3209-2009-Solutions9

# 3209-2009-Solutions9 - ECE 3209 — Electromagnetic Fields...

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Unformatted text preview: ECE 3209 — Electromagnetic Fields University of Virginia Fall 2009 Homework # 9 Solutions 1. Vector potential of small rectangular loop. Since the loop is small (compared to the distance to point P ), we can approximate the vector potential due to each side with that due to a short current filament of length ‘ , ~ A ’ μ ‘ ~ I 4 πr where r is the distance from the filament to the point P and ‘ is the length of the appropriate side. The distance from each filament to P is given by (the Law of Cosines), r 1 = p r 2 + ( a/ 2) 2 + ra sin θ sin φ ⇒ 1 r 1 ’ 1- ( a/ 2 r )sin θ sin φ r r 2 = p r 2 + ( a/ 2) 2- ra sin θ sin φ ⇒ 1 r 2 ’ 1 + ( a/ 2 r )sin θ sin φ r r 3 = p r 2 + ( b/ 2) 2 + rb sin θ cos φ ⇒ 1 r 3 ’ 1- ( b/ 2 r )sin θ cos φ r r 4 = p r 2 + ( b/ 2) 2- rb sin θ cos φ ⇒ 1 r 4 ’ 1 + ( b/ 2 r )sin θ cos φ r Above, we have used a Taylor series approximation to find the expressions on the right, with the assumption that a,b ¿ r . Thus, by superposition, the vector potential at P is, ~ A = μ Ib 4 π ‡ 1- ( a/ 2 r )sin θ sin φ r- 1 + ( a/ 2 r )sin θ sin φ r · ˆ x + μ Ia 4 π ‡ 1 + ( b/ 2 r )sin θ cos φ r- 1- ( b/ 2 r )sin θ cos φ r · ˆ y ~ A =...
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3209-2009-Solutions9 - ECE 3209 — Electromagnetic Fields...

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