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F07_hw08a - Physics 505 Homework Assignment#8 Solutions...

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Physics 505 Fall 2007 Homework Assignment #8 — Solutions Textbook problems: Ch. 5: 5.13, 5.14, 5.15, 5.16 5.13 A sphere of radius a carries a uniform surface-charge distribution σ . The sphere is rotated about a diameter with constant angular velocity ω . Find the vector potential and magnetic-flux density both inside and outside the sphere. The charge density for a uniformly charged sphere of radius a is simply ρ ( ~x ) = σδ ( | ~x | - a ) Since the sphere is rotating with constant angular velocity , the velocity at any point ~x on the sphere is given by ~v = × ~x . This allows us to write the current density as ~ J = ρ~v = σ ~ω × ~x δ ( | ~x | - a ) In Coulomb gauge, the vector potential is then given by ~ A ( ~x ) = μ 0 4 π Z ~ J ( ~x 0 ) | ~x - ~x 0 | d 3 x 0 = μ 0 σa 3 4 π × Z ˆ x 0 | ~x - ~x 0 | d Ω 0 (1) where | ~x 0 | = a . There are several ways to perform the angular integral. One quick method is to realize that the integral is a vector quantity. Then, by symmetry, once the d Ω 0 integral is performed, the only direction it can point in is given by ˆ x . This allows us to write Z ˆ x 0 | ~x - ~x 0 | d Ω 0 = f ( r x where f ( r ) is a function to be determined. In fact, by dotting both sides with ˆ x , we see that f = Z cos γ | ~x - ~x 0 | d Ω 0 = X l r l < r l +1 > Z P 1 (cos γ ) P l (cos γ ) d Ω 0 where cos γ = ˆ x · ˆ x 0 and where r < = min( r, a ) , r > = max( r, a ) Orthogonality of the Legendre polynomials then selects out l = 1, so that f = (4 π/ 3)( r < /r 2 > ) or Z ˆ x 0 | ~x - ~x 0 | d Ω 0 = 4 π 3 r < r 2 > ˆ x
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Inserting this into (1) gives ~ A ( ~x ) = μ 0 σa 3 3 r r < r 2 > × ~x More explicitly, we have ~ A = μ 0 σa 3 × ~x r < a μ 0 σa 4 3 r 3 × ~x r > a The magnetic induction is now given by ~ B in = ~ ∇ × ~ A in = μ 0 σa 3 ~ ∇ × ( × ~x ) = 2 μ 0 σa 3 r < a and ~ B out = ~ ∇ × ~ A out = μ 0 σa 4 3 ~ ∇ × × ~x r 3 = μ 0 σa 4 3 x ( ω · ˆ x ) - r 3 r > a The magnetic induction inside the sphere is uniform and parallel to the axis of rotation, while the magnetic induction outside is a dipole pattern with magnetic moment ~m = 4 π 3 σa 4 5.14 A long, hollow, right circular cylinder of inner (outer) radius a ( b ), and of relative permeability μ r , is placed in a region of initially uniform magnetic-flux density ~ B 0 at right angles to the field. Find the flux density at all points in space, and sketch the logarithm of the ratio of the magnitudes of ~ B on the cylinder axis to ~ B 0 as a function of log 10 μ r for a 2 /b 2 = 0 . 5, 0 . 1. Neglect end effects. For a long cylinder (neglecting end effects) we may think of this as a two- dimensional problem. Since there are no current sources, we use a magnetic scalar potential Φ M which must be harmonic in two dimensions. Since ~ H = - ~ Φ M , we orient the uniform magnetic field H 0 along the + x axis and write Φ M ( ρ, φ ) = ( - H 0 ρ + α ρ ) cos φ, ρ > b ( βρ + γ ρ ) cos φ, a < ρ < b δρ cos φ, ρ < a (2) Of course, the general harmonic expansion would be of the form ( A m ρ m + B m ρ - m ) cos + ( C m ρ m + D m ρ - m ) sin . However here we have already used the shortcut that all matching conditions for m 6 = 1 lead to homogeneous equa- tions admitting only a trivial (zero) solution.
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The magnetostatic boundary conditions demand that H φ and B ρ
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