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

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$1:3$

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(B) The magnetic field at the centre of a circular loop of radius $r$ with $n$ turns carrying current $I$ is given by $B = \frac{\mu_0 n I}{2r}$.For the first wire,it is bent into a single loop $(n_1 = 1)$ of radius $R$. The magnetic field is $B_1 = \frac{\mu_0 (1) I}{2R} = \frac{\mu_0 I}{2R}$.The second wire has the same length $L = 2\pi R$. It is bent into $n_2 = 3$ loops. Let the radius of each new loop be $r$. Then $L = n_2 (2\pi r) = 3(2\pi r)$.Equating the lengths: $2\pi R = 6\pi r$,which gives $r = R/3$.The current passing through the second coil is $I_2 = I/3$.The magnetic field at the centre of the second coil is $B_2 = \frac{\mu_0 n_2 I_2}{2r} = \frac{\mu_0 (3) (I/3)}{2(R/3)} = \frac{\mu_0 I}{2(R/3)} = \frac{3\mu_0 I}{2R}$.Taking the ratio: $\frac{B_1}{B_2} = \frac{\mu_0 I / 2R}{3\mu_0 I / 2R} = \frac{1}{3}$.Therefore,$B_1 : B_2 = 1 : 3$.

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