Physics - Magnetism Question with Solution | TestHub

Magnetic field due to a circular loop carrying current I is given as $B=\frac{{\mu }_0}{4\pi }\frac{2\pi I{r}^2}{{\left(r^2+x^2\right)}^{3/2}}$ where, $r=$ radius of loop $x=$ distance from centre of loop For large distances, $(x>>r)$, $\therefore B=\frac{{\mu }_0}{4\pi }\frac{2IA}{x^3}=\frac{{\mu }_0}{4\pi }\frac{2M}{x^3}$ where $M=IA=$ magnetic dipole moment of current loop which is equal to the magnetic field due to magnetic dipole. Thus, the current loop can be considered as a magnetic dipole.