Physics 1202: Lecture 17 Today’s Agenda Announcements: –Lectures posted on: –HW assignments, etc. Homework #5:Homework #5: –Due this Friday Midterm 1: –Answers today –New average = 63%

L C   R 

0 i 0 i Phasors for L,C,R i tt  i tt  i tt   Suppose: 0 i

Phasors: LCR The phasor diagram has been relabeled in terms of the reactances defined from:  L C   R The unknowns (i m,  ) can now be solved for graphically since the vector sum of the voltages V L + V C + V R must sum to the driving emf .  C = -Q/C  L = -L  I /  t  R = -RI

Phasors:LCR  

Phasors:Tips This phasor diagram was drawn as a snapshot of time t=0 with the voltages being given as the projections along the y-axis. y x  imRimR imXLimXL imXCimXC mm “Full Phasor Diagram” From this diagram, we can also create a triangle which allows us to calculate the impedance Z: Sometimes, in working problems, it is easier to draw the diagram at a time when the current is along the x-axis (when i=0). “ Impedance Triangle” Z |  R | X L -X C |

Resonance For fixed R,C,L the current i m will be a maximum at the resonant frequency  0 which makes the impedance Z purely resistive. the frequency at which this condition is obtained is given from:  Note that this resonant frequency is identical to the natural frequency of the LC circuit by itself! At this frequency, the current and the driving voltage are in phase! ie: reaches a maximum when: X L =X C

Resonance The current in an LCR circuit depends on the values of the elements and on the driving frequency through the relation Suppose you plot the current versus , the source voltage frequency, you would get: “ Impedance Triangle” Z |  R | X L -X C | 12 x imim 0 0  o  R=R o  m / R 0 R=2R o

Power in LCR Circuit The power supplied by the emf in a series LCR circuit depends on the frequency . It will turn out that the maximum power is supplied at the resonant frequency  0. The instantaneous power (for some frequency,  ) delivered at time t is given by: The most useful quantity to consider here is not the instantaneous power but rather the average power delivered in a cycle. To evaluate the average on the right, we first expand the sin(  t-  ) term. Remember what this stands for

Power in LCR Circuit Expanding, Taking the averages, Generally: sin 2  t tt 0  0 +1 Putting it all back together again, 0 1/2 (Integral of Product of even and odd function = 0) sin  tcos  t tt 0  0 +1

Power in LCR Circuit The power can be expressed in term of i max: Power delivered depends on the phase,  the  “power factor” phase depends on the values of L, C, R, and  This result is often rewritten in terms of rms values:  

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 came near a solenoidal magnet. »Light spreads out and makes interference patterns. Do we understand this?

f()x x f(x x z y

Maxwell’s Equations These equations describe all of Electricity and Magnetism. They are consistent with modern ideas such as relativity. They describe light ! (electromagnetic wave)

E & B in Electromagnetic Wave Plane Harmonic Wave: where: y x z Nothing special about (E y,B z ); eg could have (E y,-B x ) 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 17, ACT 1 Suppose the electric field in an e-m wave is given by: –In what direction is this wave traveling ? (a) + z direction (b) -z direction (c) +y direction (d) -y direction

Lecture 17, ACT 2 Suppose the electric field in an e-m wave is given by: Which of the following expressions describes the magnetic field associated with this wave? (a) B x = -(E o /c)cos(kz +  t) (b) B x = +(E o /c)cos(kz -  t) (c) B x = +(E o /c)sin(kz -  t)

Generating E-M Waves Static charges produce a constant Electric Field but no Magnetic Field. Moving charges (currents) produce both a possibly changing electric field and a static magnetic field. Accelerated charges produce EM radiation (oscillating electric and magnetic fields). Antennas are often used to produce EM waves in a controlled manner.

A Dipole Antenna V(t)=V o cos(  t) x z y time t= E time t=  /2  E time t=  /  one half cycle later

dipole radiation pattern oscillating electric dipole generates e-m radiation that is polarized in the direction of the dipole radiation pattern is doughnut shaped & outward traveling –zero amplitude directly above and below dipole –maximum amplitude in-plane proportional to sin(  t)

Receiving E-M Radiation receiving antenna One way to receive an EM signal is to use the same sort of antenna. Receiving antenna has charges which are accelerated by the E field of the EM wave. The acceleration of charges is the same thing as an EMF. Thus a voltage signal is created. Speaker y x z

Lecture 17, ACT 3 Consider an EM wave with the E field POLARIZED to lie perpendicular to the ground. y x z In which orientation should you turn your receiving dipole antenna in order to best receive this signal? C) Along E a) Along S b) Along B

Loop Antennas Magnetic Dipole Antennas The electric dipole antenna makes use of the basic electric force on a charged particle Note that you can calculate the related magnetic field using Ampere’s Law. We can also make an antenna that produces magnetic fields that look like a magnetic dipole, i.e. a loop of wire. This loop can receive signals by exploiting Faraday’s Law. For a changing B field through a fixed loop of area A:   = A B

Lecture 17, ACT 4 Consider an EM wave with the E field POLARIZED to lie perpendicular to the ground. y x z In which orientation should you turn your receiving loop antenna in order to best receive this signal? a) â Along S b) â Along B C) â Along E

Review of Waves from 1201 The one-dimensional wave equation: A specific solution for harmonic waves traveling in the +x direction is: has a general solution of the form: where h 1 represents a wave traveling in the +x direction and h 2 represents a wave traveling in the -x direction. h x A A = amplitude = wavelength f = frequency v = speed k = wave number

E & B in Electromagnetic Wave Plane Harmonic Wave: where: y x z From general properties of waves : 

Velocity of Electromagnetic Waves The wave equation for E x : (derived from Maxwell’s Eqn) Therefore, we now know the velocity of electromagnetic waves in free space: Putting in the measured values for  0 &  0, we get: This value is identical to the measured speed of light! –We identify light as an electromagnetic wave.

The EM Spectrum These EM waves can take on any wavelength from angstroms to miles (and beyond). We give these waves different names depending on the wavelength. Wavelength [m] Gamma Rays Infrared Microwaves Short Wave Radio TV and FM Radio AM Radio Long Radio Waves Ultraviolet Visible Light X Rays

Lecture 17, ACT 5 Consider your favorite radio station. I will assume that it is at 100 on your FM dial. That means that it transmits radio waves with a frequency f=100 MHz. What is the wavelength of the signal ? A) 3 cmB) 3 mC) ~0.5 mD) ~500 m

The EM Spectrum Each wavelength shows different details

The EM Spectrum Each wavelength shows different details

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 power transmitted per unit area = average energy density times wave velocity: Therefore, the total energy density in an e-m wave = u, where =

Momentum in EM Waves Electromagnetic waves contain momentum: The momentum transferred to a surface depends on the area of the surface. Thus Pressure is a more useful quantity. If a surface completely absorbs the incident light, the momentum gained by the surface p We use the above expression plus Newton’s Second Law in the form F=  p/  t to derive the following expression for the Pressure, If the surface completely reflects the light, conservation of momentum indicates the light pressure will be double that for the surface that absorbs. 