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Unformatted text preview: Physics 505 Fall 2007 Homework Assignment #9 Solutions Textbook problems: Ch. 5: 5.19, 5.21, 5.22, 5.27 5.19 A magnetically hard material is in the shape of a right circular cylinder of length L and radius a . The cylinder has a permanent magnetization M , uniform throughout its volume and parallel to its axis. a ) Determine the magnetic field ~ H and magnetic induction ~ B at all points on the axis of the cylinder, both inside and outside. We use a magnetic scalar potential and the expression M =- 1 4 Z V ~ ~ M ( ~x ) | ~x- ~x | d 3 x + 1 4 I S n ~ M ( ~x ) | ~x- ~x | da Orienting the cylinder along the z axis, we take a uniform magnetization ~ M = M z . In this case the volume integral drops out, and the surface integral only picks up contributions on the endcaps. Thus M = M 4 Z top 1 | ~x- ~x | da- Z bottom 1 | ~x- ~x | da where top and bottom denote z = L/ 2, and the integrals are restricted to < a . On axis ( = 0) we have simply M ( z ) = M 4 Z 1 p 2 + ( z- L/ 2) 2- 1 p 2 + ( z + L/ 2) 2 ! dd = M 4 Z a 2 1 p 2 + ( z- L/ 2) 2- 1 p 2 + ( z + L/ 2) 2 ! d 2 = M 2 h p a 2 + ( z- L/ 2) 2- p a 2 + ( z + L/ 2) 2- | z- L/ 2 | + | z + L/ 2 | i On axis, the field can only point in the z direction. It is given by H z =- z M =- M 2 z- L/ 2 p a 2 + ( z- L/ 2) 2- z + L/ 2 p a 2 + ( z + L/ 2) 2- sgn( z- L/ 2) + sgn( z + L/ 2) Note that the last two terms cancel when | z | > L/ 2, but add up to 2 inside the magnet. Thus we may write H z =- M 2 z- L/ 2 p a 2 + ( z- L/ 2) 2- z + L/ 2 p a 2 + ( z + L/ 2) 2 + 2( L/ 2- | z | ) where ( ) denotes the unit step function, = 1 for > 0 (and 0 otherwise). The magnetic induction is obtained by rewriting the relation ~ H = ~ B/- ~ M as ~ B = ( ~ H + ~ M ). Since the magnetization is only nonzero inside the magnet [ie M z = M ( L/ 2- | z | )], the addition ~ H + ~ M simply removes the step function term. We find B z = ( H z + M z ) =- M 2 z- L/ 2 p a 2 + ( z- L/ 2) 2- z + L/ 2 p a 2 + ( z + L/ 2) 2 (1) b ) Plot the ratios ~ B/ M and ~ H/M on the axis as functions of z for L/a = 5. The z component of the magnetic field looks like-2-1 1 2-0.4-0.2 0.2 0.4 z / L M / H while the z component of the magnetic induction looks like-2-1 1 2 0.2 0.4 0.6 0.8 1 / L / B M z Note that B z is continuous, while H z jumps at the ends of the magnet. This jump may be thought of as arising from effective magnetic surface charge....
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