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

The magnetic moment of circular current is given by AI, I being the circulating current and A is the area of cross-section, the direction is perpendicular to the plane of current. Now, for an element of torus of length rdθ, its magnetic moment is along the perpendicular to its cross-section and magnitude = current × area × number of turns in the length `r d theta`.

`dM=I(pid^(2))/4(N/(pir)r d theta)=N/4 d^(2) I d theta`
Resolving dM into components along x and y axis, we get `dMsin theta` and `dMcos theta` Components along y axis from neighbouring elements cancel out to zero, and components along x-axis are added.
so `M=int dM sin theta=int_(0)^(pi)N/4d^(2) I sin theta d theta=(Nd^(2)I)/2`
putting the given values we get `M=5Am^(2).=(10^(-7)xx0.5xx5xx10^(3))/(5^(2))(3/5)=6muT`