Electromagnetic Fields Review of Boundary Conditions Consider an electromagnetic field at the boundary between two materials with different properties. The tangent and the normal component of the fields must be examined separately, in order to understand the effects of the boundary. Medium 1 ε 1 ; µ1 G Hn1 Medium 2 ε2; µ2 G Hn2 G H1 boundary © Amanogawa, 2006 – Digital Maestro Series G H2 G H t1 G Ht 2 56 Electromagnetic Fields Tangential Magnetic Field Medium 1 ε 1 ; µ1 boundary G H n3 Medium 2 ε2; µ2 G H t1 G H n4 G Ht2 b a y . z x Ampère’s law for the boundary region in the figure can be written as ∂H y ∂H x G ∇×H⇒ − = J z + jω ε E z ∂x ∂y © Amanogawa, 2006 – Digital Maestro Series 57 Electromagnetic Fields In terms of finite differences approximation for the derivatives H n4 − H n3 H t1 − H t 2 − = J z + jω ε E z b a If one lets the boundary region shrink, with than b, a going to zero faster H n3 − H n4 H t 2 − H t1 = lim ( J z a + jωε E z a + a ) b a→ 0 for materials with finite conductivity ⇒ H t 2 − H t1 = 0 Tangential components are conserved for perfect conductors ⇒ H t 2 − H t1 = lim ( J z a) = Js a→ 0 © Amanogawa, 2006 – Digital Maestro Series (surface current) 58 Electromagnetic Fields For a general boundary geometry G G G nˆ × (H t1 − H t 2 ) = Js nˆ = unit vector normal to the surface In the case of a perfect conductor, the electromagnetic fields go immediately to zero inside the material, because the conductivity is infinite and attenuates instantly the fields. The surface current is confined to an infinitesimally thin “skin”, and it accounts for the discontinuity of the tangential magnetic field, which becomes immediately zero inside the perfect conductor. For a real medium, with finite conductivity, the fields can penetrate over a certain distance, and there is a current distributed on a thin, but not infinitesimal, skin layer. The tangential field components on the two sides of the interface are the same. Nonetheless, the perfect conductor is often a good approximation for a real metal. © Amanogawa, 2006 – Digital Maestro Series 59 Electromagnetic Fields Tangential Electric Field Medium 1 ε 1 ; µ1 boundary G E n3 Medium 2 ε2; µ2 G E t1 G Et2 b G E n4 a y . z x Faraday’s law for the same boundary region can be written as ∂ E y ∂E x G ∇×E⇒ − = jω µ H z ∂x ∂y © Amanogawa, 2006 – Digital Maestro Series 60 Electromagnetic Fields In terms of finite differences approximation for the derivatives E n4 − E n3 E t1 − E t 2 − = jωµ H z b a If one lets the boundary region shrink, with than b, a going to zero faster E n3 − E n4 E t 2 − E t1 = lim ( jωµ H z a + a ) b a→ 0 ⇒ E t 2 − E t1 = 0 Tangential components are conserved For a general boundary geometry G G nˆ × (E t1 − E t 2 ) = 0 © Amanogawa, 2006 – Digital Maestro Series 61 Electromagnetic Fields Normal components G Dn1 Medium 1 ε 1 ; µ1 ρs + boundary Medium 2 ε2; µ2 + + + + G Dn2 G Bn1 w + G Bn2 Area y . z x Consider a small box that encloses a certain area of the interface with ρ s = interface charge density © Amanogawa, 2006 – Digital Maestro Series 62 Electromagnetic Fields Integrate the divergence of the fields over the volume of the box: ∫∫∫ G G ∇ ⋅ D dr = Volume ∫∫∫ G ρ dr Volume w ∫∫ ⇓ theorem G G G D ⋅ n̂ ds = Flux of D out of the box ∫∫∫ G G ∇ ⋅ B dr = 0 Divergence Surface Volume ⇓ theorem G G G B ⋅ n̂ ds = Flux of B out of the box Divergence w ∫∫ Surface © Amanogawa, 2006 – Digital Maestro Series 63 Electromagnetic Fields If the thickness of the box tends to zero and the charge density is assumed to be uniform over the area, we have the following fluxes G D-Flux out of box = Area ⋅ (D1n − D2 n ) = = Total interface charge = Area ⋅ ρ s G B-Flux out of box = Area ⋅ (B1n − B2 n ) = 0 The resulting boundary conditions are D1n − D2 n = ρ s B1n − B2 n = 0 The discontinuity in the normal component of the displacement field D is equal to the density of surface charge. The normal components of the magnetic induction field B are continuous across the interface. © Amanogawa, 2006 – Digital Maestro Series 64 Electromagnetic Fields For isotropic and uniform values of ε and µ in the two media G G G G Dn1 − D n2 = ε1E n1 − ε 2 E n2 = ρ s G G G G Bn1 − Bn2 = µ1H n1 − µ 2H n2 = 0 Even when the interface charge is zero, the normal components of the electric field are discontinuous at the interface, if there is a change of dielectric constant . The normal components of the magnetic field have a similar discontinuity at the interface due to the change in the magnetic permeability. In many practical situations, the two media may have the same permeability as vacuum, µ0, and in such cases the normal component of the magnetic field is conserved across the interface. © Amanogawa, 2006 – Digital Maestro Series 65 Electromagnetic Fields SUMMARY If medium 2 is perfect conductor G H t1 G Ht2 G E t1 G Et2 G H n1 G H n2 G E n1 G E n2 ε 1 , µ1 ε 2 , µ2 ε 1 , µ1 ε 2 , µ2 ε 1 , µ1 ε 2 , µ2 ε 1 , µ1 ε 2 , µ2 © Amanogawa, 2006 – Digital Maestro Series G G H t1 = H t 2 G G nˆ × H t1 = J s G Ht2 = 0 G G E t1 = E t 2 G E t1 = 0 G Et2 = 0 G G µ1H n1 = µ 2 H n2 G G ε 1E n1 = ε 2 E n2 +ρ s G H n1 = 0 G H n2 = 0 G E n1 = ρ s ε 1 G E n2 = 0 66 Electromagnetic Fields Examples: An infinite current sheet generates a plane wave (free space on both sides) x Js -z +z y H G Js ( t ) = − Jso cos(ω t ) iˆx G Phasor J s = − Jso iˆx The E.M. field is transmitted on both sides of the infinitesimally thin sheet of current. © Amanogawa, 2006 – Digital Maestro Series 67 Electromagnetic Fields BOUNDARY CONDITIONS G G G nˆ × (H t1 − H t 2 ) = J s G G H t1 − H t 2 = Jso iˆx G G E t1 = E t 2 G G E t1 = η 0 H t1 G G Symmetry ⇒ H t1 = H t 2 Jso Jso H1 = H2 = − 2 2 © Amanogawa, 2006 – Digital Maestro Series 68 Electromagnetic Fields A semi-infinite perfect conductor medium in contact with free space has uniform surface current and generates a plane wave x Free Space Perfect Conductor Js -z +z y H G J s = − Jso cos(ω t ) iˆx The E.M. field is zero inside the perfect conductor. The wave is only transmitted into free space. © Amanogawa, 2006 – Digital Maestro Series 69 Electromagnetic Fields BOUNDARY CONDITIONS G G G nˆ × (H t1 − H t 2 ) = J s G G G H t1 − H t 2 = H t1 − 0 = Jso iˆx G Et2 = 0 G G Asymmetry ⇒ H t1 ≠ H t 2 H t1 = Jso Ht2 = 0 © Amanogawa, 2006 – Digital Maestro Series 70