Homework10

# Homework10 - Homework 10 Answers 95.657 Fall 2011...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 right-hand 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 '...
View Full Document

{[ snackBarMessage ]}

### Page1 / 7

Homework10 - Homework 10 Answers 95.657 Fall 2011...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online