Space Physics (I) [AP-344] Lectue by Ling-Hsiao Lyu Oct. 2 Lectue. Dipole Magnetic Field and Equations of Magnetic Field Lines.. Dipole Magnetic Field Since = we can define = A (.) whee A is called the vecto potential. We use the bold face font to denote vecto. Fo static magnetic field, we have = µ J (.2) Substituting Eq. (.) into Eq. (.2) to eliminate, it yields ( A) = 2 A + ( A) = µ J (.3) We choose the Coulomb gauge, A =, the equation (.3) can be ewitten as 2 A = µ J (.4) Note that: If we choose the Coulomb gauge: A =, the scala potential will contain no electomagnetic component. If we choose the Loentz gauge: φ(x,t) + A(x,t) =, the scala potential will contain c 2 t an electomagnetic component. Eq. (.4) is simila to the Poisson equation of the electostatic potential 2 φ = ρ c ε (.5) Geneal solution of Eq. (.5) can be witten as φ() = Special Case: ρ c (') d' (.6) ε ' The scala potential ceate by a point chage q is φ = q ε Likewise, the geneal solution of Eq. (.4) can be witten as A() = µ J(') d' (.7) '
Space Physics (I) [AP-344] Lectue by Ling-Hsiao Lyu Oct. 2 Execise.. Let f (x) = ( + x) α. Fo < x <, please detemine the appoximate polynomial expession of f (x). That is, f (x) a + a x + a 2 x 2 + a 3 x 3 whee a tem with magnitude of the ode of O(x 4 ) has been ignoed. (a) Wite down the geneal fom of a, a, a 2, a 3 fo a given α (b) Fo α = / 2, wite down the appoximate polynomial expession of f (x). (c) Fo α = / 2, wite down the appoximate polynomial expession of f (x). (d) Fo α =, wite down the appoximate polynomial expession of f (x). (e) Fo α = 2, wite down the appoximate polynomial expession of f (x). Answes of Execise. (a) a = a = α a 2 = α(α ) / 2! a 3 = α(α )(α 2) / 3! (b) f (x) = + x + 2 x + 8 x2 + 6 x3 (c) f (x) = + x 2 x + 3 8 x2 5 6 x3 (d) f (x) = + x x + x2 x 3 (e) f (x) = (+ x) 2 2x + 3x2 4x 3 Figue.. A coodinate system fo the study of the field geneate by a ing cuent J(') = ˆφ ' J δ( ' )δ(θ ' π 2 ) = ( sinφ ' ˆx'+ cosφ ' ŷ')j δ( ' )δ(θ ' π 2 ) 2
Space Physics (I) [AP-344] Lectue by Ling-Hsiao Lyu Oct. 2 Let us conside a coodinate system as illustated in Figue., whee = cosθẑ + cosφ ˆx + sinφŷ = cosθẑ + ˆx' ' = cosφ ' ˆx'+ sinφ ' ŷ' ˆφ ' = sinφ ' ˆx'+ cosφ ' ŷ' ˆφ = ŷ' Given a ing cuent J(') = ˆφ ' J δ( ' )δ(θ ' π 2 ) = ( sinφ ' ˆx'+ cosφ ' ŷ')j δ( ' )δ(θ ' π 2 ) The Geneal solution of the vecto potential A is (Jackson, section 5.5) A(x) = A φ ˆφ = Aφ ŷ' = µ J(') d' ' + Δ /2 = d ' Δ /2 (π /2)+ Δθ /2 (π /2) Δθ /2 = µ J (Δ)( Δθ) 'dθ ' µ ' 'dφ ' J δ( ' )δ[θ ' (π / 2)] ˆφ ' 2 cos 2 θ + ( cosφ ') 2 + ( sinφ ') 2 sinφ ' ˆx'+ cosφ ' ŷ' dφ ' 2 + 2 2 cosφ ' Let I = J (Δ)( Δθ), α = 2 + 2, β = 2, Eq. (.8) can be witten as (.8) A φ ŷ' = µ I = µ I ( ˆx' sinφ ' dφ '+ ŷ' ˆx'[ φ '=π d cosφ ' φ '= d cosφ ' + φ '= ] φ '=π cosφ ' dφ ') + µ I ŷ' cosφ ' α [ 2 ( β cosφ ') + O( β 2 α α 2 cos2 φ ')]dφ ' = µ I ˆx'[ dx dx + α βx ] α βx + µ I ŷ' cosφ ' α [+ β 2 α cosφ ' + O( β 2 α 2 cos2 φ ')]dφ ' = µ I ŷ' cosφ ' α [ + β 2 α cosφ ' + O( β 2 α 2 cos2 φ ')]dφ ' (.9) Fo >>, i.e., α >> β, we can ignoe the small second-ode tem O( β 2 (.9). It yields α 2 cos2 φ ') in Eq. 3
Space Physics (I) [AP-344] Lectue by Ling-Hsiao Lyu Oct. 2 A φ ŷ' µ I ŷ' cosφ ' α [ + β 2 α cosφ ' ]dφ ' = ŷ' µ I [ cosφ ' dφ '+ β cos 2 φ ' 2α dφ '] 3/2 = ŷ' µ I [ + β 2α π ] 3/2 ŷ' µ I [2 π ] 2 3 = ŷ' µ (I π 2 ) 2 (.) Using the definition of magnetic moment M = I π 2, Eq. (.) can be witten as A φ = µ M 2 (.) Execise.2. Please detemine the dipole magnetic field fom the vecto potential given in Eq. (.) Solution of Execise.2: Since = A, it yields = A = 2 ˆ ˆθ ˆφ θ φ A A θ A φ = 2 ˆ ˆθ ˆφ θ φ µ M 2 = µ M = µ M = µ M = µ M 2 ˆ ˆθ ˆφ θ φ sin 2 θ 2 [ ˆ θ θ (sin2 ) ˆθ θ (sin2 )] cosθ [ ˆ(2 ) + ˆθ( sin2 θ )] 2 2 [ ˆ(2cosθ) + ˆθ()] 3 (.2) 4
Space Physics (I) [AP-344] Lectue by Ling-Hsiao Lyu Oct. 2 Fo Eath dipole magnetic field = µ ( M E ) It can be ewitten as = 3 [ ˆ(2cosθ) + ˆθ()] (.3) ( / R E ) 3 [ ˆ(2cosθ) + ˆθ()] (.4) whee = ( = R E,θ = π / 2).35G = 35γ = 35nT is the magnitude of magnetic field on the Eath' suface at the magnetic equato. The θ is called the co-latitude. The latitude is λ = π 2 θ..2. Diffeential Equations of the Magnetic Field Line Let us conside a segment ds along the magnetic field line, whee ds = ˆd + ˆθdθ + ˆφ dφ = ˆxdx + ŷdy + ẑdz It yields ds = ds = d 2 + 2 dθ 2 + 2 sin 2 θdφ 2 = dx 2 + dy 2 + dz 2 Let = ˆ + ˆθ θ + ˆφ φ = ˆx x + ŷ y + ẑ z Since ds, it yields d = dθ θ = dφ = ds φ (.5) Eq. (.5) can be ewitten in the following system odinay diffeential equations d ds = dθ ds = θ dφ = φ ds Likewise, ds yields dx = dy = dz = ds x y z (.6) (.7) Eq. (.7) can be ewitten in the following system odinay diffeential equations 5
Space Physics (I) [AP-344] Lectue by Ling-Hsiao Lyu Oct. 2 dx ds = x dy ds = y dz ds = z Eqs. (.6) and (.8) can be solved by the 2nd ode o the fouth ode Runge-Kutta method. (.8).3. Dipole Magnetic Field Line The Eath dipole magnetic field is given in Eq. (.4). Fom Eq. (.5), the dipole magnetic field line should satisfy the following diffeential equation d 2cosθ = dθ Solving Eq. (.9), it yields d = 2cosθdθ = 2 d Integating along the field line, it yields (θ ) (.9) dln (θ) = 2 dln (.2) (θ =π /2) sin(θ =π /2) Let (θ = π / 2) = eq = LR E, Eq. (.2) yields (θ) = eq sin 2 θ = LR E sin 2 θ (.2) Execise.3. A dipole magnetic field line with a given L value will intesect with the Eath's suface at latitude λ L. Fo L =, 2, 3, 4, 5, and 6, find the coesponding latitude λ L. Execise.4. Plot the dipole magnetic field line with L =, 2, 3, 4, 5, and 6. Execise.5. Plot the dipole magnetic field lines, which intesect with the Eath's suface at λ L = 8º, 7º, 6º, 5º, 4º, 3º, 2º, and º. Estimate the coesponding L values. 6