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

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$4 B$

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(C) The magnetic field at the centre of a circular loop of radius $R$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2R}$.Since the circumference of the loop is $l = 2 \pi R$, we have $R = \frac{l}{2 \pi}$.Substituting this into the expression for $B$, we get $B = \frac{\mu_0 I}{2(l / 2 \pi)} = \frac{\mu_0 I \pi}{l}$.When the same wire of length $l$ is bent into a double loop (two turns), let the new radius be $r'$.The total length of the wire is $l = 2(2 \pi r')$, which implies $r' = \frac{l}{4 \pi} = \frac{R}{2}$.The magnetic field at the centre of a coil with $N$ turns is $B' = N \frac{\mu_0 I}{2r'}$.Here, $N = 2$ and $r' = \frac{R}{2}$.Substituting these values, we get $B' = 2 \cdot \frac{\mu_0 I}{2(R / 2)} = 2 \cdot \frac{\mu_0 I}{R} = 4 \cdot \left( \frac{\mu_0 I}{2R} \right) = 4B$.Therefore, the new magnetic field is $4B$.

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