Your SlideShare is downloading. ×
  • Like
Phy electro
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×

Now you can save presentations on your phone or tablet

Available for both IPhone and Android

Text the download link to your phone

Standard text messaging rates apply

Phy electro

  • 113 views
Published

 

  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads

Views

Total Views
113
On SlideShare
0
From Embeds
0
Number of Embeds
0

Actions

Shares
Downloads
4
Comments
0
Likes
0

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide
  • 2
  • 3
  • 10
  • 12
  • 14
  • 18
  • 19
  • 20
  • 21
  • 22
  • 24 1
  • 25 2
  • 26 2
  • 27 3
  • 29 3

Transcript

  • 1. Chapter 32The Laws of Electromagnetism Maxwell’s Equations Displacement Current Electromagnetic Radiation
  • 2. The Electromagnetic SpectrumRadio waves infra ultra -red -violet γ -rays µ -wave x-rays
  • 3. The Equations of Electromagnetism (at this point …) qGauss’ Law for Electrostatics ∫ E • dA = ε0Gauss’ Law for Magnetism ∫ B • dA = 0 dΦBFaraday’s Law of Induction ∫ E • dl = − dtAmpere’s Law ∫ B • dl = µ0 I
  • 4. The Equations of Electromagnetism Gauss’s Laws ..monopole.. q1 ∫ E • dA = ε02 ∫ B • dA = 0 ?...there’s nomagnetic monopole....!!
  • 5. The Equations of Electromagnetism Faraday’s Law .. if you change a .. if you change a dΦ B magnetic field you magnetic field you3 ∫ E • dl = − dt induce an electric induce an electric field......... field......... Ampere’s Law4 ∫ B • dl = µ0 I .......is the reverse .......is the reverse true..? true..?
  • 6. ...lets take a look at charge flowing into a capacitor......when we derived Ampere’s Law B Ewe assumed constant current... ∫ B • dl = µ0 I
  • 7. ...lets take a look at charge flowing into a capacitor......when we derived Ampere’s Law B Ewe assumed constant current... ∫ B • dl = µ0 I E .. if the loop encloses one B plate of the capacitor..there is a problem … I = 0 Side view: (Surface is now like a bag:)
  • 8. Maxwell solved this problem by realizing that....Inside the capacitor there must B Ebe an induced magnetic field...How?.
  • 9. Maxwell solved this problem by realizing that....Inside the capacitor there must B Ebe an induced magnetic field...How?. Inside the capacitor there is a changing E ⇒B A changing x x x x x E electric fieldx x x x x induces a x x magnetic field
  • 10. Maxwell solved this problem by realizing that....Inside the capacitor there must B Ebe an induced magnetic field...How?. Inside the capacitor there is a changing E ⇒B A changing dΦ E ∫ B • dl = µ 0ε 0 dt = µ 0 Id x x x x x E electric fieldx x x x x induces a where Id is called the x x magnetic field displacement current
  • 11. Maxwell solved this problem by realizing that....Inside the capacitor there must B Ebe an induced magnetic field...How?. Inside the capacitor there is a changing E ⇒B A changing dΦ E ∫ B • dl = µ 0ε 0 dt = µ 0 Id x x x x x E electric fieldx x x x x induces a where Id is called the x x magnetic field displacement current dΦ E ∫ B • dl = µ 0 I + µ 0ε 0 dtTherefore, Maxwell’s revisionof Ampere’s Law becomes....
  • 12. Derivation of Displacement Current dq d( EA )For a capacitor, q = ε0 EA and I = dt = ε0 dt . d(Φ E )Now, the electric flux is given by EA, so: I = ε 0 dt ,where this current , not being associated with charges, iscalled the “Displacement current”, Id. dΦ E Hence: I d = µ0 ε0 dt and: ∫ B • ds = µ0( I + Id ) ⇒ ∫ B • ds = µ0 I + µ0ε 0 dΦ E dt
  • 13. Derivation of Displacement Current dq d( EA )For a capacitor, q = ε0 EA and I = dt = ε0 dt . d(Φ E )Now, the electric flux is given by EA, so: I = ε 0 dt ,where this current, not being associated with charges, iscalled the “Displacement Current”, Id. dΦ E Hence: I d = µ0 ε0 dt and: ∫ B • dl = µ0( I + Id ) ⇒ ∫ B • dl = µ 0 I + µ 0ε 0 dΦ E dt
  • 14. Maxwell’s Equations of Electromagnetism qGauss’ Law for Electrostatics ∫ E • dA = ε0Gauss’ Law for Magnetism ∫ B • dA =0 dΦ BFaraday’s Law of Induction ∫ E • dl = − dt dΦ EAmpere’s Law ∫ B • dl = µ0 I + µ0ε0 dt
  • 15. Maxwell’s Equations of Electromagnetism in Vacuum (no charges, no masses) Consider these equations in a vacuum..... ......no mass, no charges. no currents..... q ∫ E • dA = ε 0 ∫ E • dA = 0 ∫ B • dA = 0 ∫ B • dA = 0 dΦB dΦ B ∫ E • dl = − dt ∫ E • dl = − dt B • dl = µ0 I + µ0ε 0 dΦ E dΦ E∫ dt ∫ B • dl = µ0ε 0 dt
  • 16. Maxwell’s Equations of Electromagnetism in Vacuum (no charges, no masses) ∫ E • dA = 0 ∫ B • dA = 0 dΦ B ∫ E • dl = − dt dΦ E ∫ B • dl = µ0ε 0 dt
  • 17. Electromagnetic WavesFaraday’s law: dB/dt electric fieldMaxwell’s modification of Ampere’s law dE/dt magnetic field dΦ E dΦB ∫ B • dl = µ0ε0 dt ∫ E • dl = − dt These two equations can be solved simultaneously. ˆ E(x, t) = EP sin (kx-ωt) j The result is: ˆ B(x, t) = BP sin (kx-ωt) z
  • 18. Electromagnetic Waves dΦ E dΦ B∫ B • dl = µ0ε 0 dt ∫ E • dl = − dt B  dE E  dB dt dt
  • 19. Electromagnetic Waves dΦ E dΦ B ∫ B • dl = µ0ε 0 dt ∫ E • dl = − dt v B  dE E  dB dt dt  Special case..PLANE WAVES... E = E y ( x ,t )  j  B = Bz ( x ,t )k ∂ 2ψ 1 ∂ 2ψsatisfy the wave equation = 2 2 ∂x 2 ν ∂t Maxwell’s Solution ψ = A sin( ωt + φ )
  • 20. Plane Electromagnetic Waves Ey Bz c x ˆE(x, t) = EP sin (kx-ωt) j ˆB(x, t) = BP sin (kx-ωt) z
  • 21. F(x) λ Static wave F(x) = FP sin (kx + φ) k = 2π / λ k = wavenumber x λ = wavelengthF(x) λ Moving wave F(x, t) = FP sin (kx - ωt ) v ω = 2π / f ω = angular frequency x f = frequency v=ω /k
  • 22. F v Moving wave x F(x, t) = FP sin (kx - ωt )What happens at x = 0 as a function of time?F(0, t) = FP sin (-ωt)For x = 0 and t = 0 ⇒ F(0, 0) = FP sin (0)For x = 0 and t = t ⇒ F (0, t) = FP sin (0 – ωt) = FP sin (– ωt)This is equivalent to: kx = - ωt ⇒ x = - (ω/k) tF(x=0) at time t is the same as F[x=-(ω/k)t] at time 0The wave moves to the right with speed ω/k
  • 23. Plane Electromagnetic Waves Ey ˆ E(x, t) = EP sin (kx-ωt) j ˆ B(x, t) = BP sin (kx-ωt) z BzNotes: Waves are in Phase, c but fields oriented at 900. k=2π/λ. x Speed of wave is c=ω/k (= fλ) c = 1 / ε0µ0 = 3 ×108 m / s At all times E=cB.