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Electromagnetism INEL 4152 Sandra Cruz-Pol, Ph. D. ECE UPRM Mayagüez, PR Electricity => Magnetism In 1820 Oersted discovered that a steady current produces a magnetic field while teaching a physics class. Cruz-Pol, Electromagnetics UPRM Would magnetism would produce electricity? Eleven years later, and at the same time, Mike Faraday in London and Joe Henry in New York discovered that a time-varying magnetic field would produce an electric current! Vemf d N dt L E dl t s B dS Cruz-Pol, Electromagnetics UPRM Electromagnetics was born! This is the principle of motors, hydro-electric generators and transformers operation. This is what Oersted discovered accidentally: D L H dl s J t dS *Mention some examples of em waves Cruz-Pol, Electromagnetics UPRM Cruz-Pol, Electromagnetics UPRM Electromagnetic Spectrum Cruz-Pol, Electromagnetics UPRM Some terms E = electric field intensity [V/m] D = electric field density H = magnetic field intensity, [A/m] B = magnetic field density, [Teslas] Cruz-Pol, Electromagnetics UPRM Maxwell Equations in General Form Differential form Integral Form D v D dS v dv s B 0 v B dS 0 s B E t D H J t Gauss’s Law for E field. Gauss’s Law for H field. Nonexistence of monopole Faraday’s Law L E dl t s B dS D H dl J L s t dS Cruz-Pol, Electromagnetics UPRM Ampere’s Circuit Law Maxwell’s Eqs. v J t Also the equation of continuity D Maxwell added the term t to Ampere’s Law so that it not only works for static conditions but also for time-varying situations. This added term is called the displacement current density, while J is the conduction current. Cruz-Pol, Electromagnetics UPRM Maxwell put them together And added Jd, the displacement current H dl J dS I L enc I S1 H dl J dS 0 L S1 I L S2 d dQ L H dl S J d dS dt S D dS dt I 2 2 S2 At low frequencies J>>Jd, but at radio frequencies both Cruz-Pol, Electromagnetics terms are comparable in magnitude. UPRM Moving loop in static field When a conducting loop is moving inside a magnet (static B field, in Teslas), the force on a charge is F Qu B F Il B Cruz-Pol, Electromagnetics UPRM Encarta® Who was NikolaTesla? Find out what inventions he made His relation to Thomas Edison Why is he not well know? Cruz-Pol, Electromagnetics UPRM Special case Consider the case of a lossless medium with no charges, i.e. . v 0 0 The wave equation can be derived from Maxwell equations as E c E 0 2 2 What is the solution for this differential equation? The equation of a wave! Cruz-Pol, Electromagnetics UPRM Phasors & complex #’s Working with harmonic fields is easier, but requires knowledge of phasor, let’s review complex numbers and phasors Cruz-Pol, Electromagnetics UPRM COMPLEX NUMBERS: Given a complex number z z x jy re j r r cos jr sin where r | z | x y is the magnitude 2 tan 1 2 y is the angle x Cruz-Pol, Electromagnetics UPRM Review: 1/ j = Addition, Subtraction, j= 1/ Multiplication, Division, Square example : Root, Complex Conjugate e j 45o / j= e j 45o - j= 3e j 90 o 2e Cruz-Pol, Electromagnetics UPRM 2e +j= + j 45o + j 45o = +10e + j 90 o = For a time varying phase Real and imaginary parts are: t j Re{re } r cos(t ) j Im{re } r sin( t ) Cruz-Pol, Electromagnetics UPRM PHASORS a sinusoidal current I (t ) I o cos(t ) j jt I e equals the real part of o e For j I e The complex term o which results from jt dropping the time factor e is called the phasor current, denoted by I s (s comes from sinusoidal) Cruz-Pol, Electromagnetics UPRM To change back to time domain The phasor is multiplied by the time factor, ejt, and taken the real part. A Re{ As e jt } Cruz-Pol, Electromagnetics UPRM Advantages of phasors Time derivative is equivalent to multiplying its phasor by j A jAs t Time integral is equivalent to dividing by the same term. As At j Cruz-Pol, Electromagnetics UPRM Time-Harmonic fields (sines and cosines) The wave equation can be derived from Maxwell equations, indicating that the changes in the fields behave as a wave, called an electromagnetic field. Since any periodic wave can be represented as a sum of sines and cosines (using Fourier), then we can deal only with harmonic fields to simplify the equations. Cruz-Pol, Electromagnetics UPRM Maxwell Equations for Harmonic fields Differential form* DE v v Gauss’s Law for E field. BH 0 Gauss’s Law for H field. No monopole 0 E jH E B t H J jE D H J t * (substituting Faraday’s Law Ampere’s Circuit Law D E andCruz-Pol, H Electromagnetics B) UPRM A wave Start taking the curl of Faraday’s law Es j H s Then apply the vectorial identity A ( A) 2 A And you’re left with ( Es ) Es j ( j ) Es 2 Es 2 Cruz-Pol, Electromagnetics UPRM A Wave E E 0 2 2 Let’s look at a special case for simplicity without loosing generality: •The electric field has only an x-component •The field travels in z direction Then we have E ( z, t ) whose general solution is E(z) Eo e z Eo' e z Cruz-Pol, Electromagnetics UPRM To change back to time domain From phasor E xs ( z ) Eo e z Eo e z ( j ) …to time domain E ( z , t ) Eo e z cos(t z ) x Cruz-Pol, Electromagnetics UPRM Ejemplo 9.23 In free space, 50 8 E cos(10 t kz) V / m Find k, Jd and H using phasors and maxwells eqs. Cruz-Pol, Electromagnetics UPRM Several Cases of Media 1. 2. 3. 4. Free space ( 0, o , o ) Lossless dielectric ( 0, r o , r o or ) Lossy dielectric ( 0, r o , r o ) Good Conductor ( , o , r o or ) Recall: Permittivity o=8.854 x 10-12[ F/m] Permeability o= 4p x 10-7 [H/m] Cruz-Pol, Electromagnetics UPRM 1. Free space There are no losses, e.g. E ( z, t ) A sin( t z ) x Let’s define The phase of the wave The angular frequency Phase constant The phase velocity of the wave The period and wavelength How does it moves? Cruz-Pol, Electromagnetics UPRM 3. General Case (Lossy Dielectrics) In general, we had E ( z , t ) Eo e z cos(t z ) x 2 j ( j ) j Re 2 2 2 2 2 2 2 2 2 2 From this we obtain 2 1 1 2 2 1 1 2 So , for a known material and frequency, we can find j Cruz-Pol, Electromagnetics UPRM Intrinsic Impedance, h If we divide E by H, we get units of ohms and the definition of the intrinsic impedance of a medium at a given frequency. given E = Eoe-g z x Find H |E| j h |H | j E(z, t) = Eoe-a z cos(w t - b z)x Eo - a z H (z, t) = e cos(w t - b z - qh ) ŷ h Cruz-Pol, Electromagnetics UPRM h h [] *Not in-phase for a lossy medium Note… E(z, t) = Eoe-a z cos(w t - b z)x Eo - a z H (z, t) = e cos(w t - b z - qh ) ŷ h E(z) = Eoe-a z e- j ( b z ) x Eo -a z - j ( b z-qh ) H (z) = e e ŷ h E and H are perpendicular to one another Travel is perpendicular to the direction of propagation The amplitude is related to the impedance And so is the phase Cruz-Pol, Electromagnetics UPRM Loss Tangent If we divide the conduction current by the displacement current J cs J ds Es tan loss tangent j Es http://fipsgold.physik.uni-kl.de/software/java/polarisation Cruz-Pol, Electromagnetics UPRM Relation between tan and c H E j E j 1 j E j c E The complex permittivi ty is c 1 j ' j ' ' " The loss tangent can be defined also as tan ' Cruz-Pol, Electromagnetics UPRM 2. Lossless dielectric ( 0, r o , r o or ) Substituting in the general equations: 0, 1 2p u o h 0 Cruz-Pol, Electromagnetics UPRM Review: 1. Free Space ( 0, o , o ) Substituting in the general equations: 0, / c 1 2p u c o o h o o 0 120p 377 o E ( z , t ) Eo cos(t z ) x V / m H ( z, t ) Eo ho cos(t z ) yˆ A/ m Cruz-Pol, Electromagnetics UPRM 4. Good Conductors ( , o , r o ) Substituting in the general equations: 2 2 u 2p Is water a good conductor??? h 45o E ( z , t ) Eo e z cos(t z ) x [V / m] H ( z, t ) Eo e z cos(t z 45o ) yˆ [ A / m] Cruz-Pol, Electromagnetics o UPRM Summary Lossless medium (=0) Any medium h uc r r c 2 1 1 2 2 1 1 2 Low-loss medium (”/’<.01) 0 2 Good conductor (”/’>100) pf pf (1 j ) j j / 1 1 4pf 2p/up/f **In free space; up up up f f f Cruz-Pol,xElectromagnetics o =8.85 10-12 F/m UPRM o=4p x 10-7 H/m Units [Np/m] [rad/m] [ohm] [m/s] [m] Skin depth, d E ( z , t ) Eo e z cos(t z ) x [V / m] We know that a wave attenuates in a lossy medium until it vanishes, but how deep does it go? Is defined as the depth at which the electric amplitude is decreased to 37% 1 e 0.37 (37%) e z e 1 at z 1 / d d 1 / [m] Cruz-Pol, Electromagnetics UPRM Short Cut … You can use Maxwell’s or use 1 H kˆ E h E h kˆ H where k is the direction of propagation of the wave, i.e., the direction in which the EM wave is traveling (a unitary vector). Cruz-Pol, Electromagnetics UPRM Waves Static charges > static electric field, E Steady current > static magnetic field, H Static magnet > static magnetic field, H Time-varying current > time varying E(t) & H(t) that are interdependent > electromagnetic wave Time-varying magnet > time varying E(t) & H(t) that are interdependent > electromagnetic wave Cruz-Pol, Electromagnetics UPRM EM waves don’t need a medium to propagate Sound waves need a medium like air or water to propagate EM wave don’t. They can travel in free space in the complete absence of matter. Look at a “wind wave”; the energy moves, the plants stay at the same place. Cruz-Pol, Electromagnetics UPRM Exercises: Wave Propagation in Lossless materials A wave in a nonmagnetic material is given by H zˆ50 cos(109 t 5 y ) [mA/m] Find: (a) direction of wave propagation, (b) wavelength in the material (c) phase velocity (d) Relative permittivity of material (e) Electric field phasor j5 y [V/m] Answer: +y, up= 2x108 m/s, 1.26m, 2.25, E xˆ12.57e Cruz-Pol, Electromagnetics UPRM Power in a wave A wave carries power and transmits it wherever it goes The power density per area carried by a wave is given by the Poynting vector. See Applet by Daniel RothCruz-Pol, at Electromagnetics UPRM http://fipsgold.physik.uni-kl.de/software/java/oszillator/index.html Poynting Vector Derivation Start with E dot Ampere’s E E H E t E H E E E Apply vector identity A B B A A B or in this case : H E E H H E And end up with: 1 E H E H E E 2 t 2 Cruz-Pol, Electromagnetics UPRM 2 E t Poynting Vector Derivation… Substitute Faraday in 1rst term H 1 E 2 2 H H E E t 2 t H H H As in derivative of square function : H t 2 t and if invert the order, it' s (-) H E E H H 2 2 t E H E 2 E 2 2 t Rearrange E 2 H 2 E 2 E H t 2Electromagnetics t 2 Cruz-Pol, UPRM Poynting Vector Derivation… Taking the integral wrt volume E H dv v 2 2 2 E H dv E dv t v 2 2 v Applying Theorem of Divergence 2 2 2 E H dS E H dv E dv S t v 2 2 v Total power across surface of volume Rate of change of stored energy in E or H Ohmic losses due to conduction current Which means that the total power coming out of a volume is either due to the electric or magnetic field energy variations or is lost in ohmic losses. Cruz-Pol, Electromagnetics UPRM Power: Poynting Vector Waves carry energy and information Poynting says that the net power flowing out of a given volume is = to the decrease in time in energy stored minus the conduction losses. P EH 2 [W/m ] Cruz-Pol, Electromagnetics UPRM Represents the instantaneous power density vector associated to the electromagnetic wave. Time Average Power The Poynting vector averaged in time is T T * 1 1 1 Pave P dt E H dt Re Es H s T0 T0 2 For the general case wave: Es Eo e z e jz xˆ [V / m] Hs Eo h e z e jz yˆ [ A / m] Pave Eo2 2z e cos h zˆ 2h Cruz-Pol, Electromagnetics UPRM [W/m 2 ] Total Power in W The total power through a surface S is Pave Pave dS [W ] S Note that the units now are in Watts Note that power nomenclature, P is not cursive. Note that the dot product indicates that the surface area needs to be perpendicular to the Poynting vector so that all the power will go thru. (give example of receiver antenna) Cruz-Pol, Electromagnetics UPRM Exercises: Power 1. At microwave frequencies, the power density considered safe for human exposure is 1 mW/cm2. A radar radiates a wave with an electric field amplitude E that decays with distance as |E(R)|=3000/R [V/m], where R is the distance in meters. What is the radius of the unsafe region? Answer: 34.64 m 2. A 5GHz wave traveling In a nonmagnetic medium with r=9 is characterized by E yˆ 3 cos(t x) zˆ 2 cos(t x)[V/m] Determine the direction of wave travel and the average power density carried by the wave 2 E Answer: P o e 2 cos aˆ xˆ 0.05 [W/m 2 ] ave 2h1 h kElectromagnetics Cruz-Pol, UPRM TEM wave x x z z y Transverse ElectroMagnetic = plane wave There are no fields parallel to the direction of propagation, only perpendicular (=transverse). If have an electric field Ex(z) …then must have a corresponding magnetic field Hx(z) The direction of propagation is aE x aH = a k Cruz-Pol, Electromagnetics UPRM PE 10.7 In free space, H=0.2 cos (t-x) z A/m. Find the total power passing through a square plate of side 10cm on plane x+z=1 x Answer; Ptot = 53mW square Hz plate at z=3 Ey Answer; Ptot = 0mW! Cruz-Pol, Electromagnetics UPRM Polarization: Why do we care?? Antenna applications – Remote Sensing and Radar Applications – Antenna can only TX or RX a polarization it is designed to support. Straight wires, square waveguides, and similar rectangular systems support linear waves (polarized in one direction) Round waveguides, helical or flat spiral antennas produce circular or elliptical waves. Many targets will reflect or absorb EM waves differently for different polarizations. Using multiple polarizations can give more information and improve results. Absorption applications – Human body, for instance, will absorb waves with E oriented from head to toe better than side-to-side, esp. in grounded cases. Also, the frequency at which maximum absorption occurs is different for these two polarizations. This has ramifications in safety guidelines and studies. Cruz-Pol, Electromagnetics UPRM Polarization of a wave IEEE Definition: The trace of the tip of the E-field vector as a function of time seen from behind. Simple cases Vertical, Ex x x y y x Ex ( z ) Eo cos(t z ) xˆ Exs ( z ) Eo e jz Horizontal, Ey yy http://fipsgold.physik.uniCruz-Pol, Electromagnetics kl.de/software/java/polarisation/ UPRM x Polarization In general, plane wave has 2 components; in x & y E ( z ) xˆE x yˆE y And y-component might be out of phase wrt to xcomponent, d is the phase difference between x and y. E x Eoxe j z x E y Eoy e j z d Ex x y y Cruz-Pol, Electromagnetics UPRM Ey Front View Several Cases polarization: ddy-dx =0o or ±180on Circular polarization: dy-dx =±90o & Eox=Eoy Elliptical polarization: dy-dx=±90o & Eox≠Eoy, or d≠0o or ≠180on even if Eox=Eoy Unpolarized- natural radiation Linear Cruz-Pol, Electromagnetics UPRM Linear polarization Front View d =0 x E x E o e j z Ex E y E o e j z y Ey @z=0 in time domain E x E xo cos(t) E y E yo cos(t) Back View: x y Cruz-Pol, Electromagnetics UPRM Circular polarization Both components have same amplitude Eox=Eoy, d =d y-d x= -90o = Right circular polarized (RCP) d =+90o = LCP E x E xo cos(t) E y E yo cos(t 90 o ) in phasor : E xˆE xo yˆ E yo e j 90 xˆE xo jE yo yˆ Cruz-Pol, Electromagnetics UPRM Elliptical polarization X and Y components have different amplitudes Eox≠Eoy, and d =±90o or d ≠±90o and Eox=Eoy Or d ≠0,180o, Or any other phase difference, for example d =56o Cruz-Pol, Electromagnetics UPRM Polarization example All light comes out Unpolarized radiation enters Nothing comes out this time. Polarizing glasses Cruz-Pol, Electromagnetics UPRM sin( 90 ) cos( ) sin( 180 ) sin( ) cos( 90 ) sin ( ) cos( 180 ) cos( ) o Example o o o Determine the polarization state of a plane wave with electric field: a. E ( z, t ) xˆ3cos(t - z 30o ) - ŷ4sin( t - z 45o ) o o b. E ( z, t ) xˆ5cos(t - z 45 ) ŷ10sin( t - z 45 ) c. E ( z, t ) xˆ 4cos(t - z 45 ) - ŷ4sin( t - z 45 ) o -jz ˆ ˆ d. Es ( z ) 14( x-jy )e Cruz-Pol, Electromagnetics UPRM o . d=105, Elliptic . d=0, linear a 30o c. +180, LP a 45o d. -90, RHCP Cell phone & brain Computer model for Cell phone Radiation inside the Human Brain SAR Specific Absorption Rate [W/Kg] FCC limit 1.6W/kg, ~.2mW/cm2 for 30mins http://www.ewg.org/cellphoneradiation/Get-a-SaferPhone/Samsung/Impression+%28SGH-a877%29/ Cruz-Pol, Electromagnetics UPRM Human absorption 30-300 MHz is where the human body absorbs RF energy most efficiently * The FCC limit in the US for public exposure from cellular telephones at the ear level is a SAR level of 1.6 watts per kilogram (1.6 W/kg) as averaged over one gram of tissue. **The ICNIRP limit in Europe for public exposure from cellular telephones at the ear level is a SAR level of 2.0 watts per kilogram (2.0 W/kg) as averaged over ten grams of tissue. http://handheld-safety.com/SAR.aspx http://www.fcc.gov/Bureaus/Engineering_Technology/Docume nts/bulletins/oet56/oet56e4.pdf Cruz-Pol, Electromagnetics UPRM Radar bands Band Name Nominal Freq Range HF, VHF, UHF 3-30 MHz0, 30-300 MHz, 3001000MHz Specific Bands 138-144 MHz 216-225, 420-450 MHz 890-942 Application TV, Radio, Clear air, soil moist L 1-2 GHz (15-30 cm) 1.215-1.4 GHz S 2-4 GHz (8-15 cm) 2.3-2.5 GHz 2.7-3.7> C 4-8 GHz (4-8 cm) 5.25-5.925 GHz TV stations, short range Weather X 8-12 GHz (2.5–4 cm) 8.5-10.68 GHz Cloud, light rain, airplane weather. Police radar. Ku 12-18 GHz 13.4-14.0 GHz, 15.7-17.7 K 18-27 GHz 24.05-24.25 GHz Ka 27-40 GHz 33.4-36.0 GHz V 40-75 GHz 59-64 GHz W 75-110 GHz millimeter 110-300 GHz 76-81 GH, 92-100 GHz Cruz-Pol, Electromagnetics UPRM Weather observations Cellular phones Weather studies Water vapor content Cloud, rain Intra-building comm. Rain, tornadoes Tornado chasers Microwave Oven Most food is lossy media at microwave frequencies, therefore EM power is lost in the food as heat. Find depth of penetration if meat which at 2.45 GHz has the complex permittivity given. The power reaches the inside as soon as the oven in turned on! c o (30 j1) j o j c 2pf ( j 30) c 4.7 j 281 [/m] d 1/ 21.3 cm Cruz-Pol, Electromagnetics UPRM Decibel Scale In many applications need comparison of two powers, a power ratio, e.g. reflected power, attenuated power, gain,… The decibel (dB) scale is logarithmic G P1 P2 V12 /R P1 V G[dB] 10 log 10 log 2 20 log 1 P2 V2 V2 /R Note that for voltages, fields, and electric currents, the log is multiplied by 20 instead of 10. Cruz-Pol, Electromagnetics UPRM Attenuation rate, A Represents the rate of decrease of the magnitude of Pave(z) as a function of propagation distance Pave(z) 10 log e 2z A 10 log Pave( 0 ) 20z log e - 8.68z - dB z [dB] where dB[dB/m ] 8.68 [ Np/m] Cruz-Pol, Electromagnetics UPRM Submarine antenna A submarine at a depth of 200m uses a wire antenna to receive signal transmissions at 1kHz. Determine the power density incident upon the submarine antenna due to the EM wave with |Eo|= 10V/m. [At 1kHz, sea water has r=81, =4]. Pave Eo2 2z e cos h zˆ 2h [W/m 2 ] At what depth the amplitude of E has decreased to 1% its initial value at z=0 (sea surface)? Cruz-Pol, Electromagnetics UPRM Exercise: Lossy media propagation For each of the following determine if the material is low-loss dielectric, good conductor, etc. (a) Glass with r=1, r=5 and =10-12 S/m at 10 GHZ (b) Animal tissue with r=1, r=12 and =0.3 S/m at 100 MHZ (c) Wood with r=1, r=3 and =10-4 S/m at 1 kHZ Answer: (a) (b) (c) low-loss, 8.4x1011 Np/m, 468 r/m, 1.34 cm, up1.34x108, hc168 general, 9.75, 12, 52 cm, up0.5x108 m/s, hc39.5j31.7 Good conductor, 6.3x104, 6.3x104, 10km, up0.1x108, hc6.281j Cruz-Pol, Electromagnetics UPRM