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Unformatted text preview: Electromagnetic Theory I Solution Set 10 Due: 7 November 2011 37. A point magnetic dipole with moment m = m z is placed at the center of a spherical shell with uniform magnetic permeability , and with inner and outer radii a,b respectively. a) Set up the boundary equations that determine the magnetic field in the three regions < r < a , a < r < b , r > b . Solution: Since J = 0, we can write H = . the potential of a dipole is pure l = 1, so only need the l = 1 part of the potential in all three regions = 1 4 ( m r 2 + A 1 r ) cos < r < a ( B 2 r 2 + A 2 r ) cos a < r < b B 3 r 2 cos b < r < (1) Continuity of H t reduces to continuity of : A 2 b + B 2 b 2 = B 3 b 2 , A 2 a + B 2 a 2 = A 1 a + m a 2 (2) Continuity of B n means continuity of /r . A 2 2 B 2 b 3 = 2 B 3 b 3 , A 2 2 B 2 a 3 = A 1 2 m a 3 (3) b) Solve the equations of part a) to find the B and H fields in all three regions. Solution: The equations at the r = b interface easily give B 2 = + 2 3 B 3 , A 2 = 2( ) 3 B 3 (4) Plugging these into the r = a interface equations then leads after a few lines of algebra to A 1 = 2 m a 3 ( )( a 3 b 3 )( + 2 ) ( + 2 )( + 2 ) b 3 2 a 3 ( ) 2 B 3 = 9 m b 3 ( + 2 )( + 2 ) b 3 2 a 3 ( ) 2 A 2 = 6 m ( ) ( + 2 )( + 2 ) b 3 2 a 3 ( ) 2 B 2 = 3 m ( + 2 ) b 3 ( + 2 )( + 2 ) b 3 2 a 3 ( ) 2 1 The fields are H = = 1 4 A 1 z + 3 m r z mr 2 z r 5 < r < a A 2 z + B 2 3 r z r 2 z r 5 a < r < b B 3 3 r z r 2 z r 5 b < r < (5) and of course B = H for a < r < b and = H in the inner and outer regions. c) Discuss the two extreme limits and 0. Determine the limiting B and H fields in each region, and discuss the qualitative differences of the two limits. Solution: first for , H 0 in and outside the shell ( r > a ). this is in accord with the electrostatic analogy of an E field and a conductor. B = 0 outside the conductor, but stays finite within the shell B 1 4 3 m b 3 a 3 bracketleftbigg z + b 3 2 3 r z r 2 z r 5 bracketrightbigg , a < r < b (6) In the inner region B = H with H m 4 bracketleftbigg r 3 a 3 a 3 r 3 z + 3 r z r 5 bracketrightbigg , < r < a (7) We see that H is normal to the interface at...
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 Spring '08
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