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UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 18 Prof. Steven Errede LECTURE NOTES 18 MAGNETIC MONOPOLES – FUNDAMENTAL / POINTLIKE MAGNETIC CHARGES No fundamental, point-like isolated separate North or South magnetic poles – i.e. N or S magnetic charges have ever been conclusively / reproducibly observed. In principle, there is no physical law, or theory, that forbids their existence. So we may well ask, why did “nature” not “choose” to have magnetic monopoles in our universe – or, if so, why are they so extremely rare, given that many people (including myself) have looked for / tried to detect their existence… If magnetic monopoles / fundamental point magnetic charges did exist in nature, they would obey a Coulomb-type force law (just as electric charges do): ⎛μ ⎞g g Fm ( r ) = g mtest Bm ( r ) = ⎜ o ⎟ m 2 m rˆ ⎝ 4π ⎠ r test src ( ) Source point S ′ r ′ Where: g msrc = magnetic charge of source particle r = r − r′ g msrc g mtest = magnetic charge of test particle g mtest ẑ g m ≡ + g ( ≡ North pole ) r′ r g m ≡ − g ( ≡ South pole ) SI units of magnetic charge g = Ampere-meters (A-m) ϑ ŷ x̂ Field / observation point P ( r ) Units check: 2 ⎛ μo ⎞ g m Fm ( Newtons ) = ⎜ ⎟ 2 ⎝ 4π ⎠ r μo = 4π × 10−7 Newtons Ampere 2 (N A ) 2 2 2 ⎛ N ⎞ ( A − m) ⎛ N ⎞ A − m =N = Newton = N = ⎜ 2 ⎟ ⎜ 2 ⎟ 2 2 ⎝A ⎠ m m A ⎝ ⎠ 2 Then (the radial) B -field of a magnetic monopole is: src N ⎛μ ⎞g ) Bm ( r ) = ⎜ o ⎟ m2 rˆ (SI units = Tesla = A−m ⎝ 4π ⎠ r ⎛ N ⎞ A2 − m N N =⎜ 2 ⎟ = Units check: Tesla = A − m ⎝ A ⎠ m2 A−m 1 Tesla ≡ 1 1T=1 Newton Ampere − meter N A−m gm = Ampere-meters (A-m) © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. 1 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 18 Prof. Steven Errede The magnetic flux associated with a magnetic monopole with magnetic charge g msrc is: ( ) ⎛ μo ⎞ g msrc src 2 2 src ˆ B r nda = i ( ) ⎜ ⎟ 2 4π r = μo g m or: Φ m = μo g m (Webers, or Tesla-m ) ∫S m ⎝ 4π ⎠ r rˆ = r , and r = r because the source charge g msrc is located at the origin ϑ : Φm ≡ SI units of magnetic flux: Tesla-meters2 ( Φ m = B i A ) ⎛ N −m⎞ 1 (T–m2) = 1 ⎜ ⎟ = 1 Weber (Wb) ⎝ A ⎠ Units check: N N − m⎫ N N −m ⎧ im 2 = T − m 2 ⎨= ⎬= 2 i A −m = A ⎭ A A ⎩ A−m gm nˆ = rˆ ẑ r area element da ŷ ϑ x̂ We know that electric charge is quantized in units of e = 1.602x 10−19 Coulombs. Similarly we would expect magnetic charges to also be quantized (if they do indeed exist in nature). We know that magnetic flux Φ m is quantized - the quantum (i.e. the smallest unit) of magnetic flux is one flux quantum: −34 ⎛ h ⎞ 6.626 ×10 Joule − sec Φ om ≡ ⎜ ⎟ = = 4.136 × 10−15Webers −19 ⎝ e ⎠ 1.602 × 10 coulombs Where: h = Planck’s constant = 6.626 x 10−34 Joule-sec ⎛ N ⎞ 2 ⎛ N −m⎞ Units check: Φ om = Webers = Tesla − m 2 = ⎜ ⎟i m = ⎜ ⎟ ⎝ A ⎠ ⎝ A− m ⎠ ⎛ h ⎞ Joule − sec Joules Newton − meters ⎛ N − m ⎞ And: ⎜ ⎟ = = = =⎜ ⎟ Amp Amp ⎝ e ⎠ coulombs ⎝ A ⎠ − g mo + g mo ⎛h⎞ 1Φ om = ⎜ ⎟ = μo g mo ⎝e⎠ Then the smallest integer unit of quantized magnetic charge g mo is: Φ om = μo g mo or: g mo = 1 μo Φ om = 1 ⎛ h ⎞ 4.136 ×10−15Wb = 3.2914 × 10−9 Ampere-meters (A-m) ⎜ ⎟= −7 2 μo ⎝ e ⎠ 4π ×10 N A Now, it’s possible that magnetic monopoles could exist with integer multiples of this smallest / quantized amount of magnetic charge g mo , i.e. g mn = ng mo where n = integer = ±1, ±2, ±3, . . . . and: + g mo = N ( North Pole) and − g mo = S ( South Pole) . ⎛h⎞ ⎛h⎞ Then if Φ om = μo g mo = ⎜ ⎟ , then Φ nm = μo ng mo = μo g mn = n ⎜ ⎟ i.e. Φ nm = nΦ om . ⎝e⎠ ⎝e⎠ 2 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I ⎛ h ⎞ ⎛h⎞ Fall Semester, 2007 Lecture Notes 18 Prof. Steven Errede 1 μo g mn = n ⎜ ⎟ ⇒ eg mn = n ⎜ ⎟ but: = ε o c 2 where c = 3 x 108 m/s (speed of light) μ μ e ⎝ ⎠ o ⎝ o⎠ 2 nε o hc n ⎛ h ⎞ ⎛ h ⎞ 2 n ⎛ 4πε o c ⎞ n = 4πε o ⎜ Defining ≡ ⎜ ⎟ then g m = ⎟c = ⎜ ⎟ ec e 2 2 ⎝ e2 ⎠ ⎝ 2π ⎠ ⎝ 2π ⎠ And: ≡ 1 α em e2 1 n⎛ 1 ⎞ α ≡ = ec (A-m) and the fine-structure constant i.e. g mn = ⎜ ⎟ em 4πε o c 137.036... 2 ⎝ α em ⎠ 2 1 (dimensionless quantity) and numerically, 2α em = 137.036 68.5 Then: g mn = 68.5n ( ec ) A-m (n = ±1, ±2, ±3, . . . .) Thus, we see that the relative strength of magnetic monopole (e.g. North-South pole) attraction is huge in comparison to that associated with electric monopole (e.g. e+-e−) attraction: (n) m Fm( n ) F = Fe Fe ⎛ μo ⎞ g mn 2 2 ⎜ ⎟ r2 ⎛ g mn ⎞ 1 2 4π ⎠ ⎝ 2 2 2 = = ε o μo ⎜ ⎟ = 2 68.5 n c = ( 68.5n ) ⎛ 1 ⎞ e2 c ⎝ e ⎠ ⎜ ⎟ r2 ⎝ 4πε o ⎠ 4700n 2 (n = ±1, ±2, ±3, . . . .) Force of attraction between two Force of magnetic attraction between N-S monopoles (opposite) electric charges. ⎛h⎞ If Φ om = ⎜ ⎟ magnetic flux quantum (= 4.136 x 10-15 Wb) ⎝e⎠ Φ om = Then: Φ oE But: ∴ ∫ ∫ S S ˆ B inda ˆ E inda = ⎛ go ⎞ 1 ⎛ go ⎞ μo g mo = ε o μo ⎜ m ⎟ = 2 ⎜ m ⎟ e εo ⎝ e ⎠ c ⎝ e ⎠ ⎛ 1 ⎞ ⎛ g mo ⎞ g = 68.5ec or: ⎜ ⎟c ⎟ = 68.5c = ⎜ ⎝ e ⎠ ⎝ 2α em ⎠ o m where: α em = e2 4πε o c = 1 137.036... Φ om 1 ⎛ g mo ⎞ 1 ⎛ 1 ⎞ 1⎛ 1 ⎞ 1 = = = = 68.5 c ⎜ ⎟ ⎜ ⎟= ⎜ ⎟ 2 2 o c Φ E c ⎝ e ⎠ c ⎝ 2α em ⎠ c ⎝ 2α em ⎠ 2α em c © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. 3 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 18 Prof. Steven Errede We can rearrange this latter relation to obtain the electric flux quantum: ⎛ e2 ⎞ ⎛ h ⎞ ⎛ eh ⎞ h ⎛h⎞ Φ oE = 2α em cΦ om = 2α em c ⎜ ⎟ = 2 ⎜⎜ e− ⎟⎟ c ⎜ ⎟ = 2 ⎜ ⎟ where ≡ 4 e πε 2 π 4 πε c e ⎝ ⎠ o ⎠ ⎝ o ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ eh ⎟ = ⎛ 4π e h ⎞ = ⎛⎜ e ⎞⎟ Φ oE = 2 ⎜ 1Φ oE = e e+ εo ⎜⎜ 4πε h ⎟⎟ ⎜⎝ 4π ε o h ⎟⎠ ⎝ ε o ⎠ o 2π ⎠ ⎝ Numerically: 2 1.602 ×10−19 Coulombs −8 N − m Φ oE = ⎛⎜ e ⎞⎟ = Electric Flux Quantum = = 1.810 × 10 ⎝ εo ⎠ Coulombs 2 Coulomb 8.85 × 10−12 2 N −m −15 2 Φ om = ( μo g mo ) = h e = Magnetic Flux Quantum = 4.136 x 10 Wb (T–m ) ( ) ( ) Φ om 1 n = ec Ampere-meters where n = ±1, ± 2, ± 3 and g mn = o Φ E 2α em c 2α em Gauss’ Law: Φ om = μo g mo = Φ oE = 4 ⎛ Fm ⎞ h ⎛ ⎛ N ⎞ 2⎞ = 4.136 × 10−15Wb ⎜ = T − m 2 = ⎜ ⎟i Area = Bm i Area ⎟−m ⎟ =⎜ e ⎝ A−m ⎠ ⎝ ⎠ ⎝ gm ⎠ ⎛⎛ N ⎞ ⎞ ⎛F = 1.810 × 10−8 ⎜ ⎜ ⎟ − m 2 ⎟ = ⎜ E εo ⎝⎝ c ⎠ ⎠ ⎝ e e ⎞ ⎟i Area = E i Area ⎠ © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 18 Prof. Steven Errede The Dirac Quantization Condition In 1931, Paul Adrian Maurice Dirac showed (see P.A.M. Dirac, Proc. Roy. Soc., London, Ser. A133, 60, (1931)) that quantization of electric charge (i.e. why e is e could be explained if magnetic monopoles existed, because then: e ∗ μo g m = eg m = nh (SI units) ε oc2 Dirac Quantization Condition e = electric charge = 1.602 x 10−19 Coulombs g m = magnetic charge (SI units of Ampere-meters) (magnetic permeability of free space) μo = 4π ×10−7 Newtons Ampere 2 ⎛ Coulombs 2 ⎞ (electric permittivity of free space) ε o = 8.85 ×10 m ⎜ Newton − meter 2 ⎟ ⎝ ⎠ c = 1 ε o μo = speed of light (in free space/vacuum) = 3 × 108 meters / second −12 F n = integer (≠ 0) n = ±1, ±2, ±3, . . . h = Planck’s Constant = 6.626 x 10−34 Joule-sec = (N-m-s) Dirac originally obtained this relation by considering the motion of an electron circling a magnetic monopole of magnetic charge gm, with radial magnetic field ⎛μ ⎞g Bm ( r ) = ⎜ o ⎟ m2 rˆ (SI units: Tesla = N / A-m) ⎝ 4π ⎠ r Quantum mechanically, the wave function ψ e ( r ) of the electron circling the magnetic monopole (assumed to be infinitely heavy) must be single-valued in ϕ , i.e. ψ e (ϕ = 2π n ) = ψ e (ϕ = 0 ) in analogy e.g. to the Bohr model of the Hydrogen atom (e− bound to proton, p) In other words for the electron-monopole system, Dirac demanded: ψ e ( r ) → ψ e′ ( r ) = ψ e ( r ,θ ) eiϕ = ψ e ( r , θ ) ei( 2π n ) The quantum physics of the e−gm system gives 2π n = 2π ( eμo g m ) h where n = ±1, ±2, ±3, . . . eμ o g m or: eμo g m = nh ⇐ Dirac Quantization Condition (in SI units) h 1 e2 and: α em ≡ = fine structure constant (dimensionless) and Using: c 2 = ε o μo 4πε o c gm n n ec or: = c 68.5nc This relation can be rewritten as: g m = 2α em 2α m e Or: n= ≡ h 2π © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. 5 UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 18 Prof. Steven Errede Classically, the motion of an electron circling a magnetic monopole is shown below, at a height zc above it (at the origin). Classical path of The “Lorentz” force acting ẑ orbiting electron on the electron is: Fem ( re ) = −eve × Bm ( re ) v is e everywhere tangent to electron’s orbit ve e− ρe ve ⎛μ ⎞g = −eve × ⎜ o ⎟ m2 rˆ ⎝ 4π ⎠ re n.b. assume g m = North re ze gm magnet pole i.e. gm > 0 re θ ŷ ϑ ⎛ eμ g ⎞ v × rˆ = −⎜ o m ⎟ e 2 ⎝ 4π ⎠ re ⎛ nh ⎞ ve × rˆ = −⎜ ⎟ 2 ⎝ 4π ⎠ re 2 ⎛ nh ⎞ me ve × rˆ ⎛ c ⎞ = −⎜ ⎜ ⎟ 2 2 ⎟ ⎝ 4π ⎠ re ⎝ me c ⎠ x̂ 2 ⎛ nh ⎞ pe × re ⎛ c ⎞ = −⎜ ⎟ 3 ⎜ 2 ⎟ ⎝ 4π ⎠ re ⎝ me c ⎠ Imaginary sphere of radius re The angular momentum of the electron is Le = re × pe : 2 ⎛ nh ⎞ Le c m = + F r ∴ e ( e) ⎜ ⎟ 3 ⎝ 4π ⎠ me c2 re (n.b. the direction of Le is not constant here..) The magnetic force acting on the electron bends it around in the orbit as shown in above figure. We can also view this from a different perspective: the electron circling the magnetic monopole in this manner creates an electric current I e e C 2πρe 2π re sin θ = = τ orbit = = I= τ orbit ( C / ve ) ve ve ve Which in turn creates an orbital magnetic dipole moment: eve πρe2 ( − zˆ ) (SI units Amp-m2) me = Ia = I πρe2 ( − zˆ ) = ρe 2πρe me = 12 eve ρe ( − zˆ ) where ρe = re sin θ me = − 12 eve ρ e zˆ = − 12 eve re sin θ zˆ ze θ re Reminder: I = conventional current, which for a circulating e− electron, flows in the direction opposite to the electron’s orbital motion (see above figure). The orbital magnetic moment of the electron then interacts with the magnetic monopole. 6 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 18 Prof. Steven Errede Quantum mechanically, the e− behaves as a wave, not a point particle and thus the wavefunction ψ e ( re ) of the electron spreads out along its orbit as a periodic wave in such a way that an integer number of deBroglie wavelengths fit around the classical circumferential path, i.e. nλn = C = 2πρe with n = 1, 2, 3, . . .. Note also that the electron is not actually bound to the magnetic monopole – its orbit is stable. ∃ no binding energy between these two particles; given an initial electron velocity, e.g. ve = −ve xˆ , with the electron initially at height ze above the magnetic monopole, the radial magnetic field of the magnetic monopole will bend the electron’s path into orbit shown. Recall also that magnetic forces do no work… The Duality Transformation for Electromagnetism Because of the intimate connection between E and B at the microscopic / fundamental / elementary particle physics level, there (obviously) exists an intimate connection between E and B at the macroscopic level. A duality transformation is a simultaneous rotation in an abstract mathematical space by an angle ϕ of all electric and magnetic phenomena, which leaves all of the laws associated with the time!!! physics of electromagnetism unchanged – it’s a “knob” that allows us to rotate space Electromagnetism is invariant under a duality transformation. By carrying out a duality transformation, we simultaneously rotate all electric and magnetic phenomena by an angle ϕ in this abstract mathematical space, thus we can change electric fields magnetic fields, electric charges magnetic charges and we would never know the difference! Note: In carrying out duality transformations on all electric and magnetic phenomena, in order for all of these quantities to transform properly, each duality transform pair must have the same physical units. ⎛ 1 ⎞ e.g. E cB , ( ec gm ) , ⎜ ε o n.b. c2 = invariant under a duality transform 2 ⎟ μo c ⎠ ⎝ ( ) Other duality transform pairs are electric and magnetic currents and/or charge densities: Je Ke Ie 1 Jm c 1 Km c 1 Im c ρe σe λe Im ≡ 1 ρm c 1 σm c 1 λm c dQm dt © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. 7 UIUC Physics 435 EM Fields & Sources I Duality Transform for E cB′ Fall Semester, 2007 1 g m′ c E′ ϕ 1 gm c 1 gm c e′ ϕ ϕ E ′ = E cos ϕ + cB sin ϕ cB′ = cB cos θ − E sin ϕ Duality Transforms for: 1 Je Jm c 1 Ke Km c 1 Ie Im c ⎛ J e′ ⎞ ⎛ J e ⎞ ⎛ Jm ⎞ ⎜ ⎟ ⎜ ⎟ ⎟ 1⎜ ⎜ K e′ ⎟ = ⎜ K e ⎟ cos ϕ + c ⎜ K m ⎟ sin ϕ ⎜ I e′ ⎟ ⎜ I e ⎟ ⎜ Im ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ J m′ ⎞ ⎛ Jm ⎞ ⎛ Je ⎞ ⎟ 1⎜ ⎟ ⎜ ⎟ 1⎜ K m′ ⎟ = ⎜ K m ⎟ cos ϕ − ⎜ K e ⎟ sin ϕ ⎜ c⎜ ⎟ c ⎜ Im ⎟ ⎜ Ie ⎟ ⎝ I m′ ⎠ ⎝ ⎠ ⎝ ⎠ 8 Prof. Steven Errede Duality Transform for e and cB cB Lecture Notes 18 E ϕ e 1 e′ = e cos ϕ + g m sin ϕ c 1 1 g m′ = g m cos ϕ − e sin ϕ c c Duality Transforms for: 1 ρe ρm c 1 σe σm c 1 λe λm c ⎛ρ ′⎞ ⎛ρ ⎞ ⎛ ρm ⎞ ⎜ e ⎟ ⎜ e⎟ 1⎜ ⎟ ⎜ σ e′ ⎟ = ⎜ σ e ⎟ cos ϕ + ⎜ σ m ⎟ sin ϕ c⎜ ⎟ ⎜⎜ λ ′ ⎟⎟ ⎜ λ ⎟ e e ⎠ ⎝ ⎝ λm ⎠ ⎝ ⎠ ⎛ρ ′⎞ ⎛ρ ⎞ ⎛ ρe ⎞ m ⎟ 1⎜ m ⎟ 1⎜ ⎜ ⎟ ⎜ σ m′ ⎟ = ⎜ σ m ⎟ cos ϕ − ⎜ σ e ⎟ sin ϕ c⎜ c⎜ ⎟ ⎜λ ⎟ ⎜ λm′ ⎟⎟ ⎝ λm ⎠ ⎝ e⎠ ⎝ ⎠ © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 1 μo′ c 2 Duality Transform for ε o and Lecture Notes 18 1 μo c 2 1 μo c 2 Prof. Steven Errede ε o′ ϕ ϕ εo ε o′ = ε o cos ϕ + 1 1 1 sin ϕ and cos ϕ − ε o sin ϕ = 2 2 μo c μo′ c μo c 2 ϕ + sin ϕ We define the 2 × 2 duality transform rotation matrices as: R (ϕ ) ≡ ( cos − sin ϕ cos ϕ ) and its inverse ϕ R −1 (ϕ ) ≡ ( cos + sin ϕ Then: -sin ϕ cos ϕ ) Then RR −1 = R −1 R = 1 = ( 10 01 ) unit matrix ( ) = R (ϕ ) ( ) E′ c B′ or: E cB ⎛ ε′ ⎞ ⎛ ⎞ ⎜ 1 o ⎟ = R (ϕ ) ⎜ 1ε o ⎟ or: ⎜ μ ′ c2 ⎟ ⎜ μ c2 ⎟ ⎝ o ⎠ ⎝ o ⎠ ( ) = R (ϕ ) ( ) e′ g m′ c e gm c or: ( )=R E cB −1 (ϕ ) ( c EB′′ ) ⎛ ε ⎞ ⎛ ⎞ ⎜ 1 o ⎟ = R −1 (ϕ ) ⎜ 1ε o′ ⎟ ⎜ μ c2 ⎟ ⎜ μ ′ c2 ⎟ ⎝ o ⎠ ⎝ o ⎠ ( ) = R (ϕ ) ( ) e gm c −1 e′ g m′ c etc…. An Example of the Use / Application of the Duality Transform ( ) = R (ϕ ) ( ) E′ c B′ E cB Convert the solenoidal magnetic field associated with the motion of an electric charge ( ve ( into the solenoidal electric field associated with the motion of a magnetic charge vgm ⎛ μ ⎞ v × rˆ Start with Be ( r ) = ⎜ o ⎟ q e 2 ⎝ 4π ⎠ r ) c ! choose: ϕ = 90o then R ( 90o ) = ( −01 01 ) 1 ⎛ μ ⎞ v × rˆ cB = ⎜ o ⎟ cq 2 E ′ = cB ε o′ = r μo c 2 ⎝ 4π ⎠ v × rˆ 1 ⎛μ c⎞ E ′ = − ⎜ o ⎟ ( −e ) 2 cB′ = − E ′ = −ε o r μo′ c 2 ⎝ 4π ⎠ ⎛ μ c ⎞ ⎛ 1 ⎞ v × rˆ ⎛ μ ⎞ v × rˆ E ′ = − ⎜ o ⎟ ⎜ g m′ ⎟ 2 = − ⎜ o ⎟ g m′ 2 r ⎝ 4π ⎠ ⎝ c ⎠ r ⎝ 4π ⎠ 1 gm c Multiply both sides by c: e′ = Change cB → E ′ : 1 g m′ = −e c 1 Change −e → g m′ : c c) 1 1 ⎛ 1 ⎞ v × rˆ : E′ = − 2 ⎜ ⎟ g m′ 2 2 ε o′ c c ⎝ 4πε o ⎠ r All EM quantities - everything electromagnetic - now duality-transformed Now, drop primes everywhere (i.e. can’t tell the difference after ϕ -rotation!!) Change μo → Em ( r ) = − 1⎛ 1 ⎜ c 2 ⎝ 4πε o ⎞ vm × rˆ ⎛ μ ⎞ v × rˆ Be ( r ) = ⎜ o ⎟ q e 2 ⎟ gm 2 r ⎝ 4π ⎠ r ⎠ This is where / how the minus sign arises!! © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. 9