Physics - Magnetism Question with Solution | TestHub

The magnetic field at the center of the wire is given as $B=\frac{{\mu }_{0}i}{2\pi r}$. The magnetic moment is given as $\mathrm{M}=\mathrm{i}\pi {\mathrm{r}}^2$ The ratio is $\frac{\frac{{\mu }_0\mathrm{i}}{2\pi \mathrm{r}}}{\mathrm{i}\pi {\mathrm{r}}^2}=x$ ....(i) $x=\frac{{\mu }_0}{2{\pi }^2{r}^3}$ When current and radius is doubled, $x_1=\frac{\frac{{\mu }_{0}2i}{2\pi 2r}}{2i\pi (2r{)}^2}$ ....(From (i)) $x_1=\frac{{\mu }_0}{16{\pi }^2{r}^3}$ ${\mathrm{x}}_1=\frac{1}{8}\left(\frac{{\mu }_0}{2{\pi }^2{\mathrm{r}}^3}\right)$ $\therefore x_1=\frac{x}{8}$