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Lecture 4

Physics 1502: Lecture 25Today’s Agenda

• Announcements:

• Midterm 2: NOT Nov. 6– Following week …

• Homework 07: due Friday next week• AC current

– Resonances

• Electromagnetic Waves– Maxwell’s Equations - Revised– Energy and Momentum in Waves

LC

∼ε

R

ω

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Lecture 4

Phasors:LCR

⇒

⇓

Phasors:Tips• This phasor diagram was drawn as asnapshot of time t=0 with the voltagesbeing given as the projections along they-axis.

y

x

φimR

imXL

imXC

εm

“Full Phasor Diagram”

From this diagram, we can also create atriangle which allows us to calculate theimpedance Z:

• Sometimes, in working problems, it iseasier to draw the diagram at a time whenthe current is along the x-axis (when i=0).

“ Impedance Triangle”

Z

| φ |

R

| XL-XC |

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Lecture 4

ResonanceThe current in an LCR circuit depends on the valuesof the elements and on the driving frequency throughthe relation

Suppose you plot the currentversus ω, the source voltagefrequency, you would get:

“ Impedance Triangle”

Z

| φ |

R

| XL-XC |

1 2x

im

00

2ωoω

R=Ro

εm / R0

R=2Ro

Power and Resonance in RLC

• Power, as well as current, peaks at ω = ω 0. Thesharpness of the resonance depends on the values of thecomponents.

• Recall:

• Therefore,

We can write this in the following manner (which we won’t try toprove):

…introducing the curious factors Q and x

Z

| φ |

R

| XL-XC |

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Lecture 4

The Q factorQ also determines the sharpness of the resonance peaks ina graph of Power delivered by the source versus frequency.

ωo

ΔωLow Q

High QPav

ω

Lecture 25, ACT 1• Consider the two circuits shown

where CII = 2 CI.– What is the relation between the

quality factors, QI and QII , of thetwo circuits?

(a) QII < QI (b) QII = QI (c) QII > QI

I II

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Lecture 4

Lecture 25, ACT 2• Consider the two circuits shown where

CII = 2 CI and LII = ½ LI.– Which circuit has the narrowest width

of the resonance peak?

(a) I (b) II (c) Both the same

I II

Power Transmission• How do we transport power from power stations to homes?

– At home, the AC voltage obtained from outlets in this country is120V at 60Hz.

– Transmission of power is typically at very high voltages ( eg~500 kV) (a “high tension” line)

– Transformers are used to raise the voltage for transmission andlower the voltage for use. We’ll describe these next.

• But why?– Calculate ohmic losses in the transmission lines:– Define efficiency of transmission:

– Note for fixed input power and line resistance, theinefficiency ∝ 1/V2

Keep Rsmall

MakeVin big

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Lecture 4

Transformers

∼ε

(primary) (secondary)

• AC voltages can be stepped upor stepped down by the use oftransformers.

iron

• The AC current in the primarycircuit creates a time-varyingmagnetic field in the iron

• The iron is used to maximize the mutual inductance. Weassume that the entire flux produced by each turn of theprimary is trapped in the iron.

V2V1

• This induces an emf on thesecondary windings due to themutual inductance of the two setsof coils.

Ideal Transformers (no load)

• The primary circuit is just an AC voltagesource in series with an inductor. The changein flux produced in each turn is given by:

• The change in flux per turn in the secondary coil is thesame as the change in flux per turn in the primary coil(ideal case). The induced voltage appearing across thesecondary coil is given by:

• Therefore,• N2 > N1 ⇒ secondary V2 is larger than primary V1 (step-up)• N1 > N2 ⇒ secondary V2 is smaller than primary V1 (step-down)

• Note: “no load” means no current in secondary. The primary current,termed “the magnetizing current” is small!

No resistance losses All flux contained in iron Nothing connected on secondary

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Lecture 4

Ideal Transformers• What happens when we connect a

resistive load to the secondary coil?– Flux produced by primary coil induces

an emf in secondaryR

with a Load

– This changing flux appears in theprimary circuit as well; the sense of itis to reduce the emf in the primary...

– However, V1 is a voltage source.– Therefore, there must be an increased

current i1 (supplied by the voltagesource) in the primary which producesa flux ∝ N1i1 which exactly cancels theflux produced by i2.

– emf in secondary produces current i2

– This current produces a flux in thesecondary coil ∝ N2i2, which opposesthe original flux -- Lenz’s law

Transformers with a Load• With a resistive load in the secondary,

the primary current is given by:R

It’s time..

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Lecture 4

Lecture 25, ACT 3• The primary coil of an ideal transformer is

connected to a battery (V1 = 12V) as shown.The secondary winding has a load of 2 Ω.There are 50 turns in the primary and 200turns in the secondary.

R

– What is the current in the secondcary ?

ε

N2N1(primary) (secondary)

iron

V2V1

(a) 24 A (b) 1.5 A (c) 6 A (d) 0 A

Lecture 25, ACT 4• The primary coil of an ideal transformer is

connected to the wall (V1 = 120V) as shown.There are 50 turns in the primary and 200turns in the secondary.

R

– If 960 W are dissipated in the resistorR, what is the current in the primary ?

(a) 8 A (b) 16 A (c) 32 A

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Lecture 4

Fields from Circuits?• We have been focusing on what happens within the circuits we have

been studying (eg currents, voltages, etc.)

• What’s happening outside the circuits??– We know that:

» charges create electric fields and» moving charges (currents) create magnetic fields.

– Can we detect these fields?– Demos:

» We saw a bulb connected to a loop glow when the loop camenear a solenoidal magnet.

» Light spreads out and makes interference patterns.Do we understand this?

f( )x

x

f( x

x

z

y

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Lecture 4

Maxwell’s Equations• These equations describe all of Electricity and

Magnetism.

• They are consistent with modern ideas such asrelativity.

• They describe light !

Maxwell’s Equations - Revised• In free space, outside the wires of a circuit, Maxwell’s equations

reduce to the following.

• These can be solved (see notes) to give the followingdifferential equations for E and B.

• These are wave equations. Just like for waves on astring. But here the field is changing instead of thedisplacement of the string.

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Lecture 4

Step 1 Assume we have a plane wave propagating in z (ie E, Bnot functions of x or y)

Plane Wave Derivation

x

z

y

z1 z2

Ex Ex

ΔZ

Δx

By

Step 2 Apply Faraday’s Law to infinitesimal loop in x-z plane

Example: does this

Plane Wave Derivation

x

z

y

z1 z2

By

ΔZ

ΔyBy

Ex

Step 3 Apply Ampere’s Law to an infinitesimal loop in the y-zplane:

Step 4 Combine results from steps 2 and 3 to eliminate By

!!

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Lecture 4

Plane Wave Derivation• We derived the wave eqn for Ex:

• By is in phase with Ex• B0 = E0 / c

• How are Ex and By related in phase and magnitude?

(Result from step 2)

• We could have also derived for By:

– Consider the harmonic solution: where

Review of Waves from last semester• The one-dimensional wave equation:

• A specific solution for harmonic waves traveling in the +xdirection is:

has a general solution of the form:

where h1 represents a wave traveling in the +x direction and h2represents a wave traveling in the -x direction.

h

x

λA

A = amplitudeλ = wavelengthf = frequencyv = speedk = wave number

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Lecture 4

E & B in Electromagnetic Wave• Plane Harmonic Wave:

where:

y

x

z

Nothing special about (Ey,Bz); eg could have (Ey,-Bx)

Note: the direction of propagation is given by the cross product

where are the unit vectors in the (E,B) directions.

Note cyclical relation:

Lecture 25, ACT 5• Suppose the electric field in an e-m wave is given by:

– In what direction is this wave traveling ?5A

(a) + z direction (b) -z direction(c) +y direction (d) -y direction

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Lecture 4

Lecture 25, ACT 5• Suppose the electric field in an e-m wave is given

by:

• Which of the following expressions describesthe magnetic field associated with this wave?(a) Bx = -(Eo/c)cos(kz + ωt) (b) Bx = +(Eo/c)cos(kz - ωt) (c) Bx = +(Eo/c)sin(kz - ωt)

5B

Velocity of Electromagnetic Waves• The wave equation for Ex: (derived from Maxwell’s Eqn)

• Therefore, we now know the velocity ofelectromagnetic waves in free space:

• Putting in the measured values for µ0 & ε0, we get:

• This value is identical to the measured speed of light!– We identify light as an electromagnetic wave.

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Lecture 4

The EM Spectrum

• These EM waves can take on any wavelength fromangstroms to miles (and beyond).

• We give these waves different names depending on thewavelength.

Wavelength [m]10-14 10-10 10-6 10-2 1 102 106 1010

Gam

ma

Ray

s

Infr

ared

Mic

row

aves

Shor

t Wav

e R

adio

TV a

nd F

M R

adio

AM

Rad

io

Long

Rad

io W

aves

Ultr

avio

let

Visi

ble

Ligh

t

X R

ays

Energy in EM Waves / review• Electromagnetic waves contain energy which is stored in E

and B fields:

• The Intensity of a wave is defined as the average powertransmitted per unit area = average energy density times wavevelocity:

• Therefore, the total energy density in an e-m wave = u, where

=