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(B) Let the length of the wire be $\ell$.For a coil of one turn with radius $r$,the circumference is $2\pi r = \ell$,so $r = \frac{\ell}{2\pi}$.The ma

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(B) Let the length of the wire be $\ell$.For a coil of one turn with radius $r$,the circumference is $2\pi r = \ell$,so $r = \frac{\ell}{2\pi}$.The magnetic induction at the centre is $B_1 = \frac{\mu_0 I}{2r} = \frac{\mu_0 I}{2(\ell / 2\pi)} = \frac{\mu_0 I \pi}{\ell}$.For a coil of two turns with radius $r'$,the total length is $2(2\pi r') = \ell$,so $r' = \frac{\ell}{4\pi}$.The magnetic induction at the centre for $N=2$ turns is $B_2 = \frac{N \mu_0 I}{2r'} = \frac{2 \mu_0 I}{2(\ell / 4\pi)} = \frac{4 \mu_0 I \pi}{\ell}$.The ratio of the magnetic induction of the two-turn coil to the one-turn coil is $\frac{B_2}{B_1} = \frac{4 \mu_0 I \pi / \ell}{\mu_0 I \pi / \ell} = \frac{4}{1}$.

$A$ direct current is passing through a wire. It is bent to form a coil of one turn. Now,it is further bent to form a coil of two turns but at a smaller radius. Find the ratio of the magnetic induction at the centre of this two-turn coil to that at the centre of the one-turn coil.

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