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Unformatted text preview: Homework 10 Answers, 95.657 Fall 2011, Electromagnetic Theory I Dr. Christopher S. Baird, UMass Lowell Problem 1 Jackson 5.1 Starting with the differential expression d B = μ I 4 π d l ' × x − x ' ∣ x − x ' ∣ 3 for the magnetic induction at the point P with coordinate x produced by an increment of current I d l ' at x ', show explicitly that for a closed loop carrying a current I the magnetic induction at P is B = μ I 4 π ∇ Ω where Ω is the solid angle subtended by the loop at the point P. This corresponds to a magnetic scalar potential, Φ M =  μ I Ω/4π. The sign convention for the solid angle Ω is positive if the point P views the “inner” side of the surface spanning the loop, that is, if a unit normal n to the surface is defined by the direction of current flow via the righthand rule, Ω is positive if n points away from the point P , and negative otherwise. This is the same convention as in Section 1.6 for the electric dipole layer. SOLUTION: Start with the differential expression and put all the constants on the other side to get them out of the way: 4 π μ I d B = d l ' × x − x ' ∣ x − x ' ∣ 3 This is a vector equation that must hold for all components. If we take one component in a general way, then it will apply to all components. Let us take the i th Cartesian component. 4 π μ I d B i =̂ x i ⋅ [ d l ' × x − x ' ∣ x − x ' ∣ 3 ] now integrate over a closed loop to get the total field due to the loop: 4 π μ I B i = ∮ ̂ x i ⋅ [ d l ' × x − x ' ∣ x − x ' ∣ 3 ] We want to try to do this integral. Use the identity x − x ' ∣ x − x ' ∣ 3 =∇ ' ( 1 ∣ x − x ' ∣ ) 4 π μ I B i = ∮ ̂ x i ⋅ [ d l ' ×∇ ' ( 1 ∣ x − x ' ∣ ) ] Use the vector identity a ⋅( b × c )= b ⋅( c × a ) 4 π μ I B i = ∮ d l ' ⋅ [ ∇ ' ( 1 ∣ x − x ' ∣ ) ×̂ x i ] Use Stoke's theorem to convert the line integral to an area integral over the surface bounded by the closed line integral. 4 π μ I B i = ∫ [ ∇ ' × [ ∇ ' ( 1 ∣ x − x ' ∣ ) ×̂ x i ] ] ⋅̂ n ' da ' Now use the identity: ∇ ' ×( a × b )= a (∇ ' ⋅ b )− b (∇ ' ⋅ a )+( b ⋅∇ ' ) a −( a ⋅∇ ' ) b and set a =∇ ' ( 1 ∣ x − x ' ∣ ) and b =̂ x i ∇ ' ×(∇ ' ( 1 ∣ x − x ' ∣ ) ×̂ x i )=−̂ x i (∇ ' 2 ( 1 ∣ x − x ' ∣ ) )+(̂ x i ⋅∇ ' )∇ ' ( 1 ∣ x − x ' ∣ ) Use this identity: 4 π μ I B i =− ∫ [ ̂ x i (∇ ' 2 ( 1 ∣ x − x ' ∣ ) ) ] ⋅̂ n ' da ' + ∫ [ (̂ x i ⋅∇ ' )∇ ' ( 1 ∣ x − x ' ∣ ) ] ⋅̂ n ' da '...
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