Maxwell's Equations PDF

The Laws of Electromagnetism Maxwell’s Equations Displacement Current Electromagnetic Radiation The Electromagnetic Spectrum  The Equations of Electromagnetism (at this point …) q Gauss’ Law for Electrostatics  E dA = 0 Gauss...

The Laws of Electromagnetism Maxwell’s Equations Displacement Current Electromagnetic Radiation The Electromagnetic Spectrum  The Equations of Electromagnetism (at this point …) q Gauss’ Law for Electrostatics  E dA = 0 Gauss’ Law for Magnetism  B dA = 0 d Faraday’s Law of Induction  E dl = − dt B Ampere’s Law  B dl = 0 I  The Equations of Electromagnetism Gauss’s Laws..monopole.. q 1  E dA = 0 2  B dA = 0 ?...there’s no magnetic monopole....!!  The Equations of Electromagnetism Faraday’s Law.. if you change a d magnetic field you 3  E dl = − dt B induce an electric field......... Ampere’s Law 4  B dl = 0 I.......is the reverse true..?...lets take a look at charge flowing into a capacitor......when we derived Ampere’s Law B E we assumed constant current...  B dl = 0 I...lets take a look at charge flowing into a capacitor......when we derived Ampere’s Law B E we assumed constant current...  B dl = 0 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:)  Maxwell solved this problem by realizing that.... Inside the capacitor there must B E be an induced magnetic field... How?.  Maxwell solved this problem by realizing that.... Inside the capacitor there must B E be an induced magnetic field... How?. Inside the capacitor there is a changing E  B A changing x x x x x E electric field x x x x x induces a x x magnetic field  Maxwell solved this problem by realizing that.... Inside the capacitor there must B E be an induced magnetic field... How?. Inside the capacitor there is a changing E  B A changing d E x x x x x E electric field  B dl = 00 dt = 0 Id x x x x x induces a where Id is called the x x magnetic field displacement current  Maxwell solved this problem by realizing that.... Inside the capacitor there must B E be an induced magnetic field... How?. Inside the capacitor there is a changing E  B A changing d E x x x x x E electric field  B dl = 00 dt = 0 Id x x x x x induces a where Id is called the x x magnetic field displacement current d     Therefore, Maxwell’s revision of Ampere’s Law becomes.... B dl = 0 I + 0 0 E dt  Derivation of Displacement Current dq d( EA ) For a capacitor, q = 0 EA and I = dt = 0 dt. d ( E ) Now, the electric flux is given by EA, so: I =  0 , dt where this current , not being associated with charges, is called the “Displacement current”, Id. d E Hence: I d = 0  0 dt and:  B ds = 0( I + Id ) d   B ds = 0 I + 00 E dt  Derivation of Displacement Current dq d( EA ) For a capacitor, q = 0 EA and I = dt = 0 dt. d ( E ) Now, the electric flux is given by EA, so: I =  0 , dt where this current, not being associated with charges, is called the “Displacement Current”, Id. d E Hence: I d = 0  0 dt and:  B dl = 0( I + Id ) d   B dl = 0 I + 00 E dt  Maxwell’s Equations of Electromagnetism q Gauss’ Law for Electrostatics  E dA =  0 Gauss’ Law for Magnetism  B dA = 0 d Faraday’s Law of Induction  E dl = − dt B d E Ampere’s Law  B dl = 0 I + 00 dt  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 =  0  E dA = 0  B dA = 0  B dA = 0 d B d B  E dl = − dt  E dl = − dt    d d E  B dl = 0 I + 0 0 dt E  B dl = 0 0 dt Maxwell’s Equations of Electromagnetism in Vacuum (no charges, no masses)  E dA = 0  B dA = 0 d B  E dl = − dt d E  B dl = 0 0 dt  Electromagnetic Waves Faraday’s law: dB/dt electric field Maxwell’s modification of Ampere’s law dE/dt magnetic field d E d B  B dl = 00 dt  E dl = − dt These two equations can be solved simultaneously. E(x, t) = EP sin (kx-t) ĵ The result is: B(x, t) = BP sin (kx-t) ẑ  Electromagnetic Waves d E d B  B dl = 0 0 dt  E dl = − dt B  dE E  dB dt dt  Electromagnetic Waves d E d B  B dl = 0 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  2 1  2 satisfy the wave equation = 2 2 x 2 t Maxwell’s Solution  = A sin( t +  )  Plane Electromagnetic Waves Ey Bz c x E(x, t) = EP sin (kx-t) ĵ B(x, t) = BP sin (kx-t) ẑ F(x)  Static wave F(x) = FP sin (kx + ) k = 2   k = wavenumber x  = wavelength F(x)  Moving wave F(x, t) = FP sin (kx - t ) v  = 2  f  = angular frequency x f = frequency v=/k  F v Moving wave x F(x, t) = FP sin (kx - t ) What happens at x = 0 as a function of time? F(0, t) = FP sin (-t) For x = 0 and t = 0  F(0, 0) = FP sin (0) For x = 0 and t = t  F (0, t) = FP sin (0 – t) = FP sin (– t) This is equivalent to: kx = - t  x = - (/k) t F(x=0) at time t is the same as F[x=-(/k)t] at time 0 The wave moves to the right with speed /k  Plane Electromagnetic Waves Ey E(x, t) = EP sin (kx-t) ĵ B(x, t) = BP sin (kx-t) ẑ Bz Notes: Waves are in Phase, c but fields oriented at 900. k=2. x Speed of wave is c=/k (= f) c = 1 / 00 = 3  108 m / s At all times E=cB.