# Jackson Eq. 5.33 (3rd Ed.)

1. Dec 7, 2008

### shehry1

1. The problem statement, all variables and given/known data
I cant 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.

2. Relevant equations

3. The attempt at a solution

2. Dec 8, 2008

### Ben Niehoff

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.

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.

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.

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 $\delta(r' - a)$.

3. Dec 8, 2008

### shehry1

Thanks a lot :)