A wire of fixed length is turned to form a coil of one turn. It is rewound and again turned to form a coil of three turns. If in both cases same amount of current is passed, then the magnetic field produced at the centre of a coil in second case will be

Magnetic field at the centre of n turn coil carrying current i

$B=\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi ni}{r}.......\left(i\right)$

For single turn n =1

$B=\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi i}{r}......\left(ii\right)$

If the same wire is turn again to form a coil of three turns i.e.n = 3 and radius of each turn $r^{\mathrm{\prime }}=\frac{r}{3}$

So new magnetic field at centre $B^{\mathrm{\prime }}=\frac{{\mu }_0}{4\pi }\cdot \frac{2\pi \left(3\right)}{r^{\mathrm{\prime }}}\Rightarrow B^{\mathrm{\prime }}=9\times \frac{{\mu }_0}{4\pi }\cdot \frac{2\pi i}{r}\dots ..\left(iii\right)$

Comparing equation (ii) and (iii) gives B' = 9B .