The magnetic moment of circular current is given by Al, being the circulating current and A rs the area of cross-section; the direction is perpendicular to the plane of current. Now, for an element of toroid of length rdq, its magnetic moment is along the direction of arrow as shown, of magnitude (perpendicular to its cross-section) = current x area x number of turns in the length rdq.

$dM=I\frac{\pi d^2}{4}\left(\frac{N}{\pi r}rd\theta \right)=\frac{N}{4}{d}^{2}Id\theta$

Resolving dM into components along x and !, we get dM sin $\theta$ and dM cos $\theta$; components along y from neighbouring elements cancel out to zero, and components along x are added. So

$M=\int dM\sin \theta ={\int }_0^{\pi }\frac{N}{4}·d^{2}I\sin \theta d\theta =\frac{N{d}^{2}I}{2}$

Putting the given values we get M = 5 Am²