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Unformatted text preview: Homework 8 solutions 5.20. (a) The force is ( ) ( ) 3 . V S d x da = ∇× × + × × ∫ ∫ F M B M n B Consider the i th component of the first integrand. This is ( ) ( ) ( ) ( ) ( ) . m m i k ijk jlm k kl im km il k k k i l l k i k k i k k k k i i k k i i M M M M B B B B x x x x B M M B B M x x x B M x ε ε δ δ δ δ ∂ ∂ ∂ ∂ ∇× × = = = ∂ ∂ ∂ ∂ ∂ ∂ ∂ = + ∂ ∂ ∂ ∂ = ⋅∇∇ ⋅ + ∂ M B B M B M The integrand of the second integral is ( ) ( ) ( ) . × × = ⋅ ⋅ M n B B M n B n M Let q be a fixed but arbitrary vector. Taking the scalar product of q and F , we get ( ) ( ) ( ) ( ) 3 . k i k i V S B q M d x d x ∂ ⋅ = ⋅ ⋅∇ ⋅∇ ⋅ + + ⋅ ⋅ ⋅ ∂ ∫ ∫ q F q B M q B M B M q q M B S Converting the surface integral to a volume integral by using the divergence theorem, we find ( ) ( ) ( ) ( ) ( )( ) 3 3 . k i k i V k i k i V B q M d x x B q M d x x ∂ ⋅ = ⋅ ⋅∇ +∇⋅ ⋅ ∂ ∂ = ⋅ ⋅∇ ⋅ ∇⋅  ⋅∇ ⋅ + ∂ ∫ ∫ q F q B M q M B q B M q M B B q M Consider ( ) ( )( ) ( )( ) 0, j i i j j i j i M M q B q B x x ∂ ∂ ⋅ ⋅∇ ⋅ ∇⋅ ⋅∇ ⋅ = = ∂ ∂ q B M q M B B q M where we have used 0. ∇⋅ = B Hence 3 . k i k i V B q M d x x ∂ ⋅ = ∂ ∫ q F Using 0, ∇× = B this becomes ( ) 3 3 3 . k i i i k k i k i i i k i k i i k k k k V V S V M q B B M M q M d x q B d x q M B dS q B d x x x x x ∂ ∂ ∂ ∂ ⋅ = = = ∂ ∂ ∂ ∂ ∫ ∫ ∫ ∫ q F Since q is arbitrary, we finally get the desired result ( ) ( ) 3 . S V dS d x = ⋅ ∇⋅ ∫ ∫ F M n B M B (b) Because the magnetization is uniform inside the sphere, there is only the surface contribution to the force. In terms of spherical polar coordinates and Cartesian unit vectors, ( ) ( ) ( ) sin cos sin sin cos , sin cos sin sin cos , 1 sin sin , 1 sin cos . x y M B B R B B R θ φ θ φ θ θ φ θ φ θ β θ φ β θ φ = + + = + + = + = + M i j k n i j k Hence ( ) ( )( ) ( ) ( )( ) 2 2 0 0 2 2 0 0 sin cos sin cos sin sin sin sin cos cos 1 sin sin sin , sin cos sin cos sin sin sin sin cos cos 1 sin cos sin . x x S y y S F B dS MR B R d d F B dS MR B R d d π π π π θ φ θ φ θ φ θ φ θ θ β θ φ θ θ φ θ φ θ φ θ φ θ φ θ θ β θ φ θ θ φ = ⋅ = + + + = ⋅ = + + + ∫ ∫ ∫ ∫ ∫ ∫ M n M n Noting that the integrals of the terms linear in sin ϕ , cos ϕ , sin 2 ϕ are zero, this simplifies to 2 2 3 2 0 0 2 2 3 2 0 0 cos cos sin sin sin sin sin . cos cos sin sin cos sin cos . x y F MR B R d d F MR B R d d π π π π θ θ θ β θ φ θ φ θ φ θ θ θ β θ φ θ φ θ φ = + = + ∫ ∫ ∫ ∫ The first terms in the integrands also contribute zero to the integrals. Hence 2 3 3 2 2 3 3 2 sin sin sin sin ....
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 Fall '11
 MacDonald
 Force, Work, Flux, Cos, Energy density, charge density, ρ cosψ

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