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# hw09a - Physics 505 Homework Assignment#9 Solutions...

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Physics 505 Fall 2005 Homework Assignment #9 — Solutions Textbook problems: Ch. 5: 5.20, 5.22, 5.26 Ch. 6: 6.1 5.20 a ) Starting from the force equation (5.12) and the fact that a magnetization M inside a volume V bounded by a surface S is equivalent to a volume current density J m = ( ∇ × M ) and a surface current density ( M × n ), show tha in the absence of macroscopic conduction currents the total magnetic force on the body can be written F = - V ( ∇ · M ) B e d 3 x + S ( M · n ) B e da where B e is the applied magnetic induction (not including that of the body in question). The force is now expressed in terms of the effective charge densities ρ M and σ M . If the distribution of magnetization is now discontinuous, the surface can be at infinity and the force given by just the volume integral. Given volume and surface current densities, we may write the force as F = V J × B d 3 x + S K × B da Using J m = ∇ × M and K m = M × ˆ n , we have F = V ( ∇ × M ) × B e d 3 x + S ( M × ˆ n ) × B e da = - V B e × ( ∇ × M ) d 3 x + S ( M × ˆ n ) × B e da All that remains now is an exercise in vector calculus. We begin with the identity ( a · b ) = ( a · ∇ ) b + ( b · ∇ ) a + a × ( ∇ × b ) + b × ( ∇ × a ) with a = M and b = B e as well as the BAC–CAB rule on the surface term to write F = V [ -∇ ( M · B e ) + ( M · ∇ ) B e + ( B e · ∇ ) M + M × ( ∇ × B e )] d 3 x + S [( B e · M n - ( B e · ˆ n ) M ] da = V [( M · ∇ ) B e + ( B e · ∇ ) M ] d 3 x - S ( B e · ˆ n ) M da (1)

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To obtain the second line, we made use of the fact that we are in a source free region for the applied B e (so that ∇ × B e = 0) and we also integrated the total divergence to cancel one of the surface terms. To proceed, we note that the volume terms may be integrated by parts. In particular V ( a · ∇ ) b i d 3 x = V [ ∇ · ( ab i ) - ( ∇ · a ) b i ] d 3 x = - V ( ∇ · a ) b i d 3 x + S n · a ) b i da or in a full vector notation V ( a · ∇ ) b d 3 x = - V ( ∇ · a ) b d 3 x + S n · a ) b da Using this partial integration on (1) results in F = - V [( ∇ · M ) B e + ( ∇ · B e ) M ] d 3 x + S n · M ) B e da = - V ( ∇ · M ) B e d 3 x + S n · M ) B e da (2) where we also used ∇ · B e = 0. b ) A sphere of radius R with uniform magnetization has its center at the origin of coordinates and its direction of magnetization making spherical angles θ 0 , φ 0 . If the external magnetic field is the same as in Problem 5.11, use the expression of part a ) to evaluate the components of the force acting on the sphere. Since the magnetization is uniform (ie constant), the volume gradient term van- ishes, and we are left with a surface integral. Explicitly, the magnetization vector may be written M = M 0 (sin θ 0 cos φ 0 , sin θ 0 sin φ 0 , cos θ 0 ) while the magnetic induction vector is B e = B 0 (1 + βy, 1 + βx, 0) = B 0 (1 + βr sin θ sin φ, 1 + βr sin θ cos φ, 0) We have used spherical coordinates where the normal vector is ˆ n = (sin θ cos φ, sin θ sin φ, cos θ ) Then F = S ( M · ˆ n ) B e da = R 2 M 0 B 0 d Ω [cos θ cos θ 0 + sin θ sin θ 0 cos( φ - φ 0 )] × (1 + βR sin θ sin φ, 1 + βR sin θ cos φ, 0)
It is straightforward to perform the φ integral. The result is F = 2 πR 2 M 0 B 0 1 - 1 d cos θ (cos θ cos θ 0 + 1 2 βR sin 2 θ sin θ 0 sin φ 0 , cos θ

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hw09a - Physics 505 Homework Assignment#9 Solutions...

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