Jackson Problema 5.1

Jackson 5.1 Homework Problem Solution
Dr. Christopher S. Baird
University of Massachusetts Lowell

PROBLEM:
Starting with the differential expression
d B=

μ0 I
x−x '
d l '×
3
4π
∣x −x'∣

for the magnetic induction at the point P with coordinate x produced by an increment of current I dl' at
x', show explicitly that for a closed loop carrying a current I the magnetic induction at Pis
B=

μ0 I
∇Ω
4π

where Ω is the solid angle subtended by the loop at the point P. This corresponds to a magnetic scalar
potential, ΦM = -μ0IΩ/4π. The sign convention for the solid angle Ωis 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, Ωis 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 expressionand put all the constants on the other side to get them out of the
way:
4π
x− x '
d B=d l ' ×
μ0 I
∣x −x '∣3
This is a vector equation that must hold for all components. If we take one componentin a general way,
then it will apply to all components. Let us take the ith Cartesian component.

[

4π
x −x '
d B i= xi⋅ d l '×
̂
μ0 I
∣x − x '∣3

]

now integrate over a closed loopto get the total field due to the loop:

[

4π
x− x '
B i=∮ x i⋅ d l '×
̂
μ0 I
∣x −x '∣3

]

We want to try to do this integral. Use the identity

(

)

x−x '
1
=∇ '
3
∣x− x '∣∣x −x '∣

[

(

4π
1
B i=∮ x i⋅ d l '× ∇ '
̂
μ0 I
∣x −x '∣

)]

(
Use the vector identity a⋅ b×c ) =b⋅( c ×a )

[( )]

4π
1
B i=∮ d l '⋅ ∇ '
× xi
̂
μ0 I
∣x − x '∣

UseStoke's theorem to convert the line integral to an area integral over the surface bounded by the
closed line integral.

[ [ ( ) ]]

4π
1
̂
̂
B i=∫ ∇ '× ∇ '
× x i ⋅n ' da '
∣x −x '∣
μ0 I...