hw08a - Physics 505 Homework Assignment#8 Solutions...

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Physics 505 Fall 2005 Homework Assignment #8 — Solutions Textbook problems: Ch. 5: 5.10, 5.14, 5.17, 5.19 5.10 A circular current loop of radius a carrying a current I lies in the x - y plane with its center at the origin. a ) Show that the only nonvanishing component of the vector potential is A φ ( ρ, z ) = μ 0 Ia π 0 dk cos kz I 1 ( < ) K 1 ( > ) where ρ < ( ρ > ) is the smaller (larger) of a and ρ . The vector potential may be obtained by A ( x ) = μ 0 4 π J ( x ) | x - x | d 3 x where (for a circular current loop) J ( x ) = ( z ) δ ( ρ - a ) ˆ φ in cylindrical coordinates. Note that to obtain the cylindrical components of A ( x ) we have to be careful to convert the basis vector ˆ φ at the point x to components at x . (This is because the basic vectors depend on position.) A bit of geometry gives ˆ φ = ˆ ρ sin( φ - φ ) + ˆ φ cos( φ - φ ) [Or, alternatively, we may choose the point x to lie at φ = 0, so that ˆ φ = ˆ y and ˆ ρ = ˆ x . Then it is straightforward to see that ˆ φ = ˆ y cos φ - ˆ x sin φ = ˆ φ cos φ - ˆ ρ sin φ . Using symmetry, we can see that only the ˆ φ component of A is nonvanishing.] The integral expression for the vector potential is then A ( x ) = μ 0 I 4 π δ ( z ) δ ( ρ - a )[ˆ ρ sin( φ - φ ) + ˆ φ cos( φ - φ )] | x - x | ρ dρ dφ dz = μ 0 Ia 4 π 2 π 0 ˆ ρ sin( φ - φ ) + ˆ φ cos( φ - φ ) | x - x | (1) where the integrand in the second line is to be evaluated at z = 0 and ρ = a . We now use the cylindrical Green’s function expressed as 1 | x - x | = 4 π 0 dk cos[ k ( z - z )] 1 2 I 0 ( < ) K 0 ( > ) + m =1 cos[ m ( φ - φ )] I m ( < ) K m ( > )
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Note that the integral over φ picks out the m = 1 term in the sum. Furthermore, the ˆ ρ component drops out because sin( φ - φ ) is orthogonal to cos( φ - φ ), a result that could have been obtained by symmetry. We end up with A ( x ) = μ 0 Ia 4 π 4 π π ˆ φ 0 dk cos( kz ) I 1 ( < ) K 1 ( > ) = μ 0 Ia π ˆ φ 0 dk cos( kz ) I 1 ( < ) K 1 ( > ) b ) Show that an alternative expression for A φ is A φ ( ρ, z ) = μ 0 Ia 2 0 dk e - k | z | J 1 ( ka ) J 1 ( ) To obtain the alternative expression, we use the alternate form of the Greens’ function 1 | x - x | = 2 0 dk e - k ( z > - z < ) 1 2 J 0 ( ) J 0 ( ) + m =1 cos[ m ( φ - φ )] J m ( ) J m ( ) Since, for z = 0, we have z > - z < = | z | , it is clear that when we stick this into (1) we end up with A ( x ) = μ 0 Ia 2 ˆ φ 0 dk e - k | z | J 1 ( ) J 1 ( ka ) c ) Write down integral expressions for the components of magnetic induction, using the expressions of parts a ) and b ). Evaluate explicitly the components of B on the z axis by performing the necessary integrations. Since B = ∇ × A and the only non-vanishing component of A is A φ , we end up with B ρ = - z A φ , B z = 1 ρ ρ ( ρA φ ) The z derivative is straightforward. For the ρ derivative, on the other hand, we may use the Bessel equation identity d dz X 1 ( z ) + 1 z X 1 ( z ) = X 0 ( z ) where X m denotes either J m , N m , I m or K m . This gives, in particular 1 ρ
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