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(A) Case-$I$: Number of turns,$N_1 = 9$. The magnetic field at the centre is given by $B_1 = \frac{\mu_0 N_1 I}{2 R} = \frac{9 \mu_0 I}{2 R}$.Case-$II

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$B_1 / 9$

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(A) Case-$I$: Number of turns,$N_1 = 9$. The magnetic field at the centre is given by $B_1 = \frac{\mu_0 N_1 I}{2 R} = \frac{9 \mu_0 I}{2 R}$.Case-$II$: Number of turns,$N_2 = 3$. Let the new radius be $R'$. Since the total length of the wire remains constant,$N_1 (2 \pi R) = N_2 (2 \pi R')$.Substituting the values: $9 (2 \pi R) = 3 (2 \pi R') \Rightarrow R' = 3 R$.The new magnetic field is $B_2 = \frac{\mu_0 N_2 I}{2 R'} = \frac{\mu_0 	imes 3 	imes I}{2 	imes 3 R} = \frac{\mu_0 I}{2 R}$.Comparing $B_2$ with $B_1$: $\frac{B_2}{B_1} = \frac{\mu_0 I / 2 R}{9 \mu_0 I / 2 R} = \frac{1}{9}$.Therefore,$B_2 = \frac{B_1}{9}$.

$A$ coil having $9$ turns carrying a current produces a magnetic field $B_1$ at the centre. Now,the coil is rewound into $3$ turns carrying the same current. Then,the magnetic field at the centre $B_2$ is:

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