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(C) The magnetic field at the centre of a circular loop of radius $R$ with $N$ turns carrying current $I$ is given by $B = \frac{\mu_0 N I}{2 R}$.For

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$4 B$

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(C) The magnetic field at the centre of a circular loop of radius $R$ with $N$ turns carrying current $I$ is given by $B = \frac{\mu_0 N I}{2 R}$.For a single loop $(N=1)$,$B = \frac{\mu_0 I}{2 R}$. Since the length of the wire $l = 2 \pi R$,we have $R = \frac{l}{2 \pi}$.Thus,$B = \frac{\mu_0 I}{2 (l / 2 \pi)} = \frac{\mu_0 I \pi}{l}$.When the same wire of length $l$ is bent into a double loop $(N=2)$,the new radius $R'$ is given by $l = 2 (2 \pi R')$,so $R' = \frac{l}{4 \pi} = \frac{R}{2}$.The new magnetic field $B'$ at the centre is $B' = \frac{\mu_0 N' I}{2 R'} = \frac{\mu_0 (2) I}{2 (R / 2)} = \frac{2 \mu_0 I}{R} = 4 \left( \frac{\mu_0 I}{2 R} \right) = 4 B$.

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