Question

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

Answer 1

Take an element of length $rd\theta$ containing $\frac{N}{\pi r}\cdot rd\theta$ turns. Its magnetic moment is
$\frac{N}{\pi }d\theta \cdot \frac{\pi }{4}{d}^{2}I$
normal to the plane of cross section. We resolve it along $OA$ and $OB$. The moment along $OA$ integrates to
${\int }_0^{\pi }\frac{N}{4}{d}^{2}Id\theta \cos \theta =0$
while that along $OB$ gives
$p_m={\int }_0^{\pi }\frac{N{d}^{2}I}{4}\sin \theta d\theta =\frac{1}{2}N{d}^{2}I$