# emsol10 - Electromagnetic Theory I Solution Set 10 Due 7...

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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 ˆ z is placed at the center of a spherical shell with uniform magnetic permeability μ , 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 = −∇ φ . the potential of a dipole is pure l = 1, so only need the l = 1 part of the potential in all three regions φ = 1 4 π ( m r 2 + A 1 r ) cos θ < r < a ( B 2 r 2 + A 2 r ) cos θ a < r < b B 3 r 2 cos θ b < r < ∞ (1) Continuity of H t reduces to continuity of φ : 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 μ∂φ/∂r . μA 2 − 2 μ B 2 b 3 = − 2 μ B 3 b 3 , μA 2 − 2 μ B 2 a 3 = μ A 1 − 2 μ 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 = μ + 2 μ 3 μ B 3 , A 2 = 2( μ − μ ) 3 μ 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 ( μ − μ )( a 3 − b 3 )( μ + 2 μ ) ( μ + 2 μ )( μ + 2 μ ) b 3 − 2 a 3 ( μ − μ ) 2 B 3 = 9 mμμ b 3 ( μ + 2 μ )( μ + 2 μ ) b 3 − 2 a 3 ( μ − μ ) 2 A 2 = 6 mμ ( μ − μ ) ( μ + 2 μ )( μ + 2 μ ) b 3 − 2 a 3 ( μ − μ ) 2 B 2 = 3 mμ ( μ + 2 μ ) b 3 ( μ + 2 μ )( μ + 2 μ ) b 3 − 2 a 3 ( μ − μ ) 2 1 The fields are H = −∇ φ = 1 4 π − A 1 ˆ z + 3 m r z − mr 2 ˆ z r 5 < r < a − A 2 ˆ z + B 2 3 r z − r 2 ˆ z r 5 a < r < b B 3 3 r z − r 2 ˆ z r 5 b < r < ∞ (5) and of course B = μ H for a < r < b and = μ H in the inner and outer regions. c) Discuss the two extreme limits μ → ∞ and μ → 0. Determine the limiting B and H fields in each region, and discuss the qualitative differences of the two limits. Solution: first for μ → ∞ , 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 π 3 μ m b 3 − a 3 bracketleftbigg − ˆ z + b 3 2 3 r z − r 2 ˆ z r 5 bracketrightbigg , a < r < b (6) In the inner region B = μ H with H → m 4 π bracketleftbigg − r 3 − a 3 a 3 r 3 ˆ z + 3 r z r 5 bracketrightbigg , < r < a (7) We see that H is normal to the interface at...
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emsol10 - Electromagnetic Theory I Solution Set 10 Due 7...

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