The ratio of magnetic field at the centre of a current carrying circular coil to its magnetic moment is \(x\), if the current and the radius both are doubled. The new ratio will become

Magnetic field at the centre of current carrying circular coil
\(B=\frac{\mu_{0} I}{2 r}\)
where, \(I=\) current and \(r=\) radius.
Magnetic dipole moment,
\(m=I A=I \cdot \pi r^{2} \)
\( \text {Given, } \frac{B}{m}=x \)
\( \Rightarrow \frac{\mu_{0} I}{2 r} \)
\( \Rightarrow \frac{\mu_{0}}{\pi I r^{2}}=x ...(i)\)
When current and radius, both are doubled, then
\(\begin{aligned} \frac{B^{\prime}}{m^{\prime}} &=\frac{\frac{\mu_{0} 2 I}{2 \cdot 2 r}}{\pi(2 I)(2 r)^{2}} \\ &=\frac{\frac{\mu_{0} I}{2 r}}{8 \pi r^{2} I}=\frac{\mu_{0}}{16 \pi r^{3}} \\ &=\frac{1}{8} \cdot \frac{\mu_{0}}{2 \pi r^{3}}=\frac{x}{8} \text { [from Eq. (i)] } \end{aligned}\)