MagneticMoment - Magnetic Moment. Ian R. Gatland,...

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Unformatted text preview: Magnetic Moment. Ian R. Gatland, Georgia Institute of Technology Physics 3122 Notes, July 22, 2010 Preliminaries. Consider a vector field, J ( r ) , that is localized (zero on and outside a surface S surrounding a volume Γ) and divergence free ( ∇⋅ J = 0 ) together with two scalar fields, f ( r ) and g( r ) . Then € ⋅ ∫Γ [ fJ ⋅ ∇g +gJ €∇f ]dτ = ∫ [ fJ ⋅ ∇g + ∇⋅ ( fgJ ) − f∇⋅ (gJ )]dτ € = ∫ ∇⋅ ( fgJ )dτ + ∫ [ fJ ⋅ ∇g − fg∇⋅ J − fJ ⋅ ∇g]dτ = ∫ fgJ ⋅ da − ∫ fg∇⋅ J dτ = 0 Γ Γ Γ S (1) Γ using the localization and divergence conditions. € Case 1: Equation (1) with f = 1 and g = x provides ˆ ∫Γ [J ⋅ ∇x +xJ ⋅ ∇1]dτ = ∫Γ [J ⋅ x +0]dτ = ∫ Γ J x dτ = 0 (2) and similar results with y and z in place of x. € Case 2: Equation (1) with f = x and g = x provides ∫Γ [ xJ ⋅ ∇x +xJ ⋅ ∇x ]dτ = ∫ 2 xJ dτ = 0 x Γ (3) and similar results with y and z in place of x. € Case 3: Equation (1) with f = x and g = y provides ∫Γ [ xJ ⋅ ∇y +yJ ⋅ ∇x ]dτ = ∫Γ [ xJy +yJx ]dτ = 0 (4) From equation (4) it follows that ∫ € Γ xJ y dτ = − ∫Γ yJ x dτ Now, from equation (5), ∫€xJ dτ = ∫ [ xJ Γ y 1 2 Γ and similar results with x, y, and z cycled. € y −yJx ]dτ = 1 2 (5) ∫ [r × J ] dτ Γ z (6) Summary. ∫ J dτ = 0 ∫ xJ dτ = ∫ yJ dτ = ∫ and zJ z dτ = 0 J ∫Γ y€z dτ = − ∫Γ zJy dτ = 12 ∫Γ [r × J ]x dτ ∫Γ zJx dτ = − ∫Γ xJz dτ = 12 ∫Γ [r × J ]y dτ Γ (7) Γ € x y Γ Γ € € ∫ xJ y dτ = − ∫Γ yJ x dτ = Γ 1 2 ∫ [r × J ] dτ z Γ (8) (9) (10) (11) Magnetic Moment. € The vector potential, A( r ) , is given in terms of the current density, J ( r ) , as µ0 J ( r+ ) A( r ) = ∫ dτ + 4 π r − r+ € € The denominator in equation (12) may be expanded as −1 r − r+ = [ r 2 − 2 r ⋅ r+ + r+ 2 ]−1 / 2 = r −1[1 − 2 r ⋅ r+ / r 2 + r+ 2 / r 2 ]−1 / 2 € = r −1[1 + r ⋅ r+ / r 2 + ] = 1 / r + r ⋅ r+ / r 3 + … Combining equations (12) and (13) yields µ µ A( r ) = 0 ∫ J ( r+ )dτ + + 0 3 € 4 πr 4 πr ∫ (r ⋅ r+ )J (r+ )dτ + + (12) (13) (14) Provided that the current density is localized and divergence free, the first term in equation (14) is zero according to equation (7). Also, from equations (8), (9), (10), and (11), the x € component of the integral in the second term of equation (14) is ∫ (r ⋅ r+ )Jx (r+ )dτ+ = x ∫ x+ Jx (r+ )dτ+ +y ∫ y+ Jx (r+ )dτ+ +z ∫ z+ Jx (r+ )dτ+ = 0 − y 1 ∫ [ r+ × J ( r+ )]z dτ+ + z 1 ∫ [ r+ × J ( r+ )]y dτ+ (15) 2 2 = − r × 1 ∫ [ r+ × J ( r+ )]z dτ+ 2 [ x and similar results with y and z in place of x. € Defining the magnetic moment as m≡1 2 € ∫ r × J (r )dτ (16) and combining equations (7), (14), and (15) allows the vector potential to be written as ˆ µ0 m × r µ0 m × r (17) A( r ) = = 3 2 4π r 4π r when higher order terms in 1/r are ignored. € For a current loop with current I, equation (16) becomes m = 1 I ∫ r × d 2 and for a plane loop it is easy to show that equation (18) reduces to ˆ m = IAn € ˆ where A is the area of the loop and n is the normal to the loop. € € (18) (19) ...
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This note was uploaded on 01/22/2012 for the course PHYS 3122 taught by Professor Staff during the Fall '08 term at Georgia Institute of Technology.

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