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F07_hw09a

# F07_hw09a - Physics 505 Homework Assignment#9 Solutions...

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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 0 , 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 Φ M = - 1 4 π Z V ~ ∇ · ~ M ( ~x 0 ) | ~x - ~x 0 | d 3 x 0 + 1 4 π I S ˆ n 0 · ~ M ( ~x 0 ) | ~x - ~x 0 | da 0 Orienting the cylinder along the z axis, we take a uniform magnetization ~ M = M 0 ˆ z . In this case the volume integral drops out, and the surface integral only picks up contributions on the endcaps. Thus Φ M = M 0 4 π Z top 1 | ~x - ~x 0 | da 0 - Z bottom 1 | ~x - ~x 0 | da 0 where ‘top’ and ‘bottom’ denote z = ± L/ 2, and the integrals are restricted to ρ < a . On axis ( ρ = 0) we have simply Φ M ( z ) = M 0 4 π Z 1 p ρ 2 + ( z - L/ 2) 2 - 1 p ρ 2 + ( z + L/ 2) 2 ! ρ dρ dφ = M 0 4 Z a 2 0 1 p ρ 2 + ( z - L/ 2) 2 - 1 p ρ 2 + ( z + L/ 2) 2 ! 2 = M 0 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 Φ M = - M 0 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 0 2 z - L/ 2 p a 2 + ( z - L/ 2) 2 - z + L/ 2 p a 2 + ( z + L/ 2) 2 + 2 Θ( L/ 2 - | z | )

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where Θ( ξ ) denotes the unit step function, Θ = 1 for ξ > 0 (and 0 otherwise). The magnetic induction is obtained by rewriting the relation ~ H = ~ B/μ 0 - ~ M as ~ B = μ 0 ( ~ H + ~ M ). Since the magnetization is only nonzero inside the magnet [ie M z = M 0 Θ( L/ 2 - | z | )], the addition ~ H + ~ M simply removes the step function term. We find B z = μ 0 ( H z + M z ) = - μ 0 M 0 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/μ 0 M 0 and ~ H/M 0 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 0 / 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 μ 0 M z 0 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. 5.21 A magnetostatic field is due entirely to a localized distribution of permanent magne- tization. a ) Show that Z ~ B · ~ H d 3 x = 0 provided the integral is taken over all space. So long as the magnetic field is due to a localized distribution of permanent magnetization, and in particular not to free currents, it satisfies the curl-free
condition ~ ∇ × ~ H = 0. As a result, we may employ a magnetic scalar potential ~ H = - ~ Φ M . This allows us to write Z ~ B · ~ H d 3 x = - Z ~ B · ~ Φ M d 3 x = Z Φ M ~

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