The ratio of magnetic field at the centre of the current carrying circular loop and magnetic moment is ' $x$…
The ratio of magnetic field at the centre of the current carrying circular loop and magnetic moment is ' $x$ '. When both the current and radius are double then the ratio will be
$2 \mathrm{x}$
$\frac{x}{2}$
$\frac{x}{4}$
$\frac{x}{8}$
Solution
The magnetic field at the center of the wire is given as $B=\frac{\mu_0 i}{2 r}$.
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,
$x_1=\frac{\frac{\mu_0 2 i}{2 \pi 2 r}}{2 i \pi(2 r)^2}$ ....(From (i))
$x_1=\frac{\mu_0}{16 \pi^2 r^3}$
$\mathrm{x}_1=\frac{1}{8}\left(\frac{\mu_0}{2 \pi^2 \mathrm{r}^3}\right)$
$\therefore \quad x_1=\frac{x}{8}$