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

Vedclass · Accepted Answer

$1: 4$

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

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