What is the expression for current density of a ring in spherical coordinates [closed]

Question

I am not understanding the following expression for the current density of a ring of radius $a$ : $$J_{\phi}= I \sin\theta'\delta (\cos\theta')\delta(r'-a)/a.$$ It is dimensionally correct. I do not know how the normalization was done to get the total current. This is Jackson 5.33 of Section 5.5. I am sorry to ask a homework-like question.

Answer 1

If your ring of current $I$ is at the parallel $r=r_0$ and $\theta=\theta_0$, then: $$ j = I\delta(\theta-\theta_0)\frac{\delta(r-r_0)}{r_0}e_\phi $$ This is dimensionally correct, like your expression. Indeed, when looking at the current flowing through a half plane $\phi=cst$, using: $$ \int_0^\pi d\theta\int_0^\infty dr r \delta(\theta-\theta_0)\frac{\delta(r-r_0)}{r_0} = 1 $$ you do get a total current $I$. You need to use the 2D area element (same as polar coordinates): $$ d\theta rdr $$

In general, if it is not located at a singular point of the coordinate system, you just need to divide by the 2D Jacobean at that point.

Hope this helps.