What are the Confusing Aspects of Jackson Eq. 5.33 (3rd Ed.)?

[shehry1]

Homework Statement

I can't seem to figure out how he writes down this equation. Specifically:
a. Isn't Theta' = 90 degrees. Then why doesn't he write it out explicitly.
b. Whats the use of adding the Sin(Theta') if he is going to use a delta function using the Cos
c. What is the radius 'a' doing there. If the wire has no thickness, shouldn't the magnitude of J equal I.
d. Shouldn't J=I/Area^2 in terms of units.

Homework Equations

The Attempt at a Solution

 

[Ben Niehoff]

Homework Statement

I can't seem to figure out how he writes down this equation. Specifically:
a. Isn't Theta' = 90 degrees. Then why doesn't he write it out explicitly.

Because it's more convenient to write the measure as

\delta(\cos \theta') d(\cos \theta')

rather than

\sin \theta' \delta(\theta') d\theta'

It saves a lot of time, trust me.

b. Whats the use of adding the Sin(Theta') if he is going to use a delta function using the Cos

The current loop is in the x-y plane, where

\sin \theta' = 1

which is the same as

\cos \theta' = 0

I suppose he could have technically left out the sin(theta) and just put in 1 (just as he puts in 1/a instead of 1/r). However, perhaps he put it in so that the formula would be easier to generalize.

c. What is the radius 'a' doing there. If the wire has no thickness, shouldn't the magnitude of J equal I.

No. The integral of J over its cross-sectional area should equal I. That is:

I = \iint_R r \; dr \; d(\cos \theta) \; I \delta(\cos \theta) \frac{\delta(r-a)}{a}

where R is some vertical region that cuts across part of the wire. So you see, the 1/a is essential to give the correct value of I.

d. Shouldn't J=I/Area^2 in terms of units.

Yes, and it is. Or rather, J = I/Area. Remember that delta functions carry reciprocal units. So you get 1/L from the factor of 1/a, and another 1/L from the factor of \delta(r' - a).

 

[shehry1]

Thanks a lot :)