Phy electro

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  • Phy electro

    1. 1. Chapter 32The Laws of Electromagnetism Maxwell’s Equations Displacement Current Electromagnetic Radiation
    2. 2. The Electromagnetic SpectrumRadio waves infra ultra -red -violet γ -rays µ -wave x-rays
    3. 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. 4. The Equations of Electromagnetism Gauss’s Laws ..monopole.. q1 ∫ E • dA = ε02 ∫ B • dA = 0 ?...there’s nomagnetic monopole....!!
    5. 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. 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. 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. 8. Maxwell solved this problem by realizing that....Inside the capacitor there must B Ebe an induced magnetic field...How?.
    9. 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. 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. 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. 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. 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. 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. 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. 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. 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. 18. Electromagnetic Waves dΦ E dΦ B∫ B • dl = µ0ε 0 dt ∫ E • dl = − dt B  dE E  dB dt dt
    19. 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. 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. 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. 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. 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.

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