Eq mazwell

253 views

Published on

0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total views
253
On SlideShare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
Downloads
2
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide
  • dS = n dA Flux = field integrated over a surface No magnetiic monopoles E .dl is an EMF (volts)
  • dS = n dA Flux = field integrated over a surface No magnetiic monopoles E .dl is an EMF (volts)
  • Eq mazwell

    1. 1. General form of Faraday’s Law b U b − Ua r r Vba = Vb − Va ≡ = − ∫ Egds q aSo the electromotive force around a closed path is: r r ε = Ñ gds ∫EAnd Faraday’s Law becomes: r r dΦ B ε = Ñ gds = − ∫E dtA changing magnetic flux produces an electric field.This electric field is necessarily non-conservative.
    2. 2. E produced by changing B r r dΦ B ε = Ñ gd l = − ∫E dt r r dB E 2πr = πr 2 dt r r rdB E = 2 dt How about outside ro ?
    3. 3. Problems with Ampere’s Law r r ÑB ⋅ d l = µo Iencl + ? ∫C r B 2πr = µ o I r µo I B= 2πr
    4. 4. But what if….. r rÑ∫C B ⋅ d l = µ o Iencl + ? r B 2πr = µo ( 0 ) r B =0 ?????
    5. 5. Maxwell’s correction to Ampere’s Law Q = CV εo A V = Ed C= d  εo A  Q=  ( Ed ) = ε o AE = ε o Φ E  d  dQ dΦ E I= = εo Called “displacement current”, Id dt dt
    6. 6. Maxwell’s Equations q Ñ ×dA = εo ∫E Gausss law ( electric ) S Ñ ×dA = 0 ∫B S Gausss law in magnetism dΦB Ñ ×ds = − dt ∫E Faradays law dΦ E Ñ ×dμ = ∫B s o + εI μ o o dt Ampere-Maxwell law•The two Gauss’s laws are symmetrical, apart from the absence of the term formagnetic monopoles in Gauss’s law for magnetism•Faraday’s law and the Ampere-Maxwell law are symmetrical in that the lineintegrals of E and B around a closed path are related to the rate of change ofthe respective fluxes
    7. 7. • Gauss’s law (electrical):• The total electric flux through any q Ñ ×dA = εo ∫E closed surface equals the net charge inside that surface divided by εo S• This relates an electric field to the charge distribution that creates it• Gauss’s law (magnetism):• The total magnetic flux through any closed surface is zero Ñ ×dA = 0 ∫B• This says the number of field lines that enter a closed volume must equal the number that leave that volume S• This implies the magnetic field lines cannot begin or end at any point• Isolated magnetic monopoles have not been observed in nature
    8. 8. • Faraday’s law of Induction:• This describes the creation of an electric field by a changing magnetic flux• The law states that the emf, which is the line dΦB Ñ ×ds = − dt ∫E integral of the electric field around any closed path, equals the rate of change of the magnetic flux through any surface bounded by that path• One consequence is the current induced in a conducting loop placed in a time-varying B• The Ampere-Maxwell law is a generalization of Ampere’s law dΦ E•• It describes the creation of a magnetic field by an electric field and electric currents The line integral of the magnetic field around any Ñ ×dμ = ∫B s o + εI μ o o dt closed path is the given sum
    9. 9. The Lorentz Force Law• Once the electric and magnetic fields are known at some point in space, the force acting on a particle of charge q can be calculated• F = qE + qv x B• This relationship is called the Lorentz force law• Maxwell’s equations, together with this force law, completely describe all classical electromagnetic interactions
    10. 10. Maxwell’s Equation’s in integral form r r Q 1ÒA E ⋅ dA = εo = εo∫∫ ∫∫∫ V ρdV Gauss’s Law r rÒ∫∫A B ⋅ dA = 0 Gauss’s Law for Magnetism r r dΦ B d r rÑ∫C E ⋅ dl = − dt = − ∫∫ B ⋅ dA dt A Faraday’s Law r r r dΦ E r dE  rÑB ⋅ d l = µo Iencl + µoεo dt = µo ∫∫A  J + εo dt  ⋅ dA∫C   Ampere’s Law
    11. 11. Maxwell’s Equation’s in free space (no charge or current) r r Ò ∫∫A E ⋅ dA = 0 Gauss’s Law r r Ò ∫∫A B ⋅ dA = 0 Gauss’s Law for Magnetism r r dΦ B d r rÑE ⋅ d l = − dt = − dt ∫∫A B ⋅ dA∫C Faraday’s Law r r dΦ E d r rÑB ×d l = µoεo dt = µoεo dt ∫∫A E ×dA∫C Ampere’s Law
    12. 12. Hertz’s Experiment• An induction coil is connected to a transmitter• The transmitter consists of two spherical electrodes separated by a narrow gap• The discharge between the electrodes exhibits an oscillatory behavior at a very high frequency• Sparks were induced across the gap of the receiving electrodes when the frequency of the receiver was adjusted to match that of the transmitter• In a series of other experiments, Hertz also showed that the radiation generated by this equipment exhibited wave properties – Interference, diffraction, reflection, refraction and polarization• He also measured the speed of the radiation
    13. 13. Implication• A magnetic field will be produced in empty space if there is a changing electric field. (correction to Ampere)• This magnetic field will be changing. (originally there was none!)• The changing magnetic field will produce an electric field. (Faraday)• This changes the electric field.• This produces a new magnetic field.• This is a change in the magnetic field.
    14. 14. An antennaHook up anAC source We have changed the magnetic field near the antenna An electric field results! This is the start of a “radiation field.”
    15. 15. Look at the cross section Called: “Electromagnetic Waves”Acceleratingelectric chargesgive rise toelectromagneticwaves E and B are perpendicular (transverse) We say that the waves are “polarized.” E and B are in phase (peaks and zeros align)
    16. 16. Angular Dependence of Intensity• This shows the angular dependence of the radiation intensity produced by a dipole antenna• The intensity and power radiated are a maximum in a plane that is perpendicular to the antenna and passing through its midpoint• The intensity varies as (sin2 θ) / r2
    17. 17. Active Figure 34.3 (SLIDESHOW MODE ONLY)
    18. 18. Harmonic Plane Wavesr At t = 0E λ = spatial period or wavelength x λ phase velocityr At x = 0 λ 2π λ ωE v = = fλ = = T T 2π k t Τ = temporal period Τ
    19. 19. Applying Faraday to radiation r r dΦ B ÑE ⋅ d l = − dt ∫C r r ÑE ⋅ d l = ( E + dE ) ∆y − E∆y = dE∆y ∫C dΦ B dB = dx∆y dt dt dB dE∆y = − dx∆y dt dE dB =− dx dt
    20. 20. Applying Ampere to radiation r r dΦ E ÑB ⋅ d l = µoεo dt ∫C r r ÑB ⋅ d l = B∆z − ( B + dB ) ∆z = −dB∆z ∫C dΦ E dE = dx∆z dt dt dE −dB∆z = µ o εo dx∆z dt dB dE = −µ o ε o dx dt
    21. 21. Fields are functions of both position (x) and time (t) dE dB Partial derivatives ∂E ∂B =− are appropriate =− dx dt ∂x ∂tdB dE ∂B ∂E = −µ o ε o = −µ o εodx dt ∂x ∂t ∂E2 ∂ ∂B ∂ ∂B ∂ 2E =− = −µ o εo 2 ∂x 2 ∂x ∂t ∂t ∂x ∂t ∂2E ∂2E This is a wave = µoεo 2 ∂x 2 ∂t equation!
    22. 22. The Trial Solution• The simplest solution to the partial differential equations is a sinusoidal wave: – E = Emax cos (kx – ωt) – B = Bmax cos (kx – ωt)• The angular wave number is k = 2π/λ – λ is the wavelength• The angular frequency is ω = 2πƒ – ƒ is the wave frequency
    23. 23. The trial solution E = E y = E o sin ( kx − ωt ) ∂2E ∂2E = µoεo 2 ∂x 2 ∂t∂2E ∂2E = − k E o sin ( kx − ωt ) 2 = −ω E o sin ( kx − ωt ) 2∂x 2 ∂t 2 −k 2 E o sin ( kx − ωt ) = −µ o ε o ω2 E o sin ( kx − ωt ) ω2 1 = k 2 µoε o
    24. 24. The speed of light(or any other electromagnetic radiation) λ 2π λ ω v = = fλ = = T T 2π k ω 1 v=c= = k µo ε o
    25. 25. 3. The speed of an electromagnetic wave traveling in a transparentnonmagnetic substance is , where κ is the dielectric constant of the substance.Determine the speed of light in water, which has a dielectric constant at opticalfrequencies of 1.78.5. Figure 34.3 shows a plane electromagnetic sinusoidal wave propagatingin the x direction. Suppose that the wavelength is 50.0 m, and the electric fieldvibrates in the xy plane with an amplitude of 22.0 V/m. Calculate (a) thefrequency of the wave and (b) the magnitude and direction of B when theelectric field has its maximum value in the negative y direction. (c) Write anexpression for B with the correct unit vector, with numerical values for Bmax, k,and ω, and with its magnitude in the form6. Write down expressions for the electric and magnetic fields of asinusoidal plane electromagnetic wave having a frequency of 3.00 GHz andtraveling in the positive x direction. The amplitude of the electric field is 300V/m.
    26. 26. The electromagnetic spectrum λ 2π λ ω v = = fλ = = T T 2π k
    27. 27. Another look dE dB =− dx dtB = Bz = Bo sin ( kx − ωt ) E = E y = E o sin ( kx − ωt ) d d E o sin ( kx − ωt ) = − Bo sin ( kx − ωt ) dx dt E o k cos ( kx − ωt ) = Bo ω cos ( kx − ωt ) Eo ω 1 = =c= Bo k µo ε o
    28. 28. Energy in Waves 1 1 2 u = ε0 E + 2 B 2 2µ 0 u = ε0 E 2Eo ω 1 2 = =c= 1 u= BBo k µoε o µ0 ε0 u= EB µ0
    29. 29. Poynting Vector r 1 r r S= µ0 E×B( ) EB E 2 c B 2 S= = = μo μo c μo r S = cu• Poynting vector points in the direction the wave moves• Poynting vector gives the energy passing through a unit area in 1 sec.• Units are Watts/m2
    30. 30. Intensity• The wave intensity, I, is the time average of S (the Poynting vector) over one or more cycles• When the average is taken, the time average of cos2(kx - ωt) = ½ is involved 2 2 Emax Bmax E cB I = Sav = = max = max = cuave 2 μo 2 μo c 2 μo
    31. 31. 11. How much electromagnetic energy per cubic meter iscontained in sunlight, if the intensity of sunlight at the Earth’ssurface under a fairly clear sky is 1 000 W/m2?16. Assuming that the antenna of a 10.0-kW radio stationradiates spherical electromagnetic waves, compute themaximum value of the magnetic field 5.00 km from theantenna, and compare this value with the surface magneticfield of the Earth.21. A lightbulb filament has a resistance of 110 Ω. Thebulb is plugged into a standard 120-V (rms) outlet, and emits1.00% of the electric power delivered to it by electromagneticradiation of frequency f. Assuming that the bulb is coveredwith a filter that absorbs all other frequencies, find theamplitude of the magnetic field 1.00 m from the bulb.
    32. 32. Radiation Pressure F 1 dp P= = A A dt ∆U (Absorption of radiationMaxwell showed: ∆p = by an object) c r 1 dU Save P= = Ac dt c What if the radiation reflects off an object?
    33. 33. Pressure and Momentum• For a perfectly reflecting surface, p = 2U/c and P = 2S/c• For a surface with a reflectivity somewhere between a perfect reflector and a perfect absorber, the momentum delivered to the surface will be somewhere in between U/c and 2U/c• For direct sunlight, the radiation pressure is about 5 x 10-6 N/m2
    34. 34. 26. A 100-mW laser beam is reflected back upon itself by amirror. Calculate the force on the mirror.27. A radio wave transmits 25.0 W/m2 of power per unitarea. A flat surface of area A is perpendicular to the direction ofpropagation of the wave. Calculate the radiation pressure on it,assuming the surface is a perfect absorber.29. A 15.0-mW helium–neon laser (λ = 632.8 nm) emits abeam of circular cross section with a diameter of 2.00 mm. (a)Find the maximum electric field in the beam. (b) What totalenergy is contained in a 1.00-m length of the beam? (c) Find themomentum carried by a 1.00-m length of the beam.
    35. 35. Background for the superior mathematics student!
    36. 36. Harmonic Plane WavesIn general, We will only be concerned with the real part of the complex phasor representation of a plane wave. Using Euler’s formula: r r i b −ωt g r E = Eo e kx b g b = Eo cos kx − ωt + i sin kx − ωt g di r r b Re E = Eo cos kx − ωt g 2π k= = propagation number λ 2π ω= = angular frequency τ (kx-ωt) = phase
    37. 37. Phase Velocity - Another View φ = kx − ωt dφ dx =k − ω = kv − ω = 0 dt dt Since the plane waves remain plane waves, the phase on a plane does not change with time ω v= k
    38. 38. Vector Calculus Theorems r r r rÒ∫∫ A F ⋅ dA = ∫∫∫ ∇ ⋅ FdV V Gauss’ Divergence Theoremz z z r r F ⋅ dl = C r r r ∇ × F ⋅ dS A Stokes Theorem And an Important Identity e j e j r r r r r r ∇ × ∇ × F = ∇ ∇⋅ F − ∇ F 2
    39. 39. Maxwell’s Equation’s In Differential Form r r ∇⋅ D = ρ Gauss’s Law r r ∇⋅ B = 0 Gauss’s Law for Magnetism r r r ∂B ∇× E = − Faraday’s Law ∂t r r r r ∂D ∇× H = J + Ampere’s Law ∂t

    ×