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Unformatted text preview: 1 PHYS809 Class 24 Notes The magnetic vector potential Since 0, ∇⋅ = B (24.1) there is a vector potential A such that . = ∇× B A (24.2) Note that A is not unique since adding the gradient of a scalar to the vector potential leaves B unchanged. The transformation ψ → +∇ A A (24.3) is called a gauge transformation . The freedom allowed by gauge transformation is exploited to have ∇⋅ A take any convenient form. In magnetostatics, a convenient choice is the Coulomb gauge 0. ∇⋅ = A (24.4) Since , μ ∇× = B J (24.5) the potential is the solution of ( ) . μ ∇× ∇× = A J (24.6) Using the identity ( ) ( ) 2 , ∇× ∇× = ∇ ∇∇ A A A i (24.7) we have ( ) 2 . μ ∇ ∇∇ = A A J i (24.8) For the Coulomb gauge this simplifies to the Poisson equation 2 . μ ∇ =  A J (24.9) By comparison with electrostatics, the solution to this equation for a localized current distribution is ( ) ( ) 3 . 4 allspace d x μ π ′ ′ = ′ ∫ J x A x x x (24.10) For a current loop ( ) . 4 loop I d μ π ′ = ′ ∫ x A x x x (24.11) 2 Vector potential for a circular loop Consider a circular loop of radius a in the x – y plane centered on the origin. First note that, because of the pattern of current flow, in spherical polar coordinates, only A φ is nonzero. Since cos sin , a a φ φ ′ ′ ′ = + x i j (24.12) the line element is ( ) sin cos . d a a d φ φ φ ′ ′ ′ ′ =  + x i j (24.13) From equation (24.11), the vector potential is From equation (24....
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 Fall '11
 MacDonald
 Magnetism, Magnetic Field, loop, µ, Vector potential

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