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(A) The magnetic field at the centre of a circular coil of $N$ turns and radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 N I}{2R}$.Let th

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$\frac{B_1}{9}$

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(A) The magnetic field at the centre of a circular coil of $N$ turns and radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 N I}{2R}$.Let the length of the wire be $L$. For the first coil,$L = N_1 (2\pi R_1)$,where $N_1 = 9$. So,$R_1 = \frac{L}{18\pi}$.The magnetic field is $B_1 = \frac{\mu_0 N_1 I}{2R_1} = \frac{\mu_0 (9) I}{2(L/18\pi)} = \frac{\mu_0 I (9^2) (2\pi)}{2L} = \frac{81 \mu_0 I \pi}{L}$.For the second coil,$N_2 = 3$. So,$R_2 = \frac{L}{2\pi N_2} = \frac{L}{6\pi}$.The magnetic field is $B_2 = \frac{\mu_0 N_2 I}{2R_2} = \frac{\mu_0 (3) I}{2(L/6\pi)} = \frac{\mu_0 I (3^2) (2\pi)}{2L} = \frac{9 \mu_0 I \pi}{L}$.Comparing $B_1$ and $B_2$,we get $\frac{B_2}{B_1} = \frac{9}{81} = \frac{1}{9}$.Therefore,$B_2 = \frac{B_1}{9}$.

$A$ coil having $9$ turns carrying current $I$ produces a magnetic field $B_1$ at the centre. If the same wire is rewound into a coil of $3$ turns carrying the same current $I$,then the magnetic field at the centre $B_2$ is:

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