emsol14 - Electromagnetic Theory I Solution Set 14 Due 5...

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Unformatted text preview: Electromagnetic Theory I Solution Set 14 Due: 5 December 2011 53. J, Problem 6.18 Solution : a) We have A = g 4 π Z-∞ dz ˆ z × ( r- z ˆ z ) | r- z ˆ z | 3 = g 4 π (ˆ z × ρ ) Z-∞ dz 1 ( ρ 2 + ( z- z ) 2 ) 3 / 2 = g 4 π ˆ ϕρ Z-∞ dz 1 ( ρ 2 + ( z- z ) 2 ) 3 / 2 = g 4 π ˆ ϕ ρ 1- z r = g 4 π ˆ ϕ r sin θ (1- cos θ ) The vector potential is well behaved for θ ∼ 0, but it shows singular behavior for θ ∼ π . This is near the negative z-axis where the Dirac string is located. b) Write A = f ( ρ, z )ˆ z × ρ . Then ∇ × A = ∂f ∂ρ ρ ˆ z + ∂f ∂z (- ρ ) + f 2ˆ z 2 f + ρ ∂f ∂ρ = g 4 π cos θ r 2 , ∂f ∂z =- g 4 πr 3 ∇ × A = g 4 π ˆ z z r 3 + ρ r 3 = g 4 π r r 3 (1) This is the magnetic field of the monopole. Note that the singular behavior in A near the negative z-axis is absent in ∇ × A a little off the axis. c) For 0 < θ < π/ 2, Φ < = 2 π Z R sin θ ρdρ g 4 π R cos θ ( R 2 cos 2 θ + ρ 2 ) 3 / 2 = g 2 (1- cos θ ) = + g 2- g 2 cos θ For π/ 2 < θ < π , by symmetry the downward flux would be numerically the same, so the upward flux is numerically the negative: Φ > =- g 2 (1- cos( π- θ )) =- g 2- g 2 cos θ (2) d) For any θ , I d l · A = 2 πR sin θ g 4 πR tan θ 2 = g 2 2 sin 2 θ = g 2 (1- cos θ ) = + g 2- g 2 cos θ This second calculation includes not only the flux from the monopole calculated in c),...
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This note was uploaded on 01/08/2012 for the course PHY 6346 taught by Professor Staff during the Spring '08 term at University of Florida.

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emsol14 - Electromagnetic Theory I Solution Set 14 Due 5...

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