Calculating the magnetic moment ( in `Am^2`) of a thin wire with a current I=8A, wound tightlly on a half a tore (see figure). The diameter of the cross section of the tore is equal to d=5cm, and the number of turns is N=500.

Explanation

Take an element of length `nd theta` containing `(N)/(pi r), rd theta` turns. Its magnetic moment is
`(N)/(pi) d theta . (pi)/(4) d^(2) I`
normal to the plane of cross section. We resolve it along `OA` and `OB`. The moment along `OA` inegrates to
`int_(0)^(pi) (N)/(4) d^(2) theta cos theta = 0`
while that along `OB` gives
`p = -int_(0)^(pi) (N d^(2) I)/(4) sin theta d theta = (1)/(2) N d^(2) I`