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Unformatted text preview: 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 surfacecharge distribution σ . The sphere is rotated about a diameter with constant angular velocity ω . Find the vector potential and magneticflux 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 ~ω , the velocity at any point ~x on the sphere is given by ~v = ~ω × ~x . This allows us to write the current density as ~ J = ρ~v = σ ~ω × ~xδ (  ~x   a ) In Coulomb gauge, the vector potential is then given by ~ A ( ~x ) = μ 4 π Z ~ J ( ~x )  ~x ~x  d 3 x = μ σa 3 4 π ~ω × Z ˆ x  ~x ~x  d Ω (1) where  ~x  = 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 integral is performed, the only direction it can point in is given by ˆ x . This allows us to write Z ˆ x  ~x ~x  d Ω = f ( r )ˆ 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  d Ω = X l r l < r l +1 > Z P 1 (cos γ ) P l (cos γ ) d Ω where cos γ = ˆ x · ˆ x 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  ~x ~x  d Ω = 4 π 3 r < r 2 > ˆ x Inserting this into (1) gives ~ A ( ~x ) = μ σa 3 3 r r < r 2 > ~ω × ~x More explicitly, we have ~ A = μ σa 3 ~ω × ~x r < a μ σa 4 3 r 3 ~ω × ~x r > a The magnetic induction is now given by ~ B in = ~ ∇ × ~ A in = μ σa 3 ~ ∇ × ( ~ω × ~x ) = 2 μ σa 3 ~ω r < a and ~ B out = ~ ∇ × ~ A out = μ σa 4 3 ~ ∇ × ~ω × ~x r 3 = μ σa 4 3 3ˆ x ( ω · ˆ x ) ~ω 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 ~ω 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 magneticflux density ~ B 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 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 along the + x axis and write Φ M ( ρ,φ ) = ( H ρ + α ρ )cos φ, ρ > b ( βρ + γ ρ )cos φ, a < ρ < b...
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This note was uploaded on 11/21/2011 for the course PHY 505 taught by Professor Stephens,p during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Stephens,P
 Charge, Magnetism, Work

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