Download I Physics in Session 2

Physics in Session 2: I
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Physics / Higher Physics 1B (PHYS1221/1231)
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Science, Advanced Science
Engineering: Electrical, Photovoltaic,Telecom
Double Degree: Science/Engineering
6 UOC
Waves
Physical Optics (light & interference)
Introduction to Quantum Physics
Solid State & Semiconductor Physics
Physics in Session 2: II
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Higher Physics 1B (Special) (PHYS1241) (6UOC)
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Advanced Science
Double Degree (Science/Engineering)
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Credit or higher in Physics 1A
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Waves: interference, diffraction, polarization
Introduction to Quantum Mechanics
Alternating Currents
The Sun and the Planets
Thermal Physics
Special Relativity
Much smaller classes!
Physics in Session 2: II
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Energy & Environmental Physics
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PHYS1211 (6UOC) / PHYS1249 (3UOC)
A possible elective course?
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Heat & Energy
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Introductory quantum theory
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Nuclear energy & radiation
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Photovoltaic energy
Two kinds of electric charges
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Topic 3
Electricity and Magnetism
Revision
Solar energy, alternative energy
Electric Charges
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Physics / Higher Physics 1A
Coulomb’s Law
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Like charges repel
Unlike charges attract
In vector form,
F12 = ke
Called positive and negative
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q1q2
rˆ
r2
rˆ is a unit vector
directed from q1 to q2
Like charges produce a
repulsive force between
them
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The Superposition Principle
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Electric Field
The resultant force on q1 is the vector
sum of all the forces exerted on it by
other charges: F1 = F21 + F31 + F41 + …
E=
n
Fe
q
= ke 2 rˆ
qo
r
Continuous charge distribution
∆qi
E = k e lim
∆qi →0
Electric Field Lines – Dipole
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The charges are
equal and opposite
The number of field
lines leaving the
positive charge
equals the number
of lines terminating
on the negative
charge
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i
2
rˆi = k e ∫
i
dq
rˆ
r2
Electric Flux
∆ΦE = E i∆Ai cosθ i = E i ⋅ ∆A i
Φ E = lim
∆Ai →0
∑ E ⋅ ∆A
i
i
=
∫
E ⋅ dA
surface
Field Due to a Plane of Charge
Gauss’s Law
ΦE =
∑r
∫ E ⋅ dA = q
in
The total charge in the surface is A
Applying Gauss’s law
σA
σ
ΦE = 2EA =
, and E =
ε0
2ε0
σ
ε0
qin is the net charge inside the surface
E represents the electric field at any point on the
surface
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Field uniform everywhere
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Properties of a Conductor in
Electrostatic Equilibrium
Electric Potential Energy
Electric field is zero everywhere inside
conductor
Charge resides on its surface of isolated
conductor
Electric field just outside a charged
conductor is perpendicular to the surface
with magnitude / o
On an irregularly shaped conductor surface
charge density is greatest where radius of
curvature is smallest
1.
2.
3.
σ
4.
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B
∆U = UB − U A = −qo ∫ E ⋅ ds
ε
A
Electric Potential, V
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Equipotential Surface
The potential energy per unit charge, U/qo, is the
electric potential
∆V =
n
B
∆U
= − ∫ E ⋅ ds
A
qo
The work performed on the charge is
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W = ∆U = q ∆V
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B
B
A
A
From ∆V = -E.ds = -Exdx
n
dV
dx
Along an equipotential surfaces ∆V = 0
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Hence E ⊥ ds
i.e. an equipotential surface is perpendicular to the electric field
lines passing through it
q
r
V Due to a Charged
Conductor
Finding E From V
Ex = −
Any surface consisting
of a continuous
distribution of points
having the same
electric potential
For a point charge
V = ke
In a uniform field
VB − VA = ∆V = − ∫ E ⋅ ds = −E ∫ ds = −Ed
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Work done by electric field is F.ds = qoE.ds
Potential energy of the charge-field system is
changed by ∆U = -qoE.ds
For a finite displacement of the charge from A
to B, the change in potential energy is
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E · ds = 0
So, potential difference
between A and B is zero
Electric field is zero inside
the conductor
So, electric potential
constant everywhere inside
conductor and equal to
value at the surface
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Definition of Capacitance
Cavity in a Conductor
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Assume an irregularly
shaped cavity is inside a
conductor
Assume no charges are
inside the cavity
The electric field inside the
conductor must be zero
C=
n
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Capacitance – Parallel Plates
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Charge density = Q/A
Electric field
E = σ/ε0 (for conductor)
σ
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Capacitors in Series
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Potential differences add
up to the battery voltage
A measure of the ability to store charge
The SI unit of capacitance is the farad (F)
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Capacitors can be replaced
with one capacitor with a
capacitance of Ceq
Ceq = C1 + C2
Energy of Capacitor
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Q = Q1 = Q2
∴
∆V ∆V1 ∆V2
=
+
Q1
Q2
Q
1 1
1
= +
C C1 C2
Work done in charging the capacitor appears as
electric potential energy U:
U=
∆V = ∆V1 + ∆V2
∴
Q
∆V
Capacitors in Parallel
Uniform between plates, zero elsewhere
Q
Q
Q
εA
=
=
= 0
C=
Q
∆V Ed
d
d
ε0 A
The capacitance, C, is ratio of the charge on either
conductor to the potential difference between the
conductors
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1
Q2 1
= Q ∆V = C ( ∆V )2
2C 2
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Energy is stored in the electric field
Energy density (energy per unit volume)
uE = U/Vol. = ½ εoE2
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Capacitors with Dielectrics
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A dielectric is a nonconducting material
that, when placed between the plates
of a capacitor, increases the
capacitance
For a parallel-plate capacitor
C = Co = o(A/d)
κ
κ
Rewiring charged capacitors
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ε
Two capacitors, C1 & C2
charged to same potential
difference, ∆Vi.
Capacitors removed from
battery and plates connected
with opposite polarity.
Switches S1 & S2 then closed.
What is final potential
difference, ∆Vf?
Q1i, Q2i before; Q1f, Q2f after.
Q1i = C1∆Vi; Q2i = -C2∆Vi
Magnetic Poles
So Q=Q1i+Q2i=(C1-C2)∆Vi
But Q= Q1f+Q2f (charge conserved)
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Every magnet has two poles
With Q1f = C1∆Vf; Q2f = C2∆Vf hence Q1f = C1/C2 Q2f
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So, Q=(C1/C2+1) Q2f
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With some algebra, find Q1f = QC1/(C1+C2) & Q2f = QC2/(C1+C2)
Called north and south poles
Poles exert forces on one another
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Like poles repel
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Unlike poles attract
So ∆V1f = Q1f / C1 = Q / (C1+C2) & ∆V2f = Q2f / C2 = Q / (C1+C2)
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i.e. ∆V1f = ∆V2f = ∆Vf, as expected
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So ∆Vf = (C1 - C2) / (C1 + C2) ∆Vi, on substituting for Q
Magnetic Field Lines
for a Bar Magnet
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Compass can be
used to trace the
field lines
The lines outside the
magnet point from
the North pole to
the South pole
N-N or S-S
N-S
Direction
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FB perpendicular to plane formed by v & B
Oppositely directed forces are exerted on
charges of different signs
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cause the particles to move in opposite directions
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Direction given by
Right-Hand Rule
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Fingers point in the
direction of v
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The Magnitude of F
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(for positive charge;
opposite direction if
negative)
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Curl fingers in the
direction of B
Then thumb points in the
direction of v x B; i.e. the
direction of FB
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F=ILxB
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Magnitude is the length L of the segment
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Equating the magnetic &
centripetal forces:
mv 2
F = qvB =
r
Solving gives r = mv/qB
The total force is
b
F = I ∫ ds × B
I is the current = nqAvD
B is the magnetic field
a
Force on
Charged Particle
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The force exerted
segment ds is
F = I ds x B
L is a vector that points in the direction of
the current (i.e. of vD)
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θ is the angle between v and B
FB is zero when v and B are parallel
FB is a maximum when perpendicular
Force on a Wire of
Arbitrary Shape
Force on a Wire
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The magnitude of the magnetic force on
a charged particle is FB = |q| vB sin θ
Biot-Savart Law
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dB is the field created by the current in
the length segment ds
Sum up contributions from all current
elements I.ds
B=
µ0
4π
∫
I ds × rˆ
r2
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B for a Long, Straight
Conductor
B for a Long, Straight
Conductor, Direction
Magnetic field lines are circles
concentric with the wire
Field lines lie in planes perpendicular
to to wire
Magnitude of B is constant on any
circle of radius a
The right-hand rule for determining
the direction of B is shown
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B=
µ0 I
2πa
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Magnetic Force
Between Two
Parallel Conductors
F1 =
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Definition of the Ampere
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µ 0 I1I 2
l
2πa
The force between two parallel wires can be used to define the
ampere
F1 µ0 I1I2
=
with µ0 = 4 π •10−7 T m A-1
l
2πa
Parallel conductors carrying currents in
the same direction attract each other
Parallel conductors carrying currents in
opposite directions repel each other
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When the magnitude of the force per unit length between two long
parallel wires that carry identical currents and are separated by 1 m
is 2 x 10-7 N/m, the current in each wire is defined to be 1 A
Field in interior
of a Solenoid
Ampere’s Law
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Grasp wire with thumb in direction of
current. Fingers wrap in direction of
B.
The line integral of B . ds around any
closed path equals µoI, where I is the
total steady current passing through
any surface bounded by the closed
path.
∫ B ⋅ ds = µ I
0
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Apply Ampere’s law
The side of length inside the
solenoid contributes to the field
ℓ
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Path 1 in the diagram
∫ B ⋅ ds = ∫ B ⋅ ds = B ∫ ds = Bl
path 1
B = µ0
path 1
N
I = µ0 nI
l
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Ampere’s vs. Gauss’s Law
∫ B ⋅ ds = µ I
∫ E ⋅ dA = q
0
n
n
ε0
Magnetic fields do not begin or end at any point
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Integrals around closed path vs. closed surface.
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Gauss’ Law in Magnetism
i.e. 2D vs. 3D geometrical figures
Gauss’ law in magnetism says:
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Integrals related to fundamental constant x
source of the field.
Concept of “Flux” – the flow of field lines through
a surface.
ΦB =
Faraday’s Law of Induction
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∫ B.dA = 0
Ways of Inducing an emf
ε=−
The emf induced in a circuit is directly
proportional to the rate of change of
the magnetic flux through that circuit
ε = −N
i.e. they form closed loops, with the number of lines
entering a surface equaling the number of lines leaving that
surface
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dΦB
dt
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d
(BAcosθ )
dt
Magnitude of B can change with time
Area enclosed, A, can change with time
Angle θ can change with time
Any combination of the above can occur
QuickTime™ and a
Cinepak decompressor
are needed to see this picture.
Sliding Conducting Bar
Motional emf
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Motional emf induced in
a conductor moving
through a constant
magnetic field
Electrons in conductor
experience a force, FB =
qv x B that is directed
along
In equilibrium, qE = qvB or
E = vB
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Magnetic flux is ΦB = Blx
The induced emf is
dΦ
d
dx
ε = − B = − (Blx ) = −Bl = −Blv
dt
dt
dt
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Thus the current is
n
ℓ
n
I=
ε
R
=
Blv
R
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Induced emf & Electric Fields
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A changing magnetic flux induces an emf and a current
in a conducting loop
An electric field is created in a conductor by a changing
magnetic flux
Faraday’s law can be written in a general form:
ε=
n
Generators
∫ E.ds = −
n
dΦ B
dt
Not an electrostatic field because the line integral of
E.ds is not zero.
Rotating Loop
Assume a loop with N turns,
all of the same area, rotating
in a magnetic field
The flux through one loop at
any time t is:
Φ B = BA cos θ = BA cos ωt
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∴ ε = −N
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Circulating currents called eddy
currents are induced in bulk pieces
of metal moving through a magnetic
field
From Lenz’s law, their direction is to
oppose the change that causes
them.
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Motors
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Motors are devices into which energy is
transferred by electrical transmission while
energy is transferred out by work
A motor is a generator operating in reverse
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A current is supplied to the coil by a battery and
the torque acting on the current-carrying coil
causes it to rotate
dΦ B
d
= −NAB (cosωt ) = NABω sin ωt
dt
dt
Eddy Currents
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Electric generators
take in energy by work
and transfer it out by
electrical transmission
The AC generator
consists of a loop of
wire rotated by some
external means in a
magnetic field
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The eddy currents are in opposite
directions as the plate enters or leaves
the field
Equations for Self-Inductance
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Induced emf proportional to the rate of
change of the current
dI
εL = −L
dt
L is a constant of proportionality called
the inductance of the coil.
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Inductance of a Solenoid
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Uniformly wound solenoid having N turns
and length . Then we have:
Energy in a Magnetic Field
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ℓ
N
I
l
NA
ΦB = BA = µ0
I
l
NΦ B µ0N 2 A
∴L =
=
I
l
Rate at which the energy is stored is
dU
dI
=LI
dt
dt
I
B = µ0 nI = µ0
n
U = L ∫ IdI = 12 LI 2
0
Magnetic energy density, uB, is
uB =
U
B2
=
Al 2µ0
RL Circuit
− Rt 
−t 


I = ε R 1− e L  = ε R 1− e τ 




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Time constant, τ = L / R, for the
circuit
τ is the time required for current
to reach 63.2% of its max value
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