¾ Bohr theory for hydrogen:
En=E1/n2 where E1=-13.6 eV
Quantum mechanics
Heisenberg Uncertainty Principle
Momentum and position ΔxΔp ≥ h/2π
Energy and time ΔEΔt ≥ h/2π
Definite circular orbits of electrons
No precise orbits of electrons, only the
probability of finding a given electron
at a given point
α, β, γ Decay
Parent nucleus and
daughter nucleus are
different
Parent nucleus and
daughter nucleus are
the same
Parent nucleus and
daughter nucleus are
different
What have we learned after all?
• Knowledge of physics
• Skills physicists apply in physics that
are transferable to other academic
disciplines such as reasoning, logical
thinking, idealization, approximation,
mathematical and graphical
representations of phenomena, etc.
I wish all of you a bright future!