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(C) The magnetic field at the centre of a circular coil with $N$ turns and radius $R$ is given by $B = \frac{\mu_0 N i}{2R}$.For the first coil,$N_1 =

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$2$

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(C) The magnetic field at the centre of a circular coil with $N$ turns and radius $R$ is given by $B = \frac{\mu_0 N i}{2R}$.For the first coil,$N_1 = 4$ and $B_1 = 32 \ T$. So,$32 = \frac{\mu_0 \cdot 4 \cdot i}{2R_1} \implies 32 = \frac{2 \mu_0 i}{R_1}$.When the wire of length $L$ is unwound and rewound into a single turn $(N_2 = 1)$,the circumference remains the same: $L = 2\pi R_1 \cdot N_1 = 2\pi R_2 \cdot N_2$.Since $N_1 = 4$ and $N_2 = 1$,we have $4(2\pi R_1) = 1(2\pi R_2)$,which gives $R_2 = 4R_1$.The new magnetic field is $B_2 = \frac{\mu_0 N_2 i}{2R_2} = \frac{\mu_0 \cdot 1 \cdot i}{2(4R_1)} = \frac{\mu_0 i}{8R_1}$.Comparing $B_1$ and $B_2$: $\frac{B_2}{B_1} = \frac{\mu_0 i / 8R_1}{2 \mu_0 i / R_1} = \frac{1}{16}$.Therefore,$B_2 = \frac{B_1}{16} = \frac{32}{16} = 2 \ T$.

The electric current in a circular coil of four turns produces a magnetic induction $32 \ T$ at its centre. The coil is unwound and is rewound into a circular coil of a single turn. If the same current flows through the new coil,the magnetic induction at the centre will be $.......... \ T$.

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