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(C) The magnetic field at the centre of a circular coil with $N$ turns,radius $r$,and current $I$ is given by $B = \frac{\mu_0 NI}{2r}$.For the first

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$1:n^2$

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(C) The magnetic field at the centre of a circular coil with $N$ turns,radius $r$,and current $I$ is given by $B = \frac{\mu_0 NI}{2r}$.For the first case,$N_1 = 1$,radius is $r_1$,so $B_1 = \frac{\mu_0 I}{2r_1}$.For the second case,$N_2 = n$,radius is $r_2$,so $B_2 = \frac{\mu_0 nI}{2r_2}$.The total length of the wire $L$ remains constant. Thus,$L = 2\pi r_1 = n(2\pi r_2)$.This implies $r_1 = n r_2$,or $\frac{r_2}{r_1} = \frac{1}{n}$.Taking the ratio of the magnetic fields:$\frac{B_1}{B_2} = \frac{\mu_0 I / 2r_1}{\mu_0 nI / 2r_2} = \frac{1}{n} \cdot \frac{r_2}{r_1} = \frac{1}{n} \cdot \frac{1}{n} = \frac{1}{n^2}$.Therefore,the ratio is $1:n^2$.

$A$ long wire is bent into a circular coil of one turn and then into a circular coil of smaller radius having $n$ turns. If the same current passes in both the cases,the ratio of magnetic fields produced at the centre for one turn to that of $n$ turns is

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