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(C) The magnetic induction $B$ at the centre of a circular coil with $n$ turns,radius $R$,and current $i$ is given by $B = \frac{\mu_0 n i}{2R}$.Since

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$1: 4$

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(C) The magnetic induction $B$ at the centre of a circular coil with $n$ turns,radius $R$,and current $i$ is given by $B = \frac{\mu_0 n i}{2R}$.Since $i$ is constant,$B \propto \frac{n}{R}$.For the first coil: $n_1 = 1$,radius $R_1$,length $l = 2\pi R_1$.For the second coil: $n_2 = 2$,radius $R_2$,length $l = 2(2\pi R_2) = 4\pi R_2$.Since the length $l$ is the same,$2\pi R_1 = 4\pi R_2 \Rightarrow R_2 = \frac{R_1}{2}$.The ratio of magnetic inductions is $\frac{B_1}{B_2} = \frac{n_1}{R_1} 	imes \frac{R_2}{n_2} = \frac{1}{R_1} 	imes \frac{R_1/2}{2} = \frac{1}{4}$.Thus,the ratio $B_1: B_2 = 1: 4$.

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