Calculate the magnetic moment of a thin wire with a current I=0.8A, wound tightly on half a tore. The diameter of the cross-section of the tore is equal to d = 5.0 cm, the number of turns is N = 500.

Magnetic moment is a vector quantity its direction is in the direction of area vector it is given by${\text{P}}_{\text{m}\pi }=\text{NIA}$

We consider an element making an angle$\text{d}\theta$  at o

The number of turns in the considered element is

$\text{dN}=\frac{\text{N}}{\pi }\left(\text{d}\theta \right)$

${\text{dP}}_{\text{m}}=\left(\text{dN}\right)\text{IA}=\left(\text{dN}\right)I\frac{\pi {\text{d}}^2}{4}$

$\text{d}{\overline{P}}_m.{\text{dP}}_{\text{m}}\cos \theta \hat{j}-{\text{dP}}_{\text{m}}\sin \theta \hat{i}$

${\text{P}}_{\text{mx}}={\displaystyle {\int }_{\theta =0}^{\pi }-{\text{dP}}_{\text{m}}\sin \theta }$

$=-{\displaystyle {\int }_0^{\pi }\frac{{\text{NId}}^{\text{2}}}{4}\sin \theta .d\theta =}\frac{{\text{NId}}^{\text{2}}}{4}{\left(\cos \theta \right)}_0^{\pi }$

$=\frac{{\text{NId}}^{\text{2}}}{4}\left(\cos \pi -\cos \theta \right)=\frac{-{\text{NId}}^{\text{2}}}{4}$

${\text{P}}_{\text{my}}=-{\displaystyle {\int }_0^{\pi }\frac{{\text{NId}}^{\text{2}}}{4}\cos \theta .\text{d}\theta =}\frac{{\text{NId}}^{\text{2}}}{4}{\left(\sin \theta \right)}_0^{\pi }=0$

${\text{P}}_{\text{m}x}-{\text{P}}_{\text{my}}=\frac{{\text{NId}}^{\text{2}}}{2}=0.5{\text{Am}}^{\text{2}}$