Physics - Magnetism Question with Solution | TestHub

Let r be the radius of the circular loop.

$\therefore A=\pi r^2$

$\Rightarrow \mathrm{r}=\sqrt{\frac{\mathrm{A}}{\pi }}$

Magnetic field at the centre of the loop is

$B=\frac{{\mu }_{0}I}{2r}$

$\Rightarrow \mathrm{B}$

$=\frac{{\mu }_0\mathrm{I}}{2\sqrt{\frac{\mathrm{A}}{\pi }}}\Rightarrow \mathrm{I}=\frac{2\mathrm{B}}{{\mu }_0}\sqrt{\frac{\mathrm{A}}{\pi }}$

Now, magnetic moment of the loop is

$\mathrm{M}=\mathrm{IA}$

$=\frac{2\mathrm{B}}{{\mu }_0}\sqrt{\frac{\mathrm{A}}{\pi }}·\mathrm{A}$

$=\frac{2\mathrm{B}}{{\mu }_0}\frac{{\mathrm{A}}^{\frac{3}{2}}}{\sqrt{\pi }}$

$\Rightarrow n=2$ and $m=1.5$

$\therefore n+m=3.5$