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Physics in Session 2: I

n Physics / Higher Physics 1B (PHYS1221/1231)n Science, Advanced Science

n Engineering: Electrical, Photovoltaic,Telecom

n Double Degree: Science/Engineering

n 6 UOC

n Waves

n Physical Optics (light & interference)

n Introduction to Quantum Physics

n Solid State & Semiconductor Physics

Physics in Session 2: IIn Higher Physics 1B (Special) (PHYS1241) (6UOC)

n Advanced Science

n Double Degree (Science/Engineering)

n Credit or higher in Physics 1A

n Waves: interference, diffraction, polarization

n Introduction to Quantum Mechanics

n Alternating Currents

n The Sun and the Planets

n Thermal Physics

n Special Relativity

Much smaller classes!

Physics in Session 2: II

n Energy & Environmental Physicsn PHYS1211 (6UOC) / PHYS1249 (3UOC)

n A possible elective course?

n Heat & Energyn Solar energy, alternative energy

n Introductory quantum theoryn Photovoltaic energy

n Nuclear energy & radiation

Physics / Higher Physics 1A

Topic 3

Electricity and Magnetism

Revision

Electric Charges

n Two kinds of electric charges

n Called positive and negative

n Like charges repel

n Unlike charges attract

Coulomb’s Law

n In vector form,

n is a unit vector directed from q1 to q2

n Like charges produce a

repulsive force between them

F12

= keq

1q

2

r2

ˆ r

ˆ r

2

The Superposition Principle

n The resultant force on q1 is the vector

sum of all the forces exerted on it by

other charges: F1 = F21 + F31 + F41 + …

Electric Field

n Continuous charge distribution

2ˆe

e

o

qk

q r= =F

E r

2 20ˆ ˆlim

i

ie i e

qi i

q dqk k

r r∆ →

∆= =∑ ∫E r r

Electric Field Lines – Dipole

n The charges are

equal and opposite

n The number of field

lines leaving the

positive charge

equals the number

of lines terminating

on the negative

charge

Electric Flux

0surface

limi

E i iA

E A d∆ →

Φ = ⋅ ∆ = ⋅∑ ∫ E A

∆ΦE = E i∆Ai cosθ i = Ei⋅ ∆A

i

Gauss’s Law

n qin is the net charge inside the surface

n E represents the electric field at any point on the

surface

ΦE = E ⋅ dA = qin ε0

∫

Field Due to a Plane of Charge

n The total charge in the surface is σ

A

n Applying Gauss’s law

n Field uniform everywhere

ΦE = 2EA = σAε

0

, and E = σ2ε

0

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Properties of a Conductor in Electrostatic Equilibrium

1. Electric field is zero everywhere inside conductor

2. Charge resides on its surface of isolated conductor

3. Electric field just outside a charged conductor is perpendicular to the surface with magnitude σ / ε o

4. On an irregularly shaped conductor surface charge density is greatest where radius of curvature is smallest

n Work done by electric field is F.ds = qoE.ds

n Potential energy of the charge-field system is

changed by ∆U = -qoE.ds

n For a finite displacement of the charge from A

to B, the change in potential energy is

Electric Potential Energy

B

B A oA

U U U q d∆ = − = − ⋅∫ E s

Electric Potential, V

n The potential energy per unit charge, U/qo, is the electric potential

n The work performed on the charge isW = ∆U = q ∆V

n In a uniform field

B

Ao

UV d

q

∆∆ = = − ⋅∫ E s

B B

B AA A

V V V d E d Ed− = ∆ = − ⋅ = − = −∫ ∫E s s

Equipotential Surface

n Any surface consisting

of a continuous

distribution of points

having the same

electric potential

n For a point charge

e

qV k

r=

n From ∆V = -E.ds = -Exdx

n Along an equipotential surfaces ∆V = 0n Hence E ⊥ dsn i.e. an equipotential surface is perpendicular to the electric field

lines passing through it

Finding E From V

x

dVE

dx= −

V Due to a Charged Conductor

n E · ds = 0

n So, potential difference between A and B is zero

n Electric field is zero inside

the conductor

n So, electric potential constant everywhere inside conductor and equal to

value at the surface

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Cavity in a Conductor

n Assume an irregularly

shaped cavity is inside a

conductor

n Assume no charges are

inside the cavity

n The electric field inside the

conductor must be zero

Definition of Capacitance

n The capacitance, C, is ratio of the charge on either conductor to the potential difference between the conductors

n A measure of the ability to store charge

n The SI unit of capacitance is the farad (F)

QC

V=

∆

Capacitance – Parallel Plates

n Charge density σ = Q/A

n Electric field E = σ/ε0 (for conductor)n Uniform between plates, zero elsewhere

C = Q∆V

= QEd

= Q

Q

ε0Ad

= ε0A

d

Capacitors in Parallel

n Capacitors can be replaced

with one capacitor with a capacitance of Ceq

n Ceq = C1 + C2

Capacitors in Series

n Potential differences add

up to the battery voltage

Q =Q1

=Q2

∆V = ∆V1

+ ∆V2

∴ ∆VQ

= ∆V1

Q1

+ ∆V2

Q2

∴ 1

C= 1

C1

+ 1

C2

Energy of Capacitor

n Work done in charging the capacitor appears as electric potential energy U:

n Energy is stored in the electric field

n Energy density (energy per unit volume)

uE = U/Vol. = ½ εoE2

221 1

( )2 2 2

QU Q V C V

C= = ∆ = ∆

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Capacitors with Dielectrics

n A dielectric is a nonconducting material that, when placed between the plates of a capacitor, increases the capacitance

n For a parallel-plate capacitor

C = κCo =

κ εo(A/d)

Rewiring charged capacitors

n Two capacitors, C1 & C2

charged to same potential difference, ∆Vi.

n Capacitors removed from battery and plates connected with opposite polarity.

n Switches S1 & S2 then closed. What is final potential difference, ∆Vf?

Q1i, Q2i before; Q1f, Q2f after.

Q1i = C1∆Vi; Q2i = -C2∆Vi

So Q=Q1i+Q2i=(C1-C2)∆Vi

But Q= Q1f+Q2f (charge conserved)

With Q1f = C1∆Vf; Q2f = C2∆Vf hence Q1f = C1/C2 Q2f

So, Q=(C1/C2+1) Q2f

With some algebra, find Q1f = QC1/(C1+C2) & Q2f = QC2/(C1+C2)

So ∆V1f = Q1f / C1 = Q / (C1+C2) & ∆V2f = Q2f / C2 = Q / (C1+C2)

i.e. ∆V1f = ∆V2f = ∆Vf, as expected

So ∆Vf = (C1 - C2) / (C1 + C2) ∆Vi, on substituting for Q

Magnetic Poles

n Every magnet has two poles

n Called north and south poles

n Poles exert forces on one another

n Like poles repel

n N-N or S-S

n Unlike poles attract

n N-S

Magnetic Field Lines for a Bar Magnet

n Compass can be

used to trace the field lines

n The lines outside the

magnet point from the North pole to the South pole

Direction

n FB perpendicular to plane formed by v & B

n Oppositely directed forces are exerted on charges of different signs

n cause the particles to move in opposite directions

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Direction given byRight-Hand Rule

n Fingers point in the

direction of v

n (for positive charge; opposite direction if negative)

n Curl fingers in the direction of B

n Then thumb points in the direction of v x B; i.e. the direction of FB

The Magnitude of F

n The magnitude of the magnetic force on a charged particle is FB = |q| vB sin θn θ is the angle between v and B

n FB is zero when v and B are parallel

n FB is a maximum when perpendicular

Force on a Wire

n F = I L x B

n L is a vector that points in the direction of the current (i.e. of vD)

n Magnitude is the length L of the segment

n I is the current = nqAvD

n B is the magnetic field

Force on a Wire of Arbitrary Shape

n The force exerted

segment ds is

F = I ds x B

n The total force is

Ib

d= ×∫aF s B

Force onCharged Particle

n Equating the magnetic & centripetal forces:

n Solving gives r = mv/qB

F = qvB = mv2

r

Biot-Savart Law

n dB is the field created by the current in

the length segment ds

n Sum up contributions from all current elements I.ds

B = µ0

4πI ds × ˆ r

r2∫

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B for a Long, Straight Conductor

B = µ0I

2πa

B for a Long, Straight Conductor, Direction

n Magnetic field lines are circles concentric with the wire

n Field lines lie in planes perpendicular to to wire

n Magnitude of B is constant on any circle of radius a

n The right-hand rule for determining the direction of B is shownn Grasp wire with thumb in direction of

current. Fingers wrap in direction of B.

Magnetic Force Between Two Parallel Conductors

n Parallel conductors carrying currents in the same direction attract each other

n Parallel conductors carrying currents in opposite directions repel each other

F1

= µ0I

1I

2

2πal

Definition of the Ampere

n The force between two parallel wires can be used to define the ampere

n When the magnitude of the force per unit length between two longparallel 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

F1

l= µ

0I1I

2

2πa with µ

0= 4π •10

−7 T m A

-1

Ampere’s Law

n 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 = µ0I∫

Field in interiorof a Solenoid

n Apply Ampere’s law

n The side of length ℓ inside the solenoid contributes to the field

n Path 1 in the diagram

lBdsBdd1path 1path

=∫ ∫ ∫=⋅=⋅ sBsB

B = µ0

N

lI = µ

0nI

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Ampere’s vs. Gauss’s Law

n Integrals around closed path vs. closed surface.n i.e. 2D vs. 3D geometrical figures

n Integrals related to fundamental constant x source of the field.

n Concept of “Flux” – the flow of field lines through a surface.

B ⋅ ds = µ0I∫ E ⋅ dA = qε

0∫

Gauss’ Law in Magnetism

n Magnetic fields do not begin or end at any pointn i.e. they form closed loops, with the number of lines

entering a surface equaling the number of lines leaving that surface

n Gauss’ law in magnetism says:

ΦB = B.dA∫ = 0

Faraday’s Law of Induction

n The emf induced in a circuit is directly proportional to the rate of change of the magnetic flux through that circuit

ε = −N dΦB

dtQuickTime™ and a

Cinepak decompressorare needed to see this picture.

Ways of Inducing an emf

n Magnitude of B can change with time

n Area enclosed, A, can change with time

n Angle θ can change with time

n Any combination of the above can occur

ε = − ddtBAcosθ( )

Motional emf

n Motional emf induced in a conductor moving through a constant magnetic field

n Electrons in conductor experience a force, FB = qv x B that is directed along ℓ

n In equilibrium, qE = qvB or E = vB

Sliding Conducting Bar

n Magnetic flux is

n The induced emf is

n Thus the current is

ε = − dΦB

dt= − ddtBlx( )= −Bl dx

dt= −Blv

ΦB = Blx

I =εR

= BlvR

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Induced emf & Electric Fields

n A changing magnetic flux induces an emf and a current in a conducting loop

n An electric field is created in a conductor by a changing magnetic flux

n Faraday’s law can be written in a general form:

n Not an electrostatic field because the line integral of E.ds is not zero.

ε = E.ds = − dΦB

dt∫

Generators

n Electric generators

take in energy by work and transfer it out by electrical transmission

n The AC generator consists of a loop of wire rotated by some external means in a magnetic field

Rotating Loop

n Assume a loop with N turns, all of the same area, rotating in a magnetic field

n The flux through one loop at any time t is:

ΦB = BA cos θ = BA cos ωt

∴ε = −N dΦB

dt= −NAB d

dtcosωt( )= NABω sinωt

Motors

n Motors are devices into which energy is

transferred by electrical transmission while energy is transferred out by work

n A motor is a generator operating in reverse

n A current is supplied to the coil by a battery and the torque acting on the current-carrying coil causes it to rotate

Eddy Currents

n Circulating currents called eddy currents are induced in bulk pieces of metal moving through a magnetic field

n From Lenz’s law, their direction is to oppose the change that causes them.

n The eddy currents are in opposite directions as the plate enters or leaves the field

Equations for Self-Inductance

n Induced emf proportional to the rate of change of the current

n L is a constant of proportionality called the inductance of the coil.

εL = −L dIdt

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Inductance of a Solenoid

n Uniformly wound solenoid having N turns and length

ℓ. Then we have:

B = µ0nI = µ

0

N

lI

ΦB = BA = µ0

NA

lI

∴L = NΦB

I= µ

0N

2A

l

Energy in a Magnetic Field

n Rate at which the energy is stored is

n Magnetic energy density, uB, is

II

dU dL

dt dt=

U = L IdI0

I

∫ = 12LI2

uB = UAl

= B2

2µ0

RL Circuit

n Time constant, τ = L / R, for the circuit

n τ is the time required for current to reach 63.2% of its max value

I = εR

1− e−RtL

= ε

R1− e− tτ