Maxwell’s Equations January 16, 2013 1. Maxwell’s Equation, integral form 2. Maxwell’s Eqaution, differential form 3. Conservation of electric charge 4. Maxwell’s equations in static case 5. Maxwell’s equations in source-free dynamic case 6. Maxwell’s Eqaution in phasor form 7. The dependency of Maxwell’s equations 1 Maxwell’s equations in integral form 1.1 Faradya’s Law of induction The induced emf (electromotive force) in a closed circuit is equal to the negative of rate of change of magnetic flux pass through it. ‹ ˛ ∂ ∂Φ B =− B̄ · dS̄ Ē · d¯l = − ∂t ∂t S ∂S 1.2 Ampère’s circuital law with Maxwell’s Correction The mmf (magnetomotive force) in a closed circuit is equal to the rate of change of electric flux and current pass through it. ) ˛ ‹ ( ∂ D̄ ¯ ¯ · dS̄ H̄ · dl = J+ ∂t ∂S S 1.3 Gauss’s Law for electricity The electric flux pass through any closed surface is proportional to the enclosed electric charge. ‹ D̄ · dS̄ = ΦD = Q = ρv V S 1 1.4 Gauss’s Law for magnetism The magnetic flux pass through any closed surface is zero ‹ B̄ · dS̄ = 0 2 Maxwell’s equations in differential form 2.1 2.1.1 Vector Identities Gauss’s Divergence Theorem ˚ ‹ V̄ · dS̄ ∇ · V̄ dV = ∂V V 2.1.2 Stokes’ Curl Theorem ¨ ˛ V̄ · d¯l ∇ × V̄ · dS̄ = S 2.2 ∂S Faraday’s Law in differential form Apply Stoke’s Theorem ˛ ‹ ∂ ¯ Ē · dl = − B̄ · dS̄ ∂t S ∂S ‹ ∂ ∇ × Ē · dS̄ = − ∂t S −→ ‹ B̄ · dS̄ S Rearrange ‹ ( S ) ‹ ( ∂ B̄ ∇ × Ē · dS̄ = − · dS̄ ∂t S ) The integral kernel thus equal to each other ∇ × Ē = − 2.3 ∂ B̄ ∂t Ampère’s circuital law in differential form Apply Stoke’s Theorem ) ˛ ‹ ( ∂ D̄ · dS̄ H̄ · d¯l = J¯ + ∂t ∂S S ) ‹ ( ∂ D̄ ∇ × H̄ · dS̄ = · dS̄ J¯ + ∂t S S ‹ −→ The integral kernel thus equal to each other ∇ × H̄ = J¯ + 2 ∂ D̄ ∂t 2.4 Gauss’s Law for electricity in differential form Apply Gauss’s Divergence Theorem ‹ D̄ · dS̄ = Q ˚ ˚ −→ ∇ · D̄dV = Q = S ρv dV V The integral kernel thus equal to each other ∇ · D̄ = ρv 2.5 Gauss’s Law for magnetism in differential form Apply Gauss’s Divergence Theorem ‹ B̄ · dS̄ = 0 ˚ ˚ −→ ∇ · B̄dV = 0 = S 0dV V The integral kernel thus equal to each other ∇ · B̄ = 0 3 Conservation of electric charge Consider the follow equations ∂ D̄ ∇ × H̄ = J¯ + ∂t ∇ · D̄ = ρv ( ) ∇ · ∇ × V̄ = 0 Faraday’s Law Gauss’s Law Div of curl is zero Take the div of Faraday’s Law ( ) ∂ D̄ ∇ · ∇ × H̄ = ∇ · J¯ + ∂t ( ) ∂ 0 = ∇ · J¯ + ∇ · D̄ ∂t ∂ρv 0 = ∇ · J¯ + ∂t i.e. The rate of current transfer out of a volume eqaul to decreasing rate of charge in that volume ∂ρv ∇ · J¯ = − ∂t 3 4 Maxwell’s equations in static case Static ⇐⇒ ∂ =0 ∂t { ∇ × Ē = 0 ∇ × H̄ = J ∇ · B̄ = 0 ∇ · D̄ = ρv ∇ · J¯ = 0 ∇ × Ē = 0 ∇ · D̄ = ρv ∇ × H̄ = J ∇ · B̄ = 0 ∇ · J¯ = 0 E & D field are generated by ρv H & B fields are generated ( by J ) J & ρv are independent ∇ · J¯ = 0 E , H are decoupled, the electrostatic field and magnetostatic field are independent. 5 Maxwell’s Equations in source-free dynamic case Source free ⇐⇒ J = ρv = 0 ∂ B̄ ∂t ∂ D̄ ∇ × H̄ = ∂t ∇ · B̄ = 0 ∇ · D̄ = 0 E & H are coupled, they are not independent. This coupling generate the phenomenon of EM wave propagation ∇ × Ē = − 6 Maxwell’s equations in phasor form 6.1 Phasor Review Apply Euler’s Eqaution to sinusoidal term [ ] [( ) ] V (t) = V0 cos (ωt + ϕ) = ℜe V0 ej(ωt+ϕ) = ℜe V0 ejϕ ejωt The phasor form is thus V0 ejϕ i.e. V (t) = V0 cos (ωt + ϕ) • Original time-domain form is real number • Phasor form is complex number 4 ←→ Ve = V0 ejϕ 6.2 Phasor Differentiation and Integration [( ) ] ∂ ∂ V (t) = V0 cos (ωt + ϕ) = −ωV0 sin (ωt + ϕ) = ℜe jωV0 ejϕ ejωt ∂t ∂t Therefore ∂ V (t) ←→ jω Ve ∂t ˆ ˆ V (t)dt = 1 V0 cos (ωt + ϕ) dt = V0 sin (ωt + ϕ) = ℜe ω [( ) ] 1 jϕ jωt V0 e e jω Therefore ˆ V (t)dt ←→ 6.3 1 e V jω Maxwell’s Equations in Phasor Form ∇ × Ē = −jω B̄ ∇ × H̄ = jω D̄ + J¯ ∇ · B̄ = 0 ∇ · D̄ = ρv ∇ · J¯ = −jωρv The equations are now complex and time-independent. 7 The dependency in Maxwell’s Equations There are 4 equations, but the other equations can be derived form the 2 curl equations with some vector identities. ∇ × Ē = −jω B̄ ∇ × H̄ = jω D̄ + J¯ ∇ · B̄ = 0 ∇ · D̄ = ρv ∇ · J¯ = −jωρv 7.1 Conservation of charge derived from Ampere’s Law Already shown previously 5 7.2 Gauss’s Law for electricity derived from Ampere’s law & Conservation of Charge Apply div of curl is zero into Ampere’s Law ∇ × H̄ = jω D̄ + J¯ ( ) ( ) ∇ · ∇ × H̄ = ∇ · jω D̄ + J¯ −→ 0 = jω∇ · D̄ + ∇ · J¯ Apply Conservation of charge ∇ · J¯ = −jωρv 0 = jω∇ · D̄ − jωρv i.e. ∇ · D̄ = ρv 7.3 Gauss’s Law for magnetism derived from Faraday’s Law Apply div of curl is zero into Faraday’s Law ∇ × Ē = jω B̄ ( ) ( ) ∇ · ∇ × Ē = ∇ · jω B̄ −→ ( ) 0 = ∇ · jω B̄ = jω∇ · B̄ i.e. ∇ · B̄ = 0 6