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Unformatted text preview: 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 ( ,z ) = Ia Z dk cos kz I 1 ( k < ) K 1 ( k > ) where < ( > ) is the smaller (larger) of a and . The vector potential may be obtained by ~ A ( ~x ) = 4 Z ~ J ( ~x )  ~x ~x  d 3 x where (for a circular current loop) ~ J ( ~x ) = I ( z ) (  a ) in cylindrical coordinates. Note that to obtain the cylindrical components of ~ A ( ~x ) we have to be careful to convert the basis vector at the point x to components at x . (This is because the basic vectors depend on position.) A bit of geometry gives = sin(  ) + cos(  ) [Or, alternatively, we may choose the point x to lie at = 0, so that = y and = x . Then it is straightforward to see that = y cos  x sin = cos  sin . Using symmetry, we can see that only the component of ~ A is nonvanishing.] The integral expression for the vector potential is then ~ A ( ~x ) = I 4 Z ( z ) (  a )[ sin(  ) + cos(  )]  ~x ~x  d d dz = Ia 4 Z 2 sin(  ) + cos(  )  ~x ~x  d (1) where the integrand in the second line is to be evaluated at z = 0 and = a . We now use the cylindrical Greens function expressed as 1  ~x ~x  = 4 Z dk cos[ k ( z z )] h 1 2 I ( k < ) K ( k > ) + X m =1 cos[ m (  )] I m ( k < ) K m ( k > ) i Note that the integral over picks out the m = 1 term in the sum. Furthermore, the component drops out because sin(  ) is orthogonal to cos(  ), a result that could have been obtained by symmetry. We end up with ~ A ( ~x ) = Ia 4 4 Z dk cos( kz ) I 1 ( k < ) K 1 ( k > ) = Ia Z dk cos( kz ) I 1 ( k < ) K 1 ( k > ) b ) Show that an alternative expression for A is A ( ,z ) = Ia 2 Z dk e k  z  J 1 ( ka ) J 1 ( k ) To obtain the alternative expression, we use the alternate form of the Greens function 1  ~x ~x  = 2 Z dk e k ( z > z < ) h 1 2 J ( k ) J ( k ) + X m =1 cos[ m (  )] J m ( k ) J m ( k ) i 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 ) = Ia 2 Z dk e k  z  J 1 ( k ) 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....
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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
 Current, Magnetism, Work

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